OFFSET
1,2
COMMENTS
Numbers k such that k^2/7 + k*(k + 1)/7 = k*(2*k + 1)/7 is a nonnegative integer. - Bruno Berselli, Feb 14 2017
LINKS
David Lovler, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).
FORMULA
a(n) = a(n-2) + 7 = a(n-1) + a(n-2) - a(n-3). - Henry Bottomley, Jan 19 2001
From Bruno Berselli, Sep 12 2011: (Start)
G.f.: x^2*(3 + 4*x)/((1 + x)*(1 - x)^2).
a(n) = (14*n - (-1)^n - 15)/4. (End)
a(n) = 2*n - 2 + floor((3*n - 3)/2). - Wesley Ivan Hurt, Jan 30 2014
E.g.f.: 4 + ((14*x - 15)*exp(x) - exp(-x))/4. - David Lovler, Aug 31 2022
MAPLE
MATHEMATICA
Table[(14n - (-1)^n - 15)/4, {n, 100}] (* Wesley Ivan Hurt, Jan 30 2014 *)
PROG
(Haskell)
a047355 n = a047355_list !! (n-1)
a047355_list = scanl (+) 0 a010702_list -- Reinhard Zumkeller, Jul 05 2012
(PARI) a(n)=n\2*7 - 4 + n%2*4 \\ Charles R Greathouse IV, Aug 01 2016
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
STATUS
approved