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A177436
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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.
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7
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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
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2,1
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COMMENTS
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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).
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LINKS
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FORMULA
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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).
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EXAMPLE
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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.
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MATHEMATICA
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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]]++]];
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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