%I #17 Jun 20 2019 11:38:03
%S 7,7,6,3,4,4,3,4,8,10,2,2,2,4,6,8,10,3,2,2,2,2,4,4,4,5,6,6,6,14,3,2,2,
%T 2,2,2,2,2,4,4,4,4,4,4,4,6,6,6,8,8,8,8,12,4,2,2,2,2,2,2,2,2,2,2,2,2,2,
%U 4,4,4,4,4,4,4,4,4,4,4,4,6,6,6,6,6,6,6
%N The number of positive integers m for which the exponents of 2 and p_n in the prime power factorization of m! are both powers of 2.
%C Or a(n) is the maximal m for which the Fermi-Dirac representation of m! (see comment in A050376) contains single power of 2 and single power of prime(n).
%H Robert Price, <a href="/A177436/b177436.txt">Table of n, a(n) for n = 2..127</a>
%H V. Shevelev, <a href="http://journals.impan.gov.pl/aa/Inf/126-3-1.html">Compact integers and factorials</a>, Acta Arithmetica 126 (2007), no. 3, 195-236.
%F a(2)=a(3)=7; a(4)=6; if p_n has the form (2^(4*k+1)+3)/5,k>=2,then a(n)=5; if p_n is a Fermat prime: p_n=2^(2^(k-1))+1, k>=3, then a(n)=4; if p_n has the form 2^k+3, k>=3, then a(n)=3; otherwise, if 2^(k-1)+3<p_n<=2^k-1, then a(n)=2*(1+floor(log_2((p_n-5)/(2^k-p_n))), where p_n=prime(n).
%e For p_5=11, we have 11=2^3+3. Therefore a(5)=3; for p_27=103, we have 103=(2^(4*2+1)+3)/5. Therefore a(27)=5; for p_31=127, a(31)=2*(1+floor(log_2((127-5)/(128-127)))=14.
%t nlim = 127; mlim = (Prime[nlim] + 1)^2/2 + 3; f = Table[0, mlim]; c = Table[0, nlim];
%t For[m = 2, m <= mlim, m++,
%t mf = FactorInteger[m];
%t For[i = 1, i <= Length[mf], i++, f[[PrimePi@First@mf[[i]]]] += Last@mf[[i]]];
%t If[! IntegerQ@Log[2, f[[1]]], Continue[]];
%t For[p = 1, p <= nlim, p++, If[IntegerQ@Log[2, f[[p]]], c[[p]]++]];
%t ]; c (* _Robert Price_, Jun 19 2019 *)
%Y Cf. A000142, A177378, A177355, A177349, A177458, A177459, A177498, A050376, A169655, A169661.
%K nonn
%O 2,1
%A _Vladimir Shevelev_, May 08 2010
%E a(32)-a(127) from _Robert Price_, Jun 19 2019