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Let $a_n$ be the sequence such that GCD$(2^x+3,3^x+2)<6.$ Here what GAP has given me:

$z:=100$;;L:=[1..z];;LL:=Filtered(L,x->GCD$(2^x+3,3^x+2)<6.$;

$[ 1, 2, 3, 4, 6, 7, 8, 9, 10, 12, 13, 14, 15, 16, 18, 19, 20, 21, 22, 24, 25, 26, 27, 28, 30, 31, 32, 33, 34, 36, 37, 38, 39, 40, 42, 43, 44, 45, 46, 48, 49, 50, 51, 52, 54, 55, 56, 57, 58, 60, 61, 62, 63, 64, 66, 67, 68, 69, 70, 72, 73, 74, 75, 76, 78, 79, 80, 81, 82, 84, 85, 86, 87, 88, 90, 91, 92, 93, 94, 96, 97, 98, 99, 100 ]$

Here what the sequence database has given me till 85: http://oeis.org/A059537. So is this sequence really equal to floor(n Zeta(3))?

1 Answers 1

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No, certainly not. The sequence concerning $\gcd(2^x+3,3^x+2)$ is exactly the set of all numbers that are not congruent to $5\pmod6$. That sequence is equal to $\{\lfloor n\alpha \rfloor\}$ for some real number $\alpha$—but that number is simply $\alpha=\frac65$. The fact that $\zeta(3) \approx 1.20206$ is close to $\frac65$ is just a coincidence. Indeed, $x=118$ is the smallest number that is in the GCD-related sequence but not in the sequence $\{\lfloor n\zeta(3)\rfloor\}$, and $x=119$ is the smallest number with the reverse properties¹, and the exceptions start to be reasonably regular after that point.

¹We note in passing that $\zeta(3)$ is a little bit larger than its 4th convergent, $\frac{119}{99}$, which explains the appearance of 119 here.