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 A236268 Smallest k such that prime and nonprime divisors alternate through exactly the first n pairs of divisors of Fibonacci(k), as {1, a prime, a nonprime, a prime, ...}. 0

%I

%S 3,25,325,275

%N Smallest k such that prime and nonprime divisors alternate through exactly the first n pairs of divisors of Fibonacci(k), as {1, a prime, a nonprime, a prime, ...}.

%C The first 2n divisors of f = Fibonacci(a(n)) are of the form {1, p(0), c(1), p(1), c(2), p(2), ..., c(n), p(n)} where c(i) are composite divisors, p(i) are prime divisors, and f = p(0)^2*p(1)*p(2)*...*p(q).

%e In the following examples, the { } contains the sequence of n couples (nonprime, prime divisors).

%e a(1) = 3 because F(3) = 2 => {1, 2};

%e a(2) = 25 because F(25) = 75025 = 5^2 * 3001 => {1, 5, 25, 3001};

%e a(3) = 325 because F(325) = 5^2 * 233 * 1301 * 3001 * 4235401 * 605416501 * 880262501 * 14736206161 * 49284706967787569058301 => {1, 5, 25, 233, 1165, 1301};

%e a(4) = 275 because F(275) = 5^2 * 89 * 661 * 3001 * 474541 * 7239101 * 15806979101 * 5527278404454199535821801 => {1, 5, 25, 89, 445, 661, 2225, 3001}.

%p with(combinat,fibonacci):with(numtheory):for n from 2 by 2 to 30 do:ii:=0:for kk from 2 to 1000 while(ii=0) do:k:=fibonacci(kk): x:=divisors(k):n1:=nops(x):it:=0:for i from 1 by 2 to n do:if n1>=n and type(x[i], prime)=false then it:=it+1:else fi:od:for j from 2 by 2 to n do:if n1>=n and type(x[j], prime)=true then it:=it+1:else fi:od:if it=n then ii:=1: printf ( "%d %d \n", n/2, k):else fi:od:od:

%Y Cf. A000045, A227337.

%K nonn,hard,more

%O 1,1

%A _Michel Lagneau_, Jan 21 2014

%E Name reworded by _Jon E. Schoenfield_, Sep 17 2017

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Last modified June 23 13:59 EDT 2021. Contains 345402 sequences. (Running on oeis4.)