A081499 Sum at 45 degrees to horizontal in triangle of A081498.

1, 2, 4, 6, 8, 11, 12, 16, 15, 20, 16, 22, 14, 21, 8, 16, -3, 6, -20, -10, -44, -33, -76, -64, -117, -104, -168, -154, -230, -215, -304, -288, -391, -374, -492, -474, -608, -589, -740, -720, -889, -868, -1056, -1034, -1242, -1219, -1448, -1424, -1675, -1650, -1924, -1898, -2196, -2169, -2492, -2464, -2813

Offset

1,2

Comments

The leading diagonal is given by A080956(n) = ((n+1)(2-n)/2).

Links

Colin Barker, Table of n, a(n) for n = 1..1000

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

Formula

a(n) = (n+floor(n/2)+1)*(n-floor(n/2))/2-binomial(ceiling(n/2)+1, ceiling(n/2)-2). - C. Ronaldo (aga_new_ac(AT)hotmail.com), Dec 20 2004

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

From Colin Barker, Nov 12 2017: (Start)

a(n) = (1/96)*(-2*n^3 + 36*n^2 + 32*n) for n even.

a(n) = (1/96)*(-2*n^3 + 30*n^2 + 50*n + 18) for n odd.

a(n) = a(n-1) + 3*a(n-2) - 3*a(n-3) - 3*a(n-4) + 3*a(n-5) + a(n-6) - a(n-7) for n>7.

(End)

Example

a(7) = 7+5+2+(-2) = 12.

Maple

seq((n+floor(n/2)+1)*(n-floor(n/2))/2-binomial(ceil(n/2)+1, ceil(n/2)-2), n=1..60); # C. Ronaldo (aga_new_ac(AT)hotmail.com), Dec 20 2004

Mathematica

LinearRecurrence[{1, 3, -3, -3, 3, 1, -1}, {1, 2, 4, 6, 8, 11, 12}, 60] (* Harvey P. Dale, Jan 17 2022 *)

Prog

(PARI) Vec(x*(1 + x - x^2 - x^3 - x^4) / ((1 - x)^4*(1 + x)^3) + O(x^60)) \\ Colin Barker, Nov 12 2017

Crossrefs

Cf. A080956, A081498.

Sequence in context: A226722 A187414 A187348 * A342527 A117638 A128403

Adjacent sequences: A081496 A081497 A081498 * A081500 A081501 A081502

Keyword

sign,easy

Author

Amarnath Murthy, Mar 25 2003

Extensions

More terms from C. Ronaldo (aga_new_ac(AT)hotmail.com), Dec 20 2004

Status

approved