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
Links
- Kelvin Voskuijl, Table of n, a(n) for n = 1..10000
- 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).
Crossrefs
Cf. A081585.
Programs
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Magma
[(8*n^3 - 12*n^2 + 7*n)/3 : n in [1..50]]; // Wesley Ivan Hurt, Jul 16 2020
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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 *)
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PARI
a(n) = (8*n^3 - 12*n^2 + 7*n)/3; \\ Michel Marcus, Jul 16 2020
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PARI
Vec(x*(1 + 3*x)^2 / (1 - x)^4 + O(x^40)) \\ Colin Barker, Jul 17 2020
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
Extensions
More terms from Michel Marcus, Jul 16 2020