A069784 Numbers m such that gcd(d((m!)^3), d(m!)) = 2^k, i.e., is a power of 2; d = A000005.

1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 13, 14, 18, 19, 30, 31, 32, 35, 38, 39

Offset: 1

Author

Labos Elemer, Apr 08 2002

Keywords

• nonn
• fini
• full

Comments

From David A. Corneth, Jul 31 2017: (Start)

Theorem: There are no further terms.

Proof:

Let e_n(n, p) be the exponent of p in n!. The prime p has exponent e_(n, p) = n/p for sqrt(n) < p < n in n!. n/4 <= p < n/3, e_(n, p) = 3 so e_(n, p) * 3 = 9. and for n/5 <= p < n/4, e = 4. The gap g_n between prime(n) and prime(n+1) is about sqrt(n) * log(n). There is a gap of n/4 - n/5 = n/20 between n/5 and n/4. primepi(1000) = 168, so for n > 5*1000, the gap between n/5 and the next prime is about sqrt(168) * log(168) ~= 66. This is much less than n/20. No 40 <= m <= 15000 is in the sequence, which completes the proof. (End)

Crossrefs

Cf. A000005, A000142, A069780, A069781, A069782, A069783.

Programs

Mathematica

Do[s=GCD[DivisorSigma[0, (n!)^3], DivisorSigma[0, n! ]]; If[IntegerQ[n/100], Print[{n}]]; If[IntegerQ[Log[2, s]], Print[n]], {n, 1, 10000}]

PARI

val(n, p) = my(r=0); while(n, r+=n\=p);r
is(n) = {my(p1 = p2 = 1); forprime(p=2, n, v = val(n, p); p1 *= (v + 1); p2 *= (3*v + 1)); g = gcd(p1, p2); g==2^(valuation(g, 2))} \\ David A. Corneth, Jul 31 2017

Extensions

Keywords fini and full added by David A. Corneth, Jul 31 2017