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A239275
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a(n) = numerator(2^n * Bernoulli(n, 1)).
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3
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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
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OFFSET
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0,3
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COMMENTS
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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.
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.
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
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LINKS
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FORMULA
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Numerators of the binomial transform of A157779(n)/(interleave A001897(n), 1)(conjectured).
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MAPLE
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seq(numer(2^n*bernoulli(n, 1)), n=0..35); # Peter Luschny, Jul 17 2017
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MATHEMATICA
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Table[Numerator[2^n*BernoulliB[n, 1]], {n, 0, 100}] (* Indranil Ghosh, Jul 18 2017 *)
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PROG
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(Python)
from sympy import bernoulli
def a(n): return (2**n * bernoulli(n, 1)).numerator()
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CROSSREFS
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KEYWORD
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sign,frac,easy
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AUTHOR
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STATUS
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approved
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