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%I #28 Jun 19 2021 16:32:53
%S 1,4,12,15,20,8,24,12,60,10,60,84,56,40,60,18,36,36,90,120,40,120,24,
%T 300,175,252,72,168,140,60,60,60,180,360,120,228,342,252,420,60,40,88,
%U 660,60,120,48,48,168,1400,900,252,189,108,180,120,72,252,406,1740
%N a(n) = least k>0 such that triangular(n) divides Fibonacci(k).
%C Triangular(n)=n*(n+1)/2 is the n-th triangular number.
%e Triangular(2)=3, least k>0 such that 3 divides Fibonacci(k) is k=4, so a(2)=4.
%t lk[n_]:=Module[{k=1,t=(n(n+1))/2},While[Mod[Fibonacci[k],t]!=0,k++];k]; Array[lk,60] (* _Harvey P. Dale_, Jun 19 2021 *)
%o (Python)
%o TOP = 333
%o prpr = y = 0
%o prev = k = 1
%o res = [-1]*TOP
%o while y<TOP-1:
%o for i in range(1,TOP):
%o if res[i]<0 and prev % (i*(i+1)/2) == 0:
%o res[i] = k
%o y += 1
%o curr = prpr+prev
%o prpr = prev
%o prev = curr
%o k += 1
%o for i in range(1,TOP):
%o print res[i],
%Y Cf. A085779 (least k such that triangular(n) divides k!).
%Y Cf. A001177 (least k such that n divides Fibonacci(k)).
%Y Cf. A132632 (least k such that n^2 divides Fibonacci(k)).
%Y Cf. A132633 (least k such that n^3 divides Fibonacci(k)).
%Y Cf. A215453 (least k such that n^n divides Fibonacci(k)).
%Y Cf. A214528 (least k such that n! divides Fibonacci(k)).
%Y Cf. A000217, A000045.
%K nonn
%O 1,2
%A _Alex Ratushnyak_, Aug 08 2012
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