A213387 a(n) = 5*2^(n-1) - 2 - 3*n.

0, 2, 9, 26, 63, 140, 297, 614, 1251, 2528, 5085, 10202, 20439, 40916, 81873, 163790, 327627, 655304, 1310661, 2621378, 5242815, 10485692, 20971449, 41942966, 83886003, 167772080, 335544237, 671088554, 1342177191

Offset: 1

Author

J. M. Bergot, Jun 28 2012

Keywords

• nonn
• easy

Comments

Create an array m(i,j) as follows: m(1,j) = j*(j-1)/2 in the top row, m(i,1) = (i-1)^2 in the left column, and m(i,j) = m(i,j-1) + m(i-1,j) recursively in the main body, j >= 1, i >= 1. The sum of the terms in an antidiagonal is one term in this sequence, a(n) = Sum_{k=1..n} m(n-k+1,k).

Examples

			For n=5, m(5,1)=16, m(4,2)=15, m(3,3)=11, m(2,4)=11, m(1,5)=10 gives the sum 63 = 2*A000295(4) + A095151(4) = 2*11 + 41.
		

Links

• Harvey P. Dale, Table of n, a(n) for n = 1..1000
• Joseph Breen and Emma Copeland, Non-orientable Nurikabe, arXiv:2506.12612 [math.CO], 2025. See pp. 1, 4.
• Index entries for linear recurrences with constant coefficients, signature (4,-5,2).

Crossrefs

Cf. A000295, A095151, A000217, A000290.

Programs

Mathematica

Table[5*2^(n-1)-2-3n,{n,30}] (* or *) LinearRecurrence[{4,-5,2},{0,2,9},30] (* Harvey P. Dale, Sep 28 2012 *)

Formula

a(n) = A095151(n-1) + 2*A000295(n-1).

G.f.: x^2*(2+x) / ( (1-2*x)*(1-x)^2 ). - R. J. Mathar, Jun 29 2012

a(n) = 4*a(n-1) - 5*a(n-2) + 2*a(n-3); a(1)=0, a(2)=2, a(3)=9. - Harvey P. Dale, Sep 28 2012