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.

A178763 Product of primitive prime factors of Fibonacci(n).

Original entry on oeis.org

1, 1, 2, 3, 5, 1, 13, 7, 17, 11, 89, 1, 233, 29, 61, 47, 1597, 19, 4181, 41, 421, 199, 28657, 23, 3001, 521, 5777, 281, 514229, 31, 1346269, 2207, 19801, 3571, 141961, 107, 24157817, 9349, 135721, 2161, 165580141, 211, 433494437, 13201, 109441, 64079
Offset: 1

Views

Author

T. D. Noe, Jun 10 2010

Keywords

Comments

Same as A001578 for the first 18 terms.
Let b(n) be the greatest divisor of Fibonacci(n) that is coprime to Fibonacci(m) for all positive integers m < n, then a(n) = b(n) for all n, provided that no Wall-Sun-Sun prime exists. Otherwise, if p is a Wall-Sun-Sun prime and A001177(p) = k (then A001177(p^2) = k), then p^2 divides b(k), but by definition a(k) is squarefree. - Jianing Song, Jul 02 2019

Crossrefs

Cf. A061446, A086597, A152012 (Indices of prime terms).

Programs

  • PARI
    a(n)=my(d=divisors(n), f=fibonacci(n), t); t=lcm(apply(fibonacci,d[1..#d-1])); while((t=gcd(t,f))>1, f/=t); f \\ Charles R Greathouse IV, Nov 30 2016

Formula

a(n) = A061446(n) / A178764(n).
a(n) = A061446(n) / gcd(A061446(n), n) if n != 5, 6, provided that no Wall-Sun-Sun prime exists. - Jianing Song, Jul 02 2019

A178762 Prime numbers that are Fibonacci integers.

Original entry on oeis.org

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 41, 47, 61, 89, 107, 199, 211, 233, 281, 421, 521, 1103, 1597, 2161, 2207, 2521, 3001, 3571, 5779, 9349, 9901, 14503, 19801, 28657, 90481, 103681, 109441, 135721, 141961, 514229, 3010349, 6376021, 11128427
Offset: 1

Views

Author

T. D. Noe, Jun 10 2010

Keywords

Comments

A Fibonacci integer is a number that can be written as the product and/or quotient of Fibonacci numbers (A000045). For example, 107 is a Fibonacci integer because Fib(36)/(Fib(18)*Fib(3)*Fib(4)^3) = 107. Observe that the prime Fibonacci numbers (A005478) are a subset of these primes. Luca, Pomerance, and Wagner conjecture that this sequence is infinite. The paper's Remark 2 and sequences A152012, A178763, and A178764 are useful in finding these primes.

A250208 Ratio of the primitive part of 2^n-1 to the product of primitive prime factors of 2^n-1.

Original entry on oeis.org

1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 5, 7, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 5, 1, 1, 1, 1, 1, 1, 1, 1, 1, 11, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
Offset: 1

Views

Author

Eric Chen, Mar 02 2015

Keywords

Comments

As with A178764, it can be shown that all terms are either 1 or prime.
a(2*3^n) = 3 (n>=1).
a(4*5^n) = 5 (n>=1).
a(3*7^n) = 7 (n>=1).
a(10*11^n) = 11 (n>=1).
a(12*13^n) = 13 (n>=1).
a(8*17^n) = 17 (n>=1).
a(18*19^n) = 19 (n>=1).
...
a(A014664(k)*prime(k)^n) = prime(k).
For other n (while Phi_n(2) is squarefree), a(n) = 1.
a(n) != 1 for n = {6, 18, 20, 21, 54, 100, 110, 136, 147, 155, 156, 162, ...}.
At least, a(A049093(n)) = 1. (In fact, since Phi_n(2) is not completely factored for n = 991, 1207, 1213, 1217, 1219, 1229, 1231, 1237, 1243, 1249, ..., so it is unknown whether they are squarefree or not, but it is likely that Phi_n(2) is squarefree for all n except 364 and 1755 (because it is likely 1093 and 3511 are the only two Wieferich primes), so a(991), a(1207), a(1213), ..., are likely to be 1.)

Examples

			a(11) = 1 since Phi_11(2) = (2^11-1)/(2-1) = 2047, and the primitive prime factors of 2^11-1 are 23 and 89, so a(11) = 2047/(23*89) = 1.
a(18) = 3 since Phi_18(2) = 2^6 - 2^3 + 1 = 57, and the only primitive prime factor of 2^18-1 is 19, so a(18) = 57/19 = 3.
		

Crossrefs

Programs

  • Mathematica
    a250208[n_] = If[n == 364, 1093, If[n == 1755, 3511, GCD[Cyclotomic[n, 2], n]]]; Table[a250208[n], {n, 0, 200}]
  • PARI
    a(n) = if (n==364, 1093, if (n==1755, 3511, gcd(polcyclo(n, 2), n)));
    
  • PARI
    isprimitive(p, n) = {for (r=1, n-1, if (((2^r-1) % p) == 0, return (0)); ); return (1); }
    ppf(n) = {my(pf = factor(2^n-1)[,1]); prod(k=1,#pf, if (isprimitive(pf[k], n), pf[k], 1));}
    a(n) = if (issquarefree(m=polcyclo(n,2)), gcd(m, n), m/ppf(n)); \\ Michel Marcus, Mar 06 2015

Formula

a(n) = A019320(n) / A064078(n) while Phi_n(2) is squarefree.
a(n) = GCD(Phi_n(2), n) while Phi_n(2) is squarefree.
Notice: a(364) = 1093, a(1755) = 3511. (See A001220.)
Showing 1-3 of 3 results.