A078412 -id:A078412 - cp's OEIS frontend

A078414 a(n) = (a(n-1)+a(n-2))/7^k, where 7^k is the highest power of 7 dividing a(n-1)+a(n-2).

1, 1, 2, 3, 5, 8, 13, 3, 16, 19, 5, 24, 29, 53, 82, 135, 31, 166, 197, 363, 80, 443, 523, 138, 661, 799, 1460, 2259, 3719, 122, 3841, 3963, 7804, 1681, 1355, 3036, 4391, 1061, 5452, 6513, 11965, 18478, 4349, 3261, 7610, 1553, 187, 1740, 1927, 3667, 5594, 27

Offset: 1

Yasutoshi Kohmoto, Dec 28 2002

From Vladimir Shevelev, Apr 01 2013; edited by Danny Rorabaugh, Feb 19 2016: (Start)
If we consider Fibonacci-like numbers {F_p(n)} without positive multiples of p, where p is a fixed prime, then {F_2(n)} has period of length 1, {F_3(n)} has period of length 3, {F_5(n)} has period of length 6. This sequence is the first which does not have a trivial period and, probably, even is non-periodic.
An open question: Is this sequence bounded?
Consider Fibonacci-like sequences without multiples of several primes, defined analogously: e.g., for {F_(p,q)(n)}, a(0)=0, a(1)=1, for n>=2, a(n)=a(n-1)+a(n-2) divided by the maximal possible powers of p and q.
Problem: For what sets of primes is the corresponding Fibonacci-like sequence without multiples of these primes periodic?
Examples: sequence {F_(7,11,13)(n)} has period of length 12: 0, 1, 1, 2, 3, 5, 8, 1, 9, 10, 19, 29, 48, 1, 1, 2, 3, 5,...; {F_(11,13,19)(n)} has period of length 9; {F_(13,19,23)(n)} has period of length 12; {F_(17,19,23,29)(n)} has period of length 15; {F_(19,23,31,53,59,89)(n)} has period of length 24; {F_(23,29,73,233)(n)} has period of length 18.
Don Reble noted that lengths of all such periods could only be multiples of 3 because every Fibonacci-like sequence considered here modulo 2 has the form 0,1,1,0,1,1,... .
(End)

Alois P. Heinz, Table of n, a(n) for n = 1..4000
B. Avila, T. Khovanova, Free Fibonacci Sequences, arXiv preprint arXiv:1403.4614 [math.NT], 2014 and J. Int. Seq. 17 (2014) # 14.8.5.

Cf. A000045, A078412, A214094, A214156, A216231, A216275, A216835.

a:= proc(n) option remember; local t, j;
      if n<3 then 1
    else t:= a(n-1)+a(n-2);
         while irem(t, 7, 'j')=0 do t:=j od; t
      fi
    end:
seq(a(n), n=1..100);  # Alois P. Heinz, Jul 25 2012

nxt[{a_,b_}]:=Module[{n=IntegerExponent[a+b,7]},{b,(a+b)/7^n}]; Transpose[ NestList[nxt,{1,1},60]][[1]] (* Harvey P. Dale, Jul 23 2012 *)

a(n) = A242603(a(n-1)+a(n-2)). - R. J. Mathar, Mar 13 2024

Corrected by Harvey P. Dale, Jul 23 2012

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).

Original entry on oeis.org

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

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 *)

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A078414 a(n) = (a(n-1)+a(n-2))/7^k, where 7^k is the highest power of 7 dividing a(n-1)+a(n-2).

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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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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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Extensions

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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Author

Keywords

Comments

Examples

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Programs

Mathematica