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A078534
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Coefficients of power series that satisfies A(x)^5 - 25x*A(x)^6 = 1, A(0)=1.
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5
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1, 5, 100, 2625, 78125, 2502500, 84150000, 2929265625, 104646953125, 3814697265625, 141323284375000, 5305403695312500, 201382633183593750, 7715985752343750000, 298023223876953125000, 11591412585295166015625, 453601640704152832031250
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OFFSET
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0,2
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COMMENTS
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If A(x) = Sum_{k>=1} a(k)x^k satisfies A(x)^n - (n^2)*x*A(x)^(n+1) = 1, then a(n-1) = n^(2n-3) and a(2n-1) = n^(4n-2) (conjecture).
If A(x) = Sum_{k>=1} a(k)x^k satisfies A(x)^n - (n^2)*x*A(x)^(n+1) = 1, then a(k) = n^(2k)*binomial(k/n+1/n+k-1,k)/(k+1) and, consequently, a(n-1) = n^(2n-3) and a(2n-1) = n^(4n-2). - Emeric Deutsch, Dec 10 2002
A generalization of the Catalan sequence (A000108) since for n = 1 the equation A(x)^n -(n^2)*x*A(x)^(n+1) = 1 reduces to A(x) = 1 + xA(x)^2. - Emeric Deutsch, Dec 10 2002
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LINKS
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FORMULA
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a(n) = 5^(2n)*binomial(6n/5 - 4/5, n)/(n+1). - Emeric Deutsch, Dec 10 2002
a(n) ~ sqrt(3) * 6^(6*n/5 - 4/5) * 5^n / (sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, Dec 03 2014
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EXAMPLE
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A(x)^5 - 25x*A(x)^6 = 1 since A(x)^5 = 1 + 25x + 750x^2 + 24375x^3 + 831250x^4 + ... and A(x)^6 = 1 + 30x + 975x^2 + 33250x^3 + ... also a(4) = 5^7, a(9) = 5^18 = 3814697265625.
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MATHEMATICA
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Table[5^(2n) Binomial[(6n-4)/5, n]/(n+1), {n, 0, 25}] (* Harvey P. Dale, Mar 27 2011 *)
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PROG
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(PARI) for(n=0, 50, print1(5^(2*n)*binomial((6*n-4)/5, n)/(n+1), ", ")) \\ G. C. Greubel, Jan 30 2017
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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