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A061446
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Primitive part of Fibonacci(n).
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28
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1, 1, 2, 3, 5, 4, 13, 7, 17, 11, 89, 6, 233, 29, 61, 47, 1597, 19, 4181, 41, 421, 199, 28657, 46, 15005, 521, 5777, 281, 514229, 31, 1346269, 2207, 19801, 3571, 141961, 321, 24157817, 9349, 135721, 2161, 165580141, 211, 433494437, 13201, 109441
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OFFSET
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1,3
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COMMENTS
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Fib(n) = A000045(n) = Product_{d|n} a(d), Lucas(n) = A000204(n) = Product_{d|2n and 2^m|d iff 2^m|2n} a(d) (e.g., Lucas(4) = 7 = a(8), Lucas(6) = 18 = a(12)*a(4)). - Len Smiley, Nov 11 2001
A 2001 Iranian Mathematical Olympiad question shows such a sequence exists whenever gcd(a(m),a(n)) = a(gcd(m,n)).
The problem of the characterization of the family of all GCD-morphic sequences F, i.e., F such that GCD(F(m),F(n)) = F(GCD(m,n)), was posed by A. K. Kwasniewski (GCD-morphic Problem). Dziemianczuk and Bajguz (2008) showed that any GCD-morphic sequence is coded by a certain natural number-valued sequence. - Maciej Dziemianczuk, Jan 15 2009
This is the LCM-transform of the Fibonacci numbers (cf. Nowicki). - N. J. A. Sloane, Jan 02 2016
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LINKS
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FORMULA
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Let r=(1+sqrt(5))/2. For n>2, the primitive part of F(n)=(r^n-(-1/r)^n)/sqrt(5) is Phi_n(-r^2)/r^phi(n) where Phi_n is n-th cyclotomic polynomial and phi is Euler's totient function A000010.
a(n) = Product_{d|n} Fib(d)^mu(n/d), where mu = A008683 (Bliss, Fulan, Lovett, Sommars, eq. (7)). - Jonathan Sondow, Jun 11 2013
a(n) = lcm(Fib(1),Fib(2),...,Fib(n))/lcm(Fib(1),Fib(2),...,Fib(n-1)). - Thomas Ordowski, Aug 03 2015
a(n) = Product_{k=1..n} Fib(gcd(n,k))^(mu(n/gcd(n,k))/phi(n/gcd(n,k))) = Product_{k=1..n} Fib(n/gcd(n,k))^(mu(gcd(n,k))/phi(n/gcd(n,k))) where mu = A008683, phi = A000010. - Richard L. Ollerton, Nov 08 2021
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MAPLE
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N:= 200; # to get a(1) to a(N)
L[0]:= 1:
for i from 1 to N do L[i]:=ilcm(L[i-1], combinat:-fibonacci(i)) od:
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MATHEMATICA
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t={1}; Do[f=Fibonacci[n]; Do[f=f/GCD[f, t[[d]]], {d, Most[Divisors[n]]}]; AppendTo[t, f], {n, 2, 100}]; t
(* Second program: *)
a[n_] := Product[Fibonacci[d]^MoebiusMu[n/d], {d, Divisors[n]}];
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PROG
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(PARI) a(n)=my(d=divisors(n)); fibonacci(n)/lcm(apply(fibonacci, d[1..#d-1])) \\ Charles R Greathouse IV, Oct 06 2016
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CROSSREFS
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Cf. A000010 (comments on product formulas).
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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