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

Author

Vladimir Shevelev, May 08 2010

Keywords

• nonn

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).

Examples

			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.
		

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.

Crossrefs

Cf. A000142, A177378, A177355, A177349, A177458, A177459, A177498, A050376, A169655, A169661.

Programs

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 *)

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).

Extensions

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