A272126 a(n) = 120*n^3 + 60*n^2 + 2*n + 1.
1, 183, 1205, 3787, 8649, 16511, 28093, 44115, 65297, 92359, 126021, 167003, 216025, 273807, 341069, 418531, 506913, 606935, 719317, 844779, 984041, 1137823, 1306845, 1491827, 1693489, 1912551, 2149733, 2405755, 2681337, 2977199, 3294061, 3632643, 3993665
Offset: 0
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..1000
- Richard P. Brent, Generalising Tuenter's binomial sums, arXiv:1407.3533 [math.CO], 2014. (page 16).
- Richard P. Brent, Generalising Tuenter's binomial sums, Journal of Integer Sequences, 18 (2015), Article 15.3.2.
- Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).
Programs
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Magma
[120*n^3 + 60*n^2 + 2*n + 1: n in [0..50]];
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Mathematica
Table[120 n^3 + 60 n^2 + 2 n + 1, {n, 0, 40}] LinearRecurrence[{4,-6,4,-1},{1,183,1205,3787},40] (* Harvey P. Dale, Nov 08 2020 *) -
PARI
a(n) = 120*n^3 + 60*n^2 + 2*n + 1; \\ Altug Alkan, Apr 30 2016
Formula
O.g.f.: (1 + 179*x + 479*x^2 + 61*x^3)/(1-x)^4.
E.g.f.: (1 + 182*x + 420*x^2 + 120*x^3)*exp(x).
a(n) = (2*n+1)*(60*n^2+1).
a(n) = (2*n+1) * A158673(n).
a(n) = 4*a(n-1)-6*a(n-2)+4*a(n-3)-a(n-4) for n>3.
See page 7 in Brent's paper:
A272127(n) = (2*n+1)^2*a(n) - 2*n*(2*n+1)*a(n-1).
From Peter Bala, Jan 22 2019: (Start)
a(n) = 1/4^n * Sum_{k = 0..n} (2*k + 1)^6 * binomial(2*n + 1, n - k).
a(n-1) = 2/4^n * binomial(2*n,n) * ( 1 + 3^6*(n - 1)/(n + 1) + 5^6*(n - 1)*(n - 2)/((n + 1)*(n + 2)) + 7^6*(n - 1)*(n - 2)*(n - 3)/((n + 1)*(n + 2)*(n + 3)) + ... ). (End)
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