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A048153 a(n) = sum_{k=1..n} (k^2 mod n). 9
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 (list; graph; refs; listen; history; text; internal format)
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}. - Moshe Levin, Nov 02 2011

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

LINKS

Moshe Levin, 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

(NB: This does not mean a(n) = n(n+1)(2n+1)/6 mod n !) - M. F. Hasler, Oct 21 2013

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}] (* Moshe Levin, 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

CROSSREFS

Cf. A048152.

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

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Last modified March 28 02:14 EDT 2017. Contains 284182 sequences.