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A214028
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Entry points for the Pell sequence: smallest k such that n divides A000129(k).
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19
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1, 2, 4, 4, 3, 4, 6, 8, 12, 6, 12, 4, 7, 6, 12, 16, 8, 12, 20, 12, 12, 12, 22, 8, 15, 14, 36, 12, 5, 12, 30, 32, 12, 8, 6, 12, 19, 20, 28, 24, 10, 12, 44, 12, 12, 22, 46, 16, 42, 30, 8, 28, 27, 36, 12, 24, 20, 10, 20, 12, 31, 30, 12, 64, 21, 12, 68, 8, 44, 6, 70, 24, 36, 38
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OFFSET
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1,2
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COMMENTS
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Conjecture: A175181(n)/A214027(n) = a(n). This says that the zeros appear somewhat uniformly in a period. The second zero in a period is exactly where n divides the first Lucas number, so this relationship is not really surprising.
The comment above is correct, since n divides A000129(k*a(n)) for all integers k and clearly a(n) divides A175181(n), so the zeros appear uniformly.
a(n) <= 4*n/3 for all n, where the equality holds iff n is a power of 3.
(End)
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LINKS
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FORMULA
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If p^2 does not divide A000129(a(p)) (that is, p is not in A238736) then a(p^e) = a(p)*p^(e - 1). If gcd(m, n) = 1 then a(mn) = lcm(a(m), a(n)). - Jianing Song, Aug 29 2018
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EXAMPLE
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11 first divides the term A000129(12) = 13860 = 2*3*5*7*11.
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MAPLE
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local a000129, k ;
a000129 := [1, 2, 5] ;
for k do
if modp(a000129[1], n) = 0 then
return k;
end if;
a000129[1] := a000129[2] ;
a000129[2] := a000129[3] ;
a000129[3] := 2*a000129[2]+a000129[1] ;
end do:
end proc:
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MATHEMATICA
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a[n_] := With[{s = Sqrt@ 2}, ((1 + s)^n - (1 - s)^n)/(2 s)] // Simplify; Table[k = 1; While[Mod[a[k], n] != 0, k++]; k, {n, 80}] (* Michael De Vlieger, Aug 25 2015, after Michael Somos at A000129 *)
Table[k = 1; While[Mod[Fibonacci[k, 2], n] != 0, k++]; k, {n, 100}] (* G. C. Greubel, Aug 10 2018 *)
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PROG
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(PARI) pell(n) = polcoeff(Vec(x/(1-2*x-x^2) + O(x^(n+1))), n);
a(n) = {k=1; while (pell(k) % n, k++); k; } \\ Michel Marcus, Aug 25 2015
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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