A242477 a(n) = floor(3*n^2/4).
0, 0, 3, 6, 12, 18, 27, 36, 48, 60, 75, 90, 108, 126, 147, 168, 192, 216, 243, 270, 300, 330, 363, 396, 432, 468, 507, 546, 588, 630, 675, 720, 768, 816, 867, 918, 972, 1026, 1083, 1140, 1200, 1260, 1323, 1386, 1452, 1518, 1587, 1656, 1728, 1800, 1875
Offset: 0
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..1000
- Craig Knecht, Maximum number of octiamonds in a hexagon.
- Robert Munafo, Sequence MCS429697.
- Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).
Crossrefs
Cf. A002620.
Programs
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Magma
[Floor(3*n^2/4): n in [0..60]];
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Mathematica
Table[Floor[3 n^2/4], {n, 0, 60}] LinearRecurrence[{2,0,-2,1},{0,0,3,6},60] (* Harvey P. Dale, Sep 07 2019 *)
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Sage
[3*floor(n^2/4) for n in (0..60)] # Bruno Berselli, May 22 2014
Formula
a(n) = a(n-2) + 3*(n-1) for n>1, a(0) = a(1) = 0.
From Bruno Berselli, May 22 2014: (Start)
G.f.: 3*x^2/((1-x)^2*(1-x^2)).
a(n) = 3*A002620(n). (End)
Sum_{n>=2} 1/a(n) = Pi^2/18 + 1/3. - Amiram Eldar, Feb 16 2023
Comments