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A274377
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E.g.f. satisfies: A(x)^A(x) = exp(2*x) * A(-x)^A(-x).
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3
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1, 1, 0, 1, 0, 16, 0, 736, 0, 67096, 0, 10163176, 0, 2306198896, 0, 732199108096, 0, 309860700130816, 0, 168568765338224896, 0, 114619705107961862656, 0, 95251358122177791486976, 0, 94984793274454431691503616, 0, 111939507886837612683516276736, 0, 153907136552991217284274400567296, 0, 244164979570216285201628515234840576, 0, 442692827509235885935744380253757341696, 0, 909667081143908558901949811564629988048896
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OFFSET
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0,6
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COMMENTS
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a(2*n+1) = 6 (mod 10) for n>1 (conjecture).
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LINKS
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FORMULA
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E.g.f.: 1 + Series_Reversion( log( sqrt( (1+x)^(1+x) / (1-x)^(1-x) ) ) ).
E.g.f.: 1 + Series_Reversion( (G(x) - G(-x))/2 ), where G(x) = Series_Reversion(x/LambertW(x) - 1) = (1+x)*log(1+x).
E.g.f.: 1 + Series_Reversion( x - Sum_{n>=1} x^(2*n+1)/(2*n*(2*n+1)) ).
If n is odd then a(n) ~ c * d^n * n^(n-1) / exp(n), where d = 1.37441749603820461..., c = 0.6508250221842049... . - Vaclav Kotesovec, Sep 22 2016
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EXAMPLE
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E.g.f.: A(x) = 1 + x + x^3/3! + 16*x^5/5! + 736*x^7/7! + 67096*x^9/9! + 10163176*x^11/11! + 2306198896*x^13/13! + 732199108096*x^15/15! + 309860700130816*x^17/17! + 168568765338224896*x^19/19! +...
such that A(x)^A(x) / A(-x)^A(-x) = exp(2*x).
RELATED SERIES.
A(x)^A(x) = 1 + x + 2*x^2/2! + 4*x^3/3! + 16*x^4/4! + 56*x^5/5! + 426*x^6/6! + 2262*x^7/7! + 26944*x^8/8! + 191536*x^9/9! + 3126160*x^10/10! +...+ A275764(n)*x^n/n! +...
Series_Reversion(A(x) - 1) = x - x^3/6 - x^5/20 - x^7/42 - x^9/72 - x^11/110 - x^13/156 - x^15/210 - x^17/272 +...+ -x^(2*n+1)/(2*n*(2*n+1)) +...
Also,
Series_Reversion(A(x) - 1) = (G(x) - G(-x))/2, where G(x) = (1+x)*log(1+x) = Series_Reversion(x/LambertW(x) - 1), and begins:
G(x) = x + x^2/2 - x^3/6 + x^4/12 - x^5/20 + x^6/30 - x^7/42 + x^8/56 - x^9/72 + x^10/90 - x^11/110 + x^12/132 +...+ (-x)^n/(n*(n-1)) +...
GENERATING METHOD.
Start with a(0)=1, a(1)=1, and set a(2*n)=0 for n>0, then use the following criterion to determine the odd-indexed terms.
Given partial sum A(x,2*n) = Sum_{k=0..2*n} a(k)*x^k/k!, and sufficiently large N, the odd-indexed term a(2*n+1) satisfies:
if t > a(2*n+1)/(2*n+1)!, then
t > [x^(2*n+1)] ( A(x,2*n) + t*x^(2*n+1) )^(1-1/N)
else if t <= a(2*n+1)/(2*n+1)! , then
t < [x^(2*n+1)] ( A(x,2*n) + t*x^(2*n+1) )^(1-1/N);
this criterion defines each term of this sequence for n>1.
Using the same method as above, but without setting even-indexed terms to zero, generates x/LambertW(x), e.g.f. of A177885.
RELATED SERIES.
log(A(x)) = x - x^2/2! + 3*x^3/3! - 10*x^4/4! + 60*x^5/5! - 346*x^6/6! + 3108*x^7/7! - 25600*x^8/8! + 306120*x^9/9! - 3283696*x^10/10! + 49021368*x^11/11! - 648526000*x^12/12! + 11606584080*x^13/13! - 182697457216*x^14/14! +...
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PROG
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(PARI) {a(n) = my(A = 1 + serreverse(x - sum(m=1, n\2+1, x^(2*m+1)/(2*m*(2*m+1)) ) +x^2*O(x^n) ) ); n!*polcoeff(A, n)}
for(n=0, 40, print1(a(n), ", "))
(PARI) /* Generating method (using sufficiently large N and precision) */
\p100
{a(n) = my(N=10^(3*n), A=[1, 1]); for(i=0, n\2, A=concat(A, [0, 0]); A[#A] = round( (#A-1)!*polcoeff( N*1.* Ser(A)^(1-1/N), #A-1) )/(#A-1)! ); n!*A[n+1]}
for(n=0, 40, 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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