A080648 -id:A080648 - cp's OEIS frontend

A064725 Sum of primes dividing Fibonacci(n) (with repetition).

0, 0, 2, 3, 5, 6, 13, 10, 19, 16, 89, 14, 233, 42, 68, 57, 1597, 42, 150, 60, 436, 288, 28657, 46, 3011, 754, 181, 326, 514229, 114, 2974, 2264, 19892, 5168, 141979, 160, 2443, 9499, 135956, 2228, 62158, 680, 433494437, 641, 109526, 29257, 2971215073

Offset: 1

Jason Earls, Oct 16 2001

nonn

			a(12) = 14 because Fibonacci(12) = 144 = 2^4*3^2 and the sum of the prime divisors with repetition is 4*2 + 2*3 = 14.

Amiram Eldar, Table of n, a(n) for n = 1..1000 (terms 1..350 from Harry J. Smith)

Cf. A000045, A001414, A080648 (without repetition).

with (numtheory):with(combinat, fibonacci):
sopfr:= proc(n) local e, j; e := ifactors(fibonacci(n))[2]:
add (e[j][1]*e[j][2], j=1..nops(e)) end:
seq (sopfr(n), n=1..100); # Michel Lagneau, Nov 13 2012
# second Maple program:
a:= n-> add(i[1]*i[2], i=ifactors((<<0|1>, <1|1>>^n)[1, 2])[2]):
seq(a(n), n=1..47);  # Alois P. Heinz, Sep 03 2019

fiboPrimeFactorSum[n_] := Plus @@ Times @@@ FactorInteger@ Fibonacci[n]; fiboPrimeFactorSum[1] = 0; Array[fiboPrimeFactorSum, 60] (* Michel Lagneau, Nov 13 2012 *)

sopfr(n)= { local(f,s=0); f=factor(n); for(i=1, matsize(f)[1], s+=f[i, 1]*f[i, 2]); return(s) }
{ for (n = 0, 350, write("b064725.txt", n, " ", sopfr(fibonacci(n))) ) } \\ Harry J. Smith, Sep 23 2009

A219187 Sum of distinct prime divisors of Lucas(n).

Original entry on oeis.org

2, 0, 3, 2, 7, 11, 5, 29, 47, 21, 44, 199, 32, 521, 284, 44, 2207, 3571, 112, 9349, 2168, 242, 353, 600, 1152, 263, 90484, 5800, 14510, 19548, 2567, 3010349, 5568, 10102, 63513, 1022, 103713, 54018521, 29134604, 1461, 4689, 370248451, 1796, 151190, 2118, 785

Offset: 0

Michel Lagneau, Nov 14 2012

nonn

			a(6) = 5 because Lucas(6) = 21 and the sum of the prime divisors {3, 7} equals 10.

Amiram Eldar, Table of n, a(n) for n = 0..1000

Cf. A000032, A008472, A080648, A219177.

with (numtheory):with(combinat,fibonacci):
sopf:= proc(n) local e, j; e := ifactors(fibonacci(n+1)+fibonacci(n-1))[2]:
add (e[j][1], j=1..nops(e)) end:
seq (sopf(n), n=0..100);

Array[If[#==1, 0, Plus@@First/@FactorInteger[LucasL[ # ]]]&, 50, 0]

a(n) = A008472(A000032(n)). - Amiram Eldar, Sep 03 2019

a(0) prepended by Amiram Eldar, Sep 03 2019

A238684 a(1) = a(2) = 1; for n > 2, a(n) is the product of prime factors of the n-th Fibonacci number.

Original entry on oeis.org

1, 1, 2, 3, 5, 2, 13, 21, 34, 55, 89, 6, 233, 377, 610, 987, 1597, 646, 4181, 6765, 10946, 17711, 28657, 966, 15005, 121393, 196418, 317811, 514229, 208010, 1346269, 2178309, 3524578, 5702887, 9227465, 207366, 24157817, 39088169, 63245986, 102334155, 165580141, 66978574, 433494437, 701408733, 1134903170

Offset: 1

Carmine Suriano, Mar 02 2014

nonn

In other words, the squarefree part of the n-th Fibonacci number.

a(gcd(m,n)) = gcd(a(m),a(n)). - Robert Israel, Nov 10 2023

			a(12) = 6 since F(12) = 144 = 2^4 * 3^2 and 2 * 3 = 6.

Robert Israel, Table of n, a(n) for n = 1..350

Cf. A000045, A080648.

f:= n -> convert(numtheory:-factorset(combinat:-fibonacci(n)),`*`):
map(f, [$1..100]); # Robert Israel, Nov 10 2023

Table[Times@@Part[Flatten[FactorInteger[Fibonacci[n]]], 1 ;; -2 ;; 2], {n, 3, 50}] (* Alonso del Arte, Mar 02 2014 *)

a(n) = my(f = factor(fibonacci(n))); prod(i=1, #f~, f[i, 1]); \\ Michel Marcus, Mar 02 2014

a(n) = A007947(A000045(n)) - Tom Edgar, Mar 03 2014

A280106 Numbers k such that the half sum of the prime factors of Fibonacci(k) is a Fibonacci number.

Original entry on oeis.org

3, 6, 8, 10, 14, 15, 22, 26, 30, 34, 94

Offset: 1

Michel Lagneau, Dec 28 2016

Or numbers k such that A080648(k)/2 is a Fibonacci number.
Is this sequence finite ?
a(12), if it exists is >5000. - Robert Price, Mar 02 2017

			10 is in the sequence because Fibonacci(10) = 5*11=> 5+11 = 2*8 = 2*F(6);
94 is in the sequence because Fibonacci(94) = 2971215073*6643838879 =>
2971215073+6643838879 = 2*4807526976 = 2*F(48).

Cf. A000045, A080648.

with(numtheory):with(combinat,fibonacci):nn:=300:
for n from 3 to nn do:
  f:=fibonacci(n):x:=factorset(f):n0:=nops(x):
  s:=sum(‘x[i]’, ‘i’=1..n0):c:=s/2:
  x1:=sqrt(5*c^2-4):x2:=sqrt(5*c^2+4):
    if x1=floor(x1) or x2=floor(x2)
     then
     print(n):
     else
    fi:
od:

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A064725 Sum of primes dividing Fibonacci(n) (with repetition).

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A219187 Sum of distinct prime divisors of Lucas(n).

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A238684 a(1) = a(2) = 1; for n > 2, a(n) is the product of prime factors of the n-th Fibonacci number.

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Maple

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A280106 Numbers k such that the half sum of the prime factors of Fibonacci(k) is a Fibonacci number.

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Maple