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Totally multiplicative sequence with a(p) = (p+3)^2 = p^2+6p+9 for prime p.
1

%I #14 Sep 20 2020 04:45:37

%S 1,25,36,625,64,900,100,15625,1296,1600,196,22500,256,2500,2304,

%T 390625,400,32400,484,40000,3600,4900,676,562500,4096,6400,46656,

%U 62500,1024,57600,1156,9765625,7056,10000,6400,810000,1600,12100,9216,1000000,1936,90000

%N Totally multiplicative sequence with a(p) = (p+3)^2 = p^2+6p+9 for prime p.

%H G. C. Greubel, <a href="/A167363/b167363.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). a(n) = A166591(n)^2.

%F Sum_{k>=1} 1/a(k) = Product_{primes p} (1 + 1/(p^2 + 6*p + 8)) = 1.1245403934230393267573507658470307221064356442604979888687782305037985824... - _Vaclav Kotesovec_, Sep 20 2020

%t b[1] = 1; b[n_] := (fi = FactorInteger[n]; Times @@ ((fi[[All, 1]] + 3)^fi[[All, 2]])); Table[b[n]^2, {n, 1, 100}] (* _G. C. Greubel_, Jun 11 2016 *)

%Y Cf. A166591.

%K nonn,mult

%O 1,2

%A _Jaroslav Krizek_, Nov 01 2009