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A074051 For each n there are uniquely determined numbers a(n) and b(n) and a polynomial p_n(x) such that for all integers m we have 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. 9
1, -1, 0, 3, -7, 0, 59, -217, 146, 2593, -15551, 32802, 160709, -1856621, 7971872, 1299951, -287113779, 2262481448, -7275903849, -36989148757, 698330745002, -4867040141851, 10231044332629, 184216198044034, -2679722886596295, 17971204188130391, -17976259717948832 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,4

COMMENTS

If a(n)=0 then Sum_{i>=1}i^n(i+1)! = b(n) in the p-adic numbers. The only known numbers n with a(n)=0 are 2 and 5.

a(n)*(-1)^n gives the alternating row sums of the Sheffer triangle A143494 (2-restricted Stirling2). - Wolfdieter Lang, Oct 06 2011

LINKS

Robert Israel, Table of n, a(n) for n = 0..545

FORMULA

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

EXAMPLE

a(2)=0 because Sum_{i=1..m}i^2(i+1)! = (m-1)(m+2)!+2.

a(3)=3 because Sum_{i=1..m}i^3(i+1)! = 3*Sum_{i=1..m}(i+1)!+(m^2-m-1)(m+2)!+2.

MAPLE

alias(S2 = combinat[stirling2]);

A074051 := proc(n) local k;

1 + add((-1)^(n+k) * (S2(n+1, k+1) - S2(n+2, k+1)), k = 0..n) end:

seq(A074051(i), i = 0..26); # Peter Luschny, Apr 17 2011

MATHEMATICA

A[a_] := Module[{p, k}, p[n_] = 0; For[k = a - 1, k >= 0, k--, p[n_] = Expand[p[n] + n^k Coefficient[n^a - (n + 2)p[n] + p[n - 1], n^(k + 1)]] ]; Expand[n^a - (n + 2)p[n] + p[n - 1]] ]

PROG

(Python)

from itertools import accumulate

def A074051_list(size):

    if size < 1: return []

    L, accu = [], [1]

    for n in range(size-1):

        accu = list(accumulate([-accu[-1]] + accu))

        L.append(-(-1)**n*accu[-2])

    return L

print(A074051_list(28)) # Peter Luschny, Apr 25 2016

CROSSREFS

Cf. A074052, A143494.

Sequence in context: A199068 A198490 A247956 * A048292 A072450 A085785

Adjacent sequences:  A074048 A074049 A074050 * A074052 A074053 A074054

KEYWORD

easy,sign

AUTHOR

Jan Fricke (fricke(AT)uni-greifswald.de), Aug 14 2002

EXTENSIONS

More terms from Vladeta Jovovic, Jan 27 2005

STATUS

approved

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Last modified December 4 05:09 EST 2016. Contains 278748 sequences.