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# User talk:Robert Israel

### From OeisWiki

I very much appreciated your help a couple weeks ago with sequence A240162. Although at first I thought your formula and Mathematica code were incorrect, it was only because I had misread/misunderstood them. Your contribution was correct, insightful, and very helpful. I eventually ended up modifying my program based on yours. Thanks so much for the help.

I am still waiting for that sequence, and the related sequences A245970 through A245974 to be approved.

I have a concern about the Sage code I am using for calculating A245970 (and the other related sequences may have the same issue), wherein it appears one needs to add an extra phi(n) to the exponent, in order to skip over bad values before the repeating cycle gets established. With the extra phi(n), I am confident the program generates correct values for n up to 10000; but I am not absolutely sure it works for all n, especially when n is very high. Basically my concern is that I can't prove there doesn't exist an n, for which we may need to add an extra 2*phi(n) or even 3*phi(n), and not merely add a single phi(n), in order to skip past bad initial values in the initial cycles. But I don't really know. Your help on answering that question about A245970 would be much appreciated if you can spare the time. More explanation of my concern is in the comments on the draft.

Thanks, Robert.

Wayne VanWeerthuizen 06:26, 19 August 2014 (UTC)

For any positive integers p, k, let T[k](p) be a tower of k p's (so T[1](p) = p, T[k+1](p) = p^(T[k](p)). Suppose p is prime. It's convenient to use the "computer scientist's" version of mod, a function rather than a relation, so X mod Y is the least nonnegative integer r such that X == r mod Y. Now the question is, can we say T[k+1](p) = p^(T[k](p) mod phi(n)) mod n for sufficiently large k? Well, not for p=2, as shown in the case n=6 where T[k](2) mod 6 = 4 while T[k](2) mod phi(6) = 0. Let's look more closely at this.

If n is not divisible by p, it's true, because a^(phi(n)) == 1 mod n for any a coprime to n. Suppose now n = p^d * m where m is not divisible by p, and d is a positive integer. Then phi(n) = phi(p^d) * phi(m) = p^(d-1) * (p-1) * phi(m). Let T[k](p) mod phi(m) = s, T[k](p) mod phi(p^d) = t and T[k](p) mod phi(n) = r. Thus r == s mod phi(m) and r == t mod phi(p^d). Now if k is large enough, T[k](p) == 0 mod p^(d-1), while T[k](p) == 1 mod p-1. As long as p > 2, this says r > 0, and (since r is divisible by p^(d-1) and p^(d-1) >= d) r >= d. Then T[k](p) == 0 == p^r mod p^d, and T[k](p) == p^s == p^r mod m, so T[k](p) == p^r mod n. Thus the only exception is for p = 2, and this exception only occurs when r = 0, i.e. T[k](2) is divisible by phi(n) for large k. And that will only happen when phi(n) is a power of 2, i.e. when n is in A003401. When n is in A003401, note that for a Fermat prime f, T[k](2) == 1 mod f for k sufficiently large, and use the Chinese Remainder Theorem.

I hope this helps.

## A266214

Dear Robert,

Please see the draft of A266214 ("Numbers n that are not coprime to the numerator of zeta(2*n)/(Pi^(2*n))"), a sequence I have proposed in response to a SeqFans thread (A046988 query). Are you content to be listed as a co-author? Feel free to modify the draft as you wish. Many thanks.

Chris Boyd 08:18, 27 December 2015 (UTC).