A048153 a(n) = Sum_{k=1..n} (k^2 mod n).

0, 1, 2, 2, 10, 13, 14, 12, 24, 45, 44, 38, 78, 77, 70, 56, 136, 129, 152, 130, 182, 209, 184, 148, 250, 325, 288, 294, 406, 365, 372, 304, 484, 561, 490, 402, 666, 665, 572, 540, 820, 805, 860, 726, 840, 897, 846, 680, 980, 1125, 1156, 1170, 1378, 1305, 1210

Offset

1,3

Comments

See A048152 for the array T[n,k] = k^2 mod n.

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

Positions of squares in A048153: 1, 2, 33, 51, 69, 105, 195, 250, 294, 1250, 4913, 9583, 13778, 48778, 65603, 83521.

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

For n > 1 also row sums of A060036. - Reinhard Zumkeller, Apr 29 2013

Links

Zak Seidov, Table of n, a(n) for n = 1..10000

Formula

a(n) == n*(n+1)*(2n+1)/6 (mod n). - Charles R Greathouse IV, Dec 28 2011

a(n) == n*(n-1)*(2n-1)/6 (mod n). - Chai Wah Wu, Jun 02 2024

a(n) mod n = A215573(n). - Alois P. Heinz, Jun 03 2024

Example

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

Mathematica

Table[Sum[PowerMod[k, 2, n], {k, n-1}], {n, 1, 10000}] (* Zak Seidov, Nov 02 2011 *)

Prog

(Haskell)
a048153 = sum . a048152_row  -- Reinhard Zumkeller, Apr 29 2013
(PARI) a(n)=sum(k=1, n, k^2%n) \\ Charles R Greathouse IV, Oct 21 2013
(Python)
def A048153(n): return sum(k**2%n for k in range(1, n)) # Chai Wah Wu, Jun 02 2024

Crossrefs

Cf. A000330, A048152, A215573.

Sequence in context: A066965 A066966 A132443 * A015623 A164124 A003609

Adjacent sequences: A048150 A048151 A048152 * A048154 A048155 A048156

Keyword

nonn

Author

Clark Kimberling

Extensions

Definition made more explicit by M. F. Hasler, Oct 21 2013

Status

approved