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Sum of squares of distinct prime divisors p of n, where each p <= sqrt(n).
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%I #19 Aug 19 2021 08:52:13

%S 0,0,0,4,0,4,0,4,9,4,0,13,0,4,9,4,0,13,0,4,9,4,0,13,25,4,9,4,0,38,0,4,

%T 9,4,25,13,0,4,9,29,0,13,0,4,34,4,0,13,49,29,9,4,0,13,25,53,9,4,0,38,

%U 0,4,58,4,25,13,0,4,9,78,0,13,0,4,34,4,49,13,0,29,9,4,0,62,25,4,9,4,0,38,49

%N Sum of squares of distinct prime divisors p of n, where each p <= sqrt(n).

%H Harvey P. Dale, <a href="/A098002/b098002.txt">Table of n, a(n) for n = 1..1000</a>

%F G.f.: Sum_{k>=1} prime(k)^2 * x^(prime(k)^2) / (1 - x^prime(k)). - _Ilya Gutkovskiy_, Aug 19 2021

%e 2 and 3 are the distinct prime divisors of 12 and both 2 and 3 are <= sqrt(12), so a(12) = 2^2 + 3^2 = 13.

%t ssdpd[n_]:=Total[Select[Transpose[FactorInteger[n]][[1]],#<=Sqrt[n]&]^2]; Join[{0},Array[ssdpd,90,2]] (* _Harvey P. Dale_, Mar 11 2013 *)

%o (PARI) a(n) = sumdiv(n, d, isprime(d)*(d^2<=n)*d^2); \\ _Michel Marcus_, Dec 22 2017

%Y Cf. A097974.

%K nonn

%O 1,4

%A _Leroy Quet_, Sep 08 2004

%E More terms from _John W. Layman_, Sep 14 2004