A022024 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(6,66).

6, 66, 727, 8009, 88232, 972018, 10708349, 117969769, 1299627646, 14317498734, 157730385799, 1737655093709, 19143078927992, 210891949829430, 2323315631208341, 25595076182769253, 281971126093205254, 3106367622527151978, 34221659288246953735, 377006879658404795777

Offset: 0

Author

R. K. Guy

Keywords

• nonn

Comments

This coincides with the linearly recurrent sequence defined by the expansion of (6 - 5*x^2)/(1 - 11*x - x^2 + 9*x^3) only up to n <= 169. - Bruno Berselli, Feb 11 2016

Links

• Colin Barker, Table of n, a(n) for n = 0..950

Crossrefs

Cf. A022018 - A022025, A022026 - A022032.

Programs

Maple

A022024 := proc(n)
    option remember;
    if n <= 1 then
        op(n+1,[6,66]) ;
    else
        a := procname(n-1)^2/procname(n-2) ;
        if type(a,'integer') then
            a+1 ;
        else
            ceil(a) ;
        fi;
    end if;
end proc: # R. J. Mathar, Feb 10 2016

Mathematica

a[n_] := a[n] = Switch[n, 0, 6, 1, 66, _, Floor[a[n-1]^2/a[n-2]]+1];
Table[a[n], {n, 0, 30}] (* Jean-François Alcover, Feb 08 2024 *)

PARI

a=[6,66];for(n=2,30,a=concat(a,a[n]^2\a[n-1]+1));a \\ M. F. Hasler, Feb 10 2016

Python

def a(n):
    if n == 0: return 6
    prev_1, prev_2 = 66, 6
    for i in range(2, n + 1):
        prev_2, prev_1 = prev_1, (prev_1 ** 2) // prev_2 + 1
    return prev_1 # Paul Muljadi, Feb 12 2024

Formula

a(n+1) = floor(a(n)^2/a(n-1))+1 for all n > 0. - M. F. Hasler, Feb 10 2016

Extensions

Double-checked and extended to 3 lines of data by M. F. Hasler, Feb 10 2016