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%I #30 Sep 11 2024 00:53:16
%S 2,5,4,10,6,11,8,19,10,17,12,24,14,23,16,36,18,29,20,38,22,35,24,49,
%T 26,41,28,52,30,47,32,69,34,53,36,66,38,59,40,79,42,65,44,80,46,71,48,
%U 98,50,77,52,94,54,83,56,109,58,89,60,108,62,95,64,134,66,101,68,122,70,107,72,139,74,113
%N Expansion of 2/(1-x) + Sum_{k>=0} t^2(3-t)/(1+t)/(1-t)^2, where t=x^2^k.
%H Seiichi Manyama, <a href="/A084432/b084432.txt">Table of n, a(n) for n = 1..10000</a>
%F a(1)=2, a(2*n) = a(n)+2*n+1, a(2*n+1) = 2*n+2.
%F Dirichlet g.f.: 2^s/(2^s-1) * (zeta(s)+zeta(s-1)). - _Ralf Stephan_, Jun 17 2007
%F From _Seiichi Manyama_, May 27 2024: (Start)
%F G.f. A(x) satisfies A(x) = 1/(1 - x)^2 - 1 + A(x^2).
%F G.f.: A(x) = Sum_{k>=0} (1/(1 - x^(2^k))^2 - 1). (End)
%F Conjecture: a(n) is the number of integer solutions (x,y,z) to the Diophantine equation 2^x*(y+z) = n, where 0 <= x,y,z <= n. - _Joseph M. Shunia_, Aug 27 2024
%F Sum_{k=1..n} a(k) ~ 2*n^2/3 + 2*n. - _Vaclav Kotesovec_, Aug 28 2024
%o (PARI) for(n=1, 100, l=ceil(log(n)/log(2)); t=polcoeff(sum(k=0, l, 1/(1-x^2^k)^2) + O(x^(n+1)), n); print1(t", "))
%Y Cf. A069895, A373184, A373185.
%K nonn,easy
%O 1,1
%A _Ralf Stephan_, Jun 27 2003