On the partition of numbers into parts of a given type and number

On the partition of n um b ers in to parts of a giv en t yp e and n um b er ∗ Leonhard Euler 1. Some time a go I had treated the problem o f the partition of num b ers, which lo oks for how man y different wa ys a given num ber can b e separa ted in to t wo, three, o r generally a ny num ber , of parts. I had b een car eful that I present nothing abo ut this solution b y induction, a s we o ft en use in solving this kind of pro blem. The metho d which I used se e ms s uc h that it could b e applied with eq ua l success to other pro blems, in par ticular the common problem which searches for how many ways a given num b er c a n b e thrown as a given num b er of dice, which indeed I hav e decided to explain here in a way that can b e easily generalized. 2. Since we are lo oking for how many ways a given num b er N can o ccur by throwing a given num b er n of dice, here the ques tion reduces to this, how many distinct wa ys can can a given nu mber N b e reso lv ed into n parts, which are altogether 1 , 2 , 3 , 4 , 5 or 6, if the sides of the dice ar e marked with these num b ers. F rom this a mor e general question presents itself, in how man y distinct wa ys can a given n umber N b e divided into n par ts , which are alto g ether α, β , γ , δ etc., the n umber of which is = m , such that b o th the num b er and t yp e o f par ts, in which a given num b er shall b e resolved, ar e g iv en. 3. Namely let the dice, not just in this par ticular case of six sides, but indeed hav e m sides or faces, so that in eac h the sides ar e marked with the n umbers α, β , γ , δ , etc., a nd one then a sks how many ways a given num b er N can b e pro duced by throwing n of these dice. It could a lso b e assumed that the dice are different from each other, so that each would hav e a particular n umber of faces, which would moreov e r be inscrib ed with particular num b ers; truly the solution of this ques tion, which I hav e g eneralized from the common problem o f dice, can found without excessive difficulty . 4. Indeed, I consider the num b ers, whic h the sides of the dice are ma r k ed with, a s e xponents of s ome quantit y x , so that for common dice we would have ∗ Presen ted to the St. Petersburg Academy on August 18, 1768. Originall y published as De p artitione numer orum in p artes t a m numer o q u am sp e cie datas , Novi Commen tarii academiae scien tiarum Petropolitanae 14 (1770) , 168–187. E394 in t he Enestr¨ om index. T ranslated from the Latin by Jordan Bell, Departmen t of Mathematics, U n i v ersity of T or on to, T oronto , Canada. Email: jordan.b e l l@uto r on to.ca 1 this expres s ion x 1 + x 2 + x 3 + x 4 + x 5 + x 6 , where I take unity a s the co efficien t of each p o wer, s ince each n umber is desig- nated by the exp onen t equa l to it. But if we take the squa re of this expressio n, each p o wer of x will receive a co efficient tha t indica tes how many wa ys this power can result from the mult iplica tion of tw o terms of the expressio n, that is, how many wa ys its exponent c a n be pr oduced from the addition of t wo num- ber s from the s equence 1 , 2 , 3 , 4 , 5 , 6. The r efore by expanding the square of our expression, if the term M x N o ccurs in it, it follows from this that the num be r N can b e thrown as tw o dice in as many wa ys a s the co efficien t M contains unities. 5. In a similar way , it is clear that if o ne takes the cube of this expr ession ( x + x 2 + x 3 + x 4 + x 5 + x 6 ) 3 , in its expa ns ion any particula r p o wer x N o ccurs just as many times as the wa ys in which the exp o nen t N can arise fro m the addition of three n umber s from the sequenc e 1 , 2 , 3 , 4 , 5 , 6; thus if M is the co efficien t of a power, and the whole term is M x N , we conclude fro m this that the the num b er N can b e pro duced from throws of three dice in as many wa ys as the co e fficien t M contains unities. Therefore in general, if n is taken as the exp onen t of our expression ( x + x 2 + x 3 + x 4 + x 5 + x 6 ) n , by expanding acco r ding to powers of x , any term M x N shows that, if the num b er of dice = n , the num ber N can b e made by throwing them tog ether in as many wa ys as the co efficien t M co n tains unities. 6. Therefor e if the n umber of dice w er e = n and we sear c h for how man y wa ys a given num b er N can b e made by thr o wing these, the question is resolved by the expansio n of this formula ( x + x 2 + x 3 + x 4 + x 5 + x 6 ) n ; now since the fir st term will b e x n , the la st indeed x 6 n , the pr ogression of the terms follows in this way x n + Ax n +1 + B x n +2 + C x n +3 + · · · + M x N + · · · + x 6 n , where for any term M x N it is cle ar that the num b er N which equals the exp o- nent can b e made in a s many wa ys a s the coefficient M contains unities; from this it is immediately c le ar that the question do es not even o ccur unless the prop osed num b er N is c o n tained within the limits n and 6 n . The who le matter th us now beco mes that this pr ogression, or the co efficients of all its terms, be assigned. 7. Therefore to find these, let the for m ula that is to b e expanded be r e pr e- sented in this way x n (1 + x + x 2 + x 3 + x 4 + x 5 ) n = V , 2 then indeed for its expansio n let V = x n (1 + Ax + B x 2 + C x 3 + D x 4 + E x 5 + F x 6 + etc.) . And by putting V x n = Z by differentiating the logar ithm it will b e for the fir s t xd Z Z dx = nx + 2 nx 2 + 3 nx 3 + 4 nx 4 + 5 nx 5 1 + x + x 2 + x 3 + x 4 + x 5 . As well, the v alue for the latter w ill follow xd Z Z dx = Ax + 2 B x 2 + 3 C x 3 + 4 D x 4 + 5 E x 5 + 6 F x 6 + etc. 1 + Ax + B x 2 + C x 3 + D x 4 + E x 5 + F x 6 + etc. , and since these tw o expres sions a re equal to ea c h other , from this the v alues of the co efficien ts ca n b e determined. 8. H aving constituted the equality o f these t wo ex pr essions, this equation follows nx + nAx 2 + nB x 3 + nC x 4 + nD x 5 + nE x 6 + nF x 7 + nGx 8 +etc. +2 n +2 nA +2 nB +2 nC +2 nD +2 nE +2 nF +3 n +3 nA +3 nB +3 nC + 3 nD +3 nE +4 n +4 nA +4 nB +4 nC +4 nD +5 n +5 nA +5 nB +5 nC = Ax +2 B x 2 +3 C x 3 +4 D x 4 +5 E x 5 +6 F x 6 +7 Gx 7 +8 H x 8 +etc. + A +2 B +3 C +4 D +5 E + 6 F +7 G + A +2 B +3 C +4 D +5 E +6 F + A +2 B +3 C +4 D +5 E + A +2 B +3 C + 4 D + A +2 B +3 C and these tw o e x pressions, since a ll their terms ar e eq ual to each o ther, will provide the v alues of a ll the co efficien ts . 9. The fo llowing determinations a re now obta ined: A = n 2 B = ( n − 1) A + 2 n, 3 C = ( n − 2) B + (2 n − 1 ) A + 3 n, 4 D = ( n − 3) C + (2 n − 2) B + (3 n − 1) A + 4 n, 5 E = ( n − 4) D + (2 n − 3) C + (3 n − 2) + (4 n − 1) A + 5 n, 6 F = ( n − 5) E + (2 n − 4) D + (3 n − 3) C + (4 n − 2) B + (5 n − 1) A, 7 G = ( n − 6) F + (2 n − 5) E + (3 n − 4) D + (4 n − 3) C + (5 n − 2 ) B, 8 H = ( n − 7) G + (2 n − 6) F + (3 n − 5) E + (4 n − 4) D + (5 n − 3) C etc. 3 Thu s any co efficien t ca n b e determined by the pr eceding five, and with these found it will be V = x n + Ax n +1 + B x n +2 + C x n +3 + D x n +4 + E x n +5 + etc. and this is the genera l s o lution to the pro blem for n fac e s. 10. If each o f the prec e ding is subtracted from the ab o ve equations, the following mu ch simpler determina tions can b e o btained: A = n, 2 B = nA + n, 3 C = nB + nA + n, 4 D = nC + nB + nA + n, 5 E = nD + n C + nB + nA + n, 6 F = nE + nD + nC + nB + nA − 5 n, 7 G = nF + nE + n D + nC + nB − (5 n − 1) A, 8 H = nG + nF + nE + nD + nC − (5 n − 2) B etc. If again the differences ar e taken, these r e la tions will beco me even simpler, in this wa y: 2 B = ( n + 1) A, 3 C = ( n + 2) B , 4 D = ( n + 3) C, 5 E = ( n + 4 ) D, 6 F = ( n + 5) E − 6 n , 7 G = ( n + 6) F − (6 n − 1) A + 5 n, 8 H = ( n + 7) G − (6 n − 2) B + (5 n − 1) A, 9 I = ( n + 8) H − (6 n − 3) C + (5 n − 2) B , 10 K = ( n + 9) I − (6 n − 4) D + (5 n − 3) C etc. 11. Then, if the num b er of dice were 2 , 3 or 4, the law for the pro gression of 4 the co efficien ts will be as follows, for tw o for three for four A = 2 3 4 2 B = 3 A 4 A 5 A 3 C = 4 B 5 B 6 B 4 D = 5 C 6 C 7 C 5 E = 6 D 7 D 8 D 6 F = 7 E − 12 8 E − 1 8 9 E − 2 4 7 G = 8 F − 1 1 A + 10 9 F − 1 7 A + 15 10 F − 23 A + 20 8 H = 9 G − 10 B + 9 A 10 G − 16 B + 14 A 11 G − 22 B + 19 A 9 I = 1 0 H − 9 C + 8 B 11 H − 15 C + 13 B 12 H − 21 C + 1 8 B 10 K = 11 I − 8 D + 7 C 12 I − 14 D + 12 C 13 I − 20 D + 17 C 11 L = 12 K − 7 E + 6 D 13 K − 13 E + 11 D 1 4 K − 19 E + 16 D 12 M = 13 L − 6 F + 5 E 14 L − 12 F + 1 0 E 15 L − 18 F + 15 E etc. etc. etc. Therefore any co efficien t can b e determined by three prece ding , where it is particularly noteworth y here that finally they shall abate to nothing, and the same happ ens for those following the first to disapp ear, which is no t en tirely clear from this law. 12. T o help us clea rly understand this law, let the for m ula ( N ) ( n ) denote the num b er of cases in which the num b er N can b e made by n dice, so that it would b e ( n ) ( n ) = 1 , ( n + 1) ( n ) = A, ( n + 2) ( n ) = B , ( n + 3) ( n ) = C, ( n + 4) ( n ) = D , . . . , ( n + 9) ( n ) = I and ( n + 10 ) ( n ) = K. It will therefore b e 10( n + 10 ) ( n ) = ( n + 9)( n + 9 ) ( n ) − (6 n − 4)( n + 4) ( n ) + (5 n − 3)( n + 3) ( n ) , and here it can b e co nc luded g enerally to b e λ ( n + λ ) ( n ) = ( n + λ − 1)( n + λ − 1) ( n ) − (6 n + 6 − λ )( n + λ − 6 ) ( n ) + (5 n + 7 − λ )( n + λ − 7 ) ( n ) . Let us now put n + λ = N , and so that λ = N − n , and it w ill b e ( N ) ( n ) = ( N − 1)( N − 1) ( n ) − (7 n + 6 − N )( N − 6) ( n ) + (6 n + 7 − N )( N − 7) ( n ) N − n , where it is noted to alwa ys b e ( P ) ( n ) = 0 when P < n . 5 13. Also, these co efficien ts can easily b e defined for any num b er of dice, if those for the num b er of dice o ne less ar e k no wn. F or if it were ( x + x 2 + x 3 + x 4 + x 5 + x 6 ) n = x n + Ax n +1 + B x n +2 + C x n +3 + D x n +4 + etc. let it b e put ( x + x 2 + x 3 + x 4 + x 5 + x 6 ) n +1 x n +1 + A ′ x n +2 + B ′ x n +3 + C ′ x n +4 + D ′ x n +5 + etc. then it will b e, becaus e this expr ession is equal to the for mer multiplied by x + x 2 + x 3 + x 4 + x 5 + x 6 , A ′ = A + 1 then taking the different ia ls B ′ = B + A + 1 B ′ = A ′ + B C ′ = C + B + A + 1 C ′ = B ′ + C D ′ = D + C + B + A + 1 D ′ = C ′ + D E ′ = E + D + C + B + A + 1 E ′ = D ′ + E F ′ = F + E + D + C + B + A F ′ = E ′ + F − 1 G ′ = G + F + E + D + C + B G ′ = F ′ + G − A etc. etc. 14. Thus if the wa y of denoting introduced ab ov e is used, fro m the equation G ′ = F ′ + G − A it is ( n + 8) ( n +1) = ( n + 7) ( n +1) + ( n + 7) ( n ) − ( n + 1) ( n ) , which will b e r epresen ted in gener al as ( n + 1 + λ ) ( n +1) = ( n + λ ) ( n +1) + ( n + λ ) ( n ) − ( n + λ − 6) ( n ) . Now if n + λ is wr itt e n for N , it will b e ( N + 1) ( n +1) = ( N ) ( n +1) + ( N ) ( n ) − ( N − 6 ) ( n ) , where it should be noted that whenever N − 6 < n it will b e ( N − 6) ( n ) = 0. It is clear immediately that all thes e num b ers are in teg ers, which was les s appa ren t from the previo us law. 15. In these series a lso the prop erties found in § 12 o ccur; th us if n = 6, it will b e ( N ) (6) = ( N − 1)( N − 1) (6) − (48 − N )( N − 6) (6) + (43 − N )( N − 7) (6) N − 6 , where, if for ex a mple N = 25, it will b e (25) (6) = 24 · (2 4) (6) − 23 · (19) (6) + 18 · (18) (6) 19 ; 6 T able 1: E xhibiting how ma n y wa ys a num b er N can b e made by n dice N n = 1 n = 2 n = 3 n = 4 n = 5 n = 6 n = 7 n = 8 1 1 0 0 0 0 0 0 0 2 1 1 0 0 0 0 0 0 3 1 2 1 0 0 0 0 0 4 1 3 3 1 0 0 0 0 5 1 4 6 4 1 0 0 0 6 1 5 10 10 5 1 0 0 7 0 6 15 20 15 6 1 0 8 0 5 21 35 35 21 7 1 9 0 4 25 56 70 56 28 8 10 0 3 27 80 126 126 84 36 11 0 2 27 104 205 252 21 0 120 12 0 1 25 125 305 456 46 2 330 13 0 0 21 140 420 756 9 1 7 792 14 0 0 15 146 540 1161 1667 170 8 15 0 0 10 140 651 1666 2807 336 8 16 0 0 6 125 735 2247 4417 6147 17 0 0 3 104 780 2856 6538 1048 0 18 0 0 1 80 780 3431 9142 1680 8 19 0 0 0 56 735 3906 12117 254 88 20 0 0 0 35 651 4221 15267 366 88 21 0 0 0 20 540 4332 1832 7 5028 8 22 0 0 0 10 420 4221 20993 658 08 23 0 0 0 4 305 3906 2296 7 8238 4 24 0 0 0 1 205 3431 240 17 988 13 25 0 0 0 0 126 2856 2401 7 113688 26 0 0 0 0 70 2247 2296 7 1258 8 27 0 0 0 0 35 1666 20993 13328 8 28 0 0 0 0 15 1161 18327 13595 4 29 0 0 0 0 5 756 15267 133288 30 0 0 0 0 1 456 1 2117 1255 88 31 0 0 0 0 0 252 91 42 113688 32 0 0 0 0 0 126 6 538 98 813 33 0 0 0 0 0 56 441 7 8238 4 34 0 0 0 0 0 21 28 07 658 08 35 0 0 0 0 0 6 16 67 502 88 36 0 0 0 0 0 1 9 1 7 36 688 7 and it is (2 4) (6) = 343 1 , (19) (6) = 3906 , (18 ) (6) = 343 1, hence (25) (6) = 24 · 3 431 − 23 · 3 906 + 18 · 3431 19 = 54264 19 = 2856 , as is obtained in the table. Similarly if N = 2 9, it will b e (29) (6) = 28 · (2 8) (6) − 19 · (23) (6) + 14 · (22) (6) 23 ; then b ecause (28 ) (6) = 1161 , (23 ) (6) = 390 6 and (22) (6) = 4221 it will be (29) (6) = 32508 − 7 4214 + 5909 4 23 = 17388 23 = 756 . 16. T r uly the expans ion of the for m ula V ( § 7 ) can b e done in a different wa y , such that any ter m can b e completely assig ned witho ut the use of the preceding. F or since it is 1 + x + x 2 + x 3 + x 4 + x 5 = 1 − x 6 1 − x , it will b e V = x n (1 − x 6 ) n (1 − x ) n and by expanding (1 − x 6 ) n = 1 − n 1 x 6 + n ( n − 1 ) 1 · 2 x 12 − n ( n − 1)( n − 2 ) 1 · 2 · 3 x 18 + n ( n − 1 )( n − 2)( n − 3) 1 · 2 · 3 · 4 x 24 − etc. , x n (1 − x ) n = x n + n 1 x n +1 + n ( n + 1 ) 1 · 2 x n +2 + n ( n + 1)( n + 2 ) 1 · 2 · 3 x n +3 + n ( n + 1 )( n + 2)( n + 3) 1 · 2 · 3 · 4 x n +4 + etc. 8 and th us it can b e concluded to be ( n ) ( n ) = 1 , ( n + 1) ( n ) = n 1 , ( n + 2) ( n ) = n ( n + 1) 1 · 2 , ( n + 3) ( n ) = n ( n + 1)( n + 2 ) 1 · 2 · 3 , ( n + 4) ( n ) = n ( n + 1) · · · ( n + 3 ) 1 · 2 · 3 · 4 , ( n + 5) ( n ) = n ( n + 1) · · · ( n + 4 ) 1 · 2 · 3 · 4 · 5 , ( n + 6) ( n ) = n ( n + 1) · · · ( n + 5 ) 1 · 2 · 3 · · · 6 − n 1 · 1 , ( n + 7) ( n ) = n ( n + 1) · · · ( n + 6 ) 1 · 2 · 3 · · · 7 − n 1 · n 1 , ( n + 8) ( n ) = n ( n + 1) · · · ( n + 7 ) 1 · 2 · 3 · · · 8 − n 1 · n ( n + 1 ) 1 · 2 , ( n + 9) ( n ) = n ( n + 1) · · · ( n + 8 ) 1 · 2 · 3 · · · 9 − n 1 · n ( n + 1 )( n + 2) 1 · 2 · 3 , ( n + 10 ) ( n ) = n ( n + 1) · · · ( n + 9 ) 1 · 2 · 3 · · · 10 − n 1 · n ( n + 1 ) · · · ( n + 3) 1 · 2 · 3 · 4 , ( n + 11 ) ( n ) = n ( n + 1) · · · ( n + 1 0) 1 · 2 · 3 · · · 11 − n 1 · n ( n + 1 ) · · · ( n + 4) 1 · 2 · 3 · 4 · 5 , ( n + 12 ) ( n ) = n ( n + 1) · · · ( n + 1 1) 1 · 2 · 3 · · · 12 − n 1 · n ( n + 1 ) · · · ( n + 5) 1 · 2 · 3 · · · 6 + n ( n − 1 ) 1 · 2 · 1 , ( n + 13 ) ( n ) = n ( n + 1) · · · ( n + 1 2) 1 · 2 · 3 · · · 13 − n 1 · n ( n + 1 ) · · · ( n + 6) 1 · 2 · 3 · · · 7 + n ( n − 1 ) 1 · 2 · n 1 etc., from which it c an concluded in g eneral ( n + λ ) ( n ) = n ( n + 1) · · · ( n + λ − 1 ) 1 · 2 · 3 · · · λ − n 1 · n ( n + 1 ) · · · ( n + λ − 7) 1 · 2 · 3 · · · ( λ − 6) + n ( n − 1 ) 1 · 2 · n ( n + 1) · · · ( n + λ − 13) 1 · 2 · 3 · · · ( λ − 12) − n ( n − 1)( n − 2 ) 1 · 2 · 3 · n ( n + 1 ) · · · ( n + λ − 19) 1 · 2 · 3 · · · ( λ − 18) + n ( n − 1 )( n − 2)( n − 3) 1 · 2 · 3 · 4 · n ( n + 1) · · · ( n + λ − 2 5 ) 1 · 2 · 3 · · · ( λ − 24) − etc. 17. This so lut io n can b e adapted to dice p ossessing any other num be r of faces. F or let the num b er of faces on a ll the dice b e m , which shall b e marked with the num b ers 1 , 2 , 3 , . . . , m , while the num b er of dice itself shall b e = n for which the num be r of throws are sought in which the given num b er N land. Or , 9 which ends up being the same thing, it is sought how many wa ys the num be r N ca n be resolved into n pa r ts, each of whic h is co mprised fro m the sequence of num b ers 1 , 2 , 3 , . . . , m ; here indeed it should b e noted that not only different partitions but also differ en t o rders o f the s a me parts are co un ted, as usually happ ens in dice, where for exa mple the throws 3 , 4 and 4 , 3 a re considered to b e t wo different cases. 18. But if therefo r e this symbo l ( N ) ( n ) denotes the num b er of c a ses in which the num b er N ca n b e pro duced by throwing n dice, each o f which hav e m sides marked with the num b ers 1 , 2 , 3 , . . . , m , it sho uld first be noted tha t ( n ) ( n ) = 1, and if N < n then ( N ) ( n ) = 0. Next, if N = mn then to o ( mn ) ( n ) = 1, and if N > mn it will b e ( N ) ( n ) = 0. Finally , if either N = n + λ or N = mn − λ , the nu mber of cases is the same, namely ( n + λ ) ( n ) = ( mn − λ ) ( n ) . The fir st for m ula ca n also b e develop ed as ( n + λ ) ( n ) = n ( n + 1) · · · ( n + λ − 1) 1 · 2 · 3 · · · λ − n 1 · n ( n + 1) · · · ( n + λ − m − 1) 1 · 2 · 3 · · · ( λ − m ) + n ( n − 1 ) 1 · 2 · n ( n + 1 ) · · · ( n + λ − 2 m − 1) 1 · 2 · 3 · · · ( λ − 2 m ) − n ( n − 1 )( n − 2) 1 · 2 · 3 · n ( n + 1) · · · ( n + λ − 3 m − 1) 1 · 2 · 3 · · · ( λ − 3 m ) + e tc. 19. As well, these num b ers can very ea sily b e determined from the preceding, where the num b e r of dice is one les s. F or in genera l it will b e, if the num ber of all the dice w e r e = m a nd ea c h were marked with the num b e rs 1 , 2 , 3 , . . . , m , ( N + 1 ) ( n +1) = ( N ) ( n +1) + ( N ) ( n ) − ( N − m ) ( n ) or ( N + 1 ) ( n ) = ( N ) ( n ) + ( N ) ( n − 1) − ( N − m ) ( n − 1) . Then, if n + λ is written for N + 1, we will o btain ( n + λ ) ( n ) = ( n + λ − 1) ( n ) + ( n + λ − 1) ( n − 1) − ( n + λ − m − 1) ( n − 1) . Finally , for the s ame n umber o f dice, this num b er n dep ends thus on pr e ceding nu mber s λ ( n + λ ) ( n ) = ( n + λ − 1)( n + λ − 1) ( n ) − ( mn + m − λ )( n + λ − m ) ( n ) +( mn − n + m + 1 − λ )( n + λ − m − 1) ( n ) . Again it is noted that the sum of all these num b ers is = m n . 20. The question can be resolved in a s imila r way e v en if not all the dice o ccur with the s ame num b er o f face s . Let us take three dice, the first a hex- ahedron b e aring the num b ers 1 , 2 , 3 , 4 , 5 , 6, the seco nd an octa hedron bearing 10 the n umbers 1 , 2 , 3 , . . . , 8, and the third a dodeca hedron b earing the n umber s 1 , 2 , 3 , . . . , 12; if we now ask for how ma n y ways a given num b er N can b e decomp osed, we should ex pa nd this pro duct ( x + x 2 + x 3 + · · · + x 6 )( x + x 2 + x 3 + · · · + x 8 )( x + x 2 + x 3 + · · · + x 12 ) = V , and the co efficien t o f the power x N will display the n umber of c ases. Now, since here it will be V = x 3 (1 − x 6 )(1 − x 8 )(1 − x 12 ) (1 − x ) 3 , by expanding the numerator it will b e V = x 3 − x 9 − x 11 − x 15 + x 17 + x 21 + x 23 − x 29 (1 − x ) 3 . 21. Here the numerator is m ultiplied b y 1 (1 − x ) 3 , o r the ser ies 1 + 3 x + 6 x 2 + 10 x 3 + 15 x 4 + 21 x 5 + 28 x 6 + 36 x 7 + etc. , whose co efficien ts are the tria ngular num b ers; and since the n th triangula r num- ber is n ( n +1) 2 , a n y term o f this series will b e n ( n + 1 ) 2 x n − 1 or ( n − 1)( n − 2 ) 2 x n − 3 . Now multiplying b y the numerator, the co efficien t of the p o wer x n turns out to be ( n − 1)( n − 2 ) 2 − ( n − 7)( n − 8 ) 2 − ( n − 9)( n − 1 0) 2 − ( n − 13 )( n − 14) 2 + ( n − 15)( n − 16) 2 + ( n − 19)( n − 20) 2 + ( n − 21 )( n − 22) 2 + ( n − 27 )( n − 28) 2 . This express ion do es not need to b e contin ued beyond the case when negativ e factors are ar riv ed at. 22. Noticing how ever for the denominator (1 − x ) 3 = 1 − 3 x + 3 x 2 − x 3 , the series that is sought will b e recurre nt, arising from the ladder relation 3 , − 3 , +1, providing that w e hav e the rule for the terms of the numerator. Then the 11 following co efficien ts for each exp o nen t ar e fo und: Exp onen ts Co efficien ts E xponents Co efficien ts 3 1 15 47 4 3 16 45 5 6 17 42 6 10 18 38 7 15 19 33 8 21 20 27 9 27 21 21 10 33 22 15 11 38 23 10 12 42 24 6 13 45 25 3 14 47 26 1 The num b ers here are not g reater tha n 26, since 26 = 6 + 8 + 12, and the sum of all the c ases is 57 6 = 6 · 8 · 12. 23. Since here we have succeeded in the res olution of num b ers into pa rts of a given n umber a nd type without the help of induction, some elegant Theorems of F ermat come to my mind; while they hav e not yet b een demonstrated, it seems that this metho d will p erhaps lead to demons trations of them. F o r F er mat ha d asserted that all num b ers are either triangles or the agg regate of t wo or three triangles, and b ecause zero also o ccurs in the or der of triangles , the theorem can th us b e s tated as that all num b ers ca n b e reso lved into three triang le s . Thus if the triangula r n umber s are taken for exp onents and this series is for med 1 + x 1 + x 3 + x 6 + x 10 + x 15 + x 21 + x 28 + etc. = S, it needs to b e demonstr ated that if the c ub e o f this ser ies is ex pa nded, then every p o wer of x o c c urs without omission; for if this could b e demons tr ated, the demonstration of this Theorem of F ermat would b e p ossessed. 24. In a similar way , if we take the fourth p ower of this series 1 + x 1 + x 4 + x 9 + x 16 + x 25 + x 36 + etc. = S, and we co uld sho w that every power of x occur s in it, then we w ould hav e a demonstration of the Theorem of F er mat, that all num b ers ar e made fro m the addition of four s quares. Indeed in g eneral if we put S = 1 + x 1 + x m + x 3 m − 3 + x 6 m − 8 + x 10 m − 15 + x 15 m − 24 + x 21 m − 35 + etc. and w e take the p ow er o f this series o f the exp onent m , it needs to b e demon- strated tha t every p ow er of x will b e pro duced in it, a nd th us tha t all num be r s are the agg regate of m or less po lygonal num b ers with a num b er of sides = m . 12 25. F rom these same principles ano th e r wa y pr esen ts itself for inv estiga ting the demonstratio ns, whic h differs from the prec e ding in that wher e there not only different parts but a ls o different orders were considered, her e the stipulation ab out the order is omitted. Namely for the reso lut io n into tria ngular num b ers let this for m ula b e constituted 1 (1 − z )(1 − xz )(1 − x 3 z )(1 − x 6 z )(1 − x 10 z )(1 − x 15 z ) etc. , which expanded pro duces this series 1 + P z + Qz 2 + Rz 3 + S z 4 + T z 5 + etc. , such that P , Q, R, S etc. ar e some functions of x . Now it will clearly be P = 1 + x + x 3 + x 6 + x 10 + x 15 + x 21 + etc.; while Q will also co n tain those p o wers o f x whose exp onen ts are the ag gregate of tw o triang les. It therefore needs to be b e demonstr ated that ev ery p ow er o f x o ccurs in the function R . 26. In a similar wa y for the reso lution of num b ers into four squares , let this fraction b e ex panded 1 (1 − z )(1 − xz )(1 − x 4 z )(1 − x 9 z )(1 − x 16 z )(1 − x 25 z ) etc. ; then if it is changed into the for m 1 + P z + Qz 2 + Rz 3 + S z 4 + e tc. , it needs to b e demonstrated that the function S co n tains every p o wer of x . F or instance, P is equal to the series 1 + x + x 4 + x 9 + x 16 + etc. and Q also contains tho s e powers of x whose exponents are the ag gregate of tw o squares, in which series therefore many p o wers are still absent. F urther, in R there are the additional p o wers whose exp onen ts a re the a ggregate o f three squa r es, and then in S those will o ccur whose ex p onents ar e the sum of four, so that in S all nu mber s should o ccur as exp onen ts. 27. F rom this principle, we can define how many solutions those pr oblems ad- mit which a re co mm o nly referred to by Arithmeticians as the R e gula Vir ginum 1 . In this wa y , the problems are reduced here to finding num b ers p, q , r, s, t etc. that satisfy these tw o co nditions ap + b q + cr + ds + etc. = n, and αp + β q + γ r + δ s + etc. = ν, 1 T ranslator: In English, R ule of the Vir gins . See V olume I I of Leonard E u gene Di c kson’s History of the The ory of Numb ers , under the indices “Co ec i rule” and “Vi rgins”. 13 and now the question is, how many solutio ns o ccur in p ositiv e integers; where indeed the num b ers a, b, c, d etc., n and α, β , γ , δ etc., ν may b e ta k en to b e int eg ers, as if this were not so, it can b e easily reduced to this. It is at once apparent that if t wo num b ers p a nd q a re g iv en to be investigated, mo re than one solution c a nnot b e given. Commonly the problem only o ccurs with pos itiv e int eg ral num b e rs for p and q . 28. Now, for defining the num b er o f all the solutions in any particular case, so that no thing is done by induction or by trials, let us cons ider this expressio n 1 (1 − x a y α )(1 − x b y β )(1 − x c y γ )(1 − x d y δ ) etc. and let it b e expanded, which y ields such a series 1 + Ax ... y ... + B x ... y ... + C x ... y ... etc.; if the term N x n y ν o ccurs in this, the co efficien t N will indicate the num b er of solutions; and if it turns out that this ter m do es not o ccur, this indicates that there will b e no solution. Th us the whole problem turns here on investigating the co efficien t of the term x n y ν . 14