A152744 7 times pentagonal numbers: a(n) = 7*n*(3*n-1)/2.
0, 7, 35, 84, 154, 245, 357, 490, 644, 819, 1015, 1232, 1470, 1729, 2009, 2310, 2632, 2975, 3339, 3724, 4130, 4557, 5005, 5474, 5964, 6475, 7007, 7560, 8134, 8729, 9345, 9982, 10640, 11319, 12019, 12740, 13482, 14245, 15029, 15834, 16660, 17507, 18375, 19264
Offset: 0
Links
- Ivan Panchenko, Table of n, a(n) for n = 0..1000
- Index entries for linear recurrences with constant coefficients, signature (3,-3,1).
Programs
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Magma
[7*n*(3*n-1)/2: n in [0..50]]; // G. C. Greubel, Sep 01 2018
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Mathematica
Table[7n (3n-1)/2,{n,0,50}] (* or *) LinearRecurrence[{3,-3,1},{0,7,35},50] (* Harvey P. Dale, Aug 08 2013 *) -
PARI
a(n)=7*n*(3*n-1)/2 \\ Charles R Greathouse IV, Jun 17 2017
Formula
a(n) = (21*n^2 - 7*n)/2 = A000326(n)*7.
a(n) = a(n-1) + 21*n - 14 (with a(0)=0). - Vincenzo Librandi, Nov 26 2010
G.f.: 7*x*(1+2*x)/(1-x)^3. - Colin Barker, Feb 14 2012
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n>2. - Harvey P. Dale, Aug 08 2013
a(n) = Sum_{i = 2..8} P(i,n), where P(i,m) = m*((i-2)*m-(i-4))/2. - Bruno Berselli, Jul 04 2018
E.g.f.: 7*x*(2+3*x)/2. - G. C. Greubel, Sep 01 2018
From Amiram Eldar, Feb 27 2022: (Start)
Sum_{n>=1} 1/a(n) = (9*log(3) - sqrt(3)*Pi)/21.
Sum_{n>=1} (-1)^(n+1)/a(n) = 2*(Pi*sqrt(3) - 6*log(2))/21. (End)