A134869 Row sums of triangle A134868.

1, 4, 7, 11, 16, 22, 29, 37, 46, 56, 67, 79, 92, 106, 121, 137, 154, 172, 191, 211, 232, 254, 277, 301, 326, 352, 379, 407, 436, 466, 497, 529, 562, 596, 631, 667, 704, 742, 781, 821, 862, 904, 947, 991, 1036, 1082, 1129, 1177, 1226, 1276, 1327, 1379, 1432

Offset: 1

Author

Gary W. Adamson, Nov 14 2007

Keywords

• nonn
• easy

Comments

Where records occur in A182703. - Omar E. Pol, Feb 14 2012

Consider quadratic polynomials x^2+cx+d. Then a(n) is the number of these polynomials with 0 <= c < n, 0 <= d < n where no polynomial can be horizontally translated into another. For example, a(3) = 7, the coefficients are as follows: (c, d) = {(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2), (2, 0)}. Two polynomials are excluded, namely x^2+2x+1 = (x+1)^2+0(x+1)+0, and x^2+2x+2 = (x+1)^2+0(x+1)+1. - Griffin N. Macris, Jul 19 2016

a(n) gives the number of regions into which the square [0,1]x[0,1] is divided by the Bernstein polynomials of degree n. - Franck Maminirina Ramaharo, Feb 28 2018

Examples

			a(4) = 11 = sum of row 4 terms of triangle A134868: (2, + 2 + 3 + 4).
a(4) = 11 = 1 + 10, where 10 = T(4).
a(4) = 11 = (1, 3, 3, 1) dot (1, 3, 0, 1) = (1 + 9 + 0 + 1).
		

Links

• Michael De Vlieger, Table of n, a(n) for n = 1..10000
• Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Crossrefs

Cf. A000217, A134868, A182703.

Essentially the same as A000124.

Programs

Maple

a:=n->sum((stirling2(j+1,n)), j=1..n):seq(a(n), n=1..50); # Zerinvary Lajos, Apr 12 2008

Mathematica

Table[(n^2 + n)/2 + Boole[n != 1], {n, 53}] (* or *)
Table[PolygonalNumber@ n + Boole[n != 1], {n, 53}] (* Version 10.4, or *)
Table[Sum[StirlingS2[k + 1, n], {k, n}], {n, 53}] (* or *)
Rest@ CoefficientList[Series[x (1 + x - 2 x^2 + x^3)/(1 - x)^3, {x, 0, 53}], x] (* Michael De Vlieger, Jul 19 2016 *)

PARI

a(n)=if(n>1, n*(n+1)/2+1, 1) \\ Charles R Greathouse IV, Aug 05 2016

Formula

a(n) = 1, then for n>1, a(n) = T(n) + 1, where A000217 = (1, 3, 6, 10, 15, ...).

Binomial transform of [1, 3, 0, 1, -1, 1, -1, 1, ...].

From R. J. Mathar, Oct 27 2008: (Start)

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

a(n) = 1 + A000217(n) = A000124(n), n > 1. (End)

Sum_{n>=1} 1/a(n) = 2*Pi*tanh(sqrt(7)*Pi/2)/sqrt(7) - 1/2. - Amiram Eldar, Jun 02 2025