A334694 a(n) = (n/4)*(n^3+2*n^2+5*n+8).

0, 4, 17, 51, 124, 260, 489, 847, 1376, 2124, 3145, 4499, 6252, 8476, 11249, 14655, 18784, 23732, 29601, 36499, 44540, 53844, 64537, 76751, 90624, 106300, 123929, 143667, 165676, 190124, 217185, 247039, 279872, 315876, 355249, 398195, 444924, 495652, 550601, 609999, 674080, 743084, 817257, 896851, 982124, 1073340

Offset: 0

Author

Scott R. Shannon and N. J. A. Sloane, May 19 2020

Keywords

• nonn
• easy

Comments

Consider a figure made up of a row of n >= 1 adjacent congruent rectangles in which all possible diagonals of the rectangles have been drawn. The number of regions formed is A306302. If we distort all these diagonals very slightly so that no three lines meet at a point, the number of regions changes to a(n).

Links

• Colin Barker, Table of n, a(n) for n = 0..1000
• Index entries for sequences related to stained glass windows
• Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Crossrefs

Cf. A306302, A331452, A333275.

Programs

Mathematica

LinearRecurrence[{5,-10,10,-5,1},{0,4,17,51,124},50] (* or *)
A334694[n_]:=n/4(n^3+2n^2+5n+8);Array[A334694,50,0] (* Paolo Xausa, Nov 08 2023 *)

PARI

concat(0, Vec(x*(4 - 3*x + 6*x^2 - x^3) / (1 - x)^5 + O(x^40))) \\ Colin Barker, May 27 2020

Formula

Satisfies the identity a(n) = A306302(n) + Sum_{k=3..(n+1)} binomial(k-1,2)*A333275(n,2*k). E.g. for n=4 we have a(4) = 104 + 8*1 + 2*3 + 1*6 = 124.

From Colin Barker, May 27 2020: (Start)

G.f.: x*(4 - 3*x + 6*x^2 - x^3) / (1 - x)^5.

a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) for n>4.

(End)