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 1-3 of 3 results.

A060385 Largest prime factor of n-th Fibonacci number.

Original entry on oeis.org

2, 3, 5, 2, 13, 7, 17, 11, 89, 3, 233, 29, 61, 47, 1597, 19, 113, 41, 421, 199, 28657, 23, 3001, 521, 109, 281, 514229, 61, 2417, 2207, 19801, 3571, 141961, 107, 2221, 9349, 135721, 2161, 59369, 421, 433494437, 307, 109441, 28657, 2971215073, 1103
Offset: 3

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Author

Labos Elemer, Apr 03 2001

Keywords

Comments

For n > 12, Fibonacci(n) is divisible by a primitive prime factor (one not dividing Fibonacci(1), ..., Fibonacci(n-1)). But all primes up to n-2 divide smaller Fibonacci numbers, see A001602, so a(n) >= n-1 for n > 12. This strengthens a more general theorem of Bravo and Luca. - Charles R Greathouse IV, Feb 01 2013

Examples

			F(82) = 2789 * 59369 * 370248451, so a(82) = 370248451.
		

Crossrefs

Programs

Formula

a(n) >= n - 1 for n > 12, see comments. It is not hard to show that a(n) > 1000 for n > 88. Similarly a(n) > 20641 for n > 120. - Charles R Greathouse IV, Feb 01 2013

A283461 Second-largest prime factor of 2^n - 1, if composite, or 1 otherwise.

Original entry on oeis.org

1, 1, 3, 1, 3, 1, 5, 7, 11, 23, 7, 1, 43, 31, 17, 1, 19, 1, 31, 127, 89, 47, 17, 601, 2731, 73, 113, 1103, 151, 1, 257, 89, 43691, 127, 73, 223, 174763, 8191, 41, 13367, 337, 9719, 683, 631, 178481, 4513, 257, 127, 1801, 11119, 2731
Offset: 2

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Author

Keywords

Comments

For clarification: if the largest prime factor occurs more than once, then that prime factor is selected.

Crossrefs

Programs

  • Mathematica
    a[n_] := If[PrimeQ[2^n-1], 1, Block[{f = FactorInteger[2^n-1]}, If[f[[-1, 2]] == 1, f[[-2, 1]], f[[-1, 1]]]]]; a /@ Range[2, 52] (* Giovanni Resta, Mar 08 2017 *)
  • PARI
    a(n)=my(f=factor(2^n-1),t=#f~); if(f[t,2]>1, f[t,1], if(t>1, f[t-1,1], 1))

Formula

a(n) = A006530(A000225(n)/A005420(n)).

A194086 Second-smallest prime factor of the n-th Lucas number (beginning with 2), if composite, or 1 otherwise.

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 3, 1, 1, 19, 41, 1, 7, 1, 281, 11, 1, 1, 3, 1, 2161, 29, 43, 461, 47, 101, 90481, 19, 14503, 19489, 3, 1, 4481, 199, 67, 29, 7, 1, 29134601, 79, 1601, 1, 3, 144481, 263, 11, 4969, 1, 769, 599786069, 41, 919, 103, 1, 3, 199, 10745088481, 229
Offset: 0

Views

Author

Jonathan Vos Post, Aug 14 2011

Keywords

Examples

			a(3) = 2 because the 3rd of the Lucas numbers (beginning at 2) is A000032(3) = 4 = 2^2, hence depending on whether one means by second-smallest prime factor (i.e., distinct or not, with multiplicity or not) a(3) would be either 2 or 1.
a(10) is unambiguously 41, because L(10) = 123 = 3 * 41, and 41 is the second-smallest prime factor, with no issues of multiplicity or distinctness.
		

Crossrefs

Programs

  • Mathematica
    Table[f = FactorInteger[LucasL[n]]; If[Length[f] > 1, f[[2, 1]], If[Length[f] == 1 && f[[1, 2]] > 1, f[[1, 1]], 1]], {n, 0, 70}] (* T. D. Noe, Aug 15 2011 *)
Showing 1-3 of 3 results.