A018923 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(16,32).

16, 32, 63, 124, 244, 480, 944, 1856, 3649, 7174, 14104, 27728, 54512, 107168, 210687, 414200, 814296, 1600864, 3147216, 6187264, 12163841, 23913482, 47012668, 92424472, 181701728, 357216192, 702268543, 1380623604, 2714234540, 5336044608, 10490387488

Offset: 0

Author

R. K. Guy

Keywords

• nonn

Comments

Not to be confused with the Pisot T(16,32) sequence, which is essentially A000079. - R. J. Mathar, Feb 13 2016

Links

• Colin Barker, Table of n, a(n) for n = 0..1000
• D. W. Boyd, Linear recurrence relations for some generalized Pisot sequences, Advances in Number Theory ( Kingston ON, 1991) 333-340, Oxford Sci. Publ., Oxford Univ. Press, New York, 1993.
• Index entries for Pisot sequences

Crossrefs

Is this the same sequence as A001949?

Programs

Mathematica

T[a_, b_, n_] := Block[{s = {a, b}, k}, Do[k = 2 Last@ s; While[k/s[[i - 1]] >= s[[i - 1]]/s[[i - 2]], k--]; AppendTo[s, k], {i, 3, n}]; s]; T[16, 32, 23] (* or *)
a = {16, 32}; Do[AppendTo[a, Ceiling[a[[n - 1]]^2/a[[n - 2]]] - 1], {n, 3, 23}]; a (* Michael De Vlieger, Feb 15 2016 *)
RecurrenceTable[{a[1] == 16, a[2] == 32, a[n] == Ceiling[a[n-1]^2/a[n-2] - 1]}, a, {n, 40}] (* Vincenzo Librandi, Feb 17 2016 *)

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
T(16, 32, 30) \\ Colin Barker, Feb 16 2016

Formula

Empirical G.f.: -(8*x^5+4*x^4+2*x^3+x^2-16) / ((x-1)*(x^5+x^4+x^3+x^2+x-1)). - Colin Barker, Dec 21 2012

a(n+1) = ceiling(a(n)^2/a(n-1))-1 for n>0. - Bruno Berselli, Feb 15 2016