

A165998


Denominators of Taylor series expansion of 1/(3*x)*log((1+x)/(1x)^2)


27



1, 6, 3, 12, 5, 18, 7, 24, 9, 30, 11, 36, 13, 42, 15, 48, 17, 54, 19, 60, 21, 66, 23, 72, 25, 78, 27, 84, 29, 90, 31, 96, 33, 102, 35, 108, 37, 114, 39, 120, 41, 126, 43, 132, 45, 138, 47, 144, 49, 150, 51, 156, 53, 162, 55, 168, 57, 174, 59, 180, 61, 186, 63, 192, 65, 198
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OFFSET

0,2


COMMENTS

Numerators are all 1.
Setting x=1/3 into 1/(3*x)*log((1+x)/(1x)^2) = Sum_{k>=0} x^k/((2(1)^k)*(k+1)),
log(3) = Sum_{k>=0} 1/((2(1)^k)*(k+1)*3^k) = Sum_{k>=0} (9/(2k+1)+1/(2k+2))/9^(k+1) is obtained.
It appears that this is also the first differences of the generalized decagonal numbers A074377.  Omar E. Pol, Sep 10 2011


LINKS



FORMULA

G.f.: (1+6*x+x^2)/(1x^2)^2.
a(n) = (2(1)^n)*(n+1) (see PARI's code by Jaume Oliver Lafont).
With offset 1 this sequence is multiplicative (in fact, a generalized totient function): a(p^e) = p^e for any odd prime p and a(2^e) = 3*2^e for e >= 1.  Charles R Greathouse IV, Mar 09 2015
With offset 1, Dirichlet g.f.: zeta(s1) * (1 + 2^(2s)).  Amiram Eldar, Oct 25 2023


MATHEMATICA

LinearRecurrence[{0, 2, 0, 1}, {1, 6, 3, 12}, 50] (* Vincenzo Librandi, Feb 22 2012 *)


PROG

(PARI) a(n)=(2(1)^n)*(n+1)


CROSSREFS



KEYWORD

frac,nonn,easy


AUTHOR



STATUS

approved



