

A112653


a(n) squared is congruent to a(n) (mod 13).


4



0, 1, 13, 14, 26, 27, 39, 40, 52, 53, 65, 66, 78, 79, 91, 92, 104, 105, 117, 118, 130, 131, 143, 144, 156, 157, 169, 170, 182, 183, 195, 196, 208, 209, 221, 222, 234, 235, 247, 248, 260, 261, 273, 274, 286, 287, 299, 300, 312, 313, 325, 326, 338, 339, 351
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

0,3


COMMENTS

Numbers that are congruent to {0,1} mod 13.  Philippe Deléham, Oct 17 2001


LINKS

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,1,1).


FORMULA

a(n) = Sum_{k>=0} A030308(n,k) * A005029(k1) with A005029(1) = 1.  Philippe Deléham, Oct 17 2011
From Colin Barker, May 14 2012: (Start)
a(n) = (11*(1+(1)^n)+26*n)/4.
a(n) = a(n1) + a(n2)  a(n3) for n > 2.
G.f.: x*(1+12*x) / ((1x)^2*(1+x)). (End)


EXAMPLE

a(3) = 14 because 14*14 = 196 = 1 (mod 13) and 14 = 1 (mod 13).


MAPLE

m:= 13; for n from 0 to 300 do if n^2 mod m = n mod m then print(n) fi od;


MATHEMATICA

Select[Range[0, 400], MemberQ[{0, 1}, Mod[#, 13]]&] (* Vincenzo Librandi, May 17 2012 *)


PROG

(MAGMA) I:=[0, 1, 13]; [n le 3 select I[n] else Self(n1)+Self(n2)Self(n3): n in [1..70]]; // Vincenzo Librandi, May 17 2012
(PARI) a(n)=(11*(1+(1)^n)+26*n)/4 \\ Charles R Greathouse IV, Oct 16 2015


CROSSREFS

Cf. A005029, A030308.
Sequence in context: A296795 A079831 A022803 * A301660 A022103 A224224
Adjacent sequences: A112650 A112651 A112652 * A112654 A112655 A112656


KEYWORD

easy,nonn


AUTHOR

Jeremy Gardiner, Dec 28 2005


STATUS

approved



