A056640 At stage 1, start with a unit square. At each successive stage add 4*(n-1) new squares around outside with edge-to-edge contacts. Sequence gives number of squares (regardless of size) at n-th stage.

1, 5, 18, 42, 83, 143, 228, 340, 485, 665, 886, 1150, 1463, 1827, 2248, 2728, 3273, 3885, 4570, 5330, 6171, 7095, 8108, 9212, 10413, 11713, 13118, 14630, 16255, 17995, 19856, 21840, 23953, 26197, 28578, 31098, 33763, 36575, 39540, 42660, 45941, 49385, 52998

Offset: 1

Author

Robert G. Wilson v, Aug 21 2000

Keywords

• nonn
• easy

Comments

Number of unit squares at n-th stage = n^2 + (n-1)^2 (A001844).

First differences are in A255840. - Wesley Ivan Hurt, Mar 13 2015

References

• Anthony Gardiner, "Mathematical Puzzling," Dover Publications, Inc., Mineola, NY., 1987, page 88.

Links

• Index entries for linear recurrences with constant coefficients, signature (3,-2,-2,3,-1).

Crossrefs

Cf. A255840.

Programs

Maple

A056640:=n->(8*n^3-2*n+3-3*(-1)^n)/12: seq(A056640(n), n=1..50);

Mathematica

Table[(8*n^3 - 2*n + 3 - 3*(-1)^n)/12, {n, 30}] (* Wesley Ivan Hurt, Mar 13 2015 *)

PARI

Vec(x*(5*x^2+2*x+1)/((x-1)^4*(x+1)) + O(x^100)) \\ Colin Barker, Sep 29 2014

Formula

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

a(n) = (8*n^3-2*n+3-3*(-1)^n)/12. - Luce ETIENNE, Aug 21 2014

a(n) = 3*a(n-1)-2*a(n-2)-2*a(n-3)+3*a(n-4)-a(n-5). - Colin Barker, Sep 29 2014

G.f.: x*(5*x^2+2*x+1) / ((x-1)^4*(x+1)). - Colin Barker, Sep 29 2014

Extensions

More terms from Colin Barker, Sep 29 2014