OFFSET
1,3
COMMENTS
a(n+1) = the remainder when n^2 is divided by 2*n+1. - J. M. Bergot, Jun 25 2013
LINKS
Reinhard Zumkeller, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0,2,0,-1).
FORMULA
a(1) = 0, a(2*n) = n and a(2*n+1) = 3*n+1.
a(n) = A123684(n) for n > 1.
a(n) = A048152(2*n-1,n).
a(n) = A060036(2*n-1,n-1) for n > 1.
From Colin Barker, May 01 2013: (Start)
a(n) = 2*a(n-2)-a(n-4) for n>5.
G.f.: -x^2*(x^3-4*x-1) / ((x-1)^2*(x+1)^2). (End)
MATHEMATICA
LinearRecurrence[{0, 2, 0, -1}, {0, 1, 4, 2, 7}, 100] (* Paolo Xausa, Mar 10 2026 *)
PROG
(Haskell)
a225126 n = a048152 (2 * n - 1) n
(PARI) a(n)=n--^2%(2*n+1) \\ Charles R Greathouse IV, Jun 25 2013
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Reinhard Zumkeller, Apr 29 2013
STATUS
approved
