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A337755
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G.f. A(x) satisfies: 1 = Sum_{n>=0} (n+1) * 2^n * ((1+x)^n - A(x))^n.
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4
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1, 1, 3, 52, 1320, 43440, 1722712, 79186272, 4118457732, 238450436416, 15189543467688, 1055122226778720, 79359687454230296, 6425540170275120528, 557306222539540276176, 51558846502494563714080, 5068865533417385007925076, 527798429103621760357553448, 58032815995114574483132149504
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OFFSET
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0,3
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COMMENTS
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In general, the following sums are equal:
(1) Sum_{n>=0} binomial(n+k-1, n) * r^n * (p + q^n)^n,
(2) Sum_{n>=0} binomial(n+k-1, n) * r^n * q^(n^2) / (1 - r*p*q^n)^(n+k),
for any fixed integer k; here, k = 2 with r = 2, p = -A(x), q = (1+x).
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LINKS
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FORMULA
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G.f. A(x) satisfies:
(1) 1 = Sum_{n>=0} (n+1) * 2^n * ((1+x)^n - A(x))^n.
(2) 1 = Sum_{n>=0} (n+1) * 2^n * (1+x)^(n^2) / (1 + 2*(1+x)^n*A(x))^(n+2).
a(n) ~ c * d^n * n! * sqrt(n), where d = (1 + 2*exp(1/r)) * r^2 = 5.9039765456700218004884947864345974878995107118996038057..., where r = 0.925556278640887084941460444526398190071550948416... is the root of the equation exp(1/r) * (1 + 1/(r*LambertW(-exp(-1/r)/r))) = -1/2 and c = 0.0300441568138... - Vaclav Kotesovec, Sep 25 2020
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EXAMPLE
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G.f.: A(x) = 1 + x + 3*x^2 + 52*x^3 + 1320*x^4 + 43440*x^5 + 1722712*x^6 + 79186272*x^7 + 4118457732*x^8 + 238450436416*x^9 + ...
where
1 = 1 + 2*2*((1+x) - A(x)) + 3*2^2*((1+x)^2 - A(x))^2 + 4*2^3*((1+x)^3 - A(x))^3 + 5*2^4*((1+x)^4 - A(x))^4 + 6*2^5*((1+x)^5 - A(x))^5 + 7*2^6*((1+x)^6 - A(x))^6 + 8*2^7*((1+x)^7 - A(x))^7 + ... + (n+1)*2^n*((1+x)^n - A(x))^n + ...
Also,
1 = 1/(1 + 2*A(x))^2 + 2*2*(1+x)/(1 + 2*(1+x)*A(x))^3 + 3*2^2*(1+x)^4/(1 + 2*(1+x)^2*A(x))^4 + 4*2^3*(1+x)^9/(1 + 2*(1+x)^3*A(x))^5 + 5*2^4*(1+x)^16/(1 + 2*(1+x)^4*A(x))^6 + 6*2^5*(1+x)^25/(1 + 2*(1+x)^5*A(x))^7 + 7*2^6*(1+x)^36/(1 + 2*(1+x)^6*A(x))^8 + ... + (n+1)*2^n*(1+x)^(n^2)/(1 + 2*(1+x)^n*A(x))^(n+2) + ...
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PROG
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(PARI) {a(n) = my(A=[1]); for(i=0, n, A=concat(A, 0); A[#A] = Vec( sum(m=0, #A, (m+1) * 2^m * ((1+x)^m - Ser(A))^m ) )[#A]/4 ); A[n+1]}
for(n=0, 30, print1(a(n), ", "))
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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