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A215011 a(n) = least k>0 such that triangular(n) divides Fibonacci(k).

%I A215011 #32 Oct 01 2024 15:44:50
%S A215011 1,4,12,15,20,8,24,12,60,10,60,84,56,40,60,18,36,36,90,120,40,120,24,
%T A215011 300,175,252,72,168,140,60,60,60,180,360,120,228,342,252,420,60,40,88,
%U A215011 660,60,120,48,48,168,1400,900,252,189,108,180,120,72,252,406,1740
%N A215011 a(n) = least k>0 such that triangular(n) divides Fibonacci(k).
%C A215011 Triangular(n)=n*(n+1)/2 is the n-th triangular number.
%e A215011 Triangular(2)=3, least k>0 such that 3 divides Fibonacci(k) is k=4, so a(2)=4.
%t A215011 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 A215011 (Python)
%o A215011 TOP = 333
%o A215011 prpr = y = 0
%o A215011 prev = k = 1
%o A215011 res = [-1]*TOP
%o A215011 while y<TOP-1:
%o A215011     for i in range(1, TOP):
%o A215011       if res[i]<0 and prev % int(i*(i+1)/2) == 0:
%o A215011         res[i] = k
%o A215011         y += 1
%o A215011     curr = prpr+prev
%o A215011     prpr = prev
%o A215011     prev = curr
%o A215011     k += 1
%o A215011 for i in range(1, TOP):
%o A215011     print(res[i], end=', ')
%Y A215011 Cf. A085779 (least k such that triangular(n) divides k!).
%Y A215011 Cf. A001177 (least k such that n divides Fibonacci(k)).
%Y A215011 Cf. A132632 (least k such that n^2 divides Fibonacci(k)).
%Y A215011 Cf. A132633 (least k such that n^3 divides Fibonacci(k)).
%Y A215011 Cf. A215453 (least k such that n^n divides Fibonacci(k)).
%Y A215011 Cf. A214528 (least k such that n! divides Fibonacci(k)).
%Y A215011 Cf. A000217, A000045.
%K A215011 nonn
%O A215011 1,2
%A A215011 _Alex Ratushnyak_, Aug 08 2012