login
a(n) = n(n+1) - A000120(n), where A000120(n) = number of 1's in binary expansion of n.
1

%I #8 Sep 08 2022 08:45:32

%S 0,1,5,10,19,28,40,53,71,88,108,129,154,179,207,236,271,304,340,377,

%T 418,459,503,548,598,647,699,752,809,866,926,987,1055,1120,1188,1257,

%U 1330,1403,1479,1556,1638,1719,1803,1888,1977,2066,2158,2251,2350,2447,2547

%N a(n) = n(n+1) - A000120(n), where A000120(n) = number of 1's in binary expansion of n.

%H Harvey P. Dale, <a href="/A134467/b134467.txt">Table of n, a(n) for n = 0..1000</a>

%F 2^a(n) = denominator of [x^n] (1+x)^(1/2^n) for n>=0 (see A134098); similarily, 2^a(n) = denominator of [x^n] 1/(1-x)^(1/2^n) for n>=0 (see A134097).

%t Table[n(n+1)-DigitCount[n,2,1],{n,0,50}] (* _Harvey P. Dale_, Jul 13 2013 *)

%o (PARI) a(n)=n*(n+1) - subst(Pol(binary(n)),x,1)

%o (Magma) [n^2 + Valuation(Factorial(n), 2): n in [0..60]]; // _Vincenzo Librandi_, Jun 12 2019

%Y Cf. A000120; A134097, A134098.

%K nonn

%O 0,3

%A _Paul D. Hanna_, Oct 27 2007