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.

Showing 1-5 of 5 results.

A213358 a(n) = maximum(continuedfraction(F(n+1)^n/F(n)^n)) - L(n) + (1-(-1)^n)/2, where F(n) is Fibonacci(n) and L(n) is the n-th Lucas number.

Original entry on oeis.org

1, 1, 0, 4, 7, 16, 0, 0, 397, 17, 0, 8047, 0, 0, 0, 0, 0, 0, 0, 6633, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
Offset: 1

Views

Author

Art DuPre, Jun 10 2012

Keywords

Comments

This is a variation on A113506 and A113500.
If we consider the sequence of quotients [1,1,1,1] of a simple continued fraction, we can express this by the continued fraction of F(4+1)/F(4), where we have chosen not to write the last quotient as 2 but as 1,1.
It is a very interesting fact that if we take the fifth power of F(n+1)/F(n), for varying n, we very soon observe extremely large values in its continued fraction expansion. That's why Benoit Cloitre, author of A113560, takes vecmax, the max of all these quotients, in order to capture the large rate of growth of these quotients. It would be better, in order to emphasize the largeness of these big quotients, to somehow quantify the ratio between the max and the average of all the quotients, or the average of all the others.
Since Cloitre found a Fibonacci-like behavior of the vecmax, I thought to replace the power 5 by the power n, thus getting the sequence vecmax(contfrac(F(n+1)^n/F(n)^n)) instead of vecmax(contfrac(F(n+1)^5/F(n)^5)). After I noticed its behavior was almost that of the Lucas sequence, I took the sequence which subtracted the almost-Lucas sequence from it, thus eventually getting 0 after the 20th term.

Crossrefs

Programs

  • Mathematica
    Table[Max[ContinuedFraction[Fibonacci[n+1]^n/Fibonacci[n]^n]] - LucasL[n] + (1 - (-1)^n)/2, {n, 200}] (* T. D. Noe, Jun 28 2012 *)
  • PARI
    F(n) = fibonacci(n);
    L(n) = F(n-1) + F(n+1);
    a(n) = vecmax(contfrac(F(n+1)^n/F(n)^n)) - L(n) + (1-(-1)^n)/2;

A113561 Least multiple of n in which the n-th digit from left is 6.

Original entry on oeis.org

6, 16, 126, 1016, 100060, 100026, 1000006, 10000016, 100000026, 10000000060, 10000000076, 100000000056, 1000000000116, 10000000000046, 1000000000000065, 1000000000000016, 10000000000000016, 100000000000000026, 1000000000000000056
Offset: 1

Views

Author

Amarnath Murthy, Nov 06 2005

Keywords

Examples

			a(4)= 10060, in which the fourth digit is 6.
		

Crossrefs

Extensions

Corrected and extended by Joshua Zucker, May 03 2006

A113562 Least multiple of n in which the n-th digit from left is 7.

Original entry on oeis.org

7, 170, 117, 10072, 100070, 1000074, 1000027, 100000072, 100000017, 10000000070, 10000000087, 1000000000176, 1000000000077, 100000000000278, 1000000000000170, 10000000000000176, 10000000000000067, 1000000000000000170
Offset: 1

Views

Author

Amarnath Murthy, Nov 06 2005

Keywords

Examples

			a(4)= 10072, in which the fourth digit is 7.
		

Crossrefs

Programs

  • Mathematica
    lmn7[n_]:=Module[{k=Ceiling[10^(n-1)/n]},While[NumberDigit[k n,IntegerLength[ k n]-n]!=7,k++];k n]; Array[lmn7,10] (* The program generates the first 10 terms of the sequence. *)  (* Harvey P. Dale, Nov 02 2022 *)

Extensions

More terms from Joshua Zucker, May 03 2006

A113563 Least multiple of n in which the n-th digit from left is 8.

Original entry on oeis.org

8, 18, 108, 1008, 100080, 100008, 1000048, 10000008, 100000008, 10000000080, 10000000098, 100000000008, 1000000000038, 10000000000018, 1000000000000080, 1000000000000048, 10000000000000118, 100000000000000008, 1000000000000000018
Offset: 1

Views

Author

Amarnath Murthy, Nov 06 2005

Keywords

Examples

			a(4)= 1008, in which the fourth digit is 8.
		

Crossrefs

Extensions

More terms from Joshua Zucker, May 03 2006

A113564 Least multiple of n in which the n-th digit from left is 9.

Original entry on oeis.org

9, 190, 129, 10092, 100090, 1000092, 1000069, 100000096, 100000089, 10000000090, 10000000109, 1000000000092, 1000000000129, 100000000000096, 1000000000000095, 10000000000000096, 10000000000000169
Offset: 1

Views

Author

Amarnath Murthy, Nov 06 2005

Keywords

Examples

			a(4)= 10092, in which the fourth digit is 9.
		

Crossrefs

Extensions

More terms from Peter Soung (soung3(AT)hotmail.com), Jan 23 2006
Showing 1-5 of 5 results.