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%I #5 Jun 14 2017 00:27:52
%S 1,1,1,1,1,1,2,2,1,1,5,6,4,1,1,16,24,20,8,1,1,69,136,136,72,16,1,1,
%T 430,1162,1360,880,272,32,1,1,4137,15702,21204,16032,6240,1056,64,1,1,
%U 64436,346768,537748,461992,214336,46784,4160,128,1,1,1676353,12836904
%N Triangle T, read by rows, where row n (shifted left) of T equals row 0 of matrix power T^n for n>=0.
%C This irregular-shaped triangle T results from inserting a left column of all 1's into triangle A129100; curiously, column k of A129100 equals column 0 of matrix power A129100^(2^k), while row n of A129100 equals row 0 of matrix power T^n (T is this triangle).
%F Column 1: T(n,1) = A129092(n) = A030067(2^n - 1) for n>=1, where A030067 is the semi-Fibonacci numbers.
%e Triangle T begins:
%e 1, 1;
%e 1, 1, 1;
%e 1, 2, 2, 1;
%e 1, 5, 6, 4, 1;
%e 1, 16, 24, 20, 8, 1;
%e 1, 69, 136, 136, 72, 16, 1; ...
%e where row 0 of matrix power T^k forms row k of T shift left,
%e as illustrated below.
%e For row 2: the matrix square T^2 begins:
%e 2, 2, 1;
%e 3, 4, 3, 1;
%e 6, 12, 12, 6, 1;
%e 17, 54, 65, 42, 12, 1;
%e 70, 362, 512, 400, 156, 24, 1;
%e 431, 3708, 6223, 5656, 2744, 600, 48, 1; ...
%e and row 0 of T^2 equals row 2 of T shift left: [2, 2, 1].
%e For row 3: the matrix cube T^3 begins:
%e 5, 6, 4, 1;
%e 11, 18, 16, 7, 1;
%e 37, 88, 96, 56, 14, 1;
%e 191, 672, 860, 609, 210, 28, 1;
%e 1525, 8038, 11956, 9856, 4256, 812, 56, 1; ...
%e and row 0 of T^3 equals row 3 of T shift left: [5, 6, 4, 1].
%e For row 4: T^4 begins:
%e 16, 24, 20, 8, 1;
%e 53, 112, 116, 64, 15, 1;
%e 292, 890, 1088, 736, 240, 30, 1;
%e 2571, 11350, 16056, 12664, 5185, 930, 60, 1; ...
%e and row 0 of T^4 equals row 4 of T shift left: [16, 24, 20, 8, 1].
%o (PARI) T(n,k)=local(A=[1,1;1,1],B);for(m=1,n+1,B=matrix(m+1,m+1); for(r=1,m,for(c=1,r+1,if(r==c-1 || c==1,B[r,c]=1, B[r,c]=(A^(r-1))[1,c-1])));A=B); return(A[n+1, k+1])
%Y Cf. A030067 (Semi-Fibonacci); A129092 (column 1), A129101 (column 2), A129102 (column 3), A129103 (column 4); variant: A129100.
%K nonn,tabf
%O 0,7
%A _Paul D. Hanna_, Apr 14 2007
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