A335301 a(n) = prime(n+1) mod (10^k) where k is the least positive integer such that floor(prime(n)/(10^k)) = floor(prime(n+1)/(10^k)) and prime(n) denotes the n-th prime number.
3, 5, 7, 11, 3, 7, 9, 23, 9, 31, 7, 41, 3, 7, 53, 9, 61, 7, 71, 3, 9, 83, 9, 97, 101, 3, 7, 9, 13, 27, 31, 7, 9, 49, 51, 7, 63, 7, 73, 9, 81, 91, 3, 7, 9, 211, 23, 7, 9, 33, 9, 41, 51, 7, 63, 9, 71, 7, 81, 3, 93, 307, 11, 3, 7, 31, 7, 47, 9, 53, 9, 67, 73, 9
Offset: 1
Examples
For n = 42: - prime(42) = 181 and prime(43) = 191, - floor(181/(10^1)) = 18 <> 19 = floor(191/(10^1)), - floor(181/(10^2)) = 1 = floor(191/(10^2)), - so a(42) = 191 mod (10^2) = 91.
Links
- Rémy Sigrist, Table of n, a(n) for n = 1..10000
Programs
-
PARI
{ base=10; p=2; forprime (q=p+1, 379, for (k=0, oo, m=base^k; if (q\m == p\m, print1 (q%m", "); p=q; break))) }
Formula
a(n) <= prime(n+1) with equality iff prime(n+1) is the least prime number with its number of digits and leading digit.
Comments