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.

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

Offset

2,1

Comments

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).

Links

Robert Price, Table of n, a(n) for n = 2..127

Vladimir Shevelev, Compact integers and factorials, Acta Arithmetica 126 (2007), no. 3, 195-236.

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^(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).

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((127-5)/(128-127)))) = 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 A334850 A199793

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