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For each n there uniquely determined numbers a(n) and b(n) and a polynomial p_n such that for all integers m: Sum_{i=1..m}i^n(i+1)! = a(n)*Sum_{i=1..m}(i+1)! + p_n(m)(m+2)! + b(n) The sequence b(n) is A074052.
Second inverse binomial transform of A000587. E.g.f.: exp(1-2*x-exp(-x)). G.f.: Sum((x/(1+2*x))^k/Product(1+l*x/(1+2*x), l = 0 .. k), k = 0 .. infinity)/(1+2*x). a(n) = Sum_{k=0..n} (-1)^(n-k)*(k^2-3*k+1)*Stirling2(n, k). - Vladeta Jovovic, Jan 27 2005
a(n) = (-1)^n*(A000587(n+2)-A000587(n+1)). - Peter Luschny, Apr 17 2011
G.f.: 1/U(0) where U(k)= x*k + 1 + x + x^2*(k+1)/U(k+1) ; (continued fraction, 1-step). - Sergei N. Gladkovskii, Sep 28 2012
G.f.: -1/U(0) where U(k)= -x*k - 1 - x + x^2*(k+1)/U(k+1) ; (continued fraction, 1-step). - Sergei N. Gladkovskii, Sep 29 2012
G.f.: 1/(U(0) - x) where U(k)= 1 + x + x*(k+1)/(1 - x/U(k+1)) ; (continued fraction, 2-step). - Sergei N. Gladkovskii, Oct 12 2012
G.f.: 1/(U(0) + x) where U(k)= 1 + x*(2*k+1) - x*(k+1)/(1 + x/U(k+1)) ; (continued fraction, 2-step). - Sergei N. Gladkovskii, Oct 13 2012
G.f.: 1/G(0) where G(k)= 1 + 2*x/(1 + 1/(1 + 2*x*(k+1)/G(k+1)));(continued fraction, 3-step). - Sergei N. Gladkovskii, Nov 21 2012
G.f.: 1 - 2*x/(G(0) + 2*x) where G(k)= 1 + 1/(1 + 2*x*(k+1)/(1 + 2*x/G(k+1)));(continued fraction, 3-step). - Sergei N. Gladkovskii, Nov 25 2012
G.f.: (G(0) - 1)/(x-1) where G(k) = 1 - 1/(1+k*x+2*x)/(1-x/(x-1/G(k+1) )); (recursively defined continued fraction). - Sergei N. Gladkovskii, Jan 17 2013
G.f.: (G(0)-2-2*x)/x^2 where G(k) = 1 + 1/(1+k*x)/(1-x/(x+1/G(k+1) )); (recursively defined continued fraction). - Sergei N. Gladkovskii, Feb 09 2013
G.f.: (S-2-2*x)/x^2 where S = sum(k>=0, (2 + x*k)*x^k/prod(i=0..k, (1+x*i)) ). - Sergei N. Gladkovskii, Feb 09 2013
G.f.: (G(0)-2)/x where G(k) = 1 + 1/(1+k*x+x)/(1-x/(x+1/G(k+1) )); (recursively defined continued fraction). - Sergei N. Gladkovskii, Feb 09 2013
G.f.: (1+x)/x/Q(0) - 1/x, where Q(k)= 1 + x - x/(1 + x*(k+1)/Q(k+1)); (continued fraction). - Sergei N. Gladkovskii, Apr 22 2013
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