OFFSET
0,5
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) = 460, a(10^4) = 3996, a(10^5) = 35962.
LINKS
M. F. Hasler, Table of n, a(n) for n = 0..10000, Nov 21 2025
FORMULA
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, the cumulated number of powers of 2 is 0, 0, 1, 1, 2, 3, 2, 2, 2, 4, 4, 8, ...
MAPLE
b:= proc(n) option remember; `if`(n<1, 1, (p->
b(n-1)*(1+p(n))/(p(n+1)-p(n)))(ithprime))
end:
a:= n-> padic[ordp](b(n), 2):
seq(a(n), n=0..68); # Alois P. Heinz, Apr 22 2026
MATHEMATICA
IntegerExponent[FoldList[Times, 1, Map[(#[[1]] + 1)/(#[[2]] - #[[1]]) &, Partition[Prime[Range[100]], 2, 1]]], 2] (* Paolo Xausa, Apr 22 2026 *)
PROG
(PARI) concat(s=0, vector(99, i, s+=valuation((prime(i)+1)/(prime(i+1)-prime(i)), 2)))
CROSSREFS
KEYWORD
nonn
AUTHOR
M. F. Hasler, Nov 21 2025
STATUS
approved
