A390240 - OEIS

A390240

a(n) is the least prime p not already listed such that m + A005117(A390138(n) + 1) is divisible by p^2 for certain m (as defined in the comments) with exactly n nonunitary prime factors.

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 61, 59, 67, 71, 73, 79, 83, 89, 97, 103, 107, 101, 113, 109, 127, 131, 139, 137, 151, 163, 149, 157, 167, 179, 173, 181, 193, 191, 197, 199, 211, 223, 227, 229, 241, 239, 233, 251, 257, 263, 269, 271, 293, 277, 281

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OFFSET

0,1

COMMENTS

This sequence is a permutation of the primes.

It is an elementary fact that we can have no more than three consecutive squarefree integers: the next integer will be divisible by 4. It is possible to generalize this in the following way. Let M_n be the set of integers with exactly n nonunitary prime factors; then M_n contains a subset M'_n of those m such that the run of squarefree integers m + A005117(i), i >= 1, has length A390138(n). For m in M'_n, m + A005117(A390138(n) + 1) is divisible by a(n)^2. For all m in M_n\M'_n, m + A005117(j) is nonsquarefree for some j < A390138(n) + 1. (See Englezou link for a proof).

Let S_n be the set of squares of primes which divide m + A005117(A390138(n) + 1) for every m in M'_n. It has been confirmed for n <= 289 that S_n is the singleton set {a(n)^2}. Nonetheless, for reasons explained in the Englezou link, we are unable to rule out the possibility that there exists an n > 289 such that |S_n| > 1. If this were the case, then it would mean A390138(r) = A390138(s) for some r <> s, and we would then require a way of ordering the corresponding primes a(r) and a(s). Let a(r) = p and a(s) = q such that p < q: we define the convention here that if A390138(r) = A390138(s) for some r <> s, then a(r) < a(s). This is equivalent to the ordering defined in the sequence name.

LINKS

Miles Englezou, Table of n, a(n) for n = 0..289

Miles Englezou, Proof of generalization

EXAMPLE

These examples follow from the proof in the Englezou link:

When n = 0, m is itself squarefree. If m + 1 = m + A005117(1) and m + 2 = m + A005117(2) are also squarefree, then m + A005117(3) is divisible by 4. Therefore a(0) = 2.

When m has n = 1 nonunitary prime factor, then for m in M'_n we find that m + A005117(12) is divisible by 9 (whereas, for other squares of primes, q^2, we find that not all m in M'_n are divisible by q^2). Therefore a(1) = 3.

When m has n = 2 nonunitary prime factors, then for m in M'_n we find that m + A005117(47) is divisible by 25. Therefore a(2) = 5.

When m has n = 16 nonunitary prime factors, then for m in M'_n we find that m + A005117(12629) is divisible by prime(18)^2 = 61^2 = 3721. Therefore a(16) = 61.

When m has n = 17 nonunitary prime factors, then for m in M'_n we find that m + A005117(12678) is divisible by prime(17)^2 = 59^2 = 3481. Therefore a(17) = 59.

PROG

(PARI)

{a(n) = my(f);

(f(n) = my(p = prime(n), N = [1..p^2 - 1], M);

for(k = 0, oo,

M = N;

N = select(x -> !issquarefree(x + k*p^2), N);

if(#N == 0,

return([vecmax(M) + k*p^2, 2*p^2, p]))));

for(m = 1, oo,

if(f(m)[2] > f(n+1)[1],

return(apply(y -> y[3], vecsort(apply(x -> f(x), [1..m])))[n+1])))}