Various analytic observations on combinations

V arious analytic observ ations on com binations ∗ Leonhard Euler 1. Let a ser ies, either finite o r infinite, of qua nt ities b e given to us, as a, b, c, d, e, f , g , h etc.; the letters denote these qua n tities, which m ay b e either equal or uneq ual to each other. A t the same time, ho wev er, I will sp eak of quantities indicated by differen t letters as b eing unequal to each other, ev en if in examples equa l nu mbers can b e s ubstituted in place of them. 2. N ow first let new series b e formed fr om these quantities by taking p ow ers , with the sums des ignated by capital letters A, B , C, D etc. as follows; th us: A = a + b + c + d + e + e tc. , B = a 2 + b 2 + c 2 + d 2 + e 2 + e tc. , C = a 3 + b 3 + c 3 + d 3 + e 3 + e tc. , D = a 4 + b 4 + c 4 + d 4 + e 4 + e tc. , E = a 5 + b 5 + c 5 + d 5 + e 5 + e tc. etc. These series will b e infinite if the n umber of quantities a, b , c, d etc. has been assumed to b e infinite; on the other hand if the num b er of these quan tities is finite a nd deter mina te, put = n , then all these series will b e comprised of that same num b er of terms. 3. Next let us now form series by ta k ing pro ducts of unequa l terms from the assumed quan tities a, b, c, d etc. in the following way . Namely , fir s t the single quantities are summed, then the pro ducts of tw o unequa l terms, from three unequal terms, fro m four unequal terms, and so on; and let us indicate these ∗ Presen ted to the St. Pe tersburg Academ y on April 6, 1741. Or iginally published as Observationes analytic ae variae de c ombinationibus , Commen tarii academiae scien tiarum Pe tropoli tanae 13 (1751), 64–93. E158 in the Enestr¨ om index. T ranslated from the Latin by Jordan Bell, Departmen t of Mathematics, University of T oronto , T oronto, Canada. Email: jordan.b ell@utoron to.ca 1 series with the Gr eek letters α, β , γ , δ etc., so that it follows: α = a + b + c + d + etc. , β = ab + ac + ad + a e + b d + etc. , γ = abc + ab d + abe + bcd + etc. , δ = abcd + abce + bcde + etc. , ǫ = abcde + etc. etc. If the n umber of a s sumed quantities a, b, c, d etc. were infinite, then these series would not only all extend infinitely , but also, the num b er of forms of these s eries would b e infinite. But if instea d the num ber of the quantities a , b, c, d etc. is finite, put = n , then the ser ies α will contain n terms , the second series β will be comprised fro m n ( n − 1) 1 · 2 terms, the third γ from n ( n − 1)( n − 2) 1 · 2 · 3 terms, the fourth δ from n ( n − 1)( n − 2)( n − 3) 1 · 2 · 3 · 4 terms, a nd so o n, until finally a series is reached that is comprised from a single term, and then all the s ubsequent ser ies would v anish, as they could not hav e a ny terms. It is als o clear that the num b er of series that o c c ur here is = n , the last of which consists of a sing le term, which is the pro duct of all the assumed quantities a, b, c, d, e etc. 4. Since here we only to ok pro ducts from unequal qua nt ities a nd for med the series tr e ated ab ov e from them, by th us rep eating the same quantities in pro ducts, ne w series of pro ducts from one, t wo, three, four etc. will be obtained, in which equal factors are not excluded as before; so the series will be o btained th us: A = a + b + c + d + e + etc. , B = a 2 + a b + b 2 + a c + bc + c 2 + e tc. , C = a 3 + a 2 b + ab 2 + b 3 + a 2 c + abc + etc. , D = a 4 + a 3 b + a 2 b 2 + a 2 bc + abcd + etc. , E = a 5 + a 4 b + a 3 b 2 + a 3 bc + a 2 bcd + etc. etc.; in other words, these series co n tain all the qua nt ities which can b e pr o duced from the multiplication of the given quantities a, b, c, d etc. Again we should note that if the num ber of q uantit ies a, b, c, d etc. w ere finite, = n , then the first series A m ust hav e n terms; while the seco nd B would ha ve n ( n +1) 1 · 2 terms, the third C would ha ve n ( n +1)( n +2) 1 · 2 · 3 terms, the fourth D indee d n ( n +1)( n +2)( n +3) 1 · 2 · 3 · 4 terms, and s o on. 5. The three or ders of series, whic h w e have co mpo s ed fro m the given quan- tities a, b, c, d etc. in three w ays, are connected to each other, such that with the order of a n y one of the ser ies known, then from it the o rders of the r e ma ining t wo can b e determined. F or the details of how this law w orks a nd a metho d for 2 inv estiga ting this, observ a tion and induction ar e t ypically a pplied for the most part; first no ticing of course that A = α = A , and for the remaining it has been chec ked that: α = A, β = αA − B 2 , γ = β A − αB + C 3 , δ = γ A − β B + αC − D 4 , ǫ = δ A − γ B + β C − αD + E 5 etc. similarly A = A, B = A A + B 2 , C = B A + A B + C 3 , D = C A + B B + A C + D 4 , E = D A + C B + B C + A D + E 5 etc. and also A = α, B = α A − β , C = α B − β A + γ , D = α C − β B + γ A − δ, E = α D − β C + γ B − δ A + ǫ etc. By means o f these rela tions, g iven the the sums of the serie s in any one class, the sums of the series contained in the other tw o classes will b e able to b e defined. 6. By car efully studying the nature and qualities o f these series, the truth of this mutual rela tion will indeed easily b e clear b y obser v ation and induction. T r uly though, how ever m uch we are convinced by the truth of this connection, it will be fruitful to c o nsider the entire pr oblem in the following way; whence at once other prop erties a r e o ffered to us in addition which induction alone do es not easily pres en t a path to. Namely , assuming as given the quantities a, b, c, d, e etc. 3 from whic h are for med the three classes of ser ies deta iled above, let us consider this expres sion P = az 1 − az + bz 1 − bz + cz 1 − cz + dz 1 − dz + ez 1 − ez + e tc. , and with all the terms r e s olved in to geo metric progre s sions in the us ual way this gives P = + z ( a + b + c + d + e + etc.) + z 2 ( a 2 + b 2 + c 2 + d 2 + e 2 + etc.) + z 3 ( a 3 + b 3 + c 3 + d 3 + e 3 + etc.) + z 4 ( a 4 + b 4 + c 4 + d 4 + e 4 + etc.) etc.; all these series ar e con tained in the first class. Then if the sums g iven ab ov e ( § 2) are written in place of these, it will be P = Az + Bz 2 + C z 3 + D z 4 + E z 5 + e tc. , and thus the sum of this series will b e, as w e ass umed, P = az 1 − az + bz 1 − bz + cz 1 − cz + dz 1 − dz + e tc. Also in a similar wa y , if Q = az 1 + az + bz 1 + bz + cz 1 + cz + dz 1 + dz + e tc. , by ser ies of the firs t class it will be Q = Az − B z 2 + C z 3 − D z 4 + E z 5 − e tc. 7. Let us next consider this expres sion R = (1 + az )(1 + bz )(1 + cz )(1 + dz )(1 + ez ) etc.; if the factors are actually m ultiplied in to each other and the terms ar e disp osed according to the expo nent s of z , the co efficient o f z will be equal to the sum of the given q uantit ies a , b , c, d etc. The co efficient of z 2 will b e the aggrega te of all the pro ducts of t wo unequal terms, the co efficient of z 3 will b e the aggrega te of all pro ducts of three unequal terms, and so on; fro m this it follows that R = 1 + αz + β z 2 + γ z 3 + δ z 4 + ǫ z 5 + e tc. according to the definitions given ab ove ( § 3). 4 While if we put S = (1 − az )(1 − bz )(1 − cz )(1 − dz )(1 − ez ) etc. , it will b e just by making z nega tive S = 1 − αz + β z 2 − γ z 3 + δ z 4 − ǫ z 5 + e tc. 8. In order to compare these series R and S with the preceding P a nd S , it should b e noted that l R = l (1 + az ) + l (1 + bz ) + l (1 + cz ) + l (1 + dz ) + etc. , which by taking the different ials will b e dR Rdz = a 1 + az + b 1 + bz + c 1 + cz + d 1 + dz + e tc. , which multiplied by z gives the same previous expre s sion which a b ove we ca lle d Q , so that it will b e Q = z dR Rdz . Also in a similar wa y it will b e dS S dz = − a 1 − az − b 1 − bz − c 1 − cz − e tc. , from which one obtains P = − z dS S dz . 9. Now, sinc e R = 1 + αz + β z 2 + γ z 3 + e tc. , it will b e z dR dz = αz + 2 β z 2 + 3 γ z 3 + 4 δ z 4 + 5 ǫz 5 + e tc. and hence Q = Az − B z 2 + C z 3 − D z 4 + E z 5 − e tc. = αz + 2 β z 2 + 3 γ z 3 + 4 δ z 4 + 5 ǫz 5 + e tc. 1 + αz + β z 2 + γ z 3 + δ z 4 + ǫ z 5 + e tc. . Then, fro m the equality of these expres sions the following relations b et ween the letters A, B , C , D etc. and α, β , γ , δ, ǫ etc. o ccur: A = α, αA − B = 2 β , β A − αB + C = 3 γ , γ A − β B − αC − D = 4 δ, δ A − γ B + β C − αD + E = 5 ǫ etc. 5 Indeed in a similar wa y , from the other equa tion P = − z dS S dz it follows that P = Az + Bz 2 + C z 3 + D z 4 + E z 5 + e tc. = αz − 2 β z 2 + 3 γ z 3 − 4 δ z 4 + 5 δ z 5 − e tc. 1 − αz + β z 2 − γ z 3 + δ z 4 − ǫ z 5 + e tc. , which even yields the same determinations which w e gave ab ov e ( § 5). 10. A lso, b y integrating the equation Q = z dR Rdz it follows that R Qdz z = l R . Indeed b ecause Q = Az − B z 2 + C z 3 − D z 4 + etc., it will b e Z Qdz z = Az − B z 2 2 + C z 3 3 − D z 4 4 + e tc. , and the v alue of this ser ie s th us expresses the lo garithm of this series R = 1 + αz + β z 2 + γ z 3 + δ z 4 + e tc. Therefore b ecaus e l (1 + αz + β z 2 + γ z 3 + e tc.) = Az − 1 2 B z 2 + 1 3 C z 3 − 1 4 D z 4 + e tc. , then from the e quation R P dz z = − l S it will b e l (1 − αz + β z 2 − γ z 3 + e tc.) = − Az − 1 2 B z 2 − 1 3 C z 3 − 1 4 D z 4 − e tc. So if k is written as the num ber who se h yp e rb olic logarithm is = 1, we will hav e 1 + αz + β z 2 + γ z 3 + δ z 4 + e tc. = k Az − 1 2 B z 2 + 1 3 C z 3 − 1 4 Dz 4 +etc. and 1 − αz + β z 2 − γ z 3 + δ z 4 − e tc. = k − Az − 1 2 B z 2 − 1 3 C z 3 − 1 4 Dz 4 − etc. . 11. Also notew orthy are the recipro cals expressions of R and S , o f co urse 1 R and 1 S . Indeed it is 1 S = 1 (1 − az )(1 − bz )(1 − cz )(1 − dz ) etc. ; to express the v alue o f this fra ction by a series whose terms pr o ceed according to powers of z , it is clear that all these g eometric pro gressio ns should be multiplied int o each other 1 1 − az = 1 + az + a 2 z 2 + a 3 z 3 + a 4 z 4 + e tc. , 1 1 − bz = 1 + bz + b 2 z 2 + b 3 z 3 + b 4 z 4 + e tc. , 1 1 − cz = 1 + cz + c 2 z 2 + c 3 z 3 + c 4 z 4 + e tc. , 1 1 − dz = 1 + dz + d 2 z 2 + d 3 z 3 + d 4 z 4 + e tc. etc. 6 In the pro duct, after the first term 1, the co efficient o f z will b e the sum of the quantities a + b + c + d + etc. , the co efficient of z 2 will be the sum of the factors from t wo, not excluding equa l factors in the same pro duct, the co efficient of z 3 will b e the sum o f the fac to rs from tw o, and so on. W e desig na ted these sums of pro ducts above ( § 4) w ith the letters of the Germanic alphab et A , B , C , D , E etc. With these letters thus introduced, we will hav e 1 S = 1 + A z + B z 2 + C z 3 + D z 4 + E z 5 + e tc. and by tr eating the v alue of R in a similar wa y it will be 1 R = 1 − A z + B z 2 − C z 3 + D z 4 − E z 5 + e tc. 12. S o these s e ries ar e the recipro cals of thos e which we defined under the letters R and S above ( § 7). And beca use of this it will b e 1 = (1 + αz + β z 2 + γ z 3 + δ z 4 + e tc.)(1 − A z + B z 2 − C z 3 + D z 4 + e tc.) and even 1 = (1 − αz + β z 2 − γ z 3 + δ z 4 − e tc.)(1 + A z + B z 2 + C z 3 + D z 4 + e tc.) . F r om either one of these follows the sa me re lation b e tween the v alues of the letters A , B , C , D etc. a nd α, β , γ , δ etc.; namely it will b e A − α = 0 , B − α A + β = 0 , C − α B + β A − γ = 0 , D − α C + β B − γ A + δ = 0 etc. , which is the same r elation w e alr e ady gav e ab ov e ( § 5). 13. But if we put 1 R = T and 1 S = V , so that T = 1 − A z + B z 2 − C z 3 + D z 4 − e tc. and V = 1 + A z + B z 2 + C z 3 + D z 4 + e tc. it will b e dR R = − dT T and dS S = − dV V and therefore it will b ecome P = z dV V d z and Q = − z dT T dz . 7 Now sinc e z dV dz = A z + 2 B z 2 + 3 C z 3 + 4 D z 4 + e tc. and − z dT dz = A z − 2 B z 2 + 3 C z 3 − 4 D z 4 + e tc. , by writing the appropria te v alues from § 6 in place of P and Q w e will hav e these equations Az + B z 2 + C z 3 + D z 4 + e tc. = A z + 2 B z 2 + 3 C z 3 + 4 D z 4 + e tc. 1 + A z + B z 2 + C z 3 + D z 4 + e tc. and Az − B z 2 + C z 3 − D z 4 + e tc. = A z − 2 B z 2 + 3 C z 3 − 4 D z 4 + e tc. 1 − A z + B z 2 − C z 3 + D z 4 − e tc. , from which the same r elation b et ween the letters A, B , C , D etc. and A , B , C , D etc. follows as we gav e ab ove ( § 5). Namely it will b e A = A, 2 B = A A + B , 3 C = B A + A B + C, 4 D = C A + B B + A C + D , 5 E = D A + C B + B C + A D + E etc. 14. F rom the equations given in § 12 it follows that l (1 + αz + β z 2 + γ z 3 + e tc.) = − l (1 − A z + B z 2 − C z 3 + e tc.) and l (1 − αz + β z 2 − γ z 3 + e tc.) = − l (1 + A z + B z 2 + C z 3 + e tc.) . Then by applying these to § 1 0 it will b e l (1 − A z + B z 2 − C z 3 + e tc.) = − Az + 1 2 B z 2 − 1 3 C z 3 + 1 4 D z 4 − e tc. and l (1 + A z + B z 2 + C z 3 + e tc.) = Az + 1 2 B z 2 + 1 3 C z 3 + 1 4 D z 4 + e tc. And hence by taking k as the num b er who s e logarithm is = 1, it will b e 1 − A z + B z 2 − C z 3 + e tc. = k − Az + 1 2 B z 2 − 1 3 C z 3 + 1 4 Dz 4 − etc. 8 and 1 + A z + B z 2 + C z 3 + e tc. = k Az + 1 2 B z 2 + 1 3 C z 3 + 1 4 Dz 4 +etc. . 15. Now if the letters R a nd S retain the v alues assumed ab ov e ( § 7), it will be 1 + αz + β z 2 + γ z 3 + δ z 4 + e tc. = R , 1 − A z + B z 2 − C z 3 + D z 4 − e tc. = 1 R and 1 − αz + β z 2 − γ z 3 + δ z 4 − e tc. = S, 1 + A z + B z 2 + C z 3 + D z 4 + e tc. = 1 S . F r om these the following conclus io ns are deduced 1 + β z 2 + δ z 4 + ζ z 6 + θ z 8 + e tc. = R + S 2 , αz + γ z 3 + ǫ z 5 + η z 7 + ιz 9 + e tc. = R − S 2 , 1 + B z 2 + D z 4 + F z 6 + H z 8 + e tc. = R + S 2 RS , A z + C z 3 + E z 5 + G z 7 + I z 9 + e tc. = R − S 2 RS and hence this prop ortion is obtained 1 + β z 2 + δ z 4 + ζ z 6 + e tc. : αz + γ z 3 + ǫ z 5 + η z 7 + e tc. =1 + B z 2 + D z 4 + F z 6 + e tc. : A z + C z 3 + E z 5 + G z 7 + e tc. Since it is also R − 1 = αz + β z 2 + γ z 3 + δ z 4 + e tc. , 1 − 1 R = A z − B z 2 + C z 3 − D z 4 + e tc. , it will b e R = αz + β z 2 + γ z 3 + δ z 4 + e tc. A z − B z 2 + C z 3 − D z 4 + e tc. , and in a simila r wa y , beca use 1 − S = αz − β z 2 + γ z 3 − δ z 4 + e tc. , 1 S − 1 = A z + B z 2 + C z 3 + D z 4 + e tc. it will b e S = αz − β z 2 + γ z 3 − δ z 4 + e tc. A z + B z 2 + C z 3 + D z 4 + e tc. . 9 16. Next indeed if a s ab ov e ( § 6) we put P = Az + B z 2 + C z 3 + D z 4 + e tc. , Q = Az − B z 2 + C z 3 − D z 4 + e tc. , it will b e from parag raph 9 αz + 2 β z 2 + 3 γ z 3 + 4 δ z 4 + e tc. = QR, αz − 2 β z 2 + 3 γ z 3 − 4 δ z 4 + e tc. = P S and in a simila r wa y from paragr aph 1 3 we will have 1 A z + 2 B z 2 + 3 C z 3 + 4 D z 4 + e tc. = P S , A z − 2 B z 2 + 3 C z 3 − 4 D z 4 + e tc. = Q R . F r om these the following cor ollaries are easily derived: αz − 2 β z 2 + 3 γ z 3 − 4 δ z 4 + e tc. Az + B z 2 + C z 3 + D z 4 + e tc. = S = Az + B z 2 + C z 3 + D z 4 + e tc. A z + 2 B z 2 + 3 C z 3 + 4 D z 4 + e tc. , αz + 2 β z 2 + 3 γ z 3 + 4 δ z 4 + e tc. Az − B z 2 + C z 3 − D z 4 + e tc. = R = Az − B z 2 + C z 3 − D z 4 + e tc. A z − 2 B z 2 + 3 C z 3 − 4 D z 4 + e tc. , F o r the letters R and S we have thes e quint uple v alues R = 1 + αz + β z 2 + γ z 3 + δ z 4 + e tc. , R = 1 1 − A z + B z 2 − C z 3 + D z 4 − e tc. , R = αz + β z 2 + γ z 3 + δ z 4 + e tc. A z − B z 2 + C z 3 − D z 4 + e tc. , R = αz + 2 β z 2 + 3 γ z 3 + 4 δ z 4 + e tc. Az − B z 2 + C z 3 − 4 z 4 + e tc. , R = Az − B z 2 + C z 3 − D z 4 + e tc. A z − 2 B z 2 + 3 C z 3 − 4 D z 4 + e tc. , in which putting − z in place of z everywhere yields the v alues for S . And from v aried combinations of these five v alues, a great num b er of prop erties can be elic ited which the three orders of o ur letters A, B , C, D etc., α, β , γ , δ etc., A , B , C , D etc. hold b etw een ea ch other, which how ever we will r efrain fro m pursuing now. 17. F ro m this it is quite clear that that we can desc end from what has bee n explained alr e a dy to more particula r cases, and first indeed this infinite geometric progr ession will b e taken for the series o f letters a, b, c, d etc. n, n 2 , n 3 , n 4 , n 5 , n 6 etc.; 1 T ranslator: The Op er a omnia , Seri es I, V olume 2, p. 168 and the origi nal ve rsion i n the Commentarii , V olume 13, p. 74 both ha ve +4 D z 4 on the second line, whic h is incorrect. 10 with these success ively intro duced int o the ab ove formulae w e will have: A = n + n 2 + n 3 + n 4 + n 5 + e tc. = n 1 − n , B = n 2 + n 4 + n 6 + n 8 + n 10 + e tc. = nn 1 − nn , C = n 3 + n 6 + n 9 + n 12 + n 15 + e tc. = n 3 1 − n 3 , D = n 4 + n 8 + n 12 + n 16 + n 20 + e tc. = n 4 1 − n 4 etc. Now fro m § 6, we w ill get tw o v alues for the letters P and Q , which will b e P = nz 1 − nz + n 2 z 1 − n 2 z + n 3 z 1 − n 3 z + n 4 z 1 − n 4 z + e tc. , Q = nz 1 + nz + n 2 z 1 + n 2 z + n 3 z 1 + n 3 z + n 4 z 1 + n 4 z + e tc. and then fr om the v a lues found for the letters A, B , C, D etc., these other v alues will aris e P = nz 1 − n + n 2 z 2 1 − nn + n 3 z 3 1 − n 3 + n 4 z 4 1 − n 4 + e tc. , Q = nz 1 − n − n 2 z 2 1 − nn + n 3 z 3 1 − n 3 − n 4 z 4 1 − n 4 + e tc. 18. Next, fr om para g raph 7 we will ha ve the following expressio ns for R and S R = (1 + nz )(1 + n 2 z )(1 + n 3 z )(1 + n 4 z ) etc. , S = (1 − nz )(1 − n 2 z )(1 − n 3 z )(1 − n 4 z ) etc. These fa ctors ac tually multiplied into each o ther a nd ar ranged acco rding to dimensions of z yield these ser ies for R and S R = 1 + αz + β z 2 + γ z 3 + δ z 4 + e tc. , S = 1 − αz + β z 2 − γ z 3 + δ z 4 − e tc. , where the letters α, β , γ , δ etc. are th us determined fr om the supp osed series n, n 2 , n 3 , n 4 , n 5 , n 6 , n 7 etc., so that it will b e: I. α = the sum of all the terms; whence it will b e α = n + n 2 + n 3 + n 4 + n 5 + n 6 + n 7 + e tc. , which is supp osed to b e a geo metric progres s ion, in whic h each pow er of n o ccurs and has the co e fficie nt +1. 11 II. β = the sum of the factors from tw o terms; whence it will b e β = n 3 + n 4 + 2 n 5 + 2 n 6 + 3 n 7 + 3 n 8 + 4 n 9 + 4 n 10 + e tc. , in which series after the third p ow er all the fo llowing p ow er s of n o ccur ; more- ov er , each p ower o ccurs as often as it ca n b e made b y m ultiplying t w o ter ms of the series α . Since ho wever the multiplication of p ow ers consists in the addition of exp onents, the co efficient of any pow er of n will app ear in the series β in as many wa ys as the exp o nent of n can be distributed into tw o unequal parts, or as many w ays as this exp onent n can b e pro duced fro m the addition of t wo unequal int egral num b ers. Th us the coefficient of the tenth p ow er n 10 is 4, because 10 can b e distributed in to tw o unequal pa rts in four ways, namely , 10 = 1 + 9 , 10 = 3 + 7 , 10 = 2 + 8 , 10 = 4 + 6 . II I. γ = the sum o f the fa ctors fro m thre e unequal terms of the s eries α ; whence it will b e γ = n 6 + n 7 + 2 n 8 + 3 n 9 + 4 n 10 + 5 n 11 + 7 n 12 + 8 n 13 + e tc. , in whic h after the sixth p ower, all the following p ow er s of n oc c ur. Moreov er the co efficient of ea ch p ower indicates how many wa ys the expo nent can b e distributed in to three unequa l par ts, or as often as the same exponent ca n b e pro duced fro m the addition of thr ee m utually unequal integral num b ers. Thus the pow e r n 12 has the co efficient 7 , b ecause the ex po nen t 12 can be partitioned int o three unequal parts in s e ven wa y s , as 12 = 1 + 2 + 9 , 12 = 1 + 5 + 6 , 12 = 1 + 3 + 8 , 12 = 2 + 3 + 7 , 12 = 1 + 4 + 7 , 12 = 2 + 4 + 6 , 12 = 3 + 4 + 5 . IV. δ = the sum of the factors fro m four mutually unequal terms of the ser ie s α ; whence it will b e δ = n 10 + n 11 + 2 n 12 + 3 n 13 + 5 n 14 + 6 n 15 + 9 n 16 + e tc. , whose first power is n 10 , who se exp onent is clea rly 1 + 2 + 3 + 4, or the fourth trigonal n um ber . Each of the following p ow er s app ears as often as its exp onent can b e made from the addition of four mutually unequal integral num b ers. Thus the sixteenth p ower n 16 has the co efficient 9, b ecause 16 can be distributed into four mutually unequal parts in nine w ays. Thes e nine partitions a re 16 = 1 + 2 + 3 + 10 , 16 = 1 + 3 + 4 + 8 , 16 = 1 + 2 + 4 + 9 , 16 = 1 + 3 + 5 + 7 , 16 = 1 + 2 + 5 + 8 , 16 = 1 + 4 + 5 + 6 , 16 = 1 + 2 + 6 + 7 , 16 = 2 + 3 + 4 + 7 , 12 16 = 2 + 3 + 5 + 6 . The sa me kind of thing happens for the v a lues of the following letters ǫ, ζ , η etc., which will be ǫ = n 15 + n 16 + 2 n 17 + 3 n 18 + 5 n 19 + 7 n 20 + 1 0 n 21 + e tc. , ζ = n 21 + n 22 + 2 n 23 + 3 n 24 + 5 n 25 + 7 n 26 + 1 1 n 27 + e tc. , η = n 28 + n 29 + 2 n 30 + 3 n 31 + 5 n 32 + 7 n 33 + 1 1 n 34 + e tc. , etc. In all these series, the co efficient of each power of n indicates how many differen t wa ys the exp onent of n can b e r e solved into as many unequal parts as the series is num ber ed fro m the first. In other words, the coefficient of any term indicates how many wa ys the e x po nen t of n can b e ma de from the addition of a s many m utually unequal integral num b ers as the p os itio n of the series from which the term is taken is num ber ed starting a t α . Thus in the seven th series the co efficient of the power n 34 is 11, b e cause the num b er 34 can b e distributed into seven unequal parts in seven w ays; these distributions are 34 = 1 + 2 + 3 + 4 + 5 + 6 + 1 3 , 34 = 1 + 2 + 3 + 4 + 5 + 7 + 1 2 , 34 = 1 + 2 + 3 + 4 + 5 + 8 + 1 1 , 34 = 1 + 2 + 3 + 4 + 5 + 9 + 1 0 , 34 = 1 + 2 + 3 + 4 + 6 + 7 + 1 1 , 34 = 1 + 2 + 3 + 4 + 6 + 8 + 1 0 , 34 = 1 + 2 + 3 + 4 + 7 + 8 + 9 , 34 = 1 + 2 + 3 + 5 + 6 + 7 + 1 0 , 34 = 1 + 2 + 3 + 5 + 6 + 8 + 9 , 34 = 1 + 2 + 4 + 5 + 6 + 7 + 9 , 34 = 1 + 3 + 4 + 5 + 6 + 7 + 8 . And indeed fr om thes e , the na tur e of the series which a ppea r for the letters α, β , γ , δ etc. is easily seen. 19. Therefore for in vestigating ho w many different w ays a n umber c a n be distributed in to a g iven num b er of unequal parts, the series expressing the letters α, β , γ , δ etc. ca n b e formed, a lthough this work w ill b e ex ceedingly tiresome. In turn, howev er, assuming these se r ies as known and already formed, a not in- elegant problem can b e s o lved, which w as thus prop osed to me by the Insigh tful Naud´ e: T o define ho w m any ways a given numb er c an b e pr o duc e d fr om the addition of sever al inte gr al numb ers, mutual ly une qual, the nu mb er of which is given. The most Insightful Pr op oser sear ched thus for how many different ways the num b er 50 ca n arise from the addition of seven unequal integral num b ers . 13 F o r r esolving this question it is clear that the appro priate series to co nsider is η , in which the co efficient o f any ter m indica tes how ma n y differen t w ays the exp onent o f n can b e re s olved into 7 unequal par ts. Hence the series η = n 28 + n 29 + 2 n 30 + 3 n 31 + 5 n 32 + 7 n 33 + 1 1 n 34 + e tc. should b e cont inu ed to the term a t which the fiftieth power of n is con ta ined, whose co efficient, which will b e 52 2 , shows that the num ber 50 can be produced in a ltogether 522 different wa ys fro m the a ddition of seven mutu ally unequa l int egral n um ber s. Thus it is clea r that if a conv enient and simple wa y were o b- tained of for ming these series α, β , γ , δ etc., Naud´ e’s problem w ould be brought to a most p er fect solution. 20. Thus since a w ay w as given ab ov e ( § 5 and 9) for finding the v alues o f the letters α, β , γ , δ etc. from the known v a lues of the letters A, B , C , D etc., for the present question we can easily o btain a solution, since from § 17 we hav e the known v alues A, B , C, D etc.; and thus it follows that α = A, β = αA − B 2 , γ = β A − αB + C 3 , δ = γ A − β B + αC − D 4 , ǫ = δ A − γ B + β C − αD + E 5 etc. W e therefore obtain fr om these α = n 1 − n , 2 β = αn 1 − n − nn 1 − nn , 3 γ = β n 1 − n − αn 2 1 − n 2 + n 3 1 − n 3 , 4 δ = γ n 1 − n − β n 2 1 − n 2 + αn 3 1 − n 3 − n 4 1 − n 4 etc. Now if ho wev er in place of α, β , γ etc. the previo us v alue s are succe ssively 14 substituted, it follows α = n 1 − n , β = n 3 (1 − n )(1 − nn ) , γ = n 6 (1 − n )(1 − nn )(1 − n 3 ) , δ = n 10 (1 − n )(1 − n 2 )(1 − n 3 )(1 − n 4 ) , ǫ = n 15 (1 − n )(1 − n 2 )(1 − n 3 )(1 − n 4 )(1 − n 5 ) etc. Thu s we c an understand tha t, in this cas e, α = A, β = AB , γ = AB C, δ = AB C D , ǫ = AB C D E etc. 21. The law according to which the v alues of the le tter s α, β , γ , δ etc. have bee n found, whose tr uth is observed b y ex pa nding several formulae, is so far not apparent except by induction. So its truth ma y thus b e secur ely confirmed, it will be conv enient to elicit the sa me law of the progressio n in a totally different wa y , in which induction plays no part. Now, let it b e pr op osed to inv estiga te the v alues o f the letters α, β , γ , δ etc. which occur in the serie s R = 1 + αz + β z 2 + γ z 3 + δ z 4 + ǫ z 5 + e tc. If as we had assumed initially R = (1 + nz )(1 + n 2 z )(1 + n 3 z )(1 + n 4 z ) · · · , it should be noted that if nz is written in place of z , the express ion whic h just now R was equal to is changed in to the for m (1 + n 2 z )(1 + n 3 z )(1 + n 4 z )(1 + n 5 z ) · · · , which m ultiplied by 1 + nz pro duces the prior expr e s sion. Therefore we can rightly conc lude that if in the ser ies 1 + αz + β z 2 + γ z 3 + δ z 4 + ǫ z 5 + e tc. 15 we wr ite nz in place of z , so that we have 1 + αnz + β n 2 z 2 + γ n 3 z 3 + δ n 4 z 4 + ǫ n 5 z 5 + e tc. , and we then multiply this expr ession by 1 + nz , then the product, whic h will be 1 + αnz + β n 2 z 2 + γ n 3 z 3 + δ n 4 z 4 + ǫn 5 z 5 +etc. + nz + αn 2 z 2 + β n 3 z 3 + γ n 4 z 4 + δ n 5 z 5 +etc. , should b e equal to that of the prior series 1 + αz + β z 2 + γ z 3 + δ z 4 + ǫ z 5 + e tc. But if we equate the co efficients of the c orresp onding terms, we will obtain the following deter mina tions for α, β , γ etc. α = n 1 − n = n 1 − n , β = αn 2 1 − n 2 = n 3 (1 − n )(1 − n 2 ) , γ = β n 3 1 − n 3 = n 6 (1 − n )(1 − n 2 )(1 − n 3 ) , δ = γ n 4 1 − n 4 = n 10 (1 − n )(1 − n 2 )(1 − n 3 )(1 − n 4 ) etc. 22. In this wa y w e have th us found con venient enough ex pressions for the sums o f these series α, β , γ , δ e tc., from which in tur n these serie s can b e formed. Namely since this series pro ceed ac c o rding to p ow er s of n , this will arise if the expressions of these sums are expanded b y divisio n in the usual w ay in to infinite series pro cee ding according to p ow er s of n . Let this div is ion be do ne, and it is clear that a ll the series α, β , γ , δ etc. b elong to a type w hich is commonly referred to by the name of recurre n t ser ies; they are such that a ny term can b e determined fro m several of the pr e c eding. So it will be clear how in each o f these series a n y term is formed fro m the preceding, let us expand the denominators of these expres s ions found for the letters α, β , γ , δ etc. by actual m ultiplication, 16 and having done this we will hav e α = n 1 − n , β = n 3 1 − n − n 2 + n 3 , γ = n 6 1 − n − n 2 + n 4 + n 5 − n 6 , δ = n 10 1 − n − n 2 + 2 n 5 − n 8 − n 9 + n 10 , ǫ = n 15 1 − n − n 2 + n 5 + n 6 + n 7 − n 8 − n 9 − n 10 + n 13 + n 14 − n 15 , ζ = n 21 1 − n − n 2 + n 5 + 2 n 7 − n 9 − n 10 − n 11 − n 12 + 2 n 14 + n 16 − n 19 − n 20 + n 21 etc. And from these denominators we unders tand how in each series any term is comp osed from the preceding, if the precept by which these recur rent s e ries ar e formed is called upo n. 23. And from the form of the expressio ns found for the letters α, β , γ , δ etc., by whic h each is the pro duct of the preceding and some new factor, a nother metho d is deduced which is suitable enoug h for finding the next series from any series whic h ha s a lready b een found. Thus, since the series α = n 1 − n is a geometric progr ession α = n + n 2 + n 3 + n 4 + n 5 + n 6 + n 7 + e tc. , from which comes the series β , if it is mult iplied b y n 2 1 − n 2 or if it is multiplied by this geometric progr ession n 2 + n 4 + n 6 + n 8 + n 10 + n 12 + n 14 + e tc. Next, with the series β ha ving b een solidly fo und, if it is multiplied b y n 3 1 − n 3 = n 3 + n 6 + n 9 + n 12 + n 15 + n 18 + e tc. , series γ is pro duced. And this multiplied by n 4 1 − n 4 = n 4 + n 8 + n 12 + n 16 + n 20 + n 24 + e tc. yields the series δ . And b y s o o n multiplying each series of this order by a certain geometr ic progression the following series comes out. In this manner, these series can b e contin ued a s far as we wan t; a nd thus the ab ov e problem prop osed by the Insightful Naud´ e is r esolved. 17 24. And each of the ser ies will be able to b e easily co ntin ued by means of the prece ding, if we consider the w ay in which the v alue of each of the letters α, β , γ , δ etc. is determined from the prece ding. Thus, since β = αn 2 1 − n 2 , it will be β = β nn + αnn ; therefor e if to the series β multiplied b y nn is added the series α m ultiplied by nn , the ser ies β will b e created. Then, since it is clear that the fir st term of the s eries β is n 3 , let us put β = a n 3 + b n 4 + c n 5 + d n 6 + e n 7 + f n 8 + g n 9 + e tc. and it will b e β n 2 = + a n 5 + b n 6 + c n 7 + d n 8 + e n 9 +etc. , αn 2 = n 3 + n 4 + n 5 + n 6 + n 7 + n 8 + n 9 +etc. Now with the terms equa ted, b ecause β = β nn + αnn we will hav e a = 1 , e = c + 1 = 3 , b = 1 , f = d + 1 = 3 , c = a + 1 = 2 , g = e + 1 = 4 , d = b + 1 = 2 , h = f + 1 = 4 etc. In a similar way , since γ = β n 3 1 − n 3 or γ = γ n 3 + β n 3 , the series γ will b e formed from the series β , and in turn the series δ is pr o duced from the ser ies γ , by means of the e quation δ = δ n 4 + γ n 4 ; and all the following will be dispatched likewise. 25. Bec ause in the expression R = 1 + αz + β z 2 + γ z 3 + δ z 4 + e tc. we have found the v alues o f the letters α, β , γ , δ etc. a nd it is R = (1 + nz )(1 + n 2 z )(1 + n 3 z )(1 + n 4 z ) . . . , this pro duct, appar ently (1 + nz )(1 + n 2 z )(1 + n 3 z )(1 + n 4 z ) . . . , will b e conv er ted from the infinite factors into this ser ies pro ceeding accor ding to p owers o f z 1+ nz 1 − n + n 3 z 3 (1 − n )(1 − n 2 ) + n 6 z 3 (1 − n )(1 − n 2 )(1 − n 3 ) + n 10 z 4 (1 − n )(1 − n 2 )(1 − n 3 )(1 − n 4 ) +etc. And from § 10 the hyperb olic log arithm of the sum of this series will be = nz 1 − n − nnz 2 2(1 − n 2 ) + n 3 z 3 3(1 − n 3 ) − n 4 z 4 4(1 − n 4 ) + e tc. 18 Or if k is written for the n um be r whose lo garithm = 1, it will be k nz 1 − n − n 2 z 2 2(1 − n 2 ) + n 3 z 3 3(1 − n 3 ) − n 4 z 4 4(1 − n 4 ) +etc. = R or this exp onential expr e ssion is equa l to the sum of se r ies whic h we ma de changed into the v alue o f R . 26. T ruly to turn the prop o sed pr oblem, whic h is to define how ma n y dif- ferent ways a given num ber m can b e partitioned into µ integral parts tha t are unequal to each other, let us indicate this n umber of wa ys which w e a r e seeking by the notation m ( µ ) i , which from no w on will indicate to us the num b er of w ays in which the num ber m can b e pro duced from the addition of µ mutually unequal integral num b ers; and for denoting the inequality of these parts we hav e adjoined the letter i ab ov e, which will be omitted if the question tak e s the form of finding altogether the num b er o f wa ys in which the g iven num b er m can b e distr ibuted into µ parts, e ither eq ual or unequa l. After this, a solution to the problem ca n e a sily be shown 27. Thus this num b er of wa ys m ( µ ) i will b e the co efficien t of the p ow er n m in the se r ies α, β , γ , δ, ǫ etc., which first at α are num ber ed as far as µ co n tains unities. The sum of this series is = n µ ( µ +1) 1 · 2 (1 − n )(1 − n 2 )(1 − n 3 )(1 − n 4 ) · · · (1 − n µ ) and hence the general term of the series whic h arises from this form is = m ( µ ) i n m . Mo r eov er , the general term o f the ser ies whic h arises from the form n µ ( µ − 1) 1 · 2 (1 − n )(1 − n 2 )(1 − n 3 )(1 − n 4 ) · · · (1 − n µ ) will b e = m ( µ ) i n m − µ or for the same p ow er of n the genera l term will be = ( m + µ ) ( µ ) i n m . The prior ex pression is subtracted from the latter, and the general term of the remaining expr ession n µ ( µ − 1) 1 · 2 (1 − n )(1 − n 2 )(1 − n 3 )(1 − n 4 ) · · · (1 − n µ − 1 ) will b e = n m (( m + µ ) ( µ ) i − m ( µ ) i ); moreover the general term of this series is m ( µ − 1) i n m , whence w e will hav e m ( µ − 1) i = ( m + µ ) ( µ ) i − m ( µ ) i , from which w e arrive at the rule that ( m + µ ) ( µ ) i = m ( µ ) i + m ( µ − 1) i , 19 by means of whic h, if the num b er of different wa y s the num b er m can be dis- tributed in to µ and µ − 1 unequal parts w er e known, by adding these t wo num- ber s would follow the n um ber of w ays in whic h the la rger num b er m + µ ca n be distributed into µ unequal parts. And thus the resolution of more difficult cases is reduced to simpler o nes, and with these known finally to the simpler ones; it is of co urse clear that if m < µµ + µ 2 then m ( µ ) i = 0, and if m = µµ + µ 2 then it will b e m ( µ ) i = 1. 28. Since the formula m ( µ ) i n m is the gener al term of the expression n µ ( µ +1) 2 (1 − n )(1 − n 2 )(1 − n 3 ) · · · (1 − n µ ) , let us see what kind of ser ies this expr ession 1 (1 − n )(1 − n 2 )(1 − n 3 ) · · · (1 − n µ ) , pro duces if expanded a nd arra nged a ccording to the dimensions of n . Let us put it to pro duce this ser ie s 1 + pn + q n 2 + r n 3 + s n 4 + t n 5 + e tc. , from whose g eneration it is clear that the coefficient o f a ny p ow er of n shows how many different wa ys the exp onent of n ca n b e pro duced by addition from the given num be r s 1 , 2 , 3 , 4 , 5 , 6 , . . . , µ ; and her e neither is a certa in num b er of parts pres crib ed fro m which it is for med, nor is the condition put that the must ar e unequal to each o ther. Therefo r e the expression m ( µ ) i will indicate altogether how man y wa y s the n um b e r m − µ ( µ +1) 2 can b e pro duce d by addition from the n umbers 1 , 2 , 3 , 4 , 5 , . . . , µ . Th us if one searches for how many different ways the num b er 50 can b e distributed into 7 unequal parts, b e c ause m = 50 and µ = 7 the q uestion is th us reduced to inv estiga ting ho w man y different wa ys the num b er 50 − 28 or 22 can arise b y addition from the seven n um be r s 1 , 2 , 3 , 4 , 5 , 6 , 7. With this under sto o d, b oth v arieties of this question can b e r esolved in a single effort. 29. With the let ters α, β , γ , δ etc. defined for the case where we have a ssumed the geometric progression n , n 2 , n 3 , n 4 , n 5 etc. in place of the letters a, b , c, d e tc., order requires that we also inquire into the v alues of the third order A , B , C , D , E , etc. But we hav e employ ed the letters A , B , C , D etc., with equa l v alues, in the series 1 R and 1 S ; for we assumed a b ove ( § 11) that 1 S = 1 + A z + B z 2 + C z 3 + D z 4 + E z 5 + e tc. and 1 R = 1 − A z + B z 2 − C z 3 + D z 4 − E z 5 + e tc. 20 where the or iginal v alues taken for R and S w er e R = (1 + nz )(1 + n 2 z )(1 + n 3 z )(1 + n 4 z ) etc. , S = (1 − nz )(1 − n 2 z )(1 − n 3 z )(1 − n 4 z ) etc. It is appa rent hence that the series 1 S = 1 + A z + B z 2 + C z 3 + D z 4 + etc. will arise if these inn umerable geometr ic progressio ns are multiplied b y each other 1 1 − nz = 1 + nz + n 2 z 2 + n 3 z 3 + n 4 z 4 + e tc. , 1 1 − n 2 z = 1 + n 2 z + n 4 z 2 + n 6 z 3 + n 8 z 4 + e tc. , 1 1 − n 3 z = 1 + n 3 z + n 6 z 2 + n 9 z 3 + n 12 z 4 + e tc. , 1 1 − n 4 z = 1 + n 4 z + n 8 z 2 + n 12 z 3 + n 16 z 4 + e tc. etc. On the other hand, b y putting − z in place of z the series 1 R follows in a similar wa y . 30. F rom the gener a tion of these se r ies it is clea r that: I. A = n + n 2 + n 3 + n 4 + n 5 + e tc. , which is a geo metric progre s sion where a ll p owers of n are multiplied by the co efficient + 1 . II. B = n 2 + n 3 + 2 n 4 + 2 n 5 + 3 n 6 + 3 n 7 + 4 n 8 + 4 n 9 + e tc. , in which the co efficient of e ach pow er of n contains as muc h unities as there are different wa ys in which the exponent of n can b e partitioned into tw o parts, either equal or unequal. Thus the co efficie nt of the pow er n 8 is 4, becaus e 8 c an be partitioned into 2 parts in fo ur wa ys 8 = 1 + 7 , 8 = 2 + 6 , 8 = 3 + 5 , 8 = 4 + 4 . II I. C = n 3 + n 4 + 2 n 5 + 3 n 6 + 4 n 7 + 5 n 8 + 7 n 9 + e tc. , in which the co efficient of eac h p ow er of n contains as many unities as there ar e different ways in which the exponent of n ca n be distributed in to three parts, either equa l or unequa l. Th us n 9 has the co efficient 7, b e cause 7 p ermits its separatio n into three parts in 9 ways: 9 = 1 + 1 + 7 , 9 = 1 + 4 + 4 , 9 = 1 + 2 + 6 , 9 = 2 + 2 + 5 , 9 = 1 + 3 + 5 , 9 = 2 + 3 + 4 , 21 9 = 3 + 3 + 3 . IV. D = n 4 + n 5 + 2 n 6 + 3 n 7 + 5 n 8 + 6 n 9 + 9 n 10 + e tc. , where the co efficient of an y power of n contains as many unities as there ar e different ways in which the expone nt of n can b e resolved into four par ts, either equal o r unequal. And there is a similar rule for the following ser ies which ar e found for the letters E , F , G etc. 31. The o ther proble m which the Insightful Naud´ e pr op osed to me alo ng with the preceding can thu s also b e resolved by means of these series. It can b e expressed thus: T o find how many differ ent way a given numb er m c an b e p artitione d into µ p art s , either e qual or une qual, or to find how many differ ent ways a given numb er m c an b e pr o duc e d by addition fr om µ inte gr al numb ers, either e qual or une qual. The difference b etw een this pro blem and the preceding is that in the pre - ceding, the partition was restr icted just to pa rts that w er e m utually unequal, while here equal pa rts a r e also allow ed. F or expres sing the n um be r of all these wa ys in this problem, the se arch will use this form m ( µ ) , which declares namely how many different wa ys the num b er m ca n b e parti- tioned into µ integral parts, with equality of some not excluded; for the letter i which w as previously affixed to the abov e sig n ( µ ), which indicated unequal parts, is omitted here. 32. The solutio n of this problem can b e reduced th us to the forma tion of the ser ies A , B , C , D , E etc.; we have alre a dy s hown ab ov e ( § 5) the v alues of these letters can b e defined from the a lready known v alues of the le tter s α, β , γ , δ etc. Thoug h this metho d is gene r al and attacks the very nature of the problem, it co uld how ever seem that the la w by which these v alues proceed is no t clear enoug h to the eyes. Therefore I shall inv es tigate the v alues of the letters A , B , C , D , E etc.in this case by a means simila r to what I used ab ov e ( § 21). Since it is 1 S = 1 (1 − nz )(1 − n 2 z )(1 − n 3 z )(1 − n 4 z ) etc. , it is clear that writing nz in place of z in this for m will pro duce this form 1 (1 − n 2 z )(1 − n 3 z )(1 − n 4 z )(1 − n 5 z ) etc. . 22 But the prio r form 1 S is turned into this if it is m ultiplied b y 1 − nz . Whence, since we ha ve assumed that 1 S = 1 + A z + B z 2 + C z 3 + D z 4 + E z 5 + e tc. , we put here nz in place of z and we will hav e 1 + A nz + B n 2 z 2 + C n 3 z 3 + D n 4 z 4 + e tc. Now let us mult iply the fir st ser ie s 1 S by 1 − nz 1 + A z + B z 2 + C z 3 + D z 4 +etc. − nz − A nz 2 − B nz 3 − C nz 4 − etc. Since this form should b e equal to the prev ious, it will b e A = n 1 − n = n 1 − n , B = A n 1 − n 2 = n 2 (1 − n )(1 − n 2 ) , C = B n 1 − n 3 = n 3 (1 − n )(1 − n 2 )(1 − n 3 ) , D = C n 1 − n 4 = n 4 (1 − n )(1 − n 2 )(1 − n 3 )(1 − n 4 ) etc. 33. Hence a new r elation is p erceived here b et ween the v alues of the letter s A , B , C , D etc. and the v alues of the letters α, β , γ , δ etc. It is noteworthy tha t these v alues do not disagree with ea ch other . F or collecting § 21 it is understo o d to b e α = A , β = n B , γ = n 3 C , δ = n 6 D , ǫ = n 10 E etc. Thu s it is clear from the rule of the co efficients that the series A , B , C , D etc. agree completely with the series α, β , γ , δ etc. except for the s ole difference being in the exponents of n . Indeed in the series A the exp onents are also equal to the exp onents in the ser ie s α , but in the series B the exp onents are one less than the exp onents in the ser ies β , in the se ries C the exp onents a re three less than the exp onents in the series γ , and the differences proce e d according to the trigonal num b ers and so on. 23 34. Therefor e b y the ser ies α, β , γ , δ etc. which we show e d ab ov e how to form and by which the first pr oblem of Naud´ e was resolved, the la tter prob- lem pro po sed by Na ud´ e can sim ultaneously b e resolved, so that its solution is reduced to the so lution of the first. Namely it will b e m (1) = m (1) i , m (2) = ( m + 1 ) (2) i , m (3) = ( m + 3 ) (3) i , m (4) = ( m + 6 ) (4) i and generally m ( µ ) =  m + µ ( µ − 1) 2  ( µ ) i and in turn m ( µ ) i =  m − µ ( µ − 1) 2  ( µ ) . F ur ther more, bec ause we hav e also found that ( m + µ ) ( µ ) i = m ( µ ) i + m ( µ − 1) i , and reducing this to the present case it will be  m − µ ( µ − 3) 2  ( µ ) =  m − µ ( µ − 1) 2  ( µ ) +  m − ( µ − 1)( µ − 2) 2  ( µ − 1) or for co n venience m ( µ ) = ( m − µ ) ( µ ) + ( m − 1) ( µ − 1) . The series for the letters A , B , C etc. ar e easily formed fro m this pro pe rty , and so the latter problem is reso lved. 35. F or a n example o f this pro blem that Insightfu l Man pr esented the ques- tion of determining in how many different w ays the n um be r 5 0 can be sepa rated int o exactly se ven pa rts, either equal or unequa l. T his question can hence be reduced to the first pro blem, with m = 50 and µ = 7, if we sea rch for how ma n y different ways the n um ber 50 + 21 or 71 can b e sepa rated in to sev en unequal parts. This in fact can b e done in 89 46 different wa ys. Indeed b esides this, the same n umber 8946 indica tes ( § 28) how many different wa ys 71 − 28 = 43 can be pro duced b y addition from the num b ers 1 , 2 , 3 , 4 , 5 , 6 , 7. And generally the nu mber of wa ys m ( µ ) in which the num b er m is res olved in to µ pa rts, either equal or unequal, also sho ws how many different wa ys the num b er m − µ can be pro duced by addition from the pa rticular num b ers 1 , 2 , 3 , 4 , 5 , . . . , µ. 36. A t the e nd of this pa p er ther e is a no teworthy obser v ation to make, which howev er I have not yet b een able to demonstrate with geometric rigor. Namely I have observed that if the infinitely many fa ctors of the pr o duct (1 − n )(1 − n 2 )(1 − n 3 )(1 − n 4 )(1 − n 5 ) etc. , 24 are expanded by actual multiplication, they pro duce this series 1 − n − n 2 + n 5 + n 7 − n 12 − n 15 + n 22 + n 26 − n 35 − n 40 + n 51 + e tc. , where only those those pow ers of n o c cur who se ex po nen ts ar e co n tained in the form 3 xx ± x 2 . And if x is an o dd num b er, the p ow ers of n , which a re n 3 xx ± x 2 , will hav e the co efficient − 1, while if x is an ev en n umber then the p ow er s n 3 xx ± x 2 will hav e the co efficient +1 . 37. It is also worth noting that the recipr o cal ser ies of this, which arises from the expa ns ion of this frac tio n 1 (1 − n )(1 − n 2 )(1 − n 3 )(1 − n 4 )(1 − n 5 ) etc. , yields namely the r ecurrent series 1 + 1 n + 2 n 2 + 3 n 3 + 5 n 4 + 7 n 5 + 1 1 n 6 + 1 5 n 7 + 2 2 n 8 + e tc. Of course this series multip lied b y the ab ov e series 1 − n − n 2 + n 5 + n 7 − n 12 − n 15 + n 22 + n 26 − e tc. pro duces unity . And in the first series the co efficient of any p ow er of n contains as many unities as there a re different ways in whic h the expo nent of n can b e distributed into parts; thus 5 can b e resolved in seven w ays in to parts, as 5 = 5 , 5 = 3 + 2 , 5 = 2 + 2 + 1 , 5 = 4 + 1 , 5 = 3 + 1 + 1 , 5 = 2 + 1 + 1 + 1 , 5 = 1 + 1 + 1 + 1 + 1 ; where namely neither the num b er o f parts or inequalities a re prescr ibed. 25