A113801 Numbers that are congruent to {1, 13} mod 14.

1, 13, 15, 27, 29, 41, 43, 55, 57, 69, 71, 83, 85, 97, 99, 111, 113, 125, 127, 139, 141, 153, 155, 167, 169, 181, 183, 195, 197, 209, 211, 223, 225, 237, 239, 251, 253, 265, 267, 279, 281, 293, 295, 307, 309, 321, 323, 335, 337, 349, 351, 363, 365, 377, 379

Offset: 1

Author

Giovanni Teofilatto, Jan 22 2006

Keywords

• nonn
• easy

Comments

If 14k+1 is a perfect square..(0,12,16,52,60,120..) then the square root of 14k+1 = a(n) - Gary Detlefs, Feb 22 2010

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 our case, a(n)^2-1==0 (mod 14). Also a(n)^2-1==0 (mod 28). - Bruno Berselli, Oct 26 2010 - Nov 17 2010

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
• Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Crossrefs

Cf. A000217, A113802, A113803, A113804, A113805, A113806, A113807, A008589, A045472 (primes), A195145 (partial sums), A005408, A047209, A007310, A047336, A047522, A056020, A090771, A175885, A091998, A175886, A175887.

Programs

Haskell

a113801 n = a113801_list !! (n-1)
a113801_list = 1 : 13 : map (+ 14) a113801_list
-- Reinhard Zumkeller, Jan 07 2012

Mathematica

LinearRecurrence[{1,1,-1},{1,13,15},60] (* or *) Select[Range[500], MemberQ[{1,13},Mod[#,14]]&] (* Harvey P. Dale, May 11 2011 *)

PARI

a(n)=n\2*14-(-1)^n \\ Charles R Greathouse IV, Sep 15 2015

Formula

a(n) = 14*(n-1)-a(n-1), n>1. - R. J. Mathar, Jan 30 2010

From Bruno Berselli, Oct 26 2010: (Start)

a(n) = -a(-n+1) = (14*n+5*(-1)^n-7)/2.

G.f.: x*(1+12*x+x^2)/((1+x)*(1-x)^2).

a(n) = a(n-2)+14 for n>2.

a(n) = 14*A000217(n-1)+1 - 2*sum[i=1..n-1] a(i) for n>1. (End)

a(0)=1, a(1)=13, a(2)=15, a(n)=a(n-1)+a(n-2)-a(n-3). - Harvey P. Dale, May 11 2011

Sum_{n>=1} (-1)^(n+1)/a(n) = (Pi/14)*cot(Pi/14). - Amiram Eldar, Dec 04 2021

E.g.f.: 1 + ((14*x - 7)*exp(x) + 5*exp(-x))/2. - David Lovler, Sep 04 2022

From Amiram Eldar, Nov 25 2024: (Start)

Product_{n>=1} (1 - (-1)^n/a(n)) = 2*cos(Pi/14).

Product_{n>=2} (1 + (-1)^n/a(n)) = (Pi/14)*cosec(Pi/14). (End)

Extensions

Corrected and extended by Giovanni Teofilatto, Nov 14 2008

Replaced the various formulas by a correct one - R. J. Mathar, Jan 30 2010