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.
1, 1, 1, 4, 5, 1, 46, 66, 21, 1, 1504, 2398, 978, 85, 1, 146821, 255113, 122914, 14962, 341, 1, 45236404, 84425001, 46001193, 7046354, 235122, 1365, 1, 46002427696, 91159696960, 54661544301, 9933169553, 432627794, 3738738, 5461, 1
Offset: 0
Examples
Triangle T begins: 1; 1,1; 4,5,1; 46,66,21,1; 1504,2398,978,85,1; 146821,255113,122914,14962,341,1; 45236404,84425001,46001193,7046354,235122,1365,1; ... Matrix third power T^3 (A113099) begins: 1; 3,1; 27,15,1; 693,513,63,1; 52812,47619,8289,255,1; ... where column 0 equals A113100. Matrix 4th power T^4 (A113101) begins: 1; 4,1; 46,20,1; 1504,894,84,1; 146821,108292,14622,340,1; 45236404,39188597,6812596,233758,1364,1; ... where adjacent sums in row n of T^4 forms row n+1 of T.
Crossrefs
Programs
-
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])}
Formula
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)].
Comments