

A177436


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.


7



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, 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, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 6, 6, 6, 6, 6, 6, 6
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OFFSET

2,1


COMMENTS

Or a(n) is the maximal m for which the FermiDirac representation of m! (see comment in A050376) contains single power of 2 and single power of prime(n).


LINKS

Robert Price, Table of n, a(n) for n = 2..127
V. Shevelev, Compact integers and factorials, Acta Arithmetica 126 (2007), no. 3, 195236.


FORMULA

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^(k1))+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^(k1)+3<p_n<=2^k1, then a(n)=2*(1+floor(log_2((p_n5)/(2^kp_n))), where p_n=prime(n).


EXAMPLE

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((1275)/(128127)))=14.


MATHEMATICA

nlim = 127; mlim = (Prime[nlim] + 1)^2/2 + 3; f = Table[0, mlim]; c = Table[0, nlim];
For[m = 2, m <= mlim, m++,
mf = FactorInteger[m];
For[i = 1, i <= Length[mf], i++, f[[PrimePi@First@mf[[i]]]] += Last@mf[[i]]];
If[! IntegerQ@Log[2, f[[1]]], Continue[]];
For[p = 1, p <= nlim, p++, If[IntegerQ@Log[2, f[[p]]], c[[p]]++]];
]; c (* Robert Price, Jun 19 2019 *)


CROSSREFS

Cf. A000142, A177378, A177355, A177349, A177458, A177459, A177498, A050376, A169655, A169661.
Sequence in context: A096151 A021567 A019619 * A318139 A199793 A202949
Adjacent sequences: A177433 A177434 A177435 * A177437 A177438 A177439


KEYWORD

nonn


AUTHOR

Vladimir Shevelev, May 08 2010


EXTENSIONS

a(32)a(127) from Robert Price, Jun 19 2019


STATUS

approved



