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 A239740 a(n) = gcd(Sum_{k=1...n} F(k), Product{j=1...n} F(j)), where F(k) is the k-th Fibonacci number. 2

%I #12 Sep 13 2017 02:20:07

%S 1,1,2,1,6,20,3,18,8,143,8,8,21,986,84,63,220,6764,55,770,144,46367,

%T 144,432,377,317810,16588,377,43428,2178308,987,53298,2584,14930351,

%U 2584,18088,6765,102334154,784740,20295,2054476,701408732,17711,1664834,46368

%N a(n) = gcd(Sum_{k=1...n} F(k), Product{j=1...n} F(j)), where F(k) is the k-th Fibonacci number.

%C The Fibonacci numbers in the sequence are 1, 2, 3, 8, 21, 55, 144, 377, 987, ... and a majority are elements of A001906 (F(2*n)= bisection of Fibonacci sequence).

%C We find consecutive values such that (1, 2), (2, 3), (20, 21), (986, 987), (6764, 6765), (46367, 46368), (317810, 317811), (14930351, 14930352), ...

%H Reinhard Zumkeller, <a href="/A239740/b239740.txt">Table of n, a(n) for n = 1..1000</a>

%e The first 8 Fibonacci numbers are 1,1,2,3,5,8,13,21 and 1+1+2+3+5+8+13+21 = 54. So a(8) = gcd(54, 1*1*2*3*5*8*13*21) = 18.

%p with(combinat,fibonacci):seq(gcd(add(fibonacci(i), i=1..n), mul(fibonacci(j), j=1..n)), n=1..60);

%t nn=60; With[{prs=Fibonacci[Range[nn]]}, Table[GCD[Total[Take[prs, n]], Times@@Take[ prs, n]], {n, nn}]]

%o (Haskell)

%o a239740 n = gcd (sum fs) (product fs)

%o where fs = take n \$ tail a000045_list

%o -- _Reinhard Zumkeller_, Mar 27 2014

%Y Cf. A000045, A001906.

%K nonn

%O 1,3

%A _Michel Lagneau_, Mar 26 2014

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Last modified October 1 15:27 EDT 2023. Contains 365826 sequences. (Running on oeis4.)