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.

A352290 Numbers m such that the greatest prime factor of m^2 + 1 is a Fibonacci number.

Original entry on oeis.org

1, 2, 3, 5, 7, 8, 18, 34, 55, 57, 89, 123, 144, 233, 239, 322, 377, 411, 500, 568, 610, 746, 788, 843, 987, 1487, 1542, 1568, 1636, 2207, 2584, 2707, 3173, 3639, 3793, 3804, 3817, 4050, 4181, 4217, 4594, 4662, 5270, 5778, 6107, 6613, 8595, 8972, 10341, 10569
Offset: 1

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Author

Michel Lagneau, Mar 11 2022

Keywords

Comments

A281618 is a subsequence.
The corresponding greatest prime Fibonacci factors of the sequence are 2, 5, 5, 13, 5, 13, 13, 89, 89, 13, 233, 89, 233, ...
The Fibonacci numbers of the sequence are 1, 2, 3, 5, 8, 34, 55, 89, 144, 233, 377, 610, 987, 2584, 4181, 10946, 17711, ... (subsequence of A000045).
The Lucas numbers of the sequence are 1, 2, 3, 7, 18, 123, 322, 843, 2207, 5778, 39603, 103682, ... (subsequence of A000032).
The prime numbers of the sequence are 2, 3, 5, 7, 89, 233, 239, 1487, 2207, 2707, 3793, 4217, 11789, 11981, 13763, ... including the prime Fibonacci numbers 2, 3, 5, 89, 233, 1066340417491710595814572169, ... (subsequence of A005478).

Examples

			18 is in the sequence because 18^2 + 1 = 5^2*13 and 13 is a Fibonacci number.
		

Crossrefs

Programs

  • Maple
    q:= n-> (t-> ormap(issqr, [t+4, t-4]))(5*max(numtheory[factorset](n^2+1))^2):
    select(q, [$1..12000])[];  # Alois P. Heinz, Mar 11 2022
  • Mathematica
    With[{f = Fibonacci[Range[21]], m = f[[-1]]}, Select[Range[m], MemberQ[f, FactorInteger[#^2 + 1][[-1, 1]]] &]] (* Amiram Eldar, Mar 11 2022 *)
  • PARI
    isfib(n) = my(k=n^2); k+=(k+1)<<2; issquare(k) || (n>0 && issquare(k-8));
    isok(m) = isfib(vecmax(factor(m^2+1)[,1])); \\ Michel Marcus, Mar 11 2022

A339315 a(n) is the smallest number k such that k^2+1 divided by its largest prime factor is equal to F(2*n-1) for n > 0, or 0 if no such k exists, where F(n) is the Fibonacci sequence.

Original entry on oeis.org

1, 3, 8, 34, 55, 144, 610, 233, 12166, 2584, 4181, 68260, 46368, 75025, 3917414, 464656, 1346269, 16349962
Offset: 1

Views

Author

Michel Lagneau, Nov 30 2020

Keywords

Comments

a(n) is the smallest number k such that A248516(k) = A001519(n) for n > 0, or 0 if no such k exists, where A001519(n) = F(2*n-1) (bisection of the Fibonacci sequence), with F(n) = A000045(n).
We observe that a(2 + 3m) = A001519(1 + 3m) = A000045(1 + 6m) for m = 2, 3, 4, 5. For n = 6, this property no longer works.
For k > 0, a(3k - 1) is odd, a(3k) and a(3k+1) are even.
We observe that a(n)^2 + 1 is the product of two prime Fibonacci numbers for n = 2, 3, 4, 6, 7.
The first 18 terms of the sequence are Fibonacci numbers, except a(9), a(12), a(15), a(16) and a(18).
The corresponding sequence b(n) = (a(n)^2+1)/ A001519(n) is 2, 5, 13, 89, 89, 233, 1597, 89, 92681, 1597, 1597, 162593, 28657, 28657, 29842993, 160373, 514229. We observe that a majority of terms of b(n) are prime Fibonacci numbers, except b(9), b(12), b(15) and b(16).

Examples

			a(4) = 34 because 34^2 + 1 = 13*89 = 1157, and 1157/89 = 13 = A248516(34) = A001519(4).
A curiosity: a(22) = 1134903170 = F(45) with F(45)^2 + 1 = F(43)*F(47) where F(43) and F(47) are prime Fibonacci numbers.
		

Crossrefs

Programs

  • Maple
    with(numtheory):with(combinat,fibonacci):
    nn:=100:n0:=20:
    for n from 1 to n0 do:
      ii:=0:
      for m from 1 to 10^10 while(ii=0) do:
       x:=m^2+1:y:=factorset(x):n1:=nops(y):
       z:=x/y[n1]:
        if z = fibonacci(2*n-1)
         then
         ii:=1:printf(`%d %d \n`,n,m):
         else
        fi:
      od:
    od:
  • PARI
    a(n) = {my(k=1, f=fibonacci(2*n-1)); while ((k^2+1)/vecmax(factor(k^2+1)[,1]) != f, k++); k;} \\ Michel Marcus, Nov 30 2020

A350707 Numbers m such that all prime factors of m^2+1 are Fibonacci numbers.

Original entry on oeis.org

0, 1, 2, 3, 5, 7, 8, 18, 34, 57, 144, 239, 322, 610, 1134903170
Offset: 1

Views

Author

Michel Lagneau, Mar 27 2022

Keywords

Comments

The Fibonacci numbers in the sequence include 1, 2, 3, 5, 8, 144, 610 and 1134903170.
The sequence includes terms of the form sqrt(f(n) - 1) and sqrt(5 * f(n) - 1), where f(n) = Fibonacci(A281087(n)) * Fibonacci(A281087(n)+2) = A140362(n). - Daniel Suteu, Mar 29 2022

Examples

			57 is in the sequence because 57^2+1 = 2*5^3*13 and 2, 5 and 13 are Fibonacci numbers;
1134903170 = Fibonacci(45) is in the sequence because 1134903170^2+1 = 433494437*2971215073 = Fibonacci(43)*Fibonacci(47).
		

Crossrefs

The sequence contains A281618 and A285282.

Programs

  • Maple
    with(numtheory):
    A005478:={2, 3, 5, 13, 89, 233, 1597, 28657, 514229, 433494437, 2971215073, 99194853094755497,1066340417491710595814572169, 19134702400093278081449423917}:
    for n from 0 to 11000 do:
       y:=factorset(n^2+1):n0:=nops(y):
       if A005478 intersect y = y
           then
           print(n):
           else
         fi:
    od:
  • PARI
    isfib(n) = my(k=n^2); k+=(k+1)<<2; issquare(k) || (n>0 && issquare(k-8));
    isok(m) = my(f=factor(m^2+1)); for (i=1, #f~, if (!isfib(f[i,1]), return(0))); return(1); \\ Michel Marcus, Mar 29 2022
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