A390843 5-valuation of Product_{1 <= k <= n} (prime(k) + 1) / (prime(k+1) - prime(k)).

0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 8, 8, 8, 8, 8, 8, 9, 8, 8, 8, 8, 10, 10, 10, 10, 11, 11, 12, 11, 11, 11, 11, 12, 12, 12, 12, 11, 11, 11, 11, 11, 11, 11, 10, 10, 12, 12, 13, 13, 13, 14

Offset

0,11

Comments

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) = 188, a(10^4) = 1705, a(10^5) = 15781.

Links

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

Formula

a(n) = A112765(A390839(n)).

Example

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, ...
Therefore, for the product of the first n = 0, 1, 2, 3, ... of these fractions contains a factor of 5 only from prime(n = 8) = 19 on, and second factor of 5 appears with prime(n = 10) = 29.

Mathematica

a[n_]:=IntegerExponent[Product[(Prime[k]+1)/(Prime[k+1]-Prime[k]), {k, 1, n}], 5]; Array[a, 76, 0] (* James C. McMahon, Nov 28 2025 *)

Prog

(PARI) concat(s=0, vector(99, i, s+=valuation((prime(i)+1)/(prime(i+1)-prime(i)), 5)))

Crossrefs

Cf. A390841, A390842, A390844 (similar for 2-, 3- and 7-valuation).

Cf. A112765 (5-valuation).

Cf. A390839.

Sequence in context: A296237 A074802 A071804 * A280953 A111894 A279012

Adjacent sequences: A390840 A390841 A390842 * A390844 A390845 A390846

Keyword

nonn

Author

M. F. Hasler, Nov 21 2025

Status

approved