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a(n) is the largest exponent k such that 3^k divides (prime(n)^2 + 2).
2

%I #16 Sep 21 2023 01:19:22

%S 1,0,3,1,1,2,1,1,2,1,2,1,2,1,1,1,4,1,2,1,1,1,1,1,1,1,4,1,1,3,1,2,1,2,

%T 2,1,3,1,3,1,1,1,1,2,1,1,3,1,1,2,1,2,1,1,2,1,1,1,1,1,2,2,1,2,1,1,1,2,

%U 2,1,1,1,1,3,1,4,1,1,2,2,2,1,1,1,1,1,1,1,1,2,1,1,1,3,2,1,2

%N a(n) is the largest exponent k such that 3^k divides (prime(n)^2 + 2).

%C Except for n=2, all a(n) > 1 because (prime(n)^2 + 2) is divisible by 3.

%H Zak Seidov, <a href="/A257567/b257567.txt">Table of n, a(n) for n = 1..1000</a>

%F a(n) = A007949(A061725(n)). - _Michel Marcus_, May 01 2015

%e a(1) = 1 because p=prime(1)=2 and p^2 + 2 = 6 = 3^1*2,

%e a(2) = 0 because p=prime(2)=3 and p^2 + 2 = 11 = 3^0*11,

%e a(3) = 3 because p=prime(3)=5 and p^2 + 2 = 27 = 3^3.

%t Table[IntegerExponent[Prime[k]^2 + 2, 3], {k, 100}]

%o (PARI) a(n) = valuation(prime(n)^2+2, 3); \\ _Michel Marcus_, May 01 2015

%Y Cf. A007949 (3-adic valuation), A061725 (p^2+2, with p prime), A257568.

%K nonn

%O 1,3

%A _Zak Seidov_, Apr 30 2015