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A291725
p-INVERT of (1,0,1,0,0,0,0,...), where p(S) = (1 - S)^2.
2
2, 3, 6, 11, 18, 30, 50, 81, 130, 208, 330, 520, 816, 1275, 1984, 3077, 4758, 7337, 11286, 17322, 26532, 40563, 61908, 94336, 143540, 218112, 331008, 501749, 759726, 1149159, 1736534, 2621751, 3954826, 5960902, 8977686, 13511461, 20320854, 30542064, 45875998
OFFSET
0,1
COMMENTS
Suppose s = (c(0), c(1), c(2), ...) is a sequence and p(S) is a polynomial. Let S(x) = c(0)*x + c(1)*x^2 + c(2)*x^3 + ... and T(x) = (-p(0) + 1/p(S(x)))/x. The p-INVERT of s is the sequence t(s) of coefficients in the Maclaurin series for T(x). Taking p(S) = 1 - S gives the "INVERT" transform of s, so that p-INVERT is a generalization of the "INVERT" transform (e.g., A033453).
See A291728 for a guide to related sequences.
LINKS
Greg Dresden, Yuechen Xiao, and Guanzhang Zhou, Self-Convolutions of Generalized Narayana Numbers, arXiv:2602.15208 [math.CO], 2026.
FORMULA
G.f.: -(-1 + x)*(1 + x^2)*(2 + x + x^2)/(-1 + x + x^3)^2.
a(n) = 2*a(n-1) - a(n-2) + 2*a(n-3) - 2*a(n-4) - a(n-6) for n >= 7.
For n>0, a(n-1) is the self-convolution of the Narayana's cows sequence. That is, a(n-1) = Sum_{i=0..n} b(i)*b(n-i) for b(n) = A000930(n). - Greg Dresden, Feb 16 2026
From Greg Dresden, Mar 19 2026: (Start)
(n+1)*a(n) = (n+2)*a(n-1) + (n+4)*a(n-3) for n >= 3.
a(n) = (1/31)*(9*(n+4)*b(n+4) - 3*(n+6)*b(n+3) - 2*(n+5)*b(n+2)) for b(n) = A000930(n). (End)
MATHEMATICA
z = 60; s = x + x^3; p = (1 - s)^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A154272 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A291725 *)
LinearRecurrence[{2, -1, 2, -2, 0, -1}, {2, 3, 6, 11, 18, 30}, 40] (* Vincenzo Librandi, Sep 10 2017 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Sep 08 2017
STATUS
approved