0,7

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).

Column 1: T(n,1) = A129092(n) = A030067(2^n - 1) for n>=1, where A030067 is the semi-Fibonacci numbers.

Triangle T begins:

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; ...

where row 0 of matrix power T^k forms row k of T shift left,

as illustrated below.

For row 2: the matrix square T^2 begins:

2, 2, 1;

3, 4, 3, 1;

6, 12, 12, 6, 1;

17, 54, 65, 42, 12, 1;

70, 362, 512, 400, 156, 24, 1;

431, 3708, 6223, 5656, 2744, 600, 48, 1; ...

and row 0 of T^2 equals row 2 of T shift left: [2, 2, 1].

For row 3: the matrix cube T^3 begins:

5, 6, 4, 1;

11, 18, 16, 7, 1;

37, 88, 96, 56, 14, 1;

191, 672, 860, 609, 210, 28, 1;

1525, 8038, 11956, 9856, 4256, 812, 56, 1; ...

and row 0 of T^3 equals row 3 of T shift left: [5, 6, 4, 1].

For row 4: T^4 begins:

16, 24, 20, 8, 1;

53, 112, 116, 64, 15, 1;

292, 890, 1088, 736, 240, 30, 1;

2571, 11350, 16056, 12664, 5185, 930, 60, 1; ...

and row 0 of T^4 equals row 4 of T shift left: [16, 24, 20, 8, 1].

(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])}

Cf. A030067 (Semi-Fibonacci); A129092 (column 1), A129101 (column 2), A129102 (column 3), A129103 (column 4); variant: A129100.

Sequence in context: A064552 A123585 A145668 this_sequence A092450 A014291 A136587

Adjacent sequences: A129101 A129102 A129103 this_sequence A129105 A129106 A129107

nonn,tabf

Paul D. Hanna (pauldhanna(AT)juno.com), Apr 14 2007