# [seqfan] Re: Do Mersenne primes ever divide A076078(n)?

ghodges14 at comcast.net ghodges14 at comcast.net
Sat Oct 16 05:18:41 CEST 2010

```Max,

Thanks very much. I've been busy today, but I checked your second number and it's a multiple of 31, all right. Thanks also for the modulo 7 rule.

> and, if I understand the inverse binomial transform correctly, A000371 (http://oeis.org/classic/A000371).

Specifically, it's my understanding that, for n>=1, A000371(n) = A076078(A002110(n)), where A002110(n) is the product of the first n primes. I also believe that Eric Weisstein's Mathworld page on the binomial transform has the binomial transform and its inverse transform reversed. If so, it's a tiny fault to point out in a great website, of course.

Thanks,
Matt

----- Original Message -----
From: Max Alekseyev <maxale at gmail.com>
To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
Sent: Thu, 14 Oct 2010 21:59:12 -0000 (UTC)
Subject: [seqfan] Re: Do Mersenne primes ever divide A076078(n)?

On Tue, Oct 12, 2010 at 11:38 PM,   wrote:

> A097416 contains the first 31 distinct values of A076078(n) which are not powers of 2

> (cf. http://oeis.org/classic/A097416).  11 larger values can be found in A076078's subsequences A097211, A097215, and, if I understand the inverse binomial transform correctly, A000371 (http://oeis.org/classic/A000371).

>

> According to Dario Alpern's factorization calculator, none of those 42 values is a multiple of 3, 7, 31, or any larger Mersenne prime. Since the formula for A076078(n) is based on Mersenne numbers (A000225), the natural conjecture is that no Mersenne prime divides any member of A076078(n).  Is this known or provable (or perhaps disprovable)?

This is true for 3 and 7 but not for 31.

In particular, A076078(2^3*3^3*5^3*7*11*13*19*23) and

A076078(2*3*5*7*11*13*19*23^3) are both divisible by 31.

> It also appears that, for any n, A076078(n) is congruent to A008836(n) mod 3.

This is true.

> I don't see any similar patterns for larger Mersenne primes.

Modulo 7, we have need to count

k1 = the number of primes p dividing n, whose exponent (i.e.,

valuation of n w.r.t. p) is == 1 (mod 3)

k2 = the number of primes dividing n, whose exponent is == 2 (mod 3)

then

A076078(n) == -(-2)^k1 (mod 7), if k1>0

A076078(n) == 2*(-1)^k2 - 1 (mod 7), if k1=0

It is easy to see that these expressions are never 0 modulo 7.

Regards,

Max
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