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A174341
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a(n) = numerator(Bernoulli(n, 1) + 1/(n+1)).
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7
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2, 1, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, -37, 1, 37, 1, -211, 1, 2311, 1, -407389, 1, 37153, 1, -1181819909, 1, 76977929, 1, -818946931, 1, 277930363757, 1, -84802531453217, 1, 90219075042851, 1, -711223555487930419, 1, 12696640293313423, 1, -6367871182840222481, 1, 35351107998094669831, 1, -83499808737903072705023, 1, 12690449182849194963361, 1
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OFFSET
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0,1
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COMMENTS
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1/(n+1) and Bernoulli(n,1) are autosequences in the sense that they remain the same (up to sign) under inverse binomial transform. This feature is kept for their sum, a(n)/A174342(n) = 2, 1, 1/2, 1/4, 1/6, 1/6, 1/6, 1/8, 7/90, 1/10, ...
Conjecture: the numerator of (A164555(n)/(n+1) + A027642(n)/(n+1)^2) is a(n) and the denominator of this fraction is equal to 1 if and only if n+1 is prime or 1. Cf. A309132. - Thomas Ordowski, Jul 09 2019
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LINKS
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MAPLE
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B := proc(n) if n <> 1 then bernoulli(n) ; else -bernoulli(n) ; end if; end proc:
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MATHEMATICA
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a[n_] := Numerator[BernoulliB[n, 1] + 1/(n + 1)];
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PROG
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(PARI)
B(n)=if(n!=1, bernfrac(n), -bernfrac(n));
a(n)=numerator(B(n) + 1/(n + 1));
(Python)
from sympy import bernoulli, Integer
def B(n): return bernoulli(n) if n != 1 else -bernoulli(n)
def a(n): return (B(n) + 1/Integer(n + 1)).numerator() # Indranil Ghosh, Jun 19 2017
(Magma) [2, 1] cat [Numerator(Bernoulli(n)+1/(n+1)): n in [2..40]]; // Vincenzo Librandi, Jul 18 2019
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CROSSREFS
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KEYWORD
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sign,frac
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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