cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-6 of 6 results.

A235265 Primes whose base-3 representation also is the base-2 representation of a prime.

Original entry on oeis.org

3, 13, 31, 37, 271, 283, 733, 757, 769, 1009, 1093, 2281, 2467, 2521, 2551, 2917, 3001, 3037, 3163, 3169, 3187, 3271, 6673, 7321, 7573, 9001, 9103, 9733, 19801, 19963, 20011, 20443, 20521, 20533, 20749, 21871, 21961, 22123, 22639, 22717, 27253, 28711, 28759, 29173, 29191, 59077, 61483, 61507, 61561, 65701, 65881
Offset: 1

Views

Author

M. F. Hasler, Jan 05 2014

Keywords

Comments

This sequence and A235383 and A229037 are winners in the contest held at the 2014 AMS/MAA Joint Mathematics Meetings. - T. D. Noe, Jan 20 2014
This sequence was motivated by work initiated by V.J. Pohjola's post to the SeqFan list, which led to a clarification of the definition and correction of some errors, in sequences A089971, A089981 and A090707 through A090721. These sequences use "rebasing" (terminology of A065361) from some base b to base 10. Sequences A065720 - A065727 follow the same idea but use rebasing in the other sense, from base 10 to base b. The observation that only (10,b) and (b,10) had been considered so far led to the definition of this and related sequences: In a systematic approach, it seems natural to start with the smallest possible pairs of different bases, (2,3) and (3,2), then (2 <-> 4), (3 <-> 4), (2 <-> 5), etc.
Among the two possibilities using the smallest possible bases, 2 and 3, the present one seems a little bit more interesting, among others because not every base-3 representation is a valid base-2 representation (in contrast to the opposite case). This is also a reason why the present sequence grows much faster than the partner sequence A235266.

Examples

			3 = 10_3 and 10_2 = 2 is prime. 13 = 111_3 and 111_2 = 7 is prime.
		

Crossrefs

Subset of A077717.
Cf. A235266, A065720 and A036952, A065721 - A065727, A235394, A235395, A089971 and A020449, A089981, A090707 - A091924, A235461 - A235482. See M. F. Hasler's OEIS wiki page for further cross-references.

Programs

  • Maple
    N:= 1000: # to get the first N terms
    count:= 0:
    for i from 1 while count < N do
       p2:= ithprime(i);
       L:= convert(p2,base,2);
       p3:= add(3^(j-1)*L[j],j=1..nops(L));
       if isprime(p3) then
          count:= count+1;
          A235265[count]:= p3;
       fi
    od:
    [seq(A235265[i], i=1..N)]; # Robert Israel, May 04 2014
  • Mathematica
    b32pQ[n_]:=Module[{idn3=IntegerDigits[n,3]},Max[idn3]<2&&PrimeQ[ FromDigits[ idn3,2]]]; Select[Prime[Range[7000]],b32pQ] (* Harvey P. Dale, Apr 24 2015 *)
  • PARI
    is(p,b=2,c=3)=vecmax(d=digits(p,c))
    				
  • Python
    from sympy import isprime, nextprime
    def agen(): # generator of terms
        p = 2
        while True:
            p3 = sum(3**i for i, bi in enumerate(bin(p)[2:][::-1]) if bi=='1')
            if isprime(p3):
                yield p3
            p = nextprime(p)
    g = agen()
    print([next(g) for n in range(1, 52)]) # Michael S. Branicky, Jan 16 2022

A229037 The "forest fire": sequence of positive integers where each is chosen to be as small as possible subject to the condition that no three terms a(j), a(j+k), a(j+2k) (for any j and k) form an arithmetic progression.

Original entry on oeis.org

1, 1, 2, 1, 1, 2, 2, 4, 4, 1, 1, 2, 1, 1, 2, 2, 4, 4, 2, 4, 4, 5, 5, 8, 5, 5, 9, 1, 1, 2, 1, 1, 2, 2, 4, 4, 1, 1, 2, 1, 1, 2, 2, 4, 4, 2, 4, 4, 5, 5, 8, 5, 5, 9, 9, 4, 4, 5, 5, 10, 5, 5, 10, 2, 10, 13, 11, 10, 8, 11, 13, 10, 12, 10, 10, 12, 10, 11, 14, 20, 13
Offset: 1

Views

Author

Jack W Grahl, Sep 11 2013

Keywords

Comments

Added name "forest fire" to make it easier to locate this sequence. - N. J. A. Sloane, Sep 03 2019
This sequence and A235383 and A235265 were winners in the best new sequence contest held at the OEIS Foundation booth at the 2014 AMS/MAA Joint Mathematics Meetings. - T. D. Noe, Jan 20 2014
See A236246 for indices n such that a(n)=1. - M. F. Hasler, Jan 20 2014
See A241673 for indices n such that a(n)=2^k. - Reinhard Zumkeller, Apr 26 2014
The graph (for up to n = 10000) has an eerie similarity (why?) to the distribution of rising smoke particles subjected to a lateral wind, and where the particles emanate from randomly distributed burning areas in a fire in a forest or field. - Daniel Forgues, Jan 21 2014
The graph (up to n = 100000) appears to have a fractal structure. The dense areas are not random but seem to repeat, approximately doubling in width and height each time. - Daniel Forgues, Jan 21 2014
a(A241752(n)) = n and a(m) != n for m < A241752(n). - Reinhard Zumkeller, Apr 28 2014
The indices n such that a(n) = 1 are given by A236313 (relative spacing) up to 19 terms, and A003278 (directly) up to 20 terms. If just placing ones, the 21st 1 would be n=91. The sequence A003278 fails at n=91 because the numbers filling the gaps create an arithmetic progression with a(27)=9, a(59)=5 and a(91)=1. Additionally, if you look at indices n starting at the first instance of 4 or 5, A003278/A236313 provide possible indices for a(n)=4 or a(n)=5, with some indexes instead filled by numbers < (4,5). A003278/A236313 also fail to predict indices for a(n)=4 or a(n)=5 by the ~20th term. - Daniel Putt, Sep 29 2022

Crossrefs

Cf. A094870, A362942 (partial sums).
For a variant see A309890.
A selection of sequences related to "no three-term arithmetic progression": A003002, A003003, A003278, A004793, A005047, A005487, A033157, A065825, A092482, A093678, A093679, A093680, A093681, A093682, A094870, A101884, A101886, A101888, A140577, A185256, A208746, A229037.

Programs

  • Haskell
    import Data.IntMap (empty, (!), insert)
    a229037 n = a229037_list !! (n-1)
    a229037_list = f 0 empty  where
       f i m = y : f (i + 1) (insert (i + 1) y m) where
         y = head [z | z <- [1..],
                       all (\k -> z + m ! (i - k) /= 2 * m ! (i - k `div` 2))
                           [1, 3 .. i - 1]]
    -- Reinhard Zumkeller, Apr 26 2014
    
  • Mathematica
    a[1] = 1; a[n_] := a[n] = Block[{z = 1}, While[Catch[ Do[If[z == 2*a[n-k] - a[n-2*k], Throw@True], {k, Floor[(n-1)/2]}]; False], z++]; z]; a /@ Range[100] (* Giovanni Resta, Jan 01 2014 *)
  • PARI
    step(v)=my(bad=List(),n=#v+1,t); for(d=1,#v\2,t=2*v[n-d]-v[n-2*d]; if(t>0, listput(bad,t))); bad=Set(bad); for(i=1,#bad, if(bad[i]!=i, return(i))); #bad+1
    first(n)=my(v=List([1])); while(n--, listput(v, step(v))); Vec(v) \\ Charles R Greathouse IV, Jan 21 2014
    
  • Python
    A229037_list = []
    for n in range(10**6):
        i, j, b = 1, 1, set()
        while n-2*i >= 0:
            b.add(2*A229037_list[n-i]-A229037_list[n-2*i])
            i += 1
            while j in b:
                b.remove(j)
                j += 1
        A229037_list.append(j) # Chai Wah Wu, Dec 21 2014

Formula

a(n) <= (n+1)/2. - Charles R Greathouse IV, Jan 21 2014

A065108 Positive numbers expressible as a product of Fibonacci numbers.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 13, 15, 16, 18, 20, 21, 24, 25, 26, 27, 30, 32, 34, 36, 39, 40, 42, 45, 48, 50, 52, 54, 55, 60, 63, 64, 65, 68, 72, 75, 78, 80, 81, 84, 89, 90, 96, 100, 102, 104, 105, 108, 110, 117, 120, 125, 126, 128, 130, 135, 136, 144, 150, 156, 160, 162
Offset: 1

Views

Author

Joseph L. Pe, Nov 21 2001

Keywords

Comments

There are infinitely many triples of consecutive terms of this sequence that are consecutive integers, see A065885. - John W. Layman, Nov 27 2001
Carmichael's theorem implies that 8 and 144 are the only Fibonacci numbers that are products of other Fibonacci numbers, cf. A235383. - Robert C. Lyons, Jan 13 2013

Examples

			52 = 2 * 2 * 13 is the product of Fibonacci numbers 2, 2 and 13.
		

Crossrefs

Cf. A000045, A065885. Complement of A065105.
Cf. A049997 and A094563: F(i)*F(j) and F(i)*F(j)*F(k) respectively.
Subsequence of A178772.

Programs

  • Maple
    with(combinat): A000045:=proc(n) options remember: RETURN(fibonacci(n)): end: mulfib:=proc(m,i) local j,q,f: f:=0: for j from i by -1 to 3 while(f=0) do if(irem(m, A000045(j))=0) then q:=iquo(m, A000045(j)): if(q=1) then RETURN(1) else f:=mulfib(q,j) fi fi od: RETURN(f): end: for i from 3 to 12 do for n from A000045(i) to A000045(i+1)-1 do m:=mulfib(n,i): if m=1 then printf("%d, ",n) fi od od: # C. Ronaldo
  • Mathematica
    nn = 1000; k = 1; fib = {}; While[k++; f = Fibonacci[k]; f <= nn, AppendTo[fib, f]]; s = fib; While[s2 = Select[Union[s, Flatten[Outer[Times, fib, s]]], # <= nn &]; Length[s2] > Length[s], s = s2]; s (* T. D. Noe, Jul 17 2012 *)
  • PARI
    list(lim)=if(lim<7, return([1..lim\1])); my(v=List([1]), F=List([2,3]), curfib, t, idx, newidx); while((t=F[#F]+F[#F-1])<=lim, listput(F,t)); F=setminus(Set(F), [8,144]); for(i=1,#F, curfib=F[i]; idx=1; while(v[idx]*curfib<=lim, newidx=#v+1; for(j=idx,#v, t=curfib*v[j]; if(t<=lim, listput(v,t))); idx=newidx)); Set(v) \\ Charles R Greathouse IV, Jun 15 2017

Formula

As Charles R Greathouse IV recently remarked, it would be good to have an asymptotic formula for this sequence. - N. J. A. Sloane, Jul 22 2012

Extensions

More terms from John W. Layman, Nov 27 2001
More terms from C. Ronaldo (aga_new_ac(AT)hotmail.com), Jan 02 2005

A227875 Fibonacci numbers which are perfect powers.

Original entry on oeis.org

0, 1, 8, 144
Offset: 1

Views

Author

Jean-François Alcover, Oct 25 2013

Keywords

Comments

Also, Fibonacci numbers which are products of Fibonacci numbers (each greater than 1 when the product is greater than 1 - see A235383). - Rick L. Shepherd, Feb 19 2014
The terms of the subsequence (1, 8, 144) are the Fibonacci numbers that are powerful numbers. - Robert C. Lyons, Jul 12 2016
Also Fibonacci numbers without any primitive divisors. See [Heuberger & Wagner]. - Michel Marcus, Aug 21 2016
It was proved (Bugeaud, Mignotte, and Siksek, 2006, p. 971) that the only perfect powers among the Fibonacci numbers and Lucas numbers are {0, 1, 8, 144} and {1, 4}, respectively. - Daniel Forgues, Apr 09 2018

Crossrefs

Programs

  • Mathematica
    perfectPowerQ[0] = True; perfectPowerQ[1] = True; perfectPowerQ[n_] := GCD @@ FactorInteger[n][[All, 2]] > 1; Union[Select[Fibonacci /@ Range[0, 20], perfectPowerQ]]

A360772 List of distinct numbers that are powers of odd-indexed Fibonacci numbers or even powers of nonzero even-indexed Fibonacci numbers.

Original entry on oeis.org

1, 2, 4, 5, 8, 9, 13, 16, 25, 32, 34, 64, 81, 89, 125, 128, 169, 233, 256, 441, 512, 610, 625, 729, 1024, 1156, 1597, 2048, 2197, 3025, 3125, 4096, 4181, 6561, 7921, 8192, 10946, 15625, 16384, 20736, 28561, 28657, 32768, 39304, 54289, 59049, 65536, 75025, 78125
Offset: 1

Views

Author

Amiram Eldar, Feb 20 2023

Keywords

Comments

Ohtsuka's (2023) problem does not include 1, and includes the even powers of 8 twice (once as powers of Fibonacci(6) = 8 and once as powers of Fibonacci(3) = 2). The sum of reciprocals in this case is (61 - 15*sqrt(5))/18.

Crossrefs

Programs

  • Mathematica
    seq[max_] := Module[{s = {1}, k = 3, f, d}, While[(f = Fibonacci[k]) <= max, If[k != 6, d = 2 - Mod[k, 2]; s = Join[s, f^Range[d, Floor[Log[f, max]], d]]]; k++]; Sort[s]]; seq[10^5]

Formula

Sum_{n>=1} 1/a(n) = 551/126 - 5*sqrt(5)/6.

A286948 a(n) is the number of distinct products p of Fibonacci numbers such that Fibonacci(n) < p <= Fibonacci(n + 1).

Original entry on oeis.org

0, 1, 1, 2, 2, 4, 5, 7, 10, 12, 17, 21, 29, 36, 48, 60, 78, 96, 124, 151, 190, 234, 290, 354, 436, 529, 648, 784, 952, 1141, 1382, 1651, 1984, 2367, 2831, 3359, 3999, 4733, 5608, 6614, 7816, 9178, 10802, 12667, 14850, 17356, 20297, 23653, 27579, 32062, 37277, 43235, 50139
Offset: 1

Views

Author

David A. Corneth, Jun 11 2017

Keywords

Comments

The products of Fibonacci numbers are not all distinct. For example, 144 is a product of Fibonacci numbers in more than 1 way but 144 still counts as one product. 8 and 144 are the only Fibonacci numbers with this property (see A235383).
a(n+1)/a(n) seems to converge to about 1.1; up to n = 70, the value drops slightly. Maybe to sqrt(5)/2?
Fibonacci(n) ~= phi^n / sqrt(5) where phi = (1 + sqrt(5)) / 2. If m is a product of k Fibonacci numbers, m is of the form Fibonacci(n_1) *...* Fibonacci(n_k). To count the numbers just once, we restrict n_i for 1 <= i <= k.
Fibonacci(1) = Fibonacci(2) = 1 isn't counted, products with factor Fibonacci(6) = 8 aren't counted and products with the factor Fibonacci(12) = 144 aren't counted. I.e., n_i >= 3, n_i != 6 and n_i != 12.
We can write m uniquely as m = Product_{i=1..k} Fibonacci(n_i) ~= Product_{i=1..k} (phi^(n_i) / sqrt(5)) = phi^(Sum_{i=1..k} n_i) / sqrt(5)^k. To determine the number of such products up to f = Fibonacci(x) of k such Fibonacci factors, we can find an upper bound on Sum_{i=1..k} n_i of about (k*log(5)/2 + log(x)) / log(phi). This somewhat relates this sequence to the partitions.

Examples

			The products of Fibonacci numbers larger than Fibonacci(7) = 13 and smaller than or equal to Fibonacci(7 + 1) = 21 are the five numbers 15, 16, 18, 20 and 21. Therefore a(7) = 5.
		

Crossrefs

Showing 1-6 of 6 results.