A246373 - OEIS

A246373

Primes p such that if 2p-1 = product_{k >= 1} A000040(k)^(c_k), then p <= product_{k >= 1} A000040(k-1)^(c_k).

2, 3, 7, 19, 29, 31, 37, 47, 67, 71, 79, 89, 97, 101, 103, 107, 109, 127, 139, 151, 157, 181, 191, 197, 199, 211, 223, 227, 229, 241, 251, 269, 271, 277, 283, 307, 317, 331, 337, 349, 359, 367, 373, 379, 397, 409, 421, 433, 439, 457, 461, 467, 487, 499, 521, 541, 547, 569, 571, 577, 601

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,1

COMMENTS

Primes p such that A064216(p) >= p, or equally, A064989(2p-1) >= p.

All primes of A005382 are present here, because if 2p-1 is prime q, Bertrand's postulate guarantees (after cases 2 and 3 which are in A048674) that there exists at least one prime r larger than p and less than q = 2p-1, for which A064989(q) = r.

LINKS

Antti Karttunen, Table of n, a(n) for n = 1..10000

Wikipedia, Bertrand's postulate

EXAMPLE

2 is present, as 2*2 - 1 = 3 = p_2, and p_{2-1} = p_1 = 2 >= 2.

3 is present, as 2*3 - 1 = 5 = p_3, and p_{3-1} = p_2 = 3 >= 3.

5 is not present, as 2*5 - 1 = 9 = p_2 * p_2, and p_1 * p_1 = 4, with 4 < 5.

7 is present, as 2*7 - 1 = 13 = p_6, and p_5 = 11 >= 7.

PROG

(PARI)

A064989(n) = {my(f); f = factor(n); if((n>1 && f[1, 1]==2), f[1, 2] = 0); for (i=1, #f~, f[i, 1] = precprime(f[i, 1]-1)); factorback(f)};

n = 0; forprime(p=2, 2^31, if((A064989((2*p)-1) >= p), n++; write("b246373.txt", n, " ", p); if(n > 9999, break)));