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a(n) = n^2 - n - 1.
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%I #151 Oct 27 2024 16:07:44

%S -1,-1,1,5,11,19,29,41,55,71,89,109,131,155,181,209,239,271,305,341,

%T 379,419,461,505,551,599,649,701,755,811,869,929,991,1055,1121,1189,

%U 1259,1331,1405,1481,1559,1639,1721,1805,1891,1979,2069,2161,2255

%N a(n) = n^2 - n - 1.

%C Previous name was: Values of Fibonacci polynomial n^2 - n - 1.

%C Shifted version of the array denoted rB(0,2) in A132382, whose e.g.f. is exp(x)(1-x)^2. Taking the derivative gives the e.g.f. of this sequence. - _Tom Copeland_, Dec 02 2013

%C The Fibonacci numbers are generated by the series x/(1 - x - x^2). - _T. D. Noe_, Dec 04 2013

%C Absolute value of expression f(k)*f(k+1) - f(k-1)*f(k+2) where f(1)=1, f(2)=n. Sign is alternately +1 and -1. - _Carmine Suriano_, Jan 28 2014 [Can anybody clarify what is meant here? - _Joerg Arndt_, Nov 24 2014]

%C Carmine's formula is a special case related to 4 consecutive terms of a Fibonacci sequence. A generalization of this formula is |a(n)| = |f(k+i)*f(k+j) - f(k)*f(k+i+j)|/F(i)*F(j), where f denotes a Fibonacci sequence with the initial values 1 and n, and F denotes the original Fibonacci sequence A000045. The same results can be obtained with the simpler formula |a(n)| = |f(k+1)^2 - f(k)^2 - f(k+1)*f(k)|. Everything said so far is also valid for Fibonacci sequences f with the initial values f(1) = n - 2, f(2) = 2*n - 3. - _Klaus Purath_, Jun 27 2022

%C a(n) is the total number of dollars won when using the Martingale method (bet $1, if win then continue to bet $1, if lose then double next bet) for n trials of a wager with exactly one loss, n-1 wins. For the case with exactly one win, n-1 losses, see A070313. - _Max Winnick_, Jun 28 2022

%C Numbers m such that 4*m+5 is a square b^2, where b = 2*n -1, for m = a(n). - _Klaus Purath_, Jul 23 2022

%H Reinhard Zumkeller, <a href="/A165900/b165900.txt">Table of n, a(n) for n = 0..10000</a>

%H J. J. Heed and L Kelly, <a href="http://www.fq.math.ca/Scanned/13-1/heed.pdf">An interesting sequence of Fibonacci sequence generators</a>, Fibonacci Quarterly, 13 (1975), pp. 29-30.

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).

%F a(n+2) = (n+1)*a(n+1) - (n+2)*a(n).

%F G.f.: (x^2+2*x-1)/(1-x)^3.

%F E.g.f.: exp(x)*(x^2-1).

%F a(n) = - A188652(2*n) for n > 0. - _Reinhard Zumkeller_, Apr 13 2011

%F a(n) = A214803(A015614(n+1)) for n > 0. - _Reinhard Zumkeller_, Jul 29 2012

%F a(n+1) = a(n) + A005843(n) = A002378(n) - 1. - _Ivan N. Ianakiev_, Feb 18 2013

%F a(n+2) = A028387(n). - _Michael B. Porter_, Sep 26 2018

%F From _Klaus Purath_, Aug 25 2022: (Start)

%F a(2*n) = n*(a(n+1) - a(n-1)) -1.

%F a(2*n+1) = (2*n+1)*(a(n+1) - a(n)) - 1.

%F a(n+2) = a(n) + 4*n + 2.

%F a(n) = A014206(n-1) - 3 = A002061(n-1) - 2.

%F a(n) = A028552(n-2) + 1 = A014209(n-2) + 2 = 2* A034856(n-2) + 3.

%F a(n) = A008865(n-1) + n = A005563(n-1) - n.

%F a(n) = A014209(n-3) + 2*n = A028387(n-1) - 2*n.

%F a(n) = A152015(n)/n, n != 0.

%F (a(n+k) - a(n-k))/(2*k) = 2*n-1, for any k.

%F (End)

%F For n > 1, 1/a(n) = Sum_{k>=1} F(k)/n^(k+1), where F(n) = A000045(n). - _Diego Rattaggi_, Nov 01 2022

%F a(n) = a(1-n) for all n in Z. - _Michael Somos_, Mar 23 2023

%e 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

%t Table[n^2 - n - 1, {n, 0, 50}] (* _Ron Knott_, Oct 27 2010 *)

%t LinearRecurrence[{3,-3,1},{-1,-1,1},60] (* _Harvey P. Dale_, Jul 05 2021 *)

%o (PARI) a(n)=n^2-n-1 \\ _Charles R Greathouse IV_, Jan 12 2012

%o (Haskell)

%o a165900 n = n * (n - 1) - 1 -- _Reinhard Zumkeller_, Jul 29 2012

%Y A028387 and A110331 are very similar sequences.

%Y Cf. A000045, A005843, A015614, A070313, A132382, A152015, A188652, A214803.

%Y Other quadratics: A002061, A002378, A005563, A008865, A014206, A014209, A028552, A034856.

%K sign,easy

%O 0,4

%A _Philippe Deléham_, Sep 29 2009

%E a(22) corrected by _Reinhard Zumkeller_, Apr 13 2011

%E Better name from _Joerg Arndt_, Oct 26 2024