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A100335
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An inverse Catalan transform of J(2n).
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6
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0, 1, 4, 11, 27, 64, 149, 341, 768, 1707, 3755, 8192, 17749, 38229, 81920, 174763, 371371, 786432, 1660245, 3495253, 7340032, 15379115, 32156331, 67108864, 139810133, 290805077, 603979776, 1252698795, 2594876075, 5368709120
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OFFSET
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0,3
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COMMENTS
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The g.f. is obtained from that of A002450 through the mapping g(x) -> g(x*(1-x)). A002450 may be retrieved through the mapping g(x) -> g(x*c(x)), where c(x) is the g.f. of A000108.
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LINKS
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FORMULA
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G.f.: x*(1-x)/(1 - 5*x + 9*x^2 - 8*x^3 + 4*x^4).
a(n) = 5*a(n-1) - 9*a(n-2) + 8*a(n-3) - 4*a(n-4).
a(n) = Sum_{k=0..floor(n/2)} binomial(n-k, k)*(-1)^k*(4^(n-k) - 1)/3.
Binomial transform of A042965: (1, 3, 4, 5, 7, 8, 9, 11, 12, 13, ...), also row sums of triangle A133110. - Gary W. Adamson, Sep 12 2007
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MATHEMATICA
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LinearRecurrence[{5, -9, 8, -4}, {0, 1, 4, 11}, 41] (* G. C. Greubel, Jan 24 2023 *)
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PROG
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(Magma) I:=[0, 1, 4, 11]; [n le 4 select I[n] else 5*Self(n-1) -9*Self(n-2) +8*Self(n-3) -4*Self(n-4): n in [1..41]]; // G. C. Greubel, Jan 24 2023
(SageMath)
def A100335(n): return (1/3)*((n+1)*2^n - chebyshev_U(n, 1/2))
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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