Original entry on oeis.org
1, 24, 2268, 461056, 160977375, 85624508376, 64363893844726, 64928246784463872, 84623205378726331245, 138408056280920732755000, 277597038523589348539241112, 670011760601512512626484887040
Offset: 0
O.g.f.: A(x) = 1 + 24*x + 2268*x^2 + 461056*x^3 + 160977375*x^4 + 85624508376*x^5 + 64363893844726*x^6 + 64928246784463872*x^7 + ...
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{a(n)=local(P=matrix(n+1,n+1,r,c,if(r>=c,(r^3)^(r-c)/(r-c)!)), D=matrix(n+1,n+1,r,c,if(r==c,r)));(P^-1*D^2*P)[n+1,1]}
for(n=0,20, print1(a(n),", "))
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/* From formula: [x^n] exp( n^3*x ) * (1 - x*A(x)) = 0 */
{a(n) = my(A=[1]); for(i=0, n, A=concat(A, 0); m=#A; A[m] = Vec( exp(x*m^3 +x^2*O(x^m)) * (1 - x*Ser(A)) )[m+1] ); A[n+1]}
for(n=0, 25, print1( a(n), ", ")) \\ Paul D. Hanna, May 12 2018
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/* From Recurrence: */
{a(n) = if(n==0,1, (n+1)^(3*n+3)/(n+1)! - sum(k=1,n, (n+1)^(3*k)/k! * a(n-k) ))}
for(n=0,25, print1( a(n),", ")) \\ Paul D. Hanna, May 12 2018
A107671
Triangular matrix T, read by rows, that satisfies: T = D + SHIFT_LEFT(T^3), where SHIFT_LEFT shifts each row 1 place to the left and D is the diagonal matrix {1, 2, 3, ...}.
Original entry on oeis.org
1, 8, 2, 513, 27, 3, 81856, 2368, 64, 4, 23846125, 469625, 7625, 125, 5, 10943504136, 160767720, 1898856, 19656, 216, 6, 7250862593527, 83548607478, 776598305, 6081733, 43561, 343, 7, 6545029128786432, 61068815111168, 465690017280, 2966844928, 16494080, 86528, 512, 8
Offset: 0
Triangle T begins:
1;
8, 2;
513, 27, 3;
81856, 2368, 64, 4;
23846125, 469625, 7625, 125, 5;
10943504136, 160767720, 1898856, 19656, 216, 6;
7250862593527, 83548607478, 776598305, 6081733, 43561, 343, 7;
...
The matrix cube T^3 shifts each row to the right 1 place, dropping the diagonal D and putting A006690 in column 0:
1;
56, 8;
7965, 513, 27;
2128064, 81856, 2368, 64;
914929500, 23846125, 469625, 7625, 125;
576689214816, 10943504136, 160767720, 1898856, 19656, 216;
...
From _Petros Hadjicostas_, Mar 11 2021: (Start)
We illustrate the above formula for T(n,k=0) with the compositions of n + 1 for n = 2. The compositions of n + 1 = 3 are 3, 1 + 2, 2 + 1, and 1 + 1 + 1. Thus the above sum has four terms with (r = 1, s_1 = 3), (r = 2, s_1 = 1, s_2 = 2), (r = 2, s_1 = 2, s_2 = 1), and (r = 3, s_1 = s_2 = s_3 = 1).
The value of the denominator Product_{j=1..r} s_j! for these four terms is 6, 2, 2, and 1, respectively.
The value of the numerator s_1^(-1)*Product_{j=1..r} (Sum_{i=1..j} s_i)^(3*s_j) for these four terms is 19683/3, 729/1, 1728/2, and 216/1.
Thus T(2,0) = (19683/3)/6 - (729/1)/2 - (1728/2)/2 + (216/1)/1 = 513. (End)
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{T(n,k)=local(P=matrix(n+1,n+1,r,c,if(r>=c,(r^3)^(r-c)/(r-c)!)), D=matrix(n+1,n+1,r,c,if(r==c,r)));if(n>=k,(P^-1*D*P)[n+1,k+1])}
Original entry on oeis.org
1, 56, 8, 7965, 513, 27, 2128064, 81856, 2368, 64, 914929500, 23846125, 469625, 7625, 125, 576689214816, 10943504136, 160767720, 1898856, 19656, 216, 500750172337212, 7250862593527, 83548607478, 776598305, 6081733, 43561, 343
Offset: 0
Triangle begins:
1;
56,8;
7965,513,27;
2128064,81856,2368,64;
914929500,23846125,469625,7625,125;
576689214816,10943504136,160767720,1898856,19656,216; ...
which equals the matrix cube of triangle A107671:
1;
8,2;
513,27,3;
81856,2368,64,4;
23846125,469625,7625,125,5;
10943504136,160767720,1898856,19656,216,6; ...
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{T(n,k)=local(P=matrix(n+1,n+1,r,c,if(r>=c,(r^3)^(r-c)/(r-c)!)), D=matrix(n+1,n+1,r,c,if(r==c,r)));if(n>=k,(P^-1*D^3*P)[n+1,k+1])}
Showing 1-3 of 3 results.
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