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A126674
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a(n) = n!*Sum_{j=0..n-1} 2^j/(j+1).
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4
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0, 1, 4, 20, 128, 1024, 9984, 115968, 1572864, 24477696, 430571520, 8452177920, 183175741440, 4343275192320, 111817607086080, 3105593229312000, 92539365359616000, 2944365169213440000, 99619235621240832000, 3571109329517936640000, 135199252993504444416000, 5390266968989421797376000
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OFFSET
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0,3
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COMMENTS
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R. J. Mathar's recurrence is correct. a(n) has a new sum term in addition to what a(n-1) has, giving a(n) = n*a(n-1) + 2^(n-1)*(n-1)!. (Cf. A000165 = 2^n*n!.) The same for a(n-1) from a(n-2), and a factor, is 2*(n-1)*(a(n-1) - (n-1)*a(n-2)) = 2^(n-1)*(n-1)! too. Substitute it leaving a(n) in terms of a(n-1) and a(n-2). The recurrence shows the o.g.f. satisfies the differential equation (2*x^2-x+1)*g + 3*x^2*(2*x-1)*g' + 2*x^4*g'' - x = 0. - Kevin Ryde, Jul 11 2019
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LINKS
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FORMULA
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O.g.f.: x*(Sum_{k>=0} k!*(2*x)^k)^2.
E.g.f.: log(1-2*x)/(x-1)/2. (End)
E.g.f.: E(x) = 1/2*log(1 - 2*x)/(x - 1) = x*(1 - x*G(0))/(x-1)/(2*x-1); G(k) = 1 + 2*x*(2*k+1)/(2*k + 3 - 2*x*(k+1)*(2*k+3)/(2*x*(k+1) + (k+2)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Dec 13 2011
Conjecture: a(n) + (-3*n+2)*a(n-1) + 2*(n-1)^2*a(n-2) = 0. - R. J. Mathar, May 23 2014
G.f.: x*(1/(1 - 2*x/(1 - 2*x/(1 - 4*x/(1 - 4*x/(1 - 6*x/(1 - 6*x/(1 - ...))))))))^2. - Ilya Gutkovskiy, May 10 2017
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MAPLE
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F:=n->add( n!*2^i/(1+i), i=0..n-1);
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MATHEMATICA
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Table[n!Sum[2^j/(j+1), {j, 0, n-1}], {n, 0, 30}] (* Harvey P. Dale, Jun 14 2017 *)
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PROG
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(Magma) [0] cat [Factorial(n)*(&+[2^j/(j+1):j in [0..n-1]]):n in [1..21]]; // Marius A. Burtea, Jul 12 2019
(PARI) a(n) = n!*sum(j=0, n-1, 2^j/(j+1)); \\ Michel Marcus, Jul 12 2019
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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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