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%I #78 Jun 19 2023 09:12:11
%S 1,-1,0,3,-7,0,59,-217,146,2593,-15551,32802,160709,-1856621,7971872,
%T 1299951,-287113779,2262481448,-7275903849,-36989148757,698330745002,
%U -4867040141851,10231044332629,184216198044034,-2679722886596295,17971204188130391,-17976259717948832
%N 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.
%C 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.
%C a(n)*(-1)^n gives the alternating row sums of the Sheffer triangle A143494 (2-restricted Stirling2). - _Wolfdieter Lang_, Oct 06 2011
%H Robert Israel, <a href="/A074051/b074051.txt">Table of n, a(n) for n = 0..545</a>
%F From _Vladeta Jovovic_, Jan 27 2005: (Start)
%F Second inverse binomial transform of A000587.
%F E.g.f.: exp(1 - 2*x - exp(-x)).
%F G.f.: Sum_{k >= 0}((x/(1+2*x))^k/Product_{l=0..k}(1 + l*x/(1+2*x)))/(1+2*x).
%F a(n) = Sum_{k=0..n} (-1)^(n-k)*(k^2-3*k+1)*Stirling2(n, k). (End)
%F a(n) = (-1)^n*(A000587(n+2)-A000587(n+1)). - _Peter Luschny_, Apr 17 2011
%F From _Sergei N. Gladkovskii_, Sep 28 2012 to Apr 22 2013: (Start)
%F Continued fractions:
%F G.f.: 1/U(0) where U(k)= x*k + 1 + x + x^2*(k+1)/U(k+1).
%F G.f.: -1/U(0) where U(k)= -x*k - 1 - x + x^2*(k+1)/U(k+1).
%F G.f.: 1/(U(0) - x) where U(k)= 1 + x + x*(k+1)/(1 - x/U(k+1)).
%F G.f.: 1/(U(0) + x) where U(k)= 1 + x*(2*k+1) - x*(k+1)/(1 + x/U(k+1)).
%F G.f.: 1/G(0) where G(k)= 1 + 2*x/(1 + 1/(1 + 2*x*(k+1)/G(k+1))).
%F 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))).
%F G.f.: (G(0) - 1)/(x-1) where G(k) = 1 - 1/(1+k*x+2*x)/(1-x/(x-1/G(k+1))).
%F G.f.: (G(0)-2-2*x)/x^2 where G(k) = 1 + 1/(1+k*x)/(1-x/(x+1/G(k+1) )).
%F 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))).
%F G.f.: (G(0)-2)/x where G(k) = 1 + 1/(1+k*x+x)/(1-x/(x+1/G(k+1))).
%F G.f.: (1+x)/x/Q(0) - 1/x, where Q(k)= 1 + x - x/(1 + x*(k+1)/Q(k+1)). (End)
%F a(n) = exp(1) * (-1)^n * Sum_{k>=0} (-1)^k * (k+2)^n / k!. - _Ilya Gutkovskiy_, Sep 02 2021
%e a(2)=0 because Sum_{i=1..m}i^2(i+1)! = (m-1)(m+2)!+2.
%e 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.
%p alias(S2 = combinat[stirling2]);
%p A074051 := proc(n) local k;
%p 1 + add((-1)^(n+k) * (S2(n+1, k+1) - S2(n+2, k+1)), k = 0..n) end:
%p seq(A074051(i), i = 0..26); # _Peter Luschny_, Apr 17 2011
%t 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]] ]
%t (* Second program: *)
%t a[n_] := (-1)^n (BellB[n+2, -1] - BellB[n+1, -1]);
%t Table[a[n], {n, 0, 30}] (* _Jean-François Alcover_, Jun 21 2018, after _Peter Luschny_ *)
%o (Python)
%o from itertools import accumulate
%o def A074051_list(size):
%o if size < 1: return []
%o L, accu = [], [1]
%o for n in range(size-1):
%o accu = list(accumulate([-accu[-1]] + accu))
%o L.append(-(-1)**n*accu[-2])
%o return L
%o print(A074051_list(28)) # _Peter Luschny_, Apr 25 2016
%Y Cf. A074052, A143494.
%K easy,sign
%O 0,4
%A _Jan Fricke_, Aug 14 2002
%E More terms from _Vladeta Jovovic_, Jan 27 2005
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