A327597 a(n) = numerator((a(n-1) + a(n-2) + 1)/a(n-1)), with a(1)=1, a(2)=2.

1, 2, 2, 5, 8, 7, 16, 3, 20, 6, 9, 16, 13, 30, 22, 53, 76, 65, 142, 104, 19, 124, 36, 161, 198, 20, 219, 80, 15, 32, 3, 12, 4, 17, 22, 20, 43, 64, 27, 92, 30, 41, 72, 19, 92, 28, 121, 150, 136, 287, 424, 89, 514, 302, 817, 1120, 969, 110, 108, 73, 182, 128, 311

Offset

1,2

Comments

As n goes to infinity, 32 < (a(n+1) + a(n))/(a(n+1) - a(n)) < 32.35 (conjectured).

This sequence is the continued fraction expansion of ~ 1.4072426692639398147... (calculated to 19 digits).

The sum of the reciprocals of this sequence is ~ 4.97804721273463... (calculated to 14 digits).

The average reduction 0.621 < a(n)/(a(n-1)+a(n-2)+1) < 0.622 (conjectured); the average reduction is the average of the individual reductions (The change from the numerator to the numerator in the simplest form):

a(3)/(a(2)+a(1)+1) -> 2/(2+1+1) -> 0.5

a(4)/(a(3)+a(2)+1) -> 5/(2+2+1) -> 1

a(5)/(a(4)+a(3)+1) -> 8/(5+2+1) -> 1

a(6)/(a(5)+a(4)+1) -> 7/(8+5+1) -> 0.5

a(7)/(a(6)+a(5)+1) -> 16/(7+8+1) -> 1

a(8)/(a(7)+a(6)+1) -> 3/(16+7+1) -> 0.125

a(9)/(a(8)+a(7)+1) -> 20/(3+16+1) -> 1

a(10)/(a(9)+a(8)+1) -> 6/(20+3+1) -> 0.25

That number is the average of these to ~ 100000 terms (There was some fluctuation to take into account).

Links

Vimal Vinod, Table of n, a(n) for n = 1..10000

Formula

It appears that this sequence's growth can be approximated by a(n) ~ (1 + 1/c)^n where 17.8 < c < 18.5.

Example

a(1) = 1.
a(2) = 2.
a(3) = numerator((1 + 2 + 1)/2) -> numerator(2/1) = 2.
a(4) = numerator((2 + 2 + 1)/2) -> numerator(5/2) = 5.
a(5) = numerator((2 + 5 + 1)/5) -> numerator(8/5) = 8.
a(6) = numerator((5 + 8 + 1)/8) -> numerator(7/4) = 7.

Mathematica

Nest[Append[#, Numerator[(#2 + #1 + 1)/#2] & @@ #[[-2 ;; -1]]] &, {1, 2}, 61] (* Michael De Vlieger, Sep 30 2019 *)

Prog

(Python)
from fractions import Fraction
num_terms = 100
S = [1, 2]
for n in range(num_terms-2):
    s = Fraction((S[n]+S[n+1]+1), S[n+1]).numerator
    S.append(s)
print(S) # Should print the sequence to the length specified.
(PARI) lista(nn) = {my(xa = 1, ya = 2, za); print1(xa, ", ", ya, ", "); for (n=3, nn, za = numerator((ya + xa + 1)/ya); print1(za, ", "); xa = ya; ya = za; ); } \\ Michel Marcus, Sep 24 2019
(Magma) a:=[1, 2]; [n le 2 select a[n] else Numerator((Self(n-1) + Self(n-2) + 1)/Self(n-1)):n in [1..64]]; // Marius A. Burtea, Sep 27 2019

Crossrefs

Sequence in context: A202396 A210804 A087910 * A284325 A358517 A035570

Adjacent sequences: A327594 A327595 A327596 * A327598 A327599 A327600

Keyword

nonn

Author

Vimal Vinod, Sep 18 2019

Status

approved