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%I #15 Nov 17 2022 06:24:50
%S 1,1,1,1,4,5,15,35,56,126,252,462,792,1716,3003,5005,8008,12376,18564,
%T 27132,38760,116280,170544,245157,346104,480700,657800,888030,1184040,
%U 1560780,2035800,2629575,3365856,4272048,18156204,23535820,30260340,38608020,48903492
%N Number of subsets of [n] in which exactly half of the elements are Fibonacci numbers.
%H Alois P. Heinz, <a href="/A357927/b357927.txt">Table of n, a(n) for n = 0..10000</a>
%F a(n) = binomial(n,A072649(n)).
%F a(n) = Sum_{j>=0} binomial(A072649(n),j)*binomial(n-A072649(n),j).
%e a(6) = 15: {}, {1,4}, {1,6}, {2,4}, {2,6}, {3,4}, {3,6}, {4,5}, {5,6}, {1,2,4,6}, {1,3,4,6}, {1,4,5,6}, {2,3,4,6}, {2,4,5,6}, {3,4,5,6}.
%p f:= proc(n) option remember; `if`(n=0, 0, f(n-1)+
%p `if`((t-> ormap(issqr, [t-4, t+4]))(5*n^2), 1, 0))
%p end:
%p a:= n-> binomial(n, f(n)):
%p seq(a(n), n=0..38);
%t f[n_] := Module[{j}, For[j = Floor@Log[GoldenRatio, n], Fibonacci[j+1] <= n, j++]; j-1];
%t a[n_] := If[n == 0, 1, Binomial[n, f[n]]];
%t Table[a[n], {n, 0, 38}] (* _Jean-François Alcover_, Nov 17 2022 *)
%Y Cf. A000045, A037031, A072649, A102366, A180272, A357812.
%K nonn
%O 0,5
%A _Alois P. Heinz_, Oct 20 2022
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