A237514 - OEIS

A237514

Numbers k such that 2^(k-1) < 3^(m-1) < 2^k < 3^m < 2^(k+1), for some m > 2, a(1) = 1.

1, 4, 7, 12, 15, 20, 23, 26, 31, 34, 39, 42, 45, 50, 53, 58, 61, 64, 69, 72, 77, 80, 85, 88, 91, 96, 99, 104, 107, 110, 115, 118, 123, 126, 129, 134, 137, 142, 145, 148, 153, 156, 161, 164, 169, 172, 175, 180, 183, 188, 191, 194, 199, 202, 207, 210, 213, 218, 221, 226, 229, 232, 237, 240

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OFFSET

1,2

COMMENTS

Exponents of A006899(n) such that A006899(n-1) and A006899(n+1) are both odd.

Probably finite? The last term?

Subsequence of primes starts 7, 23, 31, 53, 61, 107, 137, 191, 199, 229,...

Prime indices of A006899(n) such that A006899(n-1) and A006899(n+1) are both odd: 2, 7, 43, 113, 131, 139, 149, 157, 193, 211, 263, 281, 307, 317, 379,...

Let f(n) := floor( n * log(2) / log(3)), then k is in the sequence if and only if k = 1 or f(k - 1) = f(k) - 1 and f(k + 1) = f(k) + 1. - Michael Somos, Feb 24 2014

k > 1 is in the sequence if and only if the interval (k*b - b + 1, k*b + b) contains an integer, where b = log_3(2). Since b is irrational, the fractional parts of its multiples are uniformly distributed in [0,1], so the asymptotic density of this sequence is 2*b - 1 or approximately 0.26186. In particular, the sequence is infinite. Moreover, by a theorem of Vinogradov, the fractional parts of p * b for primes p are uniformly distributed in [0,1], so the subsequence of primes is infinite and has asymptotic density 2*b - 1 in the primes. - Robert Israel, Sep 10 2025

LINKS

Robert Israel, Table of n, a(n) for n = 1..10000

EXAMPLE

a(2) = 4 because k = 4 and 2^(4-1) < 3^(3-1) < 2^4 < 3^3 < 2^(4+1) for m = 3;

a(3) = 7 because k = 7 and 2^(7-1) < 3^(4-1) < 2^7 < 3^4 < 2^(7+1) for m = 4;

a(4) = 12 because k = 12 and 2^(12-1) < 3^(8-1) < 2^12 < 3^8 < 2^(12+1) for m = 8.

MAPLE

filter:= proc(k) local m;

m:= 2+ilog[3](2^(k-1));

3^m < 2^(k+1)

end proc:

filter(1):= true: