A336288 Numbers of squares formed by this procedure on n-th step: Step 1, draw a unit square. Step n, draw a unit square with center in every intersection of lines of the figure in step n-1.

1, 10, 43, 116, 245, 446, 735, 1128, 1641, 2290, 3091, 4060, 5213, 6566, 8135, 9936, 11985, 14298, 16891, 19780, 22981, 26510, 30383, 34616, 39225, 44226, 49635, 55468, 61741, 68470, 75671, 83360, 91553, 100266, 109515, 119316, 129685, 140638, 152191, 164360, 177161

Offset

1,2

Links

Table of n, a(n) for n=1..41.

Ilario Miriello, Step 1,2,3, Youtube video, Jul 16 2020.

Ilario Miriello, Illustration for a(2) and a(3)

Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Formula

a(n) = (8*n^3 - 12*n^2 + 7*n)/3.

From Colin Barker, Jul 17 2020: (Start)

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

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

(End)

E.g.f.: exp(x)*x*(3 + 12*x + 8*x^2)/3. - Stefano Spezia, Jul 23 2020

a(n+1) - a(n) = 8*n^2 + 1 = A081585(n). - Charlie Marion, Mar 21 2022

Mathematica

Table[(8*n^3 - 12*n^2 + 7*n)/3, {n, 1, 50}] (* Amiram Eldar, Jul 16 2020 *)
LinearRecurrence[{4, -6, 4, -1}, {1, 10, 43, 116}, 50] (* Harvey P. Dale, Sep 12 2021 *)

Prog

(Magma) [(8*n^3 - 12*n^2 + 7*n)/3 : n in [1..50]]; // Wesley Ivan Hurt, Jul 16 2020
(PARI) a(n) = (8*n^3 - 12*n^2 + 7*n)/3; \\ Michel Marcus, Jul 16 2020
(PARI) Vec(x*(1 + 3*x)^2 / (1 - x)^4 + O(x^40)) \\ Colin Barker, Jul 17 2020

Crossrefs

Cf. A081585.

Sequence in context: A092117 A245663 A244802 * A211070 A251936 A187673

Adjacent sequences: A336285 A336286 A336287 * A336289 A336290 A336291

Keyword

nonn,easy,nice

Author

Ilario Miriello, Jul 16 2020

Extensions

More terms from Michel Marcus, Jul 16 2020

Status

approved