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a(n) = A160799(n)/4.
1

%I #16 Oct 06 2024 20:43:58

%S 0,1,5,12,28,47,75,112,176,243,319,404,516,637,785,960,1216,1475,1743,

%T 2020,2324,2637,2977,3344,3792,4249,4733,5244,5836,6455,7155,7936,

%U 8960,9987,11023,12068,13140,14221,15329,16464,17680,18905,20157

%N a(n) = A160799(n)/4.

%F G.f.: x^2*Product_{i>=0} p(x^(2^i)) where p(x) = 1 + 5*x + 7*x^2 + 3*x^3. - _Gary W. Adamson_, Aug 25 2016 [edited by _Jason Yuen_, Oct 06 2024]

%o (PARI) a(n)=sum(i=0, n-1, (n-i)*3^hammingweight(i)) \\ _Charles R Greathouse IV_, Aug 25 2016

%Y Cf. A160410, A161411, A160799.

%Y Essentially partial sums of A130665.

%K nonn

%O 1,3

%A _Omar E. Pol_, Jun 14 2009

%E More terms from _Max Alekseyev_, Dec 12 2011