A064362 -id:A064362 - cp's OEIS frontend

A124378 Primitive elements of A064362.

5, 8, 12, 13, 17, 21, 28, 33, 37, 53, 57, 61, 69, 73, 77, 87, 89, 92, 93, 97, 109, 113, 133, 137, 141, 149, 157, 164, 172, 173, 177, 188, 193, 197, 203, 213, 217, 233, 237, 253, 257, 268, 269, 277, 287, 293, 301, 303, 309, 313, 317, 329, 332, 337, 353, 373, 381

Offset: 1

Franklin T. Adams-Watters, Oct 28 2006

nonn

Numbers n such that no Lucas number is a multiple of n, which are not divisible by any smaller number with that property.

T. D. Noe, Table of n, a(n) for n = 1..1000

Cf. A064362, A065156.

test[n_] := Module[{a, b, t}, {a, b} = {2, 1}; While[t = b; b = Mod[a + b, n]; a = t; ! (b == 0 || {a, b} == {2, 1})]; b == 0]; t = {}; n = 0; While[Length[t] < 1000, n++; If[! MemberQ[Mod[n, t], 0] && ! test[n], AppendTo[t, n]]]; t (* T. D. Noe, Mar 20 2013 *)

A065156 Numbers n such that some Lucas number (A000204) is divisible by n.

Original entry on oeis.org

1, 2, 3, 4, 6, 7, 9, 11, 14, 18, 19, 22, 23, 27, 29, 31, 38, 41, 43, 44, 46, 47, 49, 54, 58, 59, 62, 67, 71, 76, 79, 81, 82, 83, 86, 94, 98, 101, 103, 107, 116, 118, 121, 123, 124, 127, 129, 131, 134, 139, 142, 151, 158, 161, 162, 163, 166, 167, 179, 181, 191, 199

Offset: 1

Dean Hickerson, Oct 18 2001

From A.H.M. Smeets, Sep 20 2020 (Start)
For the Fibonacci numbers, each natural number divides some Fibonacci number (see A001177).
If, for some number m, m divides some Lucas number L_i (=A000204(i)), then, the smallest i satisfies i <= m. (End)

A.H.M. Smeets, Table of n, a(n) for n = 1..20000 (terms 1..1000 from T. D. Noe)
B. Avila and T. Khovanova, Free Fibonacci Sequences, arXiv preprint arXiv:1403.4614 [math.NT], 2014 and J. Int. Seq. 17 (2014) # 14.8.5

Complement of A064362. Cf. A000204.

Cf. A001177, A053032, A140409.

test[ n_ ] := For[ a=1; b=3, True, t=b; b=Mod[ a+b, n ]; a=t, If[ b==0, Return[ True ] ]; If[ a==2&&b==1, Return[ False ] ] ]; Select[ Range[ 200 ], test ]
Take[Flatten[Divisors/@LucasL[Range[200]]]//Union,70] (* Harvey P. Dale, Jun 07 2020 *)

a, n = 0, 0
while n < 1000:
    a, f0, f1, i = a+1, 1, 2, 1
    if f1%a == 0:
        n = n+1
        print(n,a)
    else:
        while f0%a != 0 and i <= a:
            f0, f1, i = f0+f1, f0, i+1
        if i <= a:
            n = n+1
            print(n,a) # A.H.M. Smeets, Sep 20 2020

Equals {1,2,4} union {p^e | p in A140409 and e > 0} union {2*p^e | p in A140409 and e > 0} union {4*p | p in A053032} union {4*p*q | p, q in A053032}. - A.H.M. Smeets, Sep 20 2020

A173491 a(n) is the least k such that the period of the decimal expansion of 1/k is A000204(n).

Original entry on oeis.org

3, 27, 101, 239, 21649, 19, 3191, 35121409, 722817036322379041, 2241, 797, 967, 1230610745978027, 3373, 60787, 509538919, 15060275578609, 5779, 37397, 423557

Offset: 1

Michel Lagneau, Feb 19 2010

Smallest k such that A007732(k) = A000204(n).

For the large numbers (p > 70), the Maple program below is very slow. So we use a process of two steps: first, factor 10^p-1 using the elliptic curve method; then, for each factor q(k), k=1,2,...,r, compute the period of 1/q(k) and keep the period q(i) such that q(i) ... [unfinished sentence? - R. J. Mathar, Feb 24 2010] Compare the Maple section of A170945!

			a(1)=3 because the period of 1/3 = 0.333... is 1, and 3 is the smallest number with that period.
a(2)=27 because the period of 1/27 = 0.037037... is 3 = A000204(2), and 27 is the smallest number with that period.
a(3)=101 because the period of 1/101 = 0.00990099... is 4 = A000204(3), and 101 is the smallest number with that period.
a(4)= 239 because the period of 1/239 = 0.00418410041841... is 7 = A000204(4), and 239 is the smallest number with that period.

V. E. Hoggatt, Jr., Fibonacci and Lucas Numbers. Houghton, Boston, MA, 1969.
Thomas Koshy, "Fibonacci and Lucas Numbers with Applications", John Wiley and Sons, 2001.
S. Vajda, Fibonacci and Lucas numbers and the Golden Section, Ellis Horwood Ltd., Chichester, 1989.

R. Javonovic, Lucas Function Calculator

Cf. A000032, A000204, A002329, A007732, A051626, A064362, A072859, A170945.

T:=array(0..100); U:=array(0..100); n0:=1: n1:=3: T[1] := 1: T[2] := 3:for i from 3 to 30 do: n2:=n0+n1: T[i]:=n2: n0:=n1: n1:=n2: od:
for q from 1 to 7 do: p0:=T[q]: indic:=0: for n from 1 to 25000 do: for p from 1 to 30 while(irem(10^p, n)<>1 or gcd(n,10)<>1 ) do: od: if irem(10^p,n) = 1 and gcd(n,10) = 1 and p=p0 and indic=0 then U[q]:=n: indic:=1: else fi: od: od:
for n from 1 to 7 do: print( U[n]): od:

(* This [slow] mma program gives all denominators < 50000 and disagrees with existing sequence for n = 11: a(11) = 797 instead of 29453 *) a204[n_] := a204[n] = Coefficient[Series[(2 - t )/(1 - t - t^2), {t, 0, n}], t^n] ; a7732[n_] := a7732[n] = MultiplicativeOrder[10, FixedPoint[Quotient[#, GCD[#, 10]] &, n]]; a[n_] := (k = 2; While[k++; k < 50000 && a7732[k] != a204[n] ]; k); Table[a[n], {n, 1, 15}](* Jean-François Alcover, Sep 02 2011 *)

References to unrelated sequences removed by R. J. Mathar, Feb 24 2010

Extended with the help of Jean-François Alcover and D. S. McNeil by T. D. Noe, Sep 07 2011

A232658 Numbers that are factors of Lucas numbers, whose multiples do not appear in some other Fibonacci-like sequence.

Original entry on oeis.org

11, 18, 19, 22, 29, 31, 38, 41, 44, 46, 47, 54, 58, 59, 62, 71, 76, 79, 82, 94, 101, 107, 116, 118, 121, 123, 124, 129, 131, 139, 142, 151, 158, 161, 162, 166, 179, 181, 191, 199, 201, 202, 209, 211, 214, 229, 236, 239, 241, 242, 246, 249, 251, 258, 262, 263, 271

Offset: 1

Brandon Avila and Tanya Khovanova, Nov 27 2013

nonn

Intersection of A230457 and A065156.

Sequence A230457 from which elements of A064362 are removed.

			Sequence A000285 is the Fibonacci-like sequence starting with 1 and 4. This sequence doesn't contain multiples of 11. On the other hand Lucas numbers contain multiples of 11. Therefore, 11 belongs to this sequence.

B. Avila and T. Khovanova, Free Fibonacci Sequences, arXiv preprint arXiv:1403.4614, 2014 and J. Int. Seq. 17 (2014) # 14.8.5

Cf. A230457, A065156, A064362.

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A124378 Primitive elements of A064362.

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A065156 Numbers n such that some Lucas number (A000204) is divisible by n.

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A173491 a(n) is the least k such that the period of the decimal expansion of 1/k is A000204(n).

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A232658 Numbers that are factors of Lucas numbers, whose multiples do not appear in some other Fibonacci-like sequence.

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