A156849 Numbers k such that k^2 == 2 (mod 23^2).

156, 373, 685, 902, 1214, 1431, 1743, 1960, 2272, 2489, 2801, 3018, 3330, 3547, 3859, 4076, 4388, 4605, 4917, 5134, 5446, 5663, 5975, 6192, 6504, 6721, 7033, 7250, 7562, 7779, 8091, 8308, 8620, 8837, 9149, 9366, 9678, 9895, 10207, 10424

Offset

1,1

Comments

Numbers k such that k = 156 or 373 (mod 529). - Charles R Greathouse IV, Dec 27 2011

Links

Vincenzo Librandi, Table of n, a(n) for n = 1..10000

Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Formula

Conjecture: a(n) = a(n-1) + a(n-2) - a(n-3) = 529*n/2 - 529/4 - 95*(-1)^n/4. - R. J. Mathar, Oct 18 2010

The conjecture is correct. - Charles R Greathouse IV, Dec 27 2011

G.f.: x*(156+217*x+156*x^2)/(1-x-x^2+x^3). - Colin Barker, Jan 16 2012

Sum_{n>=1} (-1)^(n+1)/a(n) = tan(217*Pi/1058)*Pi/529. - Amiram Eldar, Feb 26 2023

Example

156^2 - 2 == 0 (mod 23^2).
373^2 - 2 == 0 (mod 23^2).
685^2 - 2 == 0 (mod 23^2).
10424^2 - 2 == 0 (mod 23^2).

Mathematica

With[{c=23^2}, Select[Range[20000], PowerMod[#, 2, c]==2&]] (* or *) LinearRecurrence[{1, 1, -1}, {156, 373, 685}, 80] (* Harvey P. Dale, Oct 13 2014 *)

Prog

(PARI) a(n)=529*n/2-529/4-95*(-1)^n/4 \\ Charles R Greathouse IV, Dec 27 2011
(Magma) [(1058*n-529-95*(-1)^n)/4: n in [1..50]]; // Vincenzo Librandi, Jan 12 2012

Crossrefs

Cf. A156846, A156845, A156844, A156843, A156842, A156841.

Sequence in context: A037983 A298909 A366172 * A250382 A106056 A259947

Adjacent sequences: A156846 A156847 A156848 * A156850 A156851 A156852

Keyword

nonn,easy

Author

Vincenzo Librandi, Feb 17 2009

Extensions

Checked by Joerg Arndt, Jun 16 2010

Status

approved