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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. 10
 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 From Sergei N. Gladkovskii, Sep 28 2012 to Apr 22 2013. (Start) Continued fractions: G.f.: 1/U(0) where U(k)= x*k + 1 + x + x^2*(k+1)/U(k+1). G.f.: -1/U(0) where U(k)= -x*k - 1 - x + x^2*(k+1)/U(k+1). G.f.: 1/(U(0) - x) where U(k)= 1 + x + x*(k+1)/(1 - x/U(k+1)). G.f.: 1/(U(0) + x) where U(k)= 1 + x*(2*k+1) - x*(k+1)/(1 + x/U(k+1)). G.f.: 1/G(0) where G(k)= 1 + 2*x/(1 + 1/(1 + 2*x*(k+1)/G(k+1))). 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))). G.f.: (G(0) - 1)/(x-1) where G(k) =  1 - 1/(1+k*x+2*x)/(1-x/(x-1/G(k+1))). G.f.: (G(0)-2-2*x)/x^2 where G(k) = 1 + 1/(1+k*x)/(1-x/(x+1/G(k+1) )). 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))). G.f.: (G(0)-2)/x where G(k) = 1 + 1/(1+k*x+x)/(1-x/(x+1/G(k+1))). G.f.: (1+x)/x/Q(0) - 1/x, where Q(k)=  1 + x - x/(1 + x*(k+1)/Q(k+1)). (End) 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]] ] (* Second program: *) a[n_] := (-1)^n (BellB[n+2, -1] - BellB[n+1, -1]); Table[a[n], {n, 0, 30}] (* Jean-François Alcover, Jun 21 2018, after Peter Luschny *) 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: A198490 A247956 A295043 * 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 May 19 04:06 EDT 2019. Contains 323377 sequences. (Running on oeis4.)