A113422 a(n) = floor((5*n^2+1)/3).

0, 2, 7, 15, 27, 42, 60, 82, 107, 135, 167, 202, 240, 282, 327, 375, 427, 482, 540, 602, 667, 735, 807, 882, 960, 1042, 1127, 1215, 1307, 1402, 1500, 1602, 1707, 1815, 1927, 2042, 2160, 2282, 2407, 2535, 2667, 2802, 2940, 3082, 3227, 3375, 3527, 3682, 3840

Offset

0,2

Comments

If n < a(m) then all integer solutions (i,j,k) of 2*i^2 + 2*j^2 + 3*k^2 - 2*i*k = n satisfy max(|i|,|j|,|k|) < m.

Links

G. C. Greubel, Table of n, a(n) for n = 0..1000

Index entries for linear recurrences with constant coefficients, signature (2,-1,1,-2,1).

Formula

G.f.: (2*x + 3*x^2 + 3*x^3 + 2*x^4)/((1-x)^2(1-x^3)).

a(-n) = a(n).

a(3n) = 15n^2, a(3n+1) = 15n^2 + 10n + 2.

a(n) = 2*a(n-1) - a(n-2) + a(n-3) - 2*a(n-4) + a(n-5), a(0)=0, a(1)=2, a(2)=7, a(3)=15, a(4)=27. - Harvey P. Dale, Nov 28 2013

a(n) = Sum_{j=1..n} Sum_{i=1..n} ceiling((i+j-n+3)/3). - Wesley Ivan Hurt, Mar 12 2015

Maple

A113422:=n->floor((5*n^2+1)/3): seq(A113422(n), n=0..100); # Wesley Ivan Hurt, Mar 12 2015

Mathematica

Floor[(5*Range[0, 50]^2+1)/3] (* or *) LinearRecurrence[{2, -1, 1, -2, 1}, {0, 2, 7, 15, 27}, 50] (* Harvey P. Dale, Nov 28 2013 *)

Prog

(PARI) a(n)=(5*n^2+1)\3
(Magma) [Floor((5*n^2+1)/3) : n in [0..100]]; // Wesley Ivan Hurt, Mar 12 2015

Crossrefs

Sequence in context: A293401 A323038 A333637 * A061802 A263603 A003452

Adjacent sequences: A113419 A113420 A113421 * A113423 A113424 A113425

Keyword

nonn,easy

Author

Michael Somos, Oct 31 2005

Status

approved