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%I #11 Feb 18 2022 22:46:05
%S 1,1,1,1,2,0,1,3,0,1,1,4,0,3,0,1,5,0,6,0,0,1,6,0,10,0,0,0,1,7,0,15,0,
%T 0,0,1,1,8,0,21,0,0,0,4,0,1,9,0,28,0,0,0,10,0,0,1,10,0,36,0,0,0,20,0,
%U 0,0,1,11,0,45,0,0,0,35,0,0,0,0,1,12,0,55,0,0,0,56,0,0,0,0,0
%N Pascal's Fredholm-Rueppel triangle.
%C First row of the array = the Fredholm-Rueppel sequence (A036987); which becomes the right border of the triangle. Second row of the array (1, 2, 0, 3, 0, 0, 0, 4, ...) = A104117. Third row of the array (1, 3, 0, 6, 0, 0, 0, 10, ...) = A129502. Row sums of triangle A129503 = A129504: (1, 2, 3, 5, 8, 12, 17, 24, 34, ...).
%H Andrew Howroyd, <a href="/A129503/b129503.txt">Table of n, a(n) for n = 1..1275</a>
%F Antidiagonals of an array in which n-th row (n=0,1,2,...) = M^n * V, where M = A115361 as an infinite lower triangular matrix and V = the Fredholm-Rueppel sequence A036987 as a vector: [1, 1, 0, 1, 0, 0, 0, 1, ...]. The array = 1, 1, 0, 1, 0, 0, 0, 1, 0, ... 1, 2, 0, 3, 0, 0, 0, 4, 0, ... 1, 3, 0, 6, 0, 0, 0, 10, 0, ... 1, 4, 0, 10, 0, 0, 0, 20, 0, ... (n+1)-th row can be generated from A115361 * n-th row.
%F T(n, 2^e) = binomial(n + e - 2^e, e), T(n, k) = 0 otherwise. - _Andrew Howroyd_, Aug 09 2018
%e First few rows of the triangle:
%e 1;
%e 1, 1;
%e 1, 2, 0;
%e 1, 3, 0, 1;
%e 1, 4, 0, 3, 0;
%e 1, 5, 0, 6, 0, 0;
%e 1, 6, 0, 10, 0, 0, 0;
%e 1, 7, 0, 15, 0, 0, 0, 1;
%e 1, 8, 0, 21, 0, 0, 0, 4, 0;
%e 1, 9, 0, 28, 0, 0, 0, 10, 0, 0;
%e 1, 10, 0, 36, 0, 0, 0, 20, 0, 0, 0;
%e ...
%o (PARI) T(n,k)=my(e=valuation(k,2)); if(k==2^e, binomial(n-k+e, e)) \\ _Andrew Howroyd_, Aug 09 2018
%Y Row sums are A129504.
%Y Cf. A036987, A115361, A104117, A129502.
%K nonn,tabl
%O 1,5
%A _Gary W. Adamson_, Apr 18 2007
%E a(53) corrected and terms a(67) and beyond from _Andrew Howroyd_, Aug 09 2018
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