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A113801 Numbers that are congruent to {1, 13} mod 14. 24
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 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,2
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
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
MATHEMATICA
LinearRecurrence[{1, 1, -1}, {1, 13, 15}, 60] (* or *) Select[Range[500], MemberQ[{1, 13}, Mod[#, 14]]&] (* Harvey P. Dale, May 11 2011 *)
PROG
(Haskell)
a113801 n = a113801_list !! (n-1)
a113801_list = 1 : 13 : map (+ 14) a113801_list
-- Reinhard Zumkeller, Jan 07 2012
(PARI) a(n)=n\2*14-(-1)^n \\ Charles R Greathouse IV, Sep 15 2015
CROSSREFS
Sequence in context: A178724 A087814 A227449 * A108257 A217252 A299593
KEYWORD
nonn,easy
AUTHOR
Giovanni Teofilatto, Jan 22 2006
EXTENSIONS
Corrected and extended by Giovanni Teofilatto, Nov 14 2008
Replaced the various formulas by a correct one - R. J. Mathar, Jan 30 2010
STATUS
approved

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Last modified February 26 12:39 EST 2024. Contains 370352 sequences. (Running on oeis4.)