A028895 5 times triangular numbers: a(n) = 5*n*(n+1)/2.
0, 5, 15, 30, 50, 75, 105, 140, 180, 225, 275, 330, 390, 455, 525, 600, 680, 765, 855, 950, 1050, 1155, 1265, 1380, 1500, 1625, 1755, 1890, 2030, 2175, 2325, 2480, 2640, 2805, 2975, 3150, 3330, 3515, 3705, 3900, 4100, 4305, 4515, 4730, 4950, 5175, 5405, 5640
Offset: 0
References
- D. B. West, Introduction to Graph Theory, 2nd ed., Prentice-Hall, NJ, 2001, p. 205.
Links
- Ivan Panchenko, Table of n, a(n) for n = 0..1000
- Rangaswami Balakrishnan and S. Francis Raj, The Wiener number of powers of the Mycielskian, Discussiones Math. Graph Theory, Vol. 30, No. 3 (2010), pp. 489-498 (see Theorem 2.1).
- Index entries for linear recurrences with constant coefficients, signature (3,-3,1).
Crossrefs
Programs
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Magma
[5*n*(n+1)/2 : n in [0..50]]; // Wesley Ivan Hurt, Jun 09 2014
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Maple
[seq(5*binomial(n,2), n=1..45)]; # Zerinvary Lajos, Nov 24 2006
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Mathematica
Table[Sum[i + 2*n - 1, {i, 2, n}], {n, 45}] (* Zerinvary Lajos, Jul 11 2009 *) Table[5 n (n + 1)/2, {n, 0, 50}] (* Bruno Berselli, Sep 23 2016 *) 5 Accumulate[Range[0,60]] (* Harvey P. Dale, Jun 17 2025 *)
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PARI
a(n)=5*n*(n+1)/2 \\ Charles R Greathouse IV, Sep 24 2015
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Python
def A028895(n): return 5*n*(n+1)>>1 # Chai Wah Wu, Aug 07 2025
Formula
G.f.: 5*x/(1-x)^3.
a(n) = 5*n*(n+1)/2 = 5*A000217(n).
a(n+1) = 5*n+a(n). - Vincenzo Librandi, Aug 05 2010
a(n) = A005891(n) - 1. - Omar E. Pol, Oct 03 2011
a(n) = A130520(5n+4). - Philippe Deléham, Mar 26 2013
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2. - Wesley Ivan Hurt, Nov 27 2015
a(n) = Sum_{i=0..n} A001068(4i). - Wesley Ivan Hurt, May 06 2016
E.g.f.: 5*x*(2 + x)*exp(x)/2. - Ilya Gutkovskiy, May 06 2016
From Amiram Eldar, Feb 26 2022: (Start)
Sum_{n>=1} 1/a(n) = 2/5.
Sum_{n>=1} (-1)^(n+1)/a(n) = (2/5)*(2*log(2) - 1). (End)
Product_{n>=1} (1 - 1/a(n)) = -(5/(2*Pi))*cos(sqrt(13/5)*Pi/2). - Amiram Eldar, Feb 21 2023
Comments