A280340 a(n) = a(n-1) + 10^n * a(n-2) with a(0) = 1 and a(1) = 1.

1, 1, 101, 1101, 1011101, 111111101, 1011212111101, 1112122222111101, 101122323232322111101, 1112223344434333322111101, 1011224344546565545343322111101, 111223345667777878776655443322111101, 1011224455769911213121200887756443322111101

Offset: 0

Author

Seiichi Manyama, Dec 31 2016

Keywords

• nonn

Comments

The Rogers-Ramanujan continued fraction is defined by R(q) = q^(1/5)/(1+q/(1+q^2/(1+q^3/(1+ ... )))). The limit of a(n)/A015468(n+2) is 10^(-1/5) * R(10).

a(n) has A004652(n+1) digits. The last n digits are the same as the last n digits of a(n-1). - Robert Israel, Jan 12 2017

Examples

			1/1 = a(0)/A015468(2).
1/(1+10/1) = 1/11 = a(1)/A015468(3).
1/(1+10/(1+10^2/1)) = 101/111 = a(2)/A015468(4).
1/(1+10/(1+10^2/(1+10^3/1))) = 1101/11111 = a(3)/A015468(5).
		

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..62
• Eric Weisstein's World of Mathematics, Rogers-Ramanujan Continued Fraction

Crossrefs

Cf. A004652, A015468, A128915.

Cf. similar sequences with the recurrence a(n-1) + q^n * a(n-2) for n>1, a(0)=1 and a(1)=1: A280294 (q=2), A279543 (q=3), this sequence (q=10).

Programs

Maple

A[0]:= 1: A[1]:= 1:
for n from 2 to 20 do A[n]:= A[n-1]+10^n*A[n-2] od:
seq(A[i],i=0..20); # Robert Israel, Jan 12 2017

Mathematica

RecurrenceTable[{a[0]==a[1]==1,a[n]==a[n-1]+10^n a[n-2]},a,{n,15}] (* Harvey P. Dale, Jul 12 2020 *)

Formula

a(n) a(n-3) = 10 a(n-2) a(n-1) - 10 a(n-2)^2 + a(n-1) a(n-3). - Robert Israel, Jan 12 2017