cp's OEIS Frontend

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.

%I A056640 #17 Jun 13 2015 00:50:17
%S A056640 1,5,18,42,83,143,228,340,485,665,886,1150,1463,1827,2248,2728,3273,
%T A056640 3885,4570,5330,6171,7095,8108,9212,10413,11713,13118,14630,16255,
%U A056640 17995,19856,21840,23953,26197,28578,31098,33763,36575,39540,42660,45941,49385,52998
%N 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.
%C A056640 Number of unit squares at n-th stage = n^2 + (n-1)^2 (A001844).
%C A056640 First differences are in A255840. - _Wesley Ivan Hurt_, Mar 13 2015
%D A056640 Anthony Gardiner, "Mathematical Puzzling," Dover Publications, Inc., Mineola, NY., 1987, page 88.
%H A056640 <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (3,-2,-2,3,-1).
%F A056640 G.f.: x(5x^2+2x+1)/((1-x^2)(1-x)^3).
%F A056640 a(n) = (8*n^3-2*n+3-3*(-1)^n)/12. - _Luce ETIENNE_, Aug 21 2014
%F A056640 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
%F A056640 G.f.: x*(5*x^2+2*x+1) / ((x-1)^4*(x+1)). - _Colin Barker_, Sep 29 2014
%p A056640 A056640:=n->(8*n^3-2*n+3-3*(-1)^n)/12: seq(A056640(n), n=1..50);
%t A056640 Table[(8*n^3 - 2*n + 3 - 3*(-1)^n)/12, {n, 30}] (* _Wesley Ivan Hurt_, Mar 13 2015 *)
%o A056640 (PARI) Vec(x*(5*x^2+2*x+1)/((x-1)^4*(x+1)) + O(x^100)) \\ _Colin Barker_, Sep 29 2014
%Y A056640 Cf. A255840.
%K A056640 nonn,easy
%O A056640 1,2
%A A056640 _Robert G. Wilson v_, Aug 21 2000
%E A056640 More terms from _Colin Barker_, Sep 29 2014