# [seqfan] Re: Ramanujan sequence A067128

Sat May 14 21:57:10 CEST 2016

```Hi David et all,

I give a proof of existence k in new David's sequence A273038:
"Least k such that for all m >= k, A067128(m) is divisible by n."

We use other designations.

Let N in A067128 has prime power factorization (PPF)
N=2^k_1*...*p_n^k_n, k_n>=1,     (1)
where p_i=prime(i).
>From my theorem in A273015 it follows that, when N runs A067128,
p_n in (1) is unbounded and, moreover, tends to infinity, when N
tends to infinity.
Let us show that, when N runs A067128, also k_1 is unbounded.
Indeed, suppose k_1 is bounded. Consider number N_1 with
PPF N_1=2^(k_1+x)*...*p_(n-1)^k_(n-1) such that all powers
p^i , i=2,...,n-1, are the same as in (1) and 2^x<p_n^k_n. (2)
Then N_1<N.  Let us try choose x such that d(N_1)>d(N).
It should be (k_1+x+1)*...*(k_(n-1)+1)>(k_1+1)*...*(k_(n-1)+1)*
(k_n+1), or k_1+x+1>(k_1+1)*(k_n+1)=k_1*k_n+k_n+k_1+1,
or x>(k_1+1)*k_n.
So, by (2),  (k_1+1)*k_n<x<k_n*log_2(p_n).  (3)
Since, by the supposition,  k_1 is bounded, we for large n can choose
required x, which gives a contradiction. So k_1 is unbounded.
Moreover, we see that k_1 tends to infinity as log_2(p_n), n=n(N),
Let m=2^m_1*3^m_2*...*p_r^m_r.
We can choose k_1>m_1 and to fix k_1. In the same way we prove
that k_2 tends to infinity and choose and fix it >m_2,..., k_r tends to
infinity and choose and fix it >m_r.  All k_i , i=1,...,r tend to infinity,
at least, as log_p_r(p_n), n=n(N).
So there exists a large M_m such that for all N from
A067128 >M_m, m|N.

Best regards,

________________________________________
From: SeqFan [seqfan-bounces at list.seqfan.eu] on behalf of David Corneth [davidacorneth at gmail.com]
Sent: 13 May 2016 18:01
To: Sequence Fanatics Discussion list
Subject: [seqfan] Re: Ramanujan sequence A067128

I've also sumbitted a sequence relating to A067128. It's A273014, by the
name
Least k such that A067128(k) is divisible by n.

I'd like to put another sequence:
Least k sucht that A067128(m) is divisible by n for all m>=k, but I find it
a bit difficult to find data for this one.

It's based on the idea of a prime signature. I thought maybe that's a way
to get more insight in the sequence and maybe to see if A067128 = A034287.
A prime signature of a number n is a vector of size primepi(maxp(n)) where
maxp(n) is the highest prime dividing n, and primepi(p) counts the primes
upto p. The elements with index i of the vector are the exponents of the
i-th prime in the factorization of n.
To give an example, The prime signature of 264 is [3, 1, 0, 0, 1]. The
highest prime dividing 264 is 11. Primepi(11) = 5 so the signature has 5
elements.
The exponent of the first prime, 2, is 3 in the factorization of 264, so
the first element of the sig. is 3.
The exponent of the second prime, 3, is 1 in the factorization of 264, so
the second element of the sig. is 1.
The exponent of the third prime, 5, is 0 in the factorization of 264, so
the third element of the sig. is 0.
The exponent of the fourth prime, 7, is 0 in the factorization of 264, so
the fourth element of the sig. is 0.
The exponent of the fourth prime, 11, is 0 in the factorization of 264, so
the fourth element of the sig. is 1.

I put the prime signatures of the first 555 elements of A067128 in Excel
and colored the cells according to the elements in the signatures. This
gives a cool colorscheme!

Best,
David

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