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A113095 Triangle T, read by rows, that satisfies the recurrence: T(n,k) = [T^4](n-1,k-1) + [T^4](n-1,k) for n>k>=0, with T(n,n)=1 for n>=0, where T^4 is the matrix 4th power of T.

%I A113095 #15 Sep 09 2024 09:36:02
%S A113095 1,1,1,4,5,1,46,66,21,1,1504,2398,978,85,1,146821,255113,122914,14962,
%T A113095 341,1,45236404,84425001,46001193,7046354,235122,1365,1,46002427696,
%U A113095 91159696960,54661544301,9933169553,432627794,3738738,5461,1
%N A113095 Triangle T, read by rows, that satisfies the recurrence: T(n,k) = [T^4](n-1,k-1) + [T^4](n-1,k) for n>k>=0, with T(n,n)=1 for n>=0, where T^4 is the matrix 4th power of T.
%C A113095 Column 0 of the matrix power p, T^p, equals the number of 4-tournament sequences having initial term p (see A113092 for definitions).
%F A113095 Let GF[T] denote the g.f. of triangular matrix T. Then GF[T] = 1 + x*(1+y)*GF[T^4] and for all integer p>=1: GF[T^p] = 1 + x*Sum_{j=1..p} GF[T^(p+3*j)] + x*y*GF[T^(4*p)].
%e A113095 Triangle T begins:
%e A113095   1;
%e A113095   1,1;
%e A113095   4,5,1;
%e A113095   46,66,21,1;
%e A113095   1504,2398,978,85,1;
%e A113095   146821,255113,122914,14962,341,1;
%e A113095   45236404,84425001,46001193,7046354,235122,1365,1; ...
%e A113095 Matrix third power T^3 (A113099) begins:
%e A113095   1;
%e A113095   3,1;
%e A113095   27,15,1;
%e A113095   693,513,63,1;
%e A113095   52812,47619,8289,255,1; ...
%e A113095  where column 0 equals A113100.
%e A113095 Matrix 4th power T^4 (A113101) begins:
%e A113095   1;
%e A113095   4,1;
%e A113095   46,20,1;
%e A113095   1504,894,84,1;
%e A113095   146821,108292,14622,340,1;
%e A113095   45236404,39188597,6812596,233758,1364,1; ...
%e A113095  where adjacent sums in row n of T^4 forms row n+1 of T.
%o A113095 (PARI) {T(n,k)=local(M=matrix(n+1,n+1));for(r=1,n+1, for(c=1,r, M[r,c]=if(r==c,1,if(c>1,(M^4)[r-1,c-1])+(M^4)[r-1,c]))); return(M[n+1,k+1])}
%Y A113095 Cf. A097710, A113084, A113106; A113092, A113096 (column 0), A113097 (T^2), A113099 (T^3), A113101 (T^4).
%K A113095 nonn,tabl
%O A113095 0,4
%A A113095 _Paul D. Hanna_, Oct 14 2005