0,14

The given product is likely to be an integer for all n, because the cumulated numerators are extremely likely to have more of any given prime factor than the smallish denominators which are the prime gaps. But can it be proved that this is indeed always the case for all prime factors?

The sequence appears to grow slightly sub-linearly: a(10^3) = 148, a(10^4) = 1299, a(10^5) = 11688.

M. F. Hasler, Table of n, a(n) for n = 0..10000, Nov 20 2025

The first factors of the product are (2+1)/1 = 3, (3+1)/2 = 2, (5+1)/2 = 3, (7+1)/4 = 2, (11+1)/2 = 6, (13+1)/4 = 7/2, (17+1)/2 = 9, (19+1)/4 = 5, (23+1)/6 = 4, (29+1)/2 = 15, (31+1)/6 = 16/3, (37+1)/4 = 19/2, (41+1)/2 = 21, ...

Therefore, for the product of the first n = 0, 1, 2, 3, ... of these fractions contains a factor of 7 only from prime(n = 6) = 13 on, and second factor of 7 appears with prime(n = 13) = 41.

a[n_]:=IntegerExponent[Product[(Prime[k]+1)/(Prime[k+1]-Prime[k]), {k, 1, n}], 7]; Array[a, 82, 0] (* James C. McMahon, Nov 28 2025 *)

(PARI) concat(s=0, vector(99, i, s+=valuation((prime(i)+1)/(prime(i+1)-prime(i)), 7)))

Cf. A390841, A390842, A390843 (similar for 2-, 3- and 5-valuation).

Cf. A214411 (7-valuation).

Sequence in context: A337637 A156821 A025856 * A350765 A103378 A103663

Adjacent sequences: A390841 A390842 A390843 * A390845 A390846 A390847

M. F. Hasler, Nov 21 2025

approved