A129365 a(n) = A092287(n)/A129364(n).

1, 1, 1, 1, 1, 2, 2, 2, 6, 48, 48, 48, 48, 1536, 207360, 207360, 207360, 1105920, 1105920, 17694720, 30098718720, 15410543984640, 15410543984640, 481579499520, 60197437440000, 123284351877120000, 29958097506140160000

Offset

1,6

Comments

Conjectures:

A) a(n) is always an integer.

B) If p is a prime then p|a(n) if and only if p <= n/3. Let ordp(n,p) denote the exponent of the largest power of p which divides n. For example, ordp(48,2) = 4 since 48 = 3*(2^4). The precise decomposition of a(n) into primes would follow from the following two conjectures:

C) For each positive integer n and prime p, ordp(a(n*p),p) = ordp(a(n*p+1),p) = ordp(a(n*p+2),p) = . . . = ordp(a(n*p+p-1),p).

D) Let b(n) = A004125(n). Then ordp(a(n*p),p) = b(n) + b(floor(n/p)) + b(floor(n/p^2)) + b(floor(n/p^3)) + .... This is reminiscent of de Polignac's formula (also due to Legendre) for the prime factorization of n! (see the link).

Links

Table of n, a(n) for n=1..27.

Wikipedia, De Polignac's formula.

Formula

a(n) = ( Product_{j = 1..n} Product_{k = 1..n} gcd(j,k) ) / ( Product_{j = 1..n} Product_{d|j} d^(j/d) ).

a(n) = ( Product_{j = 1..n} Product_{k = 1..n} gcd(j,k) ) / ( Product_{k = 1..n} (floor(n/k)!)^k ).

Crossrefs

Cf. A004125, A092287, A129364.

Sequence in context: A371919 A098920 A270557 * A125838 A021453 A053789

Adjacent sequences: A129362 A129363 A129364 * A129366 A129367 A129368

Keyword

easy,nonn

Author

Peter Bala, Apr 13 2007

Status

approved