A019480 Define the sequence S(a(0),a(1)) by a(n+2) is the least integer such that a(n+2)/a(n+1) > a(n+1)/a(n) for n >= 0. This is S(4,12) (agrees with A019481 for n <= 19 only).

4, 12, 37, 115, 358, 1115, 3473, 10818, 33697, 104963, 326950, 1018419, 3172281, 9881362, 30779529, 95875387, 298642966, 930245227, 2897627873, 9025842914, 28114666162, 87574585658, 272786737320, 849705465331, 2646753962113, 8244393877392, 25680524664755

Offset: 0

Author

R. K. Guy

Keywords

• nonn

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..2025
• 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

Programs

Maple

a:= proc(n) option remember;
      `if`(n<2, [4, 12][n+1], floor(a(n-1)^2/a(n-2))+1)
    end:
seq(a(n), n=0..40);  # Alois P. Heinz, Sep 18 2015

Mathematica

S[a_, b_, n_] := Block[{s = {a, b}, k}, Do[k = Last@ s + 1; While[k/s[[i - 1]] <= s[[i - 1]]/s[[i - 2]], k++]; AppendTo[s, k], {i, 3, n}]; s]; S[4, 12, 14] (* Michael De Vlieger, Feb 15 2016 *)

PARI

S(a0, a1, maxn) = a=vector(maxn); a[1]=a0; a[2]=a1; for(n=3, maxn, a[n]=a[n-1]^2\a[n-2]+1); a
S(4, 12, 40) \\ Colin Barker, Feb 15 2016