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.

A018919 Define the generalized Pisot sequence T(a(0),a(1)) by: a(n+2) is the greatest integer such that a(n+2)/a(n+1) < a(n+1)/a(n). This is T(3,9).

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%I A018919 #49 Jul 13 2023 09:29:46
%S A018919 3,9,26,75,216,622,1791,5157,14849,42756,123111,354484,1020696,
%T A018919 2938977,8462447,24366645,70160958,202020427,581694636,1674922950,
%U A018919 4822748423,13886550633,39984728949,115131438424,331507764639,954538564968
%N A018919 Define the generalized Pisot sequence T(a(0),a(1)) by: a(n+2) is the greatest integer such that a(n+2)/a(n+1) < a(n+1)/a(n). This is T(3,9).
%C A018919 Let M denotes the 4 X 4 matrix = row by row (1,1,1,1)(1,1,1,0)(1,1,0,0)(1,0,0,0) and A(n) the vector (x(n),y(n),z(n),t(n))=M^n*A where A is the vector (1,1,1,1) then a(n)=y(n+1). - _Benoit Cloitre_, Apr 02 2002
%C A018919 Not to be confused with the Pisot T(3,9) sequence, which is A000244. - _R. J. Mathar_, Feb 13 2016
%H A018919 Vincenzo Librandi, <a href="/A018919/b018919.txt">Table of n, a(n) for n = 0..1000</a>
%H A018919 D. W. Boyd, <a href="http://www.researchgate.net/publication/258834801">Linear recurrence relations for some generalized Pisot sequences</a>, Advances in Number Theory ( Kingston ON, 1991) 333-340, Oxford Sci. Publ., Oxford Univ. Press, New York, 1993.
%H A018919 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,0,-1).
%H A018919 <a href="/index/Ph#Pisot">Index entries for Pisot sequences</a>
%F A018919 For n>1, a(n) = ceiling(a(n-1)^2/a(n-2)) - 1.
%F A018919 For n>2, a(n) = 3*a(n-1) - a(n-3).
%F A018919 G.f.: -(x^2-3) / (x^3-3*x+1). - _Colin Barker_, Dec 13 2012
%t A018919 CoefficientList[Series[- (x^2 - 3)/(x^3 - 3 x + 1), {x, 0, 30}], x] (* _Vincenzo Librandi_, Oct 16 2013 *)
%t A018919 RecurrenceTable[{a[1] == 3, a[2] == 9, a[n] == Ceiling[a[n-1]^2/a[n-2]] - 1}, a, {n, 30}] (* _Bruno Berselli_, Feb 17 2016 *)
%t A018919 LinearRecurrence[{3,0,-1},{3,9,26},30] (* _Harvey P. Dale_, Feb 06 2019 *)
%o A018919 (PARI) T(a0, a1, maxn) = a=vector(maxn); a[1]=a0; a[2]=a1; for(n=3, maxn, a[n]=ceil(a[n-1]^2/a[n-2])-1); a
%o A018919 T(3, 9, 30) \\ _Colin Barker_, Feb 14 2016
%o A018919 (Magma) Tiv:=[3,9]; [n le 2 select Tiv[n] else Ceiling(Self(n-1)^2/Self(n-2))-1: n in [1..40]]; // _Bruno Berselli_, Feb 17 2016
%Y A018919 Cf. A076264.
%K A018919 nonn,easy
%O A018919 0,1
%A A018919 _R. K. Guy_