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a(n) = Sum_{1 <= j <= n/2, gcd(j,n)=1} j^2.
3

%I #33 Feb 24 2022 02:03:57

%S 0,1,1,1,5,1,14,10,21,10,55,26,91,35,70,84,204,75,285,140,210,165,506,

%T 196,525,286,549,406,1015,340,1240,680,880,680,1190,654,2109,969,1482,

%U 1080,2870,966,3311,1650,2010,1771,4324,1544,4214,2050

%N a(n) = Sum_{1 <= j <= n/2, gcd(j,n)=1} j^2.

%C n does not divide a(n) iff n = (2^k)*(q^m) with k > 0, m >= 0 and q odd prime such that q == 3 (mod 4) or n = (2^k)*(3^L)*Product_{q} q^(v_q) with k >= 0, L > 0, v_q >= 0 and all q odd primes such that q == 5 (mod 6). - _René Gy_, Oct 21 2018

%H Seiichi Manyama, <a href="/A295574/b295574.txt">Table of n, a(n) for n = 1..10000</a>

%H John D. Baum, <a href="http://www.jstor.org/stable/2690056">A Number-Theoretic Sum</a>, Mathematics Magazine 55.2 (1982): 111-113.

%H René Gy, <a href="https://math.stackexchange.com/a/2946006/130022">The sum of product pairs of integers prime to n</a>, Math StackExchange.

%p R:=proc(n,k) local x,t1,S;

%p t1:={}; S:=0;

%p for x from 1 to floor(n/2) do if gcd(x,n)=1 then t1:={op(t1),x^k}; S:=S+x^k; fi; od;

%p S; end;

%p s:=k->[seq(R(n,k),n=1..50)];

%p s(2);

%t f[n_] := Plus @@ (Select[ Range[n/2], GCD[#, n] == 1 &]^2); Array[f, 50] (* _Robert G. Wilson v_, Dec 10 2017 *)

%o (PARI) a(n) = sum(j=1, n\2, (gcd(j, n)==1)*j^2); \\ _Michel Marcus_, Dec 10 2017

%Y In the Baum (1982) paper, S_1, S_2, S_3, S_4 are A023896, A053818, A053819, A053820, and S'_1, S'_2, S'_3, S'_4 are A066840, A295574, A295575, A295576.

%Y Cf. A023022.

%K nonn

%O 1,5

%A _N. J. A. Sloane_, Dec 08 2017