A250352 Number of length 3 arrays x(i), i=1..3 with x(i) in i..i+n and no value appearing more than 2 times.

8, 26, 62, 122, 212, 338, 506, 722, 992, 1322, 1718, 2186, 2732, 3362, 4082, 4898, 5816, 6842, 7982, 9242, 10628, 12146, 13802, 15602, 17552, 19658, 21926, 24362, 26972, 29762, 32738, 35906, 39272, 42842, 46622, 50618, 54836, 59282, 63962, 68882, 74048

Offset: 1

Author

R. H. Hardin, Nov 19 2014

Keywords

• nonn
• easy

Comments

a(n) = (n+1)^3 - (n-1), where (n+1)^3 is the number of ways of selecting a triple from n+1 numbers in these subintervals, and there are n-1 of these triples, (3,3,3) up to (n-2,n-2,n-2), where all values are the same, which are discarded. - R. J. Mathar, Oct 09 2020

Examples

			Some solutions for n=6:
  2  0  1  2  6  4  0  1  0  0  2  4  6  2  4  0
  4  4  7  7  2  4  2  3  1  6  1  2  3  6  5  5
  6  4  7  2  4  7  8  5  3  6  4  7  5  8  8  2
		

Links

• R. H. Hardin, Table of n, a(n) for n = 1..210
• N. J. A. Sloane, Coordination Sequences, Planing Numbers, and Other Recent Sequences (II), Experimental Mathematics Seminar, Rutgers University, Jan 31 2019, Part I, Part 2, Slides. (Mentions this sequence)

Crossrefs

Row 3 of A250351.

Formula

a(n) = n^3 + 3*n^2 + 2*n + 2 = 2*A158842(n+1).

From Colin Barker, Nov 12 2018: (Start)

G.f.: 2*x*(4 - 3*x + 3*x^2 - x^3) / (1 - x)^4.

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

(End)