A093918 a(2k-1)=(2k-1)^2+k, a(2k)=6k^2+k+1: Last term in rows of triangle A093915.

2, 8, 11, 27, 28, 58, 53, 101, 86, 156, 127, 223, 176, 302, 233, 393, 298, 496, 371, 611, 452, 738, 541, 877, 638, 1028, 743, 1191, 856, 1366, 977, 1553, 1106, 1752, 1243, 1963, 1388, 2186, 1541, 2421, 1702, 2668, 1871, 2927, 2048, 3198, 2233, 3481, 2426, 3776

Offset

1,1

Comments

Initially defined as "leading diagonal" of the triangle A093915, a(n) is the last term in row n of A093915, i.e. a(n)=A093916(n)+n-1. - M. F. Hasler, Apr 04 2009

Links

Table of n, a(n) for n=1..50.

Index entries for linear recurrences with constant coefficients, signature (0,3,0,-3,0,1).

Formula

Equals A093915 o A000217 = A093916 + A023443. - M. F. Hasler, Apr 04 2009

a(n) = (3+(-1)^n+2*n+(5+(-1)^n)*n^2)/4. a(n) = 3*a(n-2)-3*a(n-4)+a(n-6). G.f.: -x*(x^2+2*x+2)*(x^3-x^2+3*x+1) / ((x-1)^3*(x+1)^3). - Colin Barker, Dec 18 2012

Mathematica

LinearRecurrence[{0, 3, 0, -3, 0, 1}, {2, 8, 11, 27, 28, 58}, 50] (* Harvey P. Dale, Oct 22 2013 *)

Prog

(PARI) A093918(n)=if(n%2, n^2, 6*(n\2)^2)+n\2+1 \\ M. F. Hasler, Apr 04 2009

Crossrefs

Cf. A093915, A093916, A093917.

Sequence in context: A197540 A089118 A146480 * A332098 A254313 A209429

Adjacent sequences: A093915 A093916 A093917 * A093919 A093920 A093921

Keyword

nonn,easy

Author

Amarnath Murthy, Apr 25 2004

Extensions

Edited and extended beyond a(6) by M. F. Hasler, Apr 04 2009

Status

approved