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

0,1

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

Formula

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

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 *)

Prog

(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

Crossrefs

Cf. A022018 - A022025, A022026 - A022032.

Sequence in context: A213453 A186671 A131519 * A186666 A129554 A165229

Adjacent sequences: A022021 A022022 A022023 * A022025 A022026 A022027

Keyword

nonn

Author

R. K. Guy

Extensions

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

Status

approved