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Search: id:A101786
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| A101786 |
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G.f. satisfies: A(x) = 1 + x*A(x)/(1 - 2*x^2*A(x)^2). |
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1
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| 1, 1, 1, 3, 9, 25, 77, 247, 801, 2657, 8969, 30635, 105785, 368745, 1295493, 4582767, 16309953, 58357313, 209798289, 757461011, 2745281705, 9984464761, 36428252541, 133293594343, 489028250465, 1798543861537, 6629635284505
(list; graph; listen)
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OFFSET
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0,4
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COMMENT
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Formula may be derived using the Lagrange Inversion theorem (cf. A049124).
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FORMULA
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a(n) = Sum_{k=0..[(n-1)/2]} C(n-k-1, k)*C(n, 2*k)*2^k/(2*k+1) for n>0, with a(0)=1. G.f.: A(x) = [ series reversion of x*(1-2*x^2)/(1+x-2*x^2) ]/x.
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EXAMPLE
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Generated from Fibonacci polynomials (A011973) and
coefficients of odd powers of 1/(1-x):
a(1) = 1*1/1
a(2) = 1*1/1 + 0*1*2/3
a(3) = 1*1/1 + 1*3*2/3
a(4) = 1*1/1 + 2*6*2/3 + 0*1*2^2/5
a(5) = 1*1/1 + 3*10*2/3 + 1*5*2^2/5
a(6) = 1*1/1 + 4*15*2/3 + 3*15*2^2/5 + 0*1*2^3/7
a(7) = 1*1/1 + 5*21*2/3 + 6*35*2^2/5 + 1*7*2^3/7
a(8) = 1*1/1 + 6*28*2/3 + 10*70*2^2/5 + 4*28*2^3/7 + 0*1*2^4/9
This process is equivalent to the formula:
a(n) = Sum_{k=0..[(n-1)/2]} C(n-k-1,k)*C(n,2*k)*2^k/(2*k+1).
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PROGRAM
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(PARI) {a(n)=if(n==0, 1, sum(k=0, (n-1)\2, binomial(n-k-1, k)*binomial(n, 2*k)*2^k/(2*k+1)))}
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CROSSREFS
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Cf. A011973, A001045, A049124, A101785.
Sequence in context: A004665 A132835 A001189 this_sequence A012771 A120284 A074440
Adjacent sequences: A101783 A101784 A101785 this_sequence A101787 A101788 A101789
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KEYWORD
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nonn
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AUTHOR
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Paul D. Hanna (pauldhanna(AT)juno.com), Dec 16 2004
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