A046691 a(n) = (n^2 + 5*n - 2)/2.

-1, 2, 6, 11, 17, 24, 32, 41, 51, 62, 74, 87, 101, 116, 132, 149, 167, 186, 206, 227, 249, 272, 296, 321, 347, 374, 402, 431, 461, 492, 524, 557, 591, 626, 662, 699, 737, 776, 816, 857, 899, 942, 986, 1031, 1077, 1124, 1172, 1221, 1271, 1322

Offset: 0

Author

N. J. A. Sloane

Keywords

• sign
• easy

Comments

If Y_i (i=1,2,3,4) are 2-blocks of an n-set X then, for n>=8, a(n-3) is the number of (n-2)-subsets of X intersecting each Y_i (i=1,2,3,4). - Milan Janjic, Nov 09 2007

Numbers m > -3 such that 8*m + 33 is a square. - Bruno Berselli, Aug 20 2015

a(n-1) yields the second Betti number of a path graph on n vertices. - Samuel J. Bevins, Nov 27 2022

Links

• G. C. Greubel, Table of n, a(n) for n = 0..5000
• Milan Janjic, Two Enumerative Functions
• P. Di Francesco, O. Golinelli and E. Guitter, Meander, folding and arch statistics, arXiv:hep-th/9506030, 1995.
• Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Crossrefs

Triangular numbers (A000217) minus 4. Cf. A027379.

Cf. A000096, A000217, A002522, A028884.

Programs

Magma

[Binomial(n+3,2) -4: n in [0..50]]; // G. C. Greubel, Jul 31 2022

Maple

seq(binomial(n+3, 2)-4, n=0..49); # Zerinvary Lajos, Jan 13 2007

Mathematica

Table[(n^2 +5n -2)/2, {n, 0, 50}] (* Bruno Berselli, Dec 17 2014 *)

PARI

a(n)=(n^2+5*n-2)/2 \\ Charles R Greathouse IV, Oct 07 2015

SageMath

[(n^2 +5*n -2)/2 for n in (0..50)] # G. C. Greubel, Jul 31 2022

Formula

G.f.: (-1 + 5*x - 3*x^2)/(1 - x)^3.

a(n) = a(n-1) + n + 2 with a(0) = -1. - Vincenzo Librandi, Nov 18 2010

a(n) = 3*A000096(n-1) - 2*A000096(n-2), with A000096(-2)=A000096(-1)=-1. - Bruno Berselli, Dec 17 2014

a(n) = 2*A000217(n) - A000217(n-2), with A000217(-2)=1, A000217(-1)=0. - Bruno Berselli, Oct 13 2016

E.g.f.: (1/2)*(x^2 + 6*x - 2)*exp(x). - G. C. Greubel, Jul 13 2017

a(n) + a(n+1) = A028884(n). - R. J. Mathar, Mar 13 2021

a(n) = A000217(n+2) - 4. - G. C. Greubel, Jul 31 2022

Sum_{n>=0} 1/a(n) = 7/12 + 2*Pi*tan(sqrt(33)*Pi/2)/sqrt(33). - Vaclav Kotesovec, Dec 31 2022