login
A175886
Numbers that are congruent to {1, 12} mod 13.
13
1, 12, 14, 25, 27, 38, 40, 51, 53, 64, 66, 77, 79, 90, 92, 103, 105, 116, 118, 129, 131, 142, 144, 155, 157, 168, 170, 181, 183, 194, 196, 207, 209, 220, 222, 233, 235, 246, 248, 259, 261, 272, 274, 285, 287, 298, 300, 311, 313, 324, 326, 337, 339, 350
OFFSET
1,2
COMMENTS
Cf. property described by Gary Detlefs in A113801: more generally, these numbers are of the form (2*h*n+(h-4)*(-1)^n-h)/4 (h, n natural numbers), therefore ((2*h*n+(h-4)*(-1)^n-h)/4)^2-1 == 0 (mod h); in this case, a(n)^2-1 == 0 (mod 13).
FORMULA
G.f.: x*(1+11*x+x^2)/((1+x)*(1-x)^2).
a(n) = (26*n+9*(-1)^n-13)/4.
a(n) = -a(-n+1) = a(n-1)+a(n-2)-a(n-3).
a(n) = a(n-2)+13.
a(n) = 13*A000217(n-1)+1 - 2*sum(a(i), i=1..n-1) for n>1.
Sum_{n>=1} (-1)^(n+1)/a(n) = (Pi/13)*cot(Pi/13). - Amiram Eldar, Dec 04 2021
E.g.f.: 1 + ((26*x - 13)*exp(x) + 9*exp(-x))/4. - David Lovler, Sep 04 2022
MATHEMATICA
Select[Range[1, 350], MemberQ[{1, 12}, Mod[#, 13]]&] (* Bruno Berselli, Feb 29 2012 *)
CoefficientList[Series[(1 + 11 x + x^2) / ((1 + x) (1 - x)^2), {x, 0, 55}], x] (* Vincenzo Librandi, Aug 19 2013 *)
LinearRecurrence[{1, 1, -1}, {1, 12, 14}, 60] (* Harvey P. Dale, Oct 23 2015 *)
PROG
(Haskell)
a175886 n = a175886_list !! (n-1)
a175886_list = 1 : 12 : map (+ 13) a175886_list
-- Reinhard Zumkeller, Jan 07 2012
(Magma) [n: n in [1..350] | n mod 13 in [1, 12]]; // Bruno Berselli, Feb 29 2012
(Magma) [(26*n+9*(-1)^n-13)/4: n in [1..55]]; // Vincenzo Librandi, Aug 19 2013
(PARI) a(n)=(26*n+9*(-1)^n-13)/4 \\ Charles R Greathouse IV, Sep 24 2015
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Bruno Berselli, Oct 08 2010 - Nov 17 2010
STATUS
approved