cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A279543 a(n) = a(n-1) + 3^n * a(n-2) with a(0) = 1 and a(1) = 1.

Original entry on oeis.org

1, 1, 10, 37, 847, 9838, 627301, 22143007, 4137864868, 439978671649, 244776761262181, 78185678507867584, 130162592460442600405, 124783388108159412726037, 622688428086038843429228482, 1791127919536971393223950620041
Offset: 0

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Author

Seiichi Manyama, Dec 31 2016

Keywords

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)/A015460(n+2) is 3^(-1/5) * R(3).

Examples

			1/1 = a(0)/A015460(2).
1/(1+3/1) = 1/4 = a(1)/A015460(3).
1/(1+3/(1+3^2/1)) = 10/13 = a(2)/A015460(4).
1/(1+3/(1+3^2/(1+3^3/1))) = 37/121 = a(3)/A015460(5).
		

Crossrefs

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), this sequence (q=3), A280340 (q=10).

Programs

  • Mathematica
    RecurrenceTable[{a[n] == a[n - 1] + 3^n*a[n - 2], a[0] == 1, a[1] == 1}, a, {n, 15}] (* Michael De Vlieger, Dec 31 2016 *)

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

Original entry on oeis.org

1, 1, 101, 1101, 1011101, 111111101, 1011212111101, 1112122222111101, 101122323232322111101, 1112223344434333322111101, 1011224344546565545343322111101, 111223345667777878776655443322111101, 1011224455769911213121200887756443322111101
Offset: 0

Views

Author

Seiichi Manyama, Dec 31 2016

Keywords

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).
		

Crossrefs

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
Showing 1-2 of 2 results.