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n appears (n+1)/(1 + (n mod 2)) times.
6

%I #32 Jun 14 2017 09:47:37

%S 0,1,2,2,2,3,3,4,4,4,4,4,5,5,5,6,6,6,6,6,6,6,7,7,7,7,8,8,8,8,8,8,8,8,

%T 8,9,9,9,9,9,10,10,10,10,10,10,10,10,10,10,10,11,11,11,11,11,11,12,12,

%U 12,12,12,12,12,12,12,12,12,12,12,13,13,13,13,13,13

%N n appears (n+1)/(1 + (n mod 2)) times.

%C n appears A001318(n+1) - A001318(n) = A026741(n+1) times.

%C Sum_{k=0...a(n)} (-1)^ceiling(k/2)*p(n-G(k)) = 0 for n>0, where p(n)=A000041(n) is the partition function, and G(k)=A001318(k) denotes the generalized pentagonal numbers.

%C Row lengths of A238442, n >= 1. - _Omar E. Pol_, Dec 22 2016

%H G. C. Greubel, <a href="/A235963/b235963.txt">Table of n, a(n) for n = 0..5000</a>

%F Let t = (sqrt(n*8/3 + 1) - 1)/2 + 1/3 and let k = floor(t); then a(n) = 2k if t - k < 2/3, 2k+1 otherwise. - _Jon E. Schoenfield_, Jun 13 2017

%t Table[Table[n, {(n + 1)/(1 + Mod[n, 2])}], {n, 0, 14}]//Flatten (* _T. D. Noe_, Jan 29 2014 *)

%Y Cf. A000041, A001318, A026741, A238442.

%K nonn

%O 0,3

%A _Mircea Merca_, Jan 17 2014