A215011 - OEIS

A215011

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

1, 4, 12, 15, 20, 8, 24, 12, 60, 10, 60, 84, 56, 40, 60, 18, 36, 36, 90, 120, 40, 120, 24, 300, 175, 252, 72, 168, 140, 60, 60, 60, 180, 360, 120, 228, 342, 252, 420, 60, 40, 88, 660, 60, 120, 48, 48, 168, 1400, 900, 252, 189, 108, 180, 120, 72, 252, 406, 1740

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OFFSET

1,2

COMMENTS

Triangular(n)=n*(n+1)/2 is the n-th triangular number.

LINKS

Table of n, a(n) for n=1..59.

EXAMPLE

Triangular(2)=3, least k>0 such that 3 divides Fibonacci(k) is k=4, so a(2)=4.

MATHEMATICA

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 *)

PROG

(Python)

TOP = 333

prpr = y = 0

prev = k = 1

res = [-1]*TOP

while y<TOP-1:

for i in range(1, TOP):

if res[i]<0 and prev % int(i*(i+1)/2) == 0:

res[i] = k

y += 1

curr = prpr+prev

prpr = prev

prev = curr

k += 1

for i in range(1, TOP):

print(res[i], end=', ')

CROSSREFS

Cf. A085779 (least k such that triangular(n) divides k!).

Cf. A001177 (least k such that n divides Fibonacci(k)).

Cf. A132632 (least k such that n^2 divides Fibonacci(k)).

Cf. A132633 (least k such that n^3 divides Fibonacci(k)).

Cf. A215453 (least k such that n^n divides Fibonacci(k)).

Cf. A214528 (least k such that n! divides Fibonacci(k)).

Cf. A000217, A000045.

Sequence in context: A256706 A340869 A103020 * A024353 A024354 A020883

Adjacent sequences: A215008 A215009 A215010 * A215012 A215013 A215014

KEYWORD

nonn

AUTHOR

Alex Ratushnyak, Aug 08 2012

STATUS

approved