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A235623
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Numbers n for which in the prime power factorization of n!, the numbers of exponents 1 and >1 are equal.
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1
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OFFSET
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1,3
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COMMENTS
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Number n is in the sequence, if and only if pi(n) = 2*pi(n/2), where pi(x) is the number of primes<=x. Indeed, all primes from interval (n/2, n] appear in prime power factorization of n! with exponent 1, while all primes from interval (0, n/2] appear in n! with exponents >1. However, it follows from Ehrhart's link that, for n>=22, pi(n) < 2*pi(n/2). Therefore, a(9)=21 is the last term of the sequence.
m is in this sequence if and only if the number of prime divisors of [m/2]! equals the number of unitary prime divisors of m! - Peter Luschny, Apr 29 2014
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LINKS
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EXAMPLE
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21! = 2^20*3^9*5^4*7^3*11*13*17*19. Here 4 primes with exponent 1 and 4 primes with exponents >1, so 21 is in the sequence.
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MAPLE
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with(numtheory): a := proc(n) factorset(n!); factorset(iquo(n, 2)!);
`if`(nops(%% minus %) = nops(%), n, NULL) end: seq(a(n), n=0..30); # Peter Luschny, Apr 28 2014
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PROG
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(PARI) isok(n) = {f = factor(n!); sum(i=1, #f~, f[i, 2] == 1) == sum(i=1, #f~, f[i, 2] > 1); } \\ Michel Marcus, Apr 20 2014
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CROSSREFS
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Cf. A056171, A177329, A177333, A177334, A240537, A240588, A240606, A240619, A240620, A240668, A240669, A240670, A240672, A240695, A240751, A240755, A240764, A240905, A240906, A241123, A241124, A241139, A241148, A241289.
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KEYWORD
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nonn,fini,full
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AUTHOR
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STATUS
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approved
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