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%I #24 Nov 10 2022 07:42:38
%S 1,4,1,3,1,4,1,1,1,4,1,3,1,4,1,2,1,4,1,3,1,4,1,1,1,4,1,3,1,4,1,4,1,4,
%T 1,3,1,4,1,1,1,4,1,3,1,4,1,2,1,4,1,3,1,4,1,1,1,4,1,3,1,4,1,3,1,4,1,3,
%U 1,4,1,1,1,4,1,3,1,4,1,2,1,4,1,3,1,4,1,1,1,4,1,3,1,4,1,4,1,4,1,3,1,4,1,1,1
%N a(n) = ((-1)^(n+1)*A002425(n)) modulo 5.
%H Andrew Howroyd, <a href="/A089612/b089612.txt">Table of n, a(n) for n = 1..1024</a>
%F Let S(1) = {1, 4} and S(n+1) = S(n)*S'(n), where S'(n) is obtained from S(n) by changing last term using the cyclic permutation 4->3->1->2->4; sequence is S(infinity).
%F Multiplicative with a(2^e) = 2^(e + 1) mod 5, a(p^e) = 1 for odd prime p. - _Andrew Howroyd_, Aug 01 2018
%F Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 21/10. - _Amiram Eldar_, Nov 10 2022
%t Table[Mod[Numerator[2 / n (4^n - 1) BernoulliB[2 n]], 5], {n, 100}] (* _Vincenzo Librandi_, Aug 01 2018 *)
%o (PARI) a(n)=numerator(2/n*(4^n-1)*bernfrac(2*n))%5
%o (PARI) a(n)=if(n%2, 1, 2*2^valuation(n,2) % 5); \\ _Andrew Howroyd_, Aug 01 2018
%o (Magma) [Numerator(2/n*(4^n-1)*Bernoulli(2*n)) mod 5: n in [1..100]]; // _Vincenzo Librandi_, Aug 01 2018
%Y Cf. A002425, A056832.
%K nonn,mult
%O 1,2
%A _Benoit Cloitre_, Dec 30 2003
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