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A316701
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E.g.f. A(x) satisfies: A(x) = Sum_{n>=0} x^n/n! * Product_{k=1..n} (n+1-k) + k*A(x).
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4
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1, 2, 13, 157, 2819, 67621, 2036230, 73907639, 3142556933, 153268340377, 8436526507286, 517427997295353, 34994424316034815, 2587503674068863681, 207665084850599068022, 17979537469340405579571, 1670426465731302891946025, 165771247503060676475253809, 17501167047878021578046031334, 1958599892703021903310163005669
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OFFSET
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0,2
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COMMENTS
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More generally, we have the following identity. Given the biexponential series
W(x,y) = Sum_{n>=0} 1/n! * Product_{k=1..n} (n+1-k)*x + k*y,
then for fixed p and q,
Sum_{n>=0} 1/n! * Product_{k=1..n} (n+1-k + p)*x + (k + q)*y = W(x,y)^(p+q+1) / ( (1 + x*W(x,y))^q * (1 + y*W(x,y))^p ).
Further, W(x,y) satisfies the biexponential functional equation
( W(x,y)/(1 + x*W(x,y)) )^x = ( W(x,y)/(1 + y*W(x,y)) )^y.
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LINKS
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FORMULA
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E.g.f. A(x) = Sum_{n>=0} a(n)*x^n/n! satisfies:
(1) A(x) = Sum_{n>=0} x^n/n! * Product_{k=1..n} (n+1-k) + k*A(x).
(2) Sum_{n>=0} x^n/n! * Product_{k=1..n} (n+1-k + p) + (k + q)*A(x) = A(x)^(p+q+1) / ( (1 + x*A(x))^q * (1 + x*A(x)^2)^p ), for fixed p and q.
(3) A(x)/(1 + x*A(x)) = ( A(x)/(1 + x*A(x)^2) )^A(x).
a(n) ~ sqrt(((-1 + s) * s^3 * (1 + r*s) * (1 + r*s^2) * (1 + s + r*s^2)) / (-1 - (1 + 2*r)*s - 4*r*s^2 + (4 - 5*r)*r*s^3 + 7*r^2*s^4 + r^2*(1 + 2*r)*s^5 + 2*r^3*s^6 + r^4*s^7)) * n^(n-1) / (exp(n) * r^(n - 1/2)), where r = 0.15709770426545411244999697565973251074761653302546043128667... and s = 2.3042217766396380492962218073654169636152676607799337588129... are roots of the system of equations (s/(1 + r*s^2))^s = s/(1 + r*s), -1 + s - r*s^3 - r^2*s^4 + s*(1 + r*s)*(1 + r*s^2) * log(s/(1 + r*s^2)) = 0. - Vaclav Kotesovec, Oct 13 2020
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EXAMPLE
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E.g.f.: A(x) = 1 + 2*x + 13*x^2/2! + 157*x^3/3! + 2819*x^4/4! + 67621*x^5/5! + 2036230*x^6/6! + 73907639*x^7/7! + 3142556933*x^8/8! + 153268340377*x^9/9! + ...
such that A = A(x) satisfies
A(x) = 1 + (1 + A)*x + (2 + A)*(1 + 2*A)*x^2/2! + (3 + A)*(2 + 2*A)*(1 + 3*A)*x^3/3! + (4 + A)*(3 + 2*A)*(2 + 3*A)*(1 + 4*A)*x^4/4! + (5 + A)*(4 + 2*A)*(3 + 3*A)*(2 + 4*A)*(1 + 5*A)*x^5/5! + ...
Also,
A(x)^2/(1 + x*A(x)) = 1 + (1 + 2*A)*x + (2 + 2*A)*(1 + 3*A)*x^2/2! + (3 + 2*A)*(2 + 3*A)*(1 + 4*A)*x^3/3! + (4 + 2*A)*(3 + 3*A)*(2 + 4*A)*(1 + 5*A)*x^4/4! + (5 + 2*A)*(4 + 3*A)*(3 + 4*A)*(2 + 5*A)*(1 + 6*A)*x^5/5! + ...
And,
A(x)^3/((1 + x*A(x))*(1 + x*A(x)^2)) = 1 + (2 + 2*A)*x + (3 + 2*A)*(2 + 3*A)*x^2/2! + (4 + 2*A)*(3 + 3*A)*(2 + 4*A)*x^3/3! + (5 + 2*A)*(4 + 3*A)*(3 + 4*A)*(2 + 5*A)*x^4/4! + (6 + 2*A)*(5 + 3*A)*(4 + 4*A)*(3 + 5*A)*(2 + 6*A)*x^5/5! + ...
RELATED SERIES.
A(x)/(1 + x*A(x)) = 1 + x + 7*x^2/2! + 85*x^3/3! + 1527*x^4/4! + 36621*x^5/5! + 1102348*x^6/6! + 39996727*x^7/7! + 1700108469*x^8/8! + ...
A(x)/(1 + x*A(x)^2) = 1 + x + 3*x^2/3! + 22*x^3/3! + 299*x^4/4! + 6086*x^5/5! + 164782*x^6/6! + 5553185*x^7/7! + 223540669*x^8/8! + ...
where ( A(x)/(1 + x*A(x)^2) )^A(x) = A(x)/(1 + x*A(x)).
Let G(x) = A(x/G(x)) and A(x) = G(x*A(x)), where G(x) begins
G(x) = 1 + 2*x + 5*x^2/2! + 19*x^3/3! + 87*x^4/4! + 481*x^5/5! + 3058*x^6/6! + 22317*x^7/7! + 183501*x^8/8! + ... + A316700(n)*x^n/n! + ...
then G(x)/(1 + x) = ( G(x)/(1 + x*G(x)) )^G(x)
and G(x) = x/Series_Reversion( x*A(x) ).
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MATHEMATICA
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nmax = 25; aa = ConstantArray[0, nmax]; aa[[1]] = 2; Do[y = 1 + 2*x + Sum[aa[[k]]*x^k, {k, 2, j - 1}] + koef*x^j; sol = Solve[SeriesCoefficient[(1 + x*y)*(y/(1 + x*y^2))^y - y, {x, 0, j + 1}] == 0, koef][[1]]; aa[[j]] = koef /. sol[[1]], {j, 2, nmax}]; Flatten[{1, aa}] * Range[0, nmax]! (* Vaclav Kotesovec, Oct 16 2020 *)
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PROG
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(PARI) /* From Biexponential Series: */
{a(n) = my(A=1); for(i=1, n, A = sum(m=0, n, x^m/m! * prod(k=1, m, m+1-k + k*A +x*O(x^n)))); n!*polcoeff(A, n)}
for(n=0, 20, 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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