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%I #11 May 31 2024 14:37:27
%S 1,3,1,6,3,1,16,9,3,1,63,37,12,3,1,351,210,67,15,3,1,2609,1575,498,
%T 106,18,3,1,24636,14943,4701,975,154,21,3,1,284631,173109,54298,11100,
%U 1689,211,24,3,1,3909926,2381814,745734,151148,22518,2688,277,27,3,1
%N Matrix cube of triangle A107876; equals the product of triangular matrices: A107876^3 = A107862^-1*A107873.
%C Column 0 is A107885.
%C Column 1 is A107886.
%C Column 2 equals A107887.
%C Column 3 equals SHIFT_LEFT(A107878), where A107878 is column 2 of A107876.
%C Column 4 equals A107888.
%F G.f. for column k: 1 = Sum_{j>=0} T(k+j, k)*x^j*(1-x)^(3 + (k+j)*(k+j-1)/2 - k*(k-1)/2).
%e G.f. for column 0:
%e 1 = T(0,0)*(1-x)^3 + T(1,0)*x*(1-x)^3 + T(2,0)*x^2*(1-x)^4 + T(3,0)*x^3*(1-x)^6 + T(4,0)*x^4*(1-x)^9 + T(5,0)*x^5*(1-x)^13 + ...
%e = 1*(1-x)^3 + 3*x*(1-x)^3 + 6*x^2*(1-x)^4 + 16*x^3*(1-x)^6 + 63*x^4*(1-x)^9 + 351*x^5*(1-x)^13 + ...
%e G.f. for column 1:
%e 1 = T(1,1)*(1-x)^3 + T(2,1)*x*(1-x)^4 + T(3,1)*x^2*(1-x)^6 + T(4,1)*x^3*(1-x)^9 + T(5,1)*x^4*(1-x)^13 + T(6,1)*x^5*(1-x)^18 + ...
%e = 1*(1-x)^3 + 3*x*(1-x)^4 + 9*x^2*(1-x)^6 + 37*x^3*(1-x)^9 + 210*x^4*(1-x)^13 + 1575*x^5*(1-x)^18 + ...
%e Triangle begins:
%e 1;
%e 3, 1;
%e 6, 3, 1;
%e 16, 9, 3, 1;
%e 63, 37, 12, 3, 1;
%e 351, 210, 67, 15, 3, 1;
%e 2609, 1575, 498, 106, 18, 3, 1;
%e 24636, 14943, 4701, 975, 154, 21, 3, 1;
%e 284631, 173109, 54298, 11100, 1689, 211, 24, 3, 1;
%e ...
%t max = 10;
%t A107862 = Table[Binomial[If[n < k, 0, n*(n-1)/2-k*(k-1)/2 + n - k], n - k], {n, 0, max}, {k, 0, max}];
%t A107867 = Table[Binomial[If[n < k, 0, n*(n-1)/2-k*(k-1)/2 + n-k+1], n - k], {n, 0, max}, {k, 0, max}];
%t T = MatrixPower[Inverse[A107862].A107867, 3];
%t Table[T[[n+1, k+1]], {n, 0, max}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, May 31 2024 *)
%o (PARI) {T(n,k)=polcoeff(1-sum(j=0,n-k-1, T(j+k,k)*x^j*(1-x+x*O(x^n))^(3+(k+j)*(k+j-1)/2-k*(k-1)/2)),n-k)}
%Y Cf. A107862, A107870, A107873, A107867, A107876, A107880, A107884, A107885, A107886, A107887, A107888.
%K nonn,tabl,changed
%O 0,2
%A _Paul D. Hanna_, Jun 04 2005
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