1,2

Comment from Dean Hickerson, Oct 19 2007: (Start) Except for a(0)=1, all denominators in A002445 are divisible by 6 and are squarefree. To test such a number k to see if it's in the sequence, let 2n be the least common multiple of all p-1 for which p is a prime divisor of k.

Now list the primes p such that p-1 divides 2n. If all of them are divisors of k, then k is in the sequence; otherwise it's not.

For example, consider k = 78 = 2 * 3 * 13. The LCM of 2-1, 3-1 and 13-1 is 12, so 2n=12. The primes p such that p-1 divides 12 are 2, 3, 5, 7 and 13. Since 5 and 7 are not divisors of 78, 78 is not in the sequence. (End)

H. Rademacher, Topics in Analytic Number Theory, Springer, 1973, Chap. 1.

T. D. Noe, Table of n, a(n) for n=1..1001

We know from the von Staudt-Clausen theorem (see Rademacher) that the denominator of the Bernoulli number B_{2k} is the product of those distinct primes p for which p-1 divides 2k. In particular, all numbers after the first two (which are the denominators of B_0 and B_1) are divisible by 6. - N. J. A. Sloane (njas(AT)research.att.com), Feb 10, 2004

Cf. A090810, A002445 (denominators of Bernoulli numbers B_2n).

Sequence in context: A126989 A128040 A006954 this_sequence A166062 A127517 A137825

Adjacent sequences: A090798 A090799 A090800 this_sequence A090802 A090803 A090804

nonn,easy

Mohammed Bouayoun (bouyao(AT)wanadoo.fr), Feb 10 2004

Extended by Robert G. Wilson v (rgwv(AT)rgwv.com) Feb 10 2004