A390138 - OEIS

A390138

a(n) is the largest integer k such that there is an integer m with exactly n nonunitary prime factors and m + A005117(i) is squarefree for 1 <= i <= k.

2, 11, 46, 105, 253, 355, 794, 1026, 1509, 2273, 2706, 4347, 5291, 6036, 7411, 9115, 12628, 12677, 17183, 19773, 24584, 25788, 27690, 28713, 37746, 38236, 41339, 42950, 52206, 55226, 66416, 72687, 79468, 79574, 82541, 96858, 99771, 108878, 115164, 134359, 143177, 144408

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OFFSET

0,1

COMMENTS

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 a(n). For m in M'_n, m + A005117(a(n) + 1) is divisible by A390240(n)^2. For all m in M_n\M'_n, m + A005117(j) is nonsquarefree for some j < a(n) + 1. (See Englezou link for a proof).

Let S_n be the set of squares of primes which divide m + A005117(a(n) + 1) for every m in M'_n. It has been confirmed for n <= 289 that S_n is the singleton set {A390240(n)^2}. If this was the case for every n, then a(n) would be unique. 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 a(r) = a(s) for some r <> s.

LINKS

Miles Englezou, Table of n, a(n) for n = 0..222

Miles Englezou, Proof

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, m + A005117(12) is divisible by 9. For all other m with 1 nonunitary prime factor, m + A005117(j) is nonsquarefree for some j < 12. Therefore a(1) = 11.

When m has n = 2 nonunitary prime factors, then for m in M'_n, m + A005117(47) is divisible by 25. For all other m with 2 nonunitary prime factors, m + A005117(j) is nonsquarefree for some j < 47. Therefore a(2) = 46.

These first three examples may seem to suggest that m with r nonunitary prime factors has a maximum m + A005117(x) divisible by prime(r+1)^2. But this is not the case, as shown by the next two examples:

When m has n = 16 nonunitary prime factors, then for m in M'_n, m + A005117(12629) is divisible by prime(18)^2 = 61^2 = 3721. For all other m with 16 nonunitary prime factors, m + A005117(j) is nonsquarefree for some j < 12629. Therefore a(16) = 12628.

When m has n = 17 nonunitary prime factors, then for m in M'_n, m + A005117(12678) is divisible by prime(17)^2 = 59^2 = 3481. For all other m with 17 nonunitary prime factors, m + A005117(j) is nonsquarefree for some j < 12678. Therefore a(17) = 12677.

PROG

(PARI)

A100112(n) = my(v=vector(n), k); forsquarefree(m = 1, n, v[m[1]] = k++); v[#v];

{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]))));

for(m = 1, oo,

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