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A266329
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E.g.f. A(x) satisfies: A(x) = exp( Integral B(x) dx ) such that B(x) = exp(x) * exp( Integral A(x) dx ), where the constant of integration is zero.
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5
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1, 1, 3, 12, 62, 395, 2994, 26331, 263729, 2964845, 36975858, 506687604, 7568226163, 122388728056, 2130425343621, 39718373337525, 789613850257051, 16674806980716514, 372771700023167862, 8794945626017009781, 218392778569695964100, 5693513850197410142081, 155482323312112362743373, 4438621019461797437443233, 132210153223378852014571364, 4101859859297789141335079684, 132343983668857026899533814277
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OFFSET
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0,3
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COMMENTS
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Compare to: G(x) = exp( Integral G(x) dx ) when G(x) = 1/(1-x).
What is Limit (a(n)/n!)^(1/n) ? Example: (a(300)/300!)^(1/300) = 1.2409703...
Limit (a(n)/n!)^(1/n) = 1/Integral_{x=0..infinity} 1/(x + exp(x)) dx = 1.24008610649849766623949... - Vaclav Kotesovec, Aug 21 2017
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LINKS
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FORMULA
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E.g.f. A(x) satisfies:
(1) A(x) = exp( Integral A(x) + log(A(x)) dx ).
(2) A(x) = A'(x)/A(x) - log(A(x)).
(3) log(A(x)) = exp(x) * Integral exp(-x)*A(x) dx.
(4) A(x) = exp( Series_Reversion( Integral 1/(exp(x) + x) dx ) ).
a(n) ~ c^(n+1) * n!, where c = 1/Integral_{x=0..infinity} 1/(x + exp(x)) dx = 1.2400861064984976662394901721056528110217273471501174317019052800276... - Vaclav Kotesovec, Aug 21 2017
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EXAMPLE
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E.g.f.: A(x) = 1 + x + 3*x^2/2! + 12*x^3/3! + 62*x^4/4! + 395*x^5/5! + 2994*x^6/6! + 26331*x^7/7! + 263729*x^8/8! + 2964845*x^9/9! + 36975858*x^10/10! +...
such that log(A(x)) = Integral B(x) dx
where
B(x) = 1 + 2*x + 5*x^2/2! + 17*x^3/3! + 79*x^4/4! + 474*x^5/5! + 3468*x^6/6! + 29799*x^7/7! + 293528*x^8/8! + 3258373*x^9/9! + 40234231*x^10/10! +...
and A(x) and B(x) satisfy:
(1) A(x) = B'(x)/B(x) - 1,
(2) B(x) = A'(x)/A(x),
(3) B(x) = A(x) + log(A(x)),
(4) log(A(x)) = Integral B(x) dx,
(5) log(B(x)) = Integral A(x) dx + x.
The Series Reversion of log(A(x)) equals Integral 1/(exp(x) + x) dx:
Integral 1/(exp(x) + x) dx = x - 2*x^2/2! + 7*x^3/3! - 37*x^4/4! + 261*x^5/5! - 2301*x^6/6! + 24343*x^7/7! - 300455*x^8/8! + 4238153*x^9/9! - 67255273*x^10/10! +...+ (-1)^(n-1)*A072597(n-1)*x^n/n! +...
so that A( Integral 1/(exp(x) + x) dx ) = exp(x).
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MATHEMATICA
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a[ n_] := a[n] = If[ n < 1, Boole[n == 0], Sum[ Binomial[n - 1, k - 1] a[n - k] Sum[ a[k - j], {j, k}], {k, n}]]; (* Michael Somos, Aug 08 2017 *)
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PROG
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(PARI) {a(n) = my(A=1+x, B=1+x); for(i=0, n, A = exp( intformal( B + x*O(x^n) ) ); B = exp( intformal( 1 + A ) ) ); n!*polcoeff(A, n)}
for(n=0, 30, print1(a(n), ", "))
(PARI) {a(n) = n! * polcoeff( exp( serreverse( intformal( 1/(exp(x +x*O(x^n)) + x) ) )), n)}
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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