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A340542
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Number of Fibonacci divisors of Fibonacci(n)^2 + 1.
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1
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1, 2, 2, 2, 3, 3, 3, 4, 4, 3, 4, 4, 3, 5, 5, 3, 5, 5, 3, 5, 5, 3, 5, 6, 4, 5, 6, 4, 5, 5, 3, 5, 5, 5, 7, 5, 5, 7, 5, 3, 5, 5, 3, 7, 7, 3, 7, 8, 4, 5, 6, 4, 5, 7, 5, 5, 7, 5, 5, 5, 3, 7, 7, 5, 9, 7, 5, 7, 5, 3, 5, 5, 3, 7, 7, 5, 9, 7, 5, 8, 6, 3, 6, 8, 5, 5, 7
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OFFSET
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0,2
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COMMENTS
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A Fibonacci divisor of a number k is a Fibonacci number that divides k.
It is interesting to compare this sequence with A339669.
We observe that a(2n) = A339669(2n) if n = 5*k + 2 or n = 5*k + 3, with k >= 0, because Lucas(2n)^2 = 5*Fibonacci(2n)^2 + 4 (see A005248: all nonnegative integer solutions of the Pell equation a(n)^2 - 5*b(n)^2 = +4 together with b(n)= A001906(n), n>=0. - Wolfdieter Lang, Aug 31 2004).
So, Lucas(2n)^2 + 1 = 5*(Fibonacci(2n)^2 + 1). Lucas(2n)^2 + 1 and Fibonacci(2n)^2 + 1 have the same Fibonacci divisors for n = 5*k + 2 or n = 5*k + 3. For the other values of n = 5*k, 5*k + 1 or 5*k + 4, 5 is a Fibonacci divisor of Lucas(2n)^2 + 1 but not of Fibonacci(2n)^2 + 1. So for these last three values of n, a(2n) = A339669(2n) - 1 (except for m = 1 and 2, 5*F(m) is never a Fibonacci number).
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LINKS
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FORMULA
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EXAMPLE
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a(13) = 5 because the 5 Fibonacci divisors of Fibonacci(13)^2 + 1 = 233^2 + 1 are 1, 2, 5, 89 and 610.
a(16) = 5 because the 5 Fibonacci divisors of Fibonacci(16)^2 + 1 = 987^2 + 1 are 1, 2, 5, 610, and 1597.
Remark: the 5 Fibonacci divisors of Lucas(16)^2 + 1 = 2207^2 + 1 are 1, 2, 5, 610, and 1597, the index 16 = 2*8 with 8 of the form 5*k + 3.
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MAPLE
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with(combinat, fibonacci):nn:=100:F:={}:
for k from 0 to nn do:
F:=F union {fibonacci(k)}:
od:
for m from 0 to 90 do:
f:=fibonacci(m)^2+1:d:=numtheory[divisors](f):
lst:= F intersect d: n1:=nops(lst):printf(`%d, `, n1):
od:
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PROG
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(PARI) isfib(n) = my(k=n^2); k+=(k+1)<<2; issquare(k) || (n>0 && issquare(k-8)); \\ A010056
a(n) = sumdiv(fibonacci(n)^2+1, d, isfib(d)); \\ Michel Marcus, Jan 12 2021
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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