A246362 - OEIS

A246362

Numbers n such that if 2n-1 = product_{k >= 1} (p_k)^(c_k), then n < product_{k >= 1} (p_{k-1})^(c_k), where p_k indicates the k-th prime, A000040(k).

4, 6, 7, 9, 10, 12, 15, 16, 19, 20, 21, 22, 24, 27, 29, 30, 31, 34, 35, 36, 37, 40, 42, 44, 45, 46, 47, 48, 49, 51, 52, 54, 55, 56, 57, 60, 62, 64, 65, 66, 67, 69, 70, 71, 72, 75, 76, 78, 79, 80, 81, 82, 84, 85, 87, 89, 90, 91, 92, 96, 97, 99, 100, 101, 102, 103, 105, 106, 107, 108, 109, 110, 111, 112, 114, 115

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OFFSET

1,1

COMMENTS

Numbers n such that A064216(n) > n.

Numbers n such that A064989(2n-1) > n.

The sequence grows as:

a(100) = 148

a(1000) = 1449

a(10000) = 14264

a(100000) = 141259

a(1000000) = 1418197

and the powers of 10 occur at:

a(5) = 10

a(63) = 100

a(701) = 1000

a(6973) = 10000

a(70845) = 100000

a(705313) = 1000000

suggesting that the ratio a(n)/n is converging to a constant and an arbitrary natural number is more than twice as likely to be here than in the complement A246361. Compare this to the ratio present in the "inverse" case A246282.

LINKS

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

EXAMPLE

4 is present, as 2*4 - 1 = 7 = p_4, and p_{4-1} = p_3 = 5 > 4.

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

6 is present, as 2*6 - 1 = 11 = p_5, and p_{5-1} = p_4 = 7 > 6.

35 is present, as 2*35 - 1 = 69 = 3*23 = p_2 * p_9, and p_1 * p_8 = 2*19 = 38 > 35.

PROG

(PARI)

default(primelimit, 2^30);

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)};

A064216(n) = A064989((2*n)-1);