A386477 a(0) = 1; thereafter a(n) = 2*(6*n^2 - 3*n + 1).
1, 8, 38, 92, 170, 272, 398, 548, 722, 920, 1142, 1388, 1658, 1952, 2270, 2612, 2978, 3368, 3782, 4220, 4682, 5168, 5678, 6212, 6770, 7352, 7958, 8588, 9242, 9920, 10622, 11348, 12098, 12872, 13670, 14492, 15338, 16208, 17102, 18020, 18962, 19928, 20918, 21932, 22970, 24032, 25118, 26228, 27362, 28520, 29702, 30908, 32138, 33392, 34670
Offset: 0
Links
- Paolo Xausa, Table of n, a(n) for n = 0..10000
- Scott R. Shannon, Illustration for a(1) = 8 [Note that the cell counts shown on these five figures do not include the black exterior region, so the totals are off by 1]
- Scott R. Shannon, Illustration for a(2) = 38
- Scott R. Shannon, Illustration for a(3) = 92
- Scott R. Shannon, Illustration for a(8) = 722
- Scott R. Shannon, Illustration for a(10) = 1142
- N. J. A. Sloane, Sketch to illustrate a(2) = 38. The two hexagrams are colored red and black, respectively.
- N. J. A. Sloane, Sketch to illustrate a(3) = 92. The three hexagrams are colored red, blue, and black, respectively.
- N. J. A. Sloane, Analogous illustration for three pentagrams
- Index entries for linear recurrences with constant coefficients, signature (3,-3,1).
Crossrefs
Programs
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Mathematica
A386477[n_] := If[n == 0, 1, 6*n*(2*n - 1) + 2]; Array[A386477, 50, 0] (* or *) Join[{1}, 6*PolygonalNumber[6, Range[49]] + 2] (* or *) LinearRecurrence[{3, -3, 1}, {1, 8, 38, 92}, 50] (* Paolo Xausa, Jul 24 2025 *)
Formula
From Stefano Spezia, Jul 23 2025: (Start)
G.f.: (1 + 5*x + 17*x^2 + x^3)/(1 - x)^3.
E.g.f.: 2*exp(x)*(1 + 3*x + 6*x^2) - 1. (End)
a(n) = A152746(n) + 2, for n >= 1. - Paolo Xausa, Jul 24 2025
Comments