A299264 Partial sums of A299258.
1, 6, 19, 44, 85, 147, 236, 357, 514, 711, 953, 1246, 1595, 2004, 2477, 3019, 3636, 4333, 5114, 5983, 6945, 8006, 9171, 10444, 11829, 13331, 14956, 16709, 18594, 20615, 22777, 25086, 27547, 30164, 32941, 35883, 38996, 42285, 45754, 49407, 53249, 57286, 61523
Offset: 0
Examples
G.f. = 1 + 6*x + 19*x^2 + 44*x^3 + 85*x^4 + 147*x^5 + 236*x^6 + ... - _Michael Somos_, Oct 03 2018
Links
- Colin Barker, Table of n, a(n) for n = 0..1000
- Index entries for linear recurrences with constant coefficients, signature (3,-3,1,0,1,-3,3,-1).
Crossrefs
Cf. A299258.
The 28 uniform 3D tilings: cab: A299266, A299267; crs: A299268, A299269; fcu: A005901, A005902; fee: A299259, A299265; flu-e: A299272, A299273; fst: A299258, A299264; hal: A299274, A299275; hcp: A007899, A007202; hex: A005897, A005898; kag: A299256, A299262; lta: A008137, A299276; pcu: A005899, A001845; pcu-i: A299277, A299278; reo: A299279, A299280; reo-e: A299281, A299282; rho: A008137, A299276; sod: A005893, A005894; sve: A299255, A299261; svh: A299283, A299284; svj: A299254, A299260; svk: A010001, A063489; tca: A299285, A299286; tcd: A299287, A299288; tfs: A005899, A001845; tsi: A299289, A299290; ttw: A299257, A299263; ubt: A299291, A299292; bnn: A007899, A007202. See the Proserpio link in A299266 for overview.
Programs
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Mathematica
a[ n_] := (4 n^3 + 6 n^2 + 16 n + {5, 4, 7, 10, 9}[[Mod[n, 5] + 1]]) / 5; (* Michael Somos, Oct 03 2018 *) LinearRecurrence[{3,-3,1,0,1,-3,3,-1},{1,6,19,44,85,147,236,357},50] (* Harvey P. Dale, Aug 03 2025 *)
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PARI
Vec((1 + x)^3*(1 - x + x^2)*(1 + x + x^2) / ((1 - x)^4*(1 + x + x^2 + x^3 + x^4)) + O(x^60)) \\ Colin Barker, Feb 09 2018
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PARI
{a(n) = (4*n^3 + 6*n^2 + 16*n + [5, 4, 7, 10, 9][n%5+1]) / 5}; /* Michael Somos, Oct 03 2018 */
Formula
From Colin Barker, Feb 09 2018: (Start)
G.f.: (1 + x)^3*(1 - x + x^2)*(1 + x + x^2) / ((1 - x)^4*(1 + x + x^2 + x^3 + x^4)).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + a(n-5) - 3*a(n-6) + 3*a(n-7) - a(n-8) for n>7. (End)
a(n) = -a(-1-n) for all n in Z. - Michael Somos, Oct 03 2018
a(n) ~ 4*n^3/5. - Stefano Spezia, Jun 06 2024
Comments