A245236 -id:A245236 - cp's OEIS frontend

A248516 n^2+1 divided by its largest prime factor.

1, 1, 2, 1, 2, 1, 10, 5, 2, 1, 2, 5, 10, 1, 2, 1, 10, 25, 2, 1, 26, 5, 10, 1, 2, 1, 10, 5, 2, 17, 26, 25, 10, 13, 2, 1, 10, 85, 2, 1, 58, 5, 50, 13, 2, 29, 130, 5, 2, 41, 2, 5, 10, 1, 34, 1, 250, 5, 2, 13, 2, 5, 10, 17, 2, 1, 10, 125, 2, 169, 2, 85, 130, 1, 58

Offset: 1

Michel Lagneau, Jun 18 2015

a(n)=1 iff n^2+1 is prime (see A002496).

Conjecture: The Fibonacci numbers in the sequence are 1, 2, 5, 13, 34, 89, 233, 377, 610, 1597, 4181, 10946, 28657, 75025, 121393, 196418, ... including all of A001519.

Michel Lagneau, Table of n, a(n) for n = 1..10000

Cf. A001519, A002522, A014442, A245236.

A248516 := proc(n)
    n^2+1 ;
    %/A006530(%) ;
end proc:
seq(A248516(n),n=1..20) ; # R. J. Mathar, Jun 18 2015

Table[(n^2 + 1)/FactorInteger[n^2 + 1][[-1, 1]], {n, 75}] (* Michael De Vlieger, Jun 19 2015 *)

a(n) = my(x=n^2+1); my(f=factor(x)); x/f[#f~,1]; \\ Michel Marcus, Jun 18 2015

a(n) = A002522(n)/A014442(n).

a(n) = A052126(1+n^2). - R. J. Mathar, Jun 18 2015

A281618 Fibonacci numbers F such that all the prime factors of F^2 + 1 are also Fibonacci numbers.

Original entry on oeis.org

1, 2, 3, 5, 8, 34, 144, 610, 1134903170

Offset: 1

Michel Lagneau, Jan 25 2017

The corresponding indices of F are 1 or 2, 3, 4, 5, 6, 9, 12, 15, 45, ... and A245236 is in this sequence.

			a(9)^2+1 = Fibonacci(45)^2+1 = 1134903170^2+1 = 1288005205276048901 = 433494437 * 2971215073 = Fibonacci(43)*Fibonacci(47).

Cf. A000045, A245306, A245236.

with(numtheory):with(combinat,fibonacci):nn:=100:
for n from 1 to nn do:
  f:=fibonacci(n)^2+1:x:=factorset(f):n0:=nops(x):it:=0:
    for m from 1 to n0 do:
    c:=x[m]:
    x1:=sqrt(5*c^2-4):x2:=sqrt(5*c^2+4):
    if x1=floor(x1) or x2=floor(x2)
     then
     it:=it+1:
     else
    fi:
od:
if it=n0 then print(fibonacci(n)):else fi:od:

With[{s = Rest@ Fibonacci@ Range@ 120}, Select[s, Times @@ Boole@ Map[MemberQ[s, #] &, FactorInteger[#^2 + 1][[All, 1]]] > 0 &]] (* Michael De Vlieger, Jan 27 2017 *)

isfib(n) = my(k=n^2); k+=(k+1)<<2; issquare(k) || (n>0 && issquare(k-8));
isokf(n) = {my(f = factor(fibonacci(n)^2+1)); for (k=1, #f~, if (!isfib(f[k,1]), return(0));); return(1);}
for (n=2, 50, if (isokf(n), print1(fibonacci(n), ", "))) \\ Michel Marcus, Jan 28 2017

A245688 Numbers n such that n^2 + 1 is the product of three distinct Fibonacci numbers > 1.

Original entry on oeis.org

47, 99, 123, 322, 843, 2207, 5778, 15075, 15127, 39603, 103682, 271443, 710647, 1860498, 4870847, 12752043, 33385282, 87403803, 228826127, 599074578, 1568397607, 4106118243, 10749957122, 28143753123, 73681302247, 192900153618, 505019158607, 1322157322203

Offset: 1

Michel Lagneau, Jul 29 2014

Conjecture: except the numbers 99 and 15075, all the terms belong to A005248 (bisection of Lucas numbers).

The above conjecture holds for the first 1000 terms. - Jens Kruse Andersen, Aug 10 2014

			99 is in the sequence because 99^2+1 = 9802 = 2*13*377 where 2, 13 and 377 are three Fibonacci numbers, but 99 is not a Lucas number.
15075 is in the sequence because 15075^2+1 = 13*89*196418 where 13, 89 and 196418 are three Fibonacci numbers, but 15075 is not a Lucas number.

Jens Kruse Andersen, Table of n, a(n) for n = 1..1000

Cf. A000032, A000045, A005248, A245236.

with(combinat,fibonacci):with(numtheory):nn:=200:lst:={}:T:=array(1..nn):
   for n from 1 to nn do:
    T[n]:=fibonacci(n):
   od:
     for p from 1 to nn-1 do:
       for q from p+1 to nn-1 do:
          for r from q+1 to nn-1 do:
           f:=T[p]*T[q]*T[r]-1:x:=sqrt(f):
           if x=floor(x)and T[p]<>1
           then
           lst:=lst union {x}:
           else
           fi:
          od:
       od:
     od:
     print(lst):

Empirical g.f.: x*(9297*x^9-24320*x^8-52*x^7+52*x^3-127*x^2-42*x+47) / (x^2-3*x+1). - Colin Barker, Aug 13 2014

Doubtful link, formula and PARI code deleted by Colin Barker, Jul 31 2014

A360107 Numbers k such that sigma_2(Fibonacci(k)^2 + 1) == 0 (mod Fibonacci(k)).

Original entry on oeis.org

1, 2, 3, 5, 7, 9, 11, 13, 15, 19, 21, 25, 27, 31, 41, 45, 49, 81, 85, 129, 133, 135, 139, 357, 361, 429, 431, 433, 435, 447, 451, 507, 511, 567, 569, 571, 573

Offset: 1

Michel Lagneau, Jan 26 2023

			7 is in the sequence because the divisors of Fibonacci(7)^2 + 1 = 13^2 + 1 = 170 are {1, 2, 5, 10, 17, 34, 85, 170}, and 1^2 + 2^2 + 5^2 + 10^2 + 17^2 + 34^2 + 85^2 + 170^2 = 37700 = 13*2900 == 0 (mod 13).

Cf. A000045, A001157, A067719, A245236, A245306, A338762, A360105.

Select[Range[140],Divisible[DivisorSigma[2,Fibonacci[#]^2+1],Fibonacci[#]]&]

isok(k) = my(f=fibonacci(k)); sigma(f^2 + 1, 2) % f == 0; \\ Michel Marcus, Jan 26 2023

a(24)-a(37) from Daniel Suteu, Jan 27 2023

A245046 Smallest non-Fibonacci number k such that k^2 + F(n)^2 = f1*f2 where F(n) = A000045(n) and f1, f2 are distinct Fibonacci numbers.

Original entry on oeis.org

70, 70, 6, 991, 27, 183, 443, 38, 27, 373

Offset: 1

Michel Lagneau, Jul 16 2014

The sequence is probably finite.
Conjecture : if k exists, the number of solutions non-Fibonacci k of the equation k^2 + F(n)^2 = f1*f2 is finite.
The primes of the sequence are 373, 443 and 991.
If f1<=f2, the new sequence is 70, 70, 6, 4, 12, 183, 443, 38, 27, 373

			a(1) = a(2) = 70 because 70^2+1 = F(7)*F(14) = 13*377. The number 70 is probably unique.
a(3) = 6 because 6^2+2^2 = F(5)*F(6) = 5*8. But there exists also k = 10 such that 10^2+4 = F(6)*F(7) = 8*13.
a(7) = 443 because 443^2 + 13^2 = F(1)*F(27) = 1*196418.

Cf. A000045, A245236.

with(combinat,fibonacci):with(numtheory):nn:=200:T:=array(1..nn):
  for i from 1 to nn do:
    T[i]:=fibonacci(i):
  od:
   for n from 1 to 10 do:
     ff:=fibonacci(n):ii:=0:
       for p from 1 to nn-1 while(ii=0)do:
         for q from p+1 to nn-1 while(ii=0)do:
           f:=T[p]*T[q]-ff^2:x:=sqrt(f):x1:=sqrt(5*f+4):x2:=sqrt(5*f-4):
           if f>0 and x=floor(x)
           and x1<>floor(x1) and x2<>floor(x2)
           then
           ii:=1:printf ( "%d %d %d %d \n",n,x,T[p],T[q]):
           else
           fi:
         od:
       od:
    od:

isfib(n) = {my(k=n^2); k+=(k+1)<<2; issquare(k) || (n>0 && issquare(k-8));} \\ from A010056
isprod(pf) = {sqrpf = sqrtint(pf); ifib = 1; while((fif = fibonacci(ifib)) < sqrpf, if (pf % fif == 0, if (isfib(pf/fif), return (1));); ifib ++;); return (0);}
a(n) = {k = 1; fsq = fibonacci(n)^2; ok = 0; while (!ok, if (! isfib(k), pf = k^2 + fsq; ok = isprod(pf);); if (! ok, k++);); k;} \\ Michel Marcus, Jul 17 2014

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

Michel Lagneau, Nov 30 2020

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).

			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.

Cf. A000045, A001519, A001906, A002522, A005478, A014442, A245236, A245306, A248516, A281618.

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:

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

Michel Lagneau, Mar 27 2022

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

			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).

The sequence contains A281618 and A285282.

Cf. A000032, A000045, A005478, A352290, A245236, A339461.

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:

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

cp's OEIS Frontend

A248516 n^2+1 divided by its largest prime factor.

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A281618 Fibonacci numbers F such that all the prime factors of F^2 + 1 are also Fibonacci numbers.

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A245688 Numbers n such that n^2 + 1 is the product of three distinct Fibonacci numbers > 1.

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A360107 Numbers k such that sigma_2(Fibonacci(k)^2 + 1) == 0 (mod Fibonacci(k)).

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A245046 Smallest non-Fibonacci number k such that k^2 + F(n)^2 = f1*f2 where F(n) = A000045(n) and f1, f2 are distinct Fibonacci numbers.

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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.

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

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