A078414 -id:A078414 - cp's OEIS frontend

A214684 a(1)=1, a(2)=1, and, for n>2, a(n)=(a(n-1)+a(n-2))/5^k, where 5^k is the highest power of 5 dividing a(n-1)+a(n-2).

1, 1, 2, 3, 1, 4, 1, 1, 2, 3, 1, 4, 1, 1, 2, 3, 1, 4, 1, 1, 2, 3, 1, 4, 1, 1, 2, 3, 1, 4, 1, 1, 2, 3, 1, 4, 1, 1, 2, 3, 1, 4, 1, 1, 2, 3, 1, 4, 1, 1, 2, 3, 1, 4, 1, 1, 2, 3, 1, 4, 1, 1, 2, 3, 1, 4, 1, 1, 2, 3, 1, 4, 1, 1, 2, 3, 1, 4, 1, 1

Offset: 1

John W. Layman, Jul 25 2012

This sequence is periodic with period 1,1,2,3,1,4 of length 6.

It appears that for most choices of a(1), a(2), and divisor b^k (replacing 5^k), the resulting sequence is not periodic.

G. C. Greubel, Table of n, a(n) for n = 1..10000
B. Avila and T. Khovanova, Free Fibonacci Sequences, arXiv preprint arXiv:1403.4614, 2014 and J. Int. Seq. 17 (2014) # 14.8.5
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,1).

Cf. A010892, A049347, A078412, A078414, A132739.

I:=[1,1,2,3,1,4]; [n le 6 select I[n] else Self(n-6): n in [1..100]]; // G. C. Greubel, Mar 08 2024

CoefficientList[Series[(4*x^5 + x^4 + 3*x^3 + 2*x^2 + x + 1)/((1 - x)*(x + 1)*(x^2 - x + 1)*(x^2 + x + 1)), {x, 0, 100}], x] (* Wesley Ivan Hurt, Jul 08 2014 *)
LinearRecurrence[{0, 0, 0, 0, 0, 1},{1, 1, 2, 3, 1, 4},80] (* Ray Chandler, Aug 25 2015 *)

lista(nn) = {va = vector(nn); va[1] = 1; va[2] = 1; for (n=3, nn, sump = va[n-1] + va[n-2]; va[n] = sump/5^(valuation(sump, 5));); va;} \\ Michel Marcus, Jul 08 2014

Vec(-x*(4*x^5+x^4+3*x^3+2*x^2+x+1)/((x-1)*(x+1)*(x^2-x+1)*(x^2+x+1)) + O(x^100)) \\ Colin Barker, Jul 08 2014

def A214684_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( x*(1+x+2*x^2+3*x^3+x^4+4*x^5)/(1-x^6) ).list()
a=A214684_list(100); a[1:] # G. C. Greubel, Mar 08 2024

a(n) = A132739(a(n-1) + a(n-2)), for n>2, and a(1)=1, a(2)=1. - Michel Marcus, Jul 08 2014
G.f.: x*(1+x+2*x^2+3*x^3+x^4+4*x^5)/((1-x)*(1+x)*(1-x+x^2)*(1+x+x^2)) . - Colin Barker, Jul 08 2014
a(n) = -4*[n=0] + (1/3)*(2 + 2*(-1)^n + A010892(n) - 2*A010892(n-1) + 3*A049347(n) + 3*A049347(n-1)). - G. C. Greubel, Mar 08 2024

A224382 Fibonacci-like numbers without positive multiples of 4: a(0) = 0, a(1) = 1, for n>=2, a(n) = a(n-1) + a(n-2) divided by maximal possible power of 4.

Original entry on oeis.org

0, 1, 1, 2, 3, 5, 2, 7, 9, 1, 10, 11, 21, 2, 23, 25, 3, 7, 10, 17, 27, 11, 38, 49, 87, 34, 121, 155, 69, 14, 83, 97, 45, 142, 187, 329, 129, 458, 587, 1045, 102, 1147, 1249, 599, 462, 1061, 1523, 646, 2169, 2815, 1246, 4061, 5307, 2342, 7649, 9991, 4410

Offset: 0

Vladimir Shevelev, Apr 05 2013

nonn

Peter J. C. Moses, Table of n, a(n) for n = 0..9999
B. Avila and T. Khovanova, Free Fibonacci Sequences, arXiv preprint arXiv:1403.4614, 2014 and J. Int. Seq. 17 (2014) # 14.8.5

Cf. A000045, A078414, A214094, A214156, A216231, A216275, A216835.

a[0]:=0; a[1]:=1; a[n_]:=a[n]=#/4^IntegerExponent[#,4]&[(a[n-1]+a[n-2])]; Map[a,Range[0,99]] (* Peter J. C. Moses, Apr 05 2013 *)

A078412 a(0) = 5, a(1) = 8; for n >1, a(n)=(a(n-1)+a(n-2))/3^n, where 3^n is the highest power of 3 dividing a(n-1)+a(n-2).

Original entry on oeis.org

5, 8, 13, 7, 20, 1, 7, 8, 5, 13, 2, 5, 7, 4, 11, 5, 16, 7, 23, 10, 11, 7, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2

Offset: 0

Yasutoshi Kohmoto, Dec 28 2002

Index entries for linear recurrences with constant coefficients, signature (0,0,1).

Cf. A038502, A078414.

Join[{5, 8, 13, 7, 20, 1, 7, 8, 5, 13, 2, 5, 7, 4, 11, 5, 16, 7, 23, 10, 11, 7},LinearRecurrence[{0, 0, 1},{2, 1, 1},79]] (* Ray Chandler, Aug 25 2015 *)

a(3n-1)=2, a(3n)=1, a(3n+1)=1 for n>=8. - Sascha Kurz, Jan 04 2003

More terms from Sascha Kurz, Jan 04 2003

A224523 Let {f_n(k)} be the n-th sequence of Fibonacci-like numbers defined by recursion f_n(0) = 0, f_n(1) = 1 and, for k>=2, f_n(k) = f_n(k-1) + f_n(k-2) divided by maximal possible powers of primes >= prime(n). a(n) is length of the smallest period of {f_n(k)}, and a(n)=0, if {f_n(k)} is not eventually periodic.

Original entry on oeis.org

1, 3, 6, 9, 9, 12, 15, 27, 12, 12, 15, 15, 15, 15, 15, 15, 15, 42, 42, 42, 42, 42, 42, 90, 72, 36, 36, 36, 36, 36, 36, 36, 36, 36, 54, 54, 66, 102, 102, 102, 102, 102, 102, 102, 102, 36, 36, 36, 36, 36, 36, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24

Offset: 1

Vladimir Shevelev, Apr 09 2013

nonn

Conjecture. For n>=1, a(n)>0.

If Conjecture is true, then we have an answer in affirmative on question in A078414.

			Let n=9, prime(9)=23. Sequence {f_9(k)} begins 0,1,1,2,3,5,8,13,21,34,55. Now 34 + 55 = 89 is prime >=23, so the following terms are 89/89 = 1,56,57. Further, since 56 + 57 = 113 is prime >=23, then the following term is 113/113 = 1 and, since 57 + 1 = 58=29*2, then the following term is 58/29 = 2. Now we have period {1,2,3,5,8,13,21,34,55,1,56,57} with length 12. Thus a(9)=12.

Peter J. C. Moses, Table of n, a(n) for n = 1..4000

seqPosition[{list_,seqtofind_}]:=Position[Partition[list,Length[#],1],Flatten[{_,#,_}]]&[seqtofind];Table[Clear[a];a[0]:=0;a[1]:=1;a[n_]:=a[n]=#/(Product[Prime[i]^IntegerExponent[#,Prime[i]],{i,z,PrimePi[#]+1}])&[(a[n-1]+a[n-2])];NestWhile[#+2&,24,Length[diff=Flatten[seqPosition[{#,Take[#,-2]}]]&[Map[a,Range[0,#]]]]<=1&];(#[[2]]-#[[1]])&[diff],{z,1,50}] (* Peter J. C. Moses, Apr 10 2013 *)

For n >= 9 the terms were calculated by Peter J. C. Moses, Apr 09 2013

A232666 6-free Fibonacci numbers.

Original entry on oeis.org

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 4, 93, 97, 190, 287, 477, 764, 1241, 2005, 541, 2546, 3087, 5633, 8720, 14353, 23073, 37426, 60499, 97925, 26404, 124329, 150733, 275062, 425795, 700857, 1126652, 1827509, 2954161, 796945, 3751106, 4548051, 8299157, 12847208, 21146365, 33993573

Offset: 0

Brandon Avila and Tanya Khovanova, Nov 27 2013

nonn

The sequences of n-free Fibonacci numbers were suggested by John H. Conway.
a(n) is the sum of the two previous terms divided by the largest possible power of 6.
4-free Fibonacci numbers are A224382.
The sequence coincides with the Fibonacci sequence until the first multiple of 6 in the Fibonacci sequence: 144, which in this sequence is divided by 36 to produce 4.
7-free Fibonacci numbers is A078414.

B. Avila, T. Khovanova, Free Fibonacci Sequences, J. Int. Seq. 17 (2014) # 14.8.5.

Cf. A224382, A214684.

sixPower[n_] := (a = Transpose[FactorInteger[n]]; a2 = Position[a[[1]], 2]; a3 = Position[a[[1]], 3]; If[Length[a2] == 0 || Length[a3] == 0 , res = 0, res = Min[a[[2]][[a2[[1]][[1]]]], a[[2]][[a3[[1]][[1]]]]]]; res); sixFree[n_] := n/6^sixPower[n]; appendNext6Free[list_] := Append[list, sixFree[list[[-1]] + list[[-2]]]]; Nest[appendNext6Free, {0, 1}, 50]

A219328 Number of different prime divisors >= prime(n) of sums of two consecutive terms of sequence {f_n(k)} defined in A224523.

Original entry on oeis.org

1, 1, 1, 3, 3, 3, 4, 7, 3, 3, 3, 3, 3, 3, 3, 3, 3, 9, 9, 9, 9, 9, 9, 18, 14, 6, 6, 6, 6, 6, 6, 6, 6, 6, 9, 9, 12, 19, 19, 19, 19, 19, 19, 19, 19, 5, 5, 5, 5, 5, 5, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 6

Offset: 1

Vladimir Shevelev, Apr 11 2013

nonn

a(n) shows that it is sufficient to choose a(n) primes >= prime(n) such that Fibonacci-like sequence without multiples of these primes is periodic (see comment in A078414).

			1) In case n=4, the sequence {f_4(k)} has period {1,1,2,3,5,8,1,9,10}. We see that only sums of consecutive terms 5+8=13, 9+10=19, 10+1=11 have divisors >= prime(4)=7. Thus {f_4(k)} is the Fibonacci-like sequence without multiples of 11,13,19. So a(4)=3.
2) In cases 52 <= n <= 120, prime(n) >= prime(52) = 239, every sequence {f_n(k)} has period {1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1, 988, 989, 3, 992, 995, 1, 996}. It is Fibonacci-like sequence without multiples of 659, 997, 1597, or 1987. Since 659 = prime(120), then in the considered interval every a(n)=4.

Peter J. C. Moses, Table of n, a(n) for n = 1..4000

Cf. A078414, A078412, A214684, A219255, A224382, A224523.

Corrections and terms beginning a(37) were calculated by Peter J. C. Moses, Apr 19 2013

A219255 Numbers n for which prime(n) divides at least one sum of two consecutive terms in sequence {f_m(k)}, defined in A224523.

Original entry on oeis.org

1, 2, 3, 5, 6, 7, 8, 10, 17, 23, 24, 25, 34, 36, 37, 45, 51, 120, 124, 219, 231, 244, 251, 2034, 2057, 3121, 4176, 5185, 9492

Offset: 1

Vladimir Shevelev, Apr 11 2013

Places of the last elements of runs of the same terms in A224523 are in the sequence. Whether these sequences coincide one to another?

			7 is in the sequence, since prime(7)=17 and sequence {f_7(k)} is periodic with period {1,1,2,3,5,8,13,21,2,1,3,4,7,11,18}, such that sum of terms 13+21=34 is divisible by 17.

Cf. A078414, A078412, A214684, A219328, A224523.

Rest[Flatten[Position[Differences[A224523], Except[0]]]] (* Peter J. C. Moses, Apr 13 2013 *)

cp's OEIS Frontend

A214684 a(1)=1, a(2)=1, and, for n>2, a(n)=(a(n-1)+a(n-2))/5^k, where 5^k is the highest power of 5 dividing a(n-1)+a(n-2).

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A224382 Fibonacci-like numbers without positive multiples of 4: a(0) = 0, a(1) = 1, for n>=2, a(n) = a(n-1) + a(n-2) divided by maximal possible power of 4.

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A078412 a(0) = 5, a(1) = 8; for n >1, a(n)=(a(n-1)+a(n-2))/3^n, where 3^n is the highest power of 3 dividing a(n-1)+a(n-2).

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A232666 6-free Fibonacci numbers.

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A219328 Number of different prime divisors >= prime(n) of sums of two consecutive terms of sequence {f_n(k)} defined in A224523.

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A219255 Numbers n for which prime(n) divides at least one sum of two consecutive terms in sequence {f_m(k)}, defined in A224523.

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