A239275 a(n) = numerator(2^n * Bernoulli(n, 1)).

1, 1, 2, 0, -8, 0, 32, 0, -128, 0, 2560, 0, -1415168, 0, 57344, 0, -118521856, 0, 5749735424, 0, -91546451968, 0, 1792043646976, 0, -1982765704675328, 0, 286994513002496, 0, -3187598700536922112, 0, 4625594563496048066560, 0, -16555640873195841519616, 0, 22142170101965089931264, 0

Offset: 0

Author

Paul Curtz, Mar 13 2014

Keywords

• sign
• frac
• easy

Comments

Difference table of f(n) = 2^n *A164555(n)/A027642(n) = a(n)/A141459(n):

1, 1, 2/3, 0, -8/15, 0, 32/21, 0,...

0, -1/3, -2/3, -8/15, 8/15, 32/21, -32/21,...

-1/3, -1/3, 2/15, 16/15, 104/105, -64/21,...

0, 7/15, 14/15, -8/105, -424/105,...

7/15, 7/15, -106/105, -416/105,...

0, -31/21, -62/31,

-31/21, -31/21,...

0,... etc.

Main diagonal: A212196(n)/A181131(n). See A190339(n).

First upper diagonal: A229023(n)/A181131(n).

The inverse binomial transform of f(n) is g(n). Reciprocally, the inverse binomial transform of g(n) is f(n) with -1 instead of f(1)=1, i.e., f(n) signed.

Sum of the antidiagonals: 1,1,0,-1,0,3,0,-17,... = (-1)^n*A036968(n) = -A226158(n+1).

Following A211163(n+2), f(n) is the coefficients of a polynomial in Pi^n.

Bernoulli numbers, twice, and Genocchi numbers, twice, are linked to Pi.

f(n) - g(n) = -A226158(n).

Also the numerators of the centralized Bernoulli polynomials 2^n*Bernoulli(n, x/2+1/2) evaluated at x=1. The denominators are A141459. - Peter Luschny, Nov 22 2015

(-1)^n*a(n) = 2^n*numerator(A027641(n)/A027642(n)) (that is the present sequence with a(1) = -1 instead of +1). - Wolfdieter Lang, Jul 05 2017

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..200
• Wolfdieter Lang, On Sums of Powers of Arithmetic Progressions, and Generalized Stirling, Eulerian and Bernoulli Numbers, arXiv:math/1707.04451 [math.NT], July 2017. See B(2;n), eq. (53).

Crossrefs

Cf. A141459 (denominators), A001896/A001897, A027641/A027642.

Programs

Maple

seq(numer(2^n*bernoulli(n, 1)), n=0..35); # Peter Luschny, Jul 17 2017

Mathematica

Table[Numerator[2^n*BernoulliB[n, 1]], {n, 0, 100}] (* Indranil Ghosh, Jul 18 2017 *)

Python

from sympy import bernoulli
def a(n): return (2**n * bernoulli(n, 1)).numerator
print([a(n) for n in range(51)]) # Indranil Ghosh, Jul 18 2017

Formula

a(n) = numerators of 2^n * A164555(n)/A027642(n).

Numerators of the binomial transform of A157779(n)/(interleave A001897(n), 1)(conjectured).