A080827 Rounded up staircase on natural numbers.

1, 3, 5, 9, 13, 19, 25, 33, 41, 51, 61, 73, 85, 99, 113, 129, 145, 163, 181, 201, 221, 243, 265, 289, 313, 339, 365, 393, 421, 451, 481, 513, 545, 579, 613, 649, 685, 723, 761, 801, 841, 883, 925, 969, 1013, 1059, 1105, 1153, 1201, 1251, 1301, 1353, 1405, 1459

Offset: 1

Author

Paul Barry, Feb 28 2003

Keywords

• nonn
• easy

Comments

Represents the 'rounded up' staircase diagonal on A000027, arranged as a square array. A000982 is the 'rounded down' staircase.

Partial sums of A131055. - Paul Barry, Jun 14 2008

The same sequence arises in the triangular array of integers >= 1 according to a simple "zig-zag" rule for selection of terms. a(n-1) lies in the (n-1)-th row of the array and the second row of that subarray (with apex a(n-1)) contains just two numbers, one odd one even. The one with the same (odd) parity as a(n-1) is a(n). - David James Sycamore, Jul 29 2018

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..1000
• J. C. F. de Winter, Using the Student's t-test with extremely small sample sizes, Practical Assessment, Research & Evaluation, 18(10), 2013.
• Girtrude Hamm, Classification of lattice triangles by their two smallest widths, arXiv:2304.03007 [math.CO], 2023.
• David James Sycamore, Triangular array.
• Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).

Crossrefs

Apart from leading term identical to A099392.

Cf. A000027, A000982, A007590, A131055.

Programs

GAP

List([1..10],n->Int(n^2/2)+1); # Muniru A Asiru, Aug 02 2018

Magma

[n*(n+1)/2-Floor((n-1)/2) : n in [1..60]]; // Vincenzo Librandi, Aug 05 2013

Maple

A080827:=n->(n^2+2-(1-(-1)^n)/2)/2: seq(A080827(n), n=1..100); # Wesley Ivan Hurt, Sep 08 2015

Mathematica

CoefficientList[Series[(1 + x - x^2 + x^3) / ((1 + x) (1 - x)^3), {x, 0, 40}], x] (* Vincenzo Librandi, Aug 05 2013 *)
Ceiling[(Range[100]^2 + 1)/2] (* Paolo Xausa, Jul 19 2025 *)

Formula

a(n) = ceiling((n^2+1)/2).

a(1) = 1, a(2n) = a(2n-1) + 2n, a(2n+1) = a(2n) + 2n. - Amarnath Murthy, May 07 2003

From Paul Barry, Apr 12 2008: (Start)

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

a(n) = n*(n+1)/2-floor((n-1)/2). [corrected by R. J. Mathar, Jul 14 2013] (End)

From Wesley Ivan Hurt, Sep 08 2015: (Start)

a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4), n > 4.

a(n) = (n^2 + 2 - (1 - (-1)^n)/2)/2.

a(n) = floor(n^2/2) + 1 = A007590(n-1) + 1. (End)

Sum_{n>=1} 1/a(n) = tanh(Pi/2)*Pi/2 + coth(Pi/sqrt(2))*Pi/(2*sqrt(2)) - 1/2. - Amiram Eldar, Sep 15 2022

E.g.f.: ((2 + x + x^2)*cosh(x) + (1 + x + x^2)*sinh(x) - 2)/2. - Stefano Spezia, Jan 27 2024