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A133146 Antidiagonal sums of the triangle A133128. 2
2, 5, 7, 14, 18, 29, 35, 50, 58, 77, 87, 110, 122, 149, 163, 194, 210, 245, 263, 302, 322, 365, 387, 434, 458, 509, 535, 590, 618, 677, 707, 770, 802, 869, 903, 974, 1010, 1085, 1123, 1202, 1242, 1325, 1367, 1454, 1498, 1589, 1635, 1730, 1778, 1877, 1927, 2030 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,1

LINKS

Vincenzo Librandi, Table of n, a(n) for n = 0..10000

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

FORMULA

First differences: a(n+1) - a(n) = A059029(n+1).

Bisections: a(2n+1) = A005918(n+1). a(2n) = A141631(n+1).

G.f.: (1+2*x)(2 - x + x^3)/((1-x)^3*(1+x)^2). - R. J. Mathar, Oct 15 2008

a(n) = 19/8 + 5*n/4 + 3*n^2/4 - (-1)^n*(n/4 + 3/8). - R. J. Mathar, Oct 15 2008

From Harvey P. Dale, Aug 26 2013: (Start)

a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5); a(0)=2, a(1)=5, a(2)=7, a(3)=14, a(4)=18. (End)

EXAMPLE

a(2) = A133128(2,0) + A133128(1,1) = 10 - 3 = 7.

a(3) = A133128(3,0) + A133128(2,1) = 17 - 3 = 14.

MATHEMATICA

CoefficientList[Series[(1+2x)(2-x+x^3)/((1-x)^3(1+x)^2), {x, 0, 60}], x] (* or *) LinearRecurrence[{1, 2, -2, -1, 1}, {2, 5, 7, 14, 18}, 60] (* Harvey P. Dale, Aug 26 2013 *)

PROG

(MAGMA) [19/8 +5*n/4 +3*n^2/4 -(-1)^n*(n/4+3/8): n in [0..60]]; // Vincenzo Librandi, Aug 10 2011

CROSSREFS

Sequence in context: A031457 A044990 A266259 * A217753 A022771 A132603

Adjacent sequences:  A133143 A133144 A133145 * A133147 A133148 A133149

KEYWORD

nonn,less

AUTHOR

Paul Curtz, Aug 27 2008

EXTENSIONS

Edited and extended by R. J. Mathar, Oct 15 2008

STATUS

approved

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Last modified April 6 08:53 EDT 2020. Contains 333268 sequences. (Running on oeis4.)