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%I #30 Sep 11 2019 22:43:50
%S 0,1,2,2,10,13,14,12,24,45,44,38,78,77,70,56,136,129,152,130,182,209,
%T 184,148,250,325,288,294,406,365,372,304,484,561,490,402,666,665,572,
%U 540,820,805,860,726,840,897,846,680,980,1125
%N a(n) = Sum_{k=1..n} (k^2 mod n).
%C See A048152 for the array T[n,k] = k^2 mod n.
%C Starting with a(2)=1 each 4th term is odd: a(n=2+4*k)= 1, 13, 45, 77, 129, 209, 325, 365,... - _Zak Seidov_, Apr 22 2009
%C Positions of squares in A048153: 1, 2, 33, 51, 69, 105, 195, 250, 294, 1250, 4913, 9583, 13778, 48778, 65603, 83521.
%C Corresponding values of squares are: {0, 1, 22, 34, 46, 70, 130, 175, 203, 875, 3468, 6734, 9711, 34481, 46308, 58956}^2 = {0, 1, 484, 1156, 2116, 4900, 16900, 30625, 41209, 765625, 12027024, 45346756, 94303521, 1188939361, 2144430864, 3475809936}. - _Zak Seidov_, Nov 02 2011
%C For n > 1 also row sums of A060036. - _Reinhard Zumkeller_, Apr 29 2013
%H Zak Seidov, <a href="/A048153/b048153.txt">Table of n, a(n) for n = 1..10000</a>
%F a(n) = n(n+1)(2n+1)/6 (mod n). - _Charles R Greathouse IV_, Dec 28 2011
%F (Note: This does not mean a(n) = n(n+1)(2n+1)/6 mod n !) - _M. F. Hasler_, Oct 21 2013
%e a(5) = 1^2 + 2^2 + (3^2 mod 5) + (4^2 mod 5) + (5^2 mod 5) = 1 + 4 + 4 + 1 + 0 = 10. (It is easily seen that the last term, n^2 mod n, is always zero and would not need to be included.) - _M. F. Hasler_, Oct 21 2013
%t Table[Sum[PowerMod[k,2,n], {k,n-1}], {n,1,10000}] (* Zak Seidov, Nov 02 2011 *)
%o (Haskell)
%o a048153 = sum . a048152_row -- _Reinhard Zumkeller_, Apr 29 2013
%o (PARI) a(n)=sum(k=1,n,k^2%n) \\ _Charles R Greathouse IV_, Oct 21 2013
%Y Cf. A048152.
%K nonn
%O 1,3
%A _Clark Kimberling_
%E Definition made more explicit by _M. F. Hasler_, Oct 21 2013
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