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Smallest prime(k) such that for every r = 1 to n, there exist i,j, k >= j > i, such that prime(j)-prime(i) == 0 ( mod r).
3

%I #7 Sep 23 2023 03:13:38

%S 3,5,5,7,7,11,17,17,17,17,17,17,29,29,29,29,29,29,41,41,41,41,53,53,

%T 53,53,53,53,53,53,67,67,71,71,71,71,79,79,79,79,79,79,89,89,89,89,97,

%U 97,101,101,101,101,109,109,113,113,113,113,113,113,127,127,131,131,131

%N Smallest prime(k) such that for every r = 1 to n, there exist i,j, k >= j > i, such that prime(j)-prime(i) == 0 ( mod r).

%e a(11) = a(12) = 17, but a(13) = 29 as 29-3 == 0 (mod 13). And we have

%e 5-3 = 2, 11-5 = 2*3, 7-3 = 4, 7-2 = 5, 11-5 = 6, 17-3 = 2*7, 11-3 = 8, 23-5 = 2*9, 13-3 = 10, 13 - 2 = 11, 17-5 = 12, 29-3 = 2*13.

%o (PARI) found = vector(500); x = 1; forprime(p = 3, 500, oldX = x; forprime (q = 2, p - 1, v = divisors(p - q); for (i = 1, length(v), found[v[i]] = 1; if (v[i] == x, while (found[x], x++; write1("A088081.txt", p" "))))); if (oldX != x, write1("A088083.txt", p" "); write1("A088082.txt", x - 1" "))); \\ _David Wasserman_, Jul 11 2005

%Y Cf. A088082.

%Y Cf. A088083.

%K nonn

%O 1,1

%A _Amarnath Murthy_, Sep 22 2003

%E More terms from _David Wasserman_, Jul 11 2005