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Search: id:A078018
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| A078018 |
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a(0)=1, for n>=1 a(n)=sum(k=0,n,6^k*N(n,k)) where N(n,k) =1/n*C(n,k)*C(n,k+1) are the Narayana numbers (A001263). |
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+0
8
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| 1, 1, 7, 55, 469, 4237, 39907, 387739, 3858505, 39130777, 402972031, 4202705311, 44299426717, 471189693925, 5051001609115, 54513542257795, 591858123926545, 6459813793353265, 70837427884259575, 780073647992404615
(list; graph; listen)
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OFFSET
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0,3
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COMMENT
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More generally coefficients of (1+m*x-sqrt(m^2*x^2-(2*m+4)*x+1))/((2*m+2)*x) are given by : a(n)=sum(k=0,n,(m+1)^k*N(n,k))
The Hankel transform of this sequence is 6^C(n+1,2) . - Philippe DELEHAM (kolotoko(AT)wanadoo.fr), Oct 29 2007
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REFERENCES
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Paul Barry, On Integer-Sequence-Based Constructions of Generalized Pascal Triangles, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.4.
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FORMULA
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G.f. (1+5*x-sqrt(25*x^2-14*x+1))/(12*x)
a(n) = Sum_{k=0..n} A088617(n, k)*6^k*(-5)^(n-k) . - DELEHAM Philippe (kolotoko(AT)wanadoo.fr), Jan 21 2004
a(n) = [7(2n-1)a(n-1) - 25(n-2)a(n-2)] / (n+1) for n>=2, a(0) = a(1) = 1 . - Philippe DELEHAM (kolotoko(AT)wanadoo.fr), Aug 19 2005
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PROGRAM
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(PARI) a(n)=if(n<1, 1, sum(k=0, n, 6^k/n*binomial(n, k)*binomial(n, k+1)))
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CROSSREFS
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Cf. A001003, A007564, A059231.
Adjacent sequences: A078015 A078016 A078017 this_sequence A078019 A078020 A078021
Sequence in context: A049028 A096951 A113714 this_sequence A108628 A116862 A096307
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KEYWORD
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nonn
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AUTHOR
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Benoit Cloitre (benoit7848c(AT)orange.fr), May 10 2003
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