A226203 a(5n) = a(5n+3) = a(5n+4) = 2n+1, a(5n+1) = 2n-3, a(5n+2) = 2n-1.

1, -3, -1, 1, 1, 3, -1, 1, 3, 3, 5, 1, 3, 5, 5, 7, 3, 5, 7, 7, 9, 5, 7, 9, 9, 11, 7, 9, 11, 11, 13, 9, 11, 13, 13, 15, 11, 13, 15, 15, 17, 13, 15, 17, 17, 19, 15, 17, 19, 19, 21, 17, 19, 21, 21, 23, 19, 21, 23, 23, 25, 21, 23, 25, 25

Offset

0,2

Comments

Given the numerators of A225948/A226008 ordered according to A226096: 0, -15, -3, 2, 3, 6, -7, 5, 12, 15, 20, 9, 21, 30, 35,... = t(n), then (a(n) + t(n)/a(n))^2 = A226096(n).

First six differences (of period 5):

...-4, 2, 2, 0, 2, -4, 2, 2, 0, 2, ...

....6, 0, -2, 2, -6, 6, 0, -2, 2, -6, ...

...-6, -2, 4, -8, 12, -6, -2, 4, -8, 12, ...

....4, 6, -12, 20, -18, 4, 6, -12, 20, -18, ...

....2, -18, 32, -38, 22, 2, -18, 32, -38, 22, ...

..-20, 50, -70, 60, -20, -20, 50, -70, 60, -20, ...

Links

Table of n, a(n) for n=0..64.

Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,1,-1).

Formula

a(n+5) = a(n) + 2.

G.f.: (1-4*x+2*x^2+2*x^3+x^5)/((1-x)^2*(1+x+x^2+x^3+x^4)). [Bruno Berselli, Jun 01 2013]

a(n) = a(n-1)+a(n-5)-a(n-6) with a(0)=a(3)=a(4)=1, a(1)=-3, a(2)=-1, a(5)=3. [Bruno Berselli, Jun 01 2013]

Mathematica

a[n_] := 2 Quotient[n, 5] + Switch[Mod[n, 5], 0, 1, 1, -3, 2, -1, 3, 1, 4, 1]; Table[a[n], {n, 0, 64}] (* Jean-François Alcover, Jun 22 2017 *)

Prog

(Haskell)
import Data.List (transpose)
a226203 n = a226203_list !! n
a226203_list = concat $ transpose
               [[1, 3 ..], [-3, -1 ..], [-1, 1 ..], [1, 3 ..], [1, 3 ..]]
-- Reinhard Zumkeller, Jun 02 2013

Crossrefs

Cf. A007395, A005408, A130497, A226096.

Sequence in context: A087612 A260626 A155828 * A327791 A051997 A155744

Adjacent sequences: A226200 A226201 A226202 * A226204 A226205 A226206

Keyword

sign,easy

Author

Paul Curtz, May 31 2013

Extensions

Edited by Bruno Berselli, Jun 01 2013

Status

approved