cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A028560 a(n) = n*(n + 6).

Original entry on oeis.org

0, 7, 16, 27, 40, 55, 72, 91, 112, 135, 160, 187, 216, 247, 280, 315, 352, 391, 432, 475, 520, 567, 616, 667, 720, 775, 832, 891, 952, 1015, 1080, 1147, 1216, 1287, 1360, 1435, 1512, 1591, 1672, 1755, 1840, 1927, 2016, 2107, 2200, 2295, 2392
Offset: 0

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Nonnegative X values of solutions to the equation X + (X + 3)^2 + (X + 6)^3 = Y^2. To prove that X = n^2 + 6n: Y^2 = X + (X + 3)^2 + (X + 6)^3 = X^3 + 19*X^2 + 115X + 225 = (X + 9)*(X^2 + 10X + 25) = (X + 9)*(X + 5)^2 it means: (X + 9) must be a perfect square, so X = k^2 - 9 with k>=3. we can put: k = n + 3, which gives: X = n^2 + 6n and Y = (n + 3)*(n^2 + 6n + 5). - Mohamed Bouhamida, Nov 12 2007
a(m) where m is a positive integer are the only positive integer values of t for which the Binet-de Moivre Formula of the recurrence b(n)=6*b(n-1)+t*b(n-2) with b(0)=0 and b(1)=1 has a root which is a square. In particular, sqrt(6^2+4*t) is an integer since 6^2+4*t=6^2+4*a(m)=(2*m+6)^2. Thus, the charcteristic roots are k1=6+m and k2=-m. - Felix P. Muga II, Mar 27 2014
Also, numbers k such that k + 9 is a perfect square.

Crossrefs

a(n-3), n>=4, third column (used for the Paschen series of the hydrogen atom) of triangle A120070.
Cf. A005563.

Programs

Formula

a(n) = (n+3)^2 - 3^2 = n*(n+6).
G.f.: x*(7-5*x)/(1-x)^3.
a(n) = 2*n + a(n-1) + 5. - Vincenzo Librandi, Aug 05 2010
Sum_{n>=1} 1/a(n) = 49/120 = 0.4083333... - R. J. Mathar, Mar 22 2011
a(n) = A028884(n) - 1. - Reinhard Zumkeller, Apr 07 2013
E.g.f.: x*(x+7)*exp(x). - G. C. Greubel, Aug 19 2017
Sum_{n>=1} (-1)^(n+1)/a(n) = 37/360. - Amiram Eldar, Nov 04 2020
a(n) = A056220(n+1) - A000290(n-1). - Leo Tavares, Sep 29 2022
From Amiram Eldar, Feb 05 2024: (Start)
Product_{n>=1} (1 - 1/a(n)) = -4*sqrt(10)*sin(sqrt(10)*Pi)/(3*Pi).
Product_{n>=1} (1 + 1/a(n)) = 45*sqrt(2)*sin(2*sqrt(2)*Pi)/(7*Pi). (End)

Extensions

Edited by Robert G. Wilson v, Feb 06 2002