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A177436 The number of positive integers m for which the exponents of 2 and prime(n) in the prime power factorization of m! are both powers of 2.

%I A177436 #28 May 11 2025 01:16:35
%S A177436 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 A177436 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 A177436 4,4,4,4,4,4,4,4,4,4,4,4,6,6,6,6,6,6,6
%N A177436 The number of positive integers m for which the exponents of 2 and prime(n) in the prime power factorization of m! are both powers of 2.
%C A177436 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 A177436 Robert Price, <a href="/A177436/b177436.txt">Table of n, a(n) for n = 2..127</a>
%H A177436 Vladimir Shevelev, <a href="https://eudml.org/doc/277854">Compact integers and factorials</a>, Acta Arithmetica 126 (2007), no. 3, 195-236.
%F A177436 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 A177436 For p_5 = 11, we have 11 = 2^3+3. Therefore a(5) = 3.
%e A177436 For p_27 = 103, we have 103 = (2^(4*2+1)+3)/5. Therefore a(27) = 5.
%e A177436 For p_31 = 127, a(31) = 2*(1+floor(log_2((127-5)/(128-127)))) = 14.
%t A177436 nlim = 127; mlim = (Prime[nlim] + 1)^2/2 + 3; f = Table[0, mlim]; c = Table[0, nlim];
%t A177436 For[m = 2, m <= mlim, m++,
%t A177436   mf = FactorInteger[m];
%t A177436   For[i = 1, i <= Length[mf], i++, f[[PrimePi@First@mf[[i]]]] += Last@mf[[i]]];
%t A177436   If[! IntegerQ@Log[2, f[[1]]], Continue[]];
%t A177436   For[p = 1, p <= nlim, p++, If[IntegerQ@Log[2, f[[p]]], c[[p]]++]];
%t A177436 ]; c (* _Robert Price_, Jun 19 2019 *)
%Y A177436 Cf. A000142, A177378, A177355, A177349, A177458, A177459, A177498, A050376, A169655, A169661.
%K A177436 nonn
%O A177436 2,1
%A A177436 _Vladimir Shevelev_, May 08 2010
%E A177436 a(32)-a(127) from _Robert Price_, Jun 19 2019