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%I #25 Oct 07 2016 05:37:07
%S 0,1,1,4,5,10,14,26,35,56,77,116,157,226,302,424,560,762,998,1334,
%T 1727,2270,2914,3779,4809,6163,7781,9875,12378,15565,19383,24191,
%U 29934,37093,45643,56201,68789,84212,102564,124903,151424,183499,221508
%N Total number of parts which are positive powers of 2 in all partitions of n.
%H Alois P. Heinz and Vaclav Kotesovec, <a href="/A073119/b073119.txt">Table of n, a(n) for n = 1..10000</a> (terms 1..2000 from Alois P. Heinz)
%F a(n) = Sum_{k=1..n} A007814(k)*A000041(n-k).
%F G.f.: g(x) = (Sum_{i>0} x^(h(i))/(1-x^(h(i))))/Product_{i>0}(1 - x^i), where h(i) = 2^i. - _Emeric Deutsch_, Sep 19 2016.
%F Conjecture: a(n) ~ exp(sqrt(2*n/3)*Pi)/(2*Pi*sqrt(2*n)) ~ p(n) * sqrt(6*n)/Pi, where p(n) is the partition function A000041. - _Vaclav Kotesovec_, Oct 07 2016
%e a(5) = 5 because in the partitions [1,1,1,1,1], [1,1,1,2'], [1,2'2'], [1,1,3], [2',3],[1,4'], [5] we have 5 positive powers of 2 (they are marked). - _Emeric Deutsch_, Sep 19 2016.
%p p2:= proc(n) p2(n):= is(n=2^ilog2(n)) end: p2(1):= false:
%p b:= proc(n, i) option remember; local t, l;
%p if n<0 then [0, 0]
%p elif n=0 then [1, 0]
%p elif i<1 then [0, 0]
%p else t:= b(n, i-1);
%p l:= b(n-i, i);
%p [t[1]+l[1], t[2]+l[2]+ `if`(p2(i), l[1], 0)]
%p fi
%p end:
%p a:= n-> b(n, n)[2]:
%p seq(a(n), n=1..50); # _Alois P. Heinz_, Sep 29 2011
%t Needs["DiscreteMath`Combinatorica`"]; f[n_] := Length[ Select[ Log[2, Flatten[ Partitions[n]]], IntegerQ[ # ] && # > 0 & ]]; Table[ f[n], {n, 1, 45}]
%t a[n_] := Sum[IntegerExponent[k, 2]*PartitionsP[n-k], {k, 1, n}]; Table[a[n], {n, 1, 50}] (* _Jean-François Alcover_, Jan 28 2014 *)
%Y Cf. A024786, A000070.
%K easy,nonn
%O 1,4
%A _Vladeta Jovovic_, Aug 24 2002
%E Edited and extended by _Robert G. Wilson v_, Aug 26 2002
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