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%I #17 Dec 15 2022 06:15:02
%S 1,1,0,1,4,0,16,1,0,4,64,0,100,16,0,1,196,0,256,4,0,64,400,0,16,100,0,
%T 16,676,0,784,1,0,196,64,0,1156,256,0,4,1444,0,1600,64,0,400,1936,0,
%U 256,16,0,100,2500,0,256,16,0,676,3136,0,3364,784,0,1,400
%N Totally multiplicative sequence with a(p) = (p-3)^2 = p^2-6p+9 for prime p.
%H G. C. Greubel, <a href="/A167361/b167361.txt">Table of n, a(n) for n = 1..1000</a>
%F Multiplicative with a(p^e) = ((p-3)^2)^e. If n = Product p(k)^e(k) then a(n) = Product ((p(k)-3)^2)^e(k).
%F a(3k) = 0 for k >= 1.
%F a(n) = A166589(n)^2.
%F Sum_{k=1..n} a(k) ~ c * n^3, where c = (2/Pi^2) / Product_{p prime} (1 + 5/p^2 - 3/p^3 - 9/p^4) = 0.07909568395... . - _Amiram Eldar_, Dec 15 2022
%t a[1] = 1; a[n_] := (fi = FactorInteger[n]; Times @@ ((fi[[All, 1]] - 3)^fi[[All, 2]])); Table[a[n]^2, {n, 1, 100}] (* _G. C. Greubel_, Jun 11 2016 *)
%Y Cf. A166589.
%K nonn,mult
%O 1,5
%A _Jaroslav Krizek_, Nov 01 2009