login
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
OFFSET
0,3
COMMENTS
Numbers that are congruent to {0,1} mod 13. - Philippe Deléham, Oct 17 2001
FORMULA
a(n) = Sum_{k>=0} A030308(n,k) * A005029(k-1) 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(n-1) + a(n-2) - a(n-3) for n > 2.
G.f.: x*(1+12*x) / ((1-x)^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 *)
Select[Range[0, 400], Mod[#, 13]==PowerMod[#, 2, 13]&] (* or *) LinearRecurrence[ {1, 1, -1}, {0, 1, 13}, 60] (* Harvey P. Dale, Feb 07 2023 *)
PROG
(Magma) I:=[0, 1, 13]; [n le 3 select I[n] else Self(n-1)+Self(n-2)-Self(n-3): 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
Sequence in context: A296795 A079831 A022803 * A301660 A022103 A224224
KEYWORD
easy,nonn
AUTHOR
Jeremy Gardiner, Dec 28 2005
STATUS
approved