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A179929
a(n) = 2^n*A(n, -1/2), A(n, x) the Eulerian polynomials.
5
1, 2, 2, -6, -30, 42, 882, 954, -39870, -203958, 2300562, 29677914, -120958110, -4657703958, -7059175758, 807984602874, 6667870853250, -145556787011958, -2827006784652078, 21703953751815834, 1108558810703202210
OFFSET
0,2
COMMENTS
Conjecture: the sequence {a(n)} reduced modulo a positive integer k is eventually periodic with period a divisor of phi(k), where phi() = A000010(k). For example, modulo 11 the sequence becomes [1, 2, 2, 5, 3, 9, 2, 8, 5, 4, 0, 2, 2, 5, 3, 9, 2, 8, 5, 4, 0, 2, 2, 5, 3, 9, 2, 8, 5, 4, 0, ...], with an apparent period of 10 = phi(11) starting at n = 1. - Peter Bala, Feb 20 2026
FORMULA
From Philippe Deléham, Dec 22 2011: (Start)
a(n) = Sum_{k = 0..n} A123125(n,k)*(-1)^(n-k)*2^k
a(n) = Sum_{k = 0..n} A173018(n,k)*2^(n-k)*(-1)^k. (End)
From Peter Bala, Mar 12 2013: (Start)
E.g.f.: 3/(1 + 2*exp(-3*x)) = 1 + 2*x + 2*x^2/2! - 6*x^3/3! - 30*x^4/4! + ....
Recurrence equation: a(n+1) = 3*a(n) - Sum_{k = 0..n} binomial(n, k)*a(k)*a(n-k).
(-1)^n*a(n) are the coefficients of a delta operator associated with a sequence of polynomials of binomial type - see A195205. (End)
a(n) ~ n! * 2*3^(n+1)/(Pi^2+(log(2))^2)^(n/2+1) * (Pi*sin(n*arctan(Pi/log(2))) - log(2)*cos(n*arctan(Pi/log(2)))). - Vaclav Kotesovec, Oct 09 2013
From Stanislav Sykora, May 15 2014: (Start)
a(n) = -2*A212846(n) for n > 0.
a(n) = -3^(n+1)*Li(-n, -1/2), with Li(-n, x) = Sum_{k>=0} ((k^n)*(x^k)) the polylogarithm.
a(n) = Sum_{k = 0..n} 3^(n-k)*(-1)^k*k!*S(n+1, k+1), S(m, l) the Stirling number of second kind. (End)
MAPLE
# uses Graves' method to compute the series coefficients (but with offset 1)
d := proc (n, x) option remember; if n = 0 then x else simplify( (3 - 2*exp(-x))*
diff( d(n-1, x), x) ) end if end proc:
seq(eval(d(n, x), x = 0), n = 1..20);
MATHEMATICA
a[n_] := Sum[3^(n-k) (-1)^k k! StirlingS2[n+1, k+1], {k, 0, n}];
Table[a[n], {n, 0, 20}] (* Jean-François Alcover, Jun 17 2019 *)
PROG
(PARI) A179929(n) = {my(s, k, term);
term = 3^n; s = term*stirling(n+1, 1, 2);
for (k=1, n, term *= -k/3; s += term*stirling(n+1, k+1, 2); );
return(s); } \\ Stanislav Sykora, May 15 2014
CROSSREFS
Cf. A000629 = 2^n*A(n, 1/2).
Sequence in context: A326907 A270487 A058250 * A278258 A067644 A184312
KEYWORD
sign,easy
AUTHOR
Peter Luschny, Aug 03 2010
STATUS
approved