|
%I #18 Apr 30 2022 08:21:31
%S 1,1,1,1,1,1,3,1,1,1,1,1,3,3,1,1,1,1,3,1,3,1,1,1,1,3,1,3,1,1,3,1,1,1,
%T 3,1,3,3,3,1,1,3,3,1,1,1,1,1,3,1,1,3,1,1,1,3,3,1,1,1,3,3,3,1,3,1,3,1,
%U 1,3,1,1,3,3,1,3,3,3,3,1,1,1,1,3,1,3,1,1,1,1,9,1,3,1,3,1,3,3,1
%N a(n) = 3^b(n), where b(n) is #{primes p=1 mod 3 dividing n}.
%H Amiram Eldar, <a href="/A115069/b115069.txt">Table of n, a(n) for n = 1..10000</a>
%H Steven Finch and Pascal Sebah, <a href="https://arxiv.org/abs/math/0604465">Squares and Cubes Modulo n</a>, arXiv:math/0604465 [math.NT], 2006-2016.
%F a(n) = 3^A005088(n). - _R. J. Mathar_, May 19 2020
%p a:= n-> 3^add(`if`(irem(i[1], 3)=1, 1, 0), i=ifactors(n)[2](n)):
%p seq(a(n), n=1..100); # _Alois P. Heinz_, Feb 17 2019
%t b[n_] := Count[FactorInteger[n][[All, 1]], p_ /; Mod[p, 3] == 1];
%t a[1] = 1; a[n_] := 3^b[n];
%t Table[a[n], {n, 1, 99}] (* _Jean-François Alcover_, Feb 17 2019 *)
%Y Cf. A005088, A060839.
%K nonn
%O 1,7
%A _Steven Finch_, Mar 01 2006
%E a(1)=1 prepended by _Alois P. Heinz_, Feb 17 2019
|
