A165900 a(n) = n^2 - n - 1.
-1, -1, 1, 5, 11, 19, 29, 41, 55, 71, 89, 109, 131, 155, 181, 209, 239, 271, 305, 341, 379, 419, 461, 505, 551, 599, 649, 701, 755, 811, 869, 929, 991, 1055, 1121, 1189, 1259, 1331, 1405, 1481, 1559, 1639, 1721, 1805, 1891, 1979, 2069, 2161, 2255
Offset: 0
Examples
G.f. = -1 - x + x^2 + 5*x^3 + 11*x^4 + 19*x^5 + 29*x^6 + 41*x^7 + ... - _Michael Somos_, Mar 23 2023
Links
- Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
- Joseph J. Heed and Lucille Kelly, An interesting sequence of Fibonacci sequence generators, Fibonacci Quarterly, 13 (1975), pp. 29-30.
- Index entries for linear recurrences with constant coefficients, signature (3,-3,1).
Crossrefs
Programs
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Haskell
a165900 n = n * (n - 1) - 1 -- Reinhard Zumkeller, Jul 29 2012
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Mathematica
Table[n^2 - n - 1, {n, 0, 50}] (* Ron Knott, Oct 27 2010 *) LinearRecurrence[{3,-3,1},{-1,-1,1},60] (* Harvey P. Dale, Jul 05 2021 *)
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PARI
a(n)=n^2-n-1 \\ Charles R Greathouse IV, Jan 12 2012
Formula
a(n+2) = (n+1)*a(n+1) - (n+2)*a(n).
G.f.: (x^2+2*x-1)/(1-x)^3.
E.g.f.: exp(x)*(x^2-1).
a(n) = - A188652(2*n) for n > 0. - Reinhard Zumkeller, Apr 13 2011
a(n+2) = A028387(n). - Michael B. Porter, Sep 26 2018
From Klaus Purath, Aug 25 2022: (Start)
a(2*n) = n*(a(n+1) - a(n-1)) -1.
a(2*n+1) = (2*n+1)*(a(n+1) - a(n)) - 1.
a(n+2) = a(n) + 4*n + 2.
a(n) = A152015(n)/n, n != 0.
(a(n+k) - a(n-k))/(2*k) = 2*n-1, for any k.
(End)
For n > 1, 1/a(n) = Sum_{k>=1} F(k)/n^(k+1), where F(n) = A000045(n). - Diego Rattaggi, Nov 01 2022
a(n) = a(1-n) for all n in Z. - Michael Somos, Mar 23 2023
For n > 1, 1/a(n) = Sum_{k>=1} F(2k)/((n+1)^(k+1)), where F(2n) = A001906(n). - Diego Rattaggi, Jan 20 2025
From Amiram Eldar, May 11 2025: (Start)
Sum_{n>=1} 1/a(n) = tan(sqrt(5)*Pi/2)*Pi/sqrt(5).
Product_{n>=3} 1 - 1/a(n) = -sec(sqrt(5)*Pi/2)*Pi/6.
Product_{n>=2} 1 + 1/a(n) = -sec(sqrt(5)*Pi/2)*Pi. (End)
Extensions
a(22) corrected by Reinhard Zumkeller, Apr 13 2011
Better name from Joerg Arndt, Oct 26 2024
Comments