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 1-5 of 5 results.

A112310 Number of terms in lazy Fibonacci representation of n.

Original entry on oeis.org

0, 1, 1, 2, 2, 2, 3, 2, 3, 3, 3, 4, 3, 3, 4, 3, 4, 4, 4, 5, 3, 4, 4, 4, 5, 4, 4, 5, 4, 5, 5, 5, 6, 4, 4, 5, 4, 5, 5, 5, 6, 4, 5, 5, 5, 6, 5, 5, 6, 5, 6, 6, 6, 7, 4, 5, 5, 5, 6, 5, 5, 6, 5, 6, 6, 6, 7, 5, 5, 6, 5, 6, 6, 6, 7, 5, 6, 6, 6, 7, 6, 6, 7, 6, 7, 7, 7, 8, 5, 5, 6, 5, 6, 6, 6, 7, 5, 6, 6, 6, 7, 6, 6, 7, 6
Offset: 0

Views

Author

N. J. A. Sloane, Dec 01 2005

Keywords

Comments

Equivalently, the number of ones in the maximal Fibonacci bit-representation (A104326) of n.
Conjecture: if we split the sequence in groups that contain Fibonacci(k) terms like (0), (1), (1, 2), (2, 2, 3), (2, 3, 3, 3, 4), (3, 3, 4, 3, 4, 4, 4, 5) etc, the sums in the groups are the terms of A023610. - Gary W. Adamson, Nov 02 2010
Equivalently, the number of periods in the length-n prefix of the infinite Fibonacci word (A003849). An integer p, 1 <= p <= n, is a period of a length-n word x if x[i] = x[i+p] for 1 <= i <= n-p. - Jeffrey Shallit, May 23 2020

Examples

			a(10) = 3 because A104326(10) = 1110 contains three ones.
		

Crossrefs

Number of terms in row n of A112309.
Record positions are in A001911. - Ray Chandler, Dec 01 2005

Programs

  • Haskell
    a112310 n = a112310_list !! n
    a112310_list = concat fss where
       fss = [0] : [1] : (map (map (+ 1))) (zipWith (++) fss $ tail fss)
    -- Reinhard Zumkeller, Oct 26 2013
  • Maple
    A112310 := proc(n)
        convert(A104326(n),base,10) ;
        add(d,d=%) ;
    end proc:
    seq(A112310(n),n=0..120) ; # R. J. Mathar, Aug 28 2025
  • Mathematica
    DeleteCases[IntegerDigits[Range[200], 2], {_, 0, 0, _}]
    A112309 = Map[DeleteCases[Reverse[#] Fibonacci[Range[Length[#]] + 1], 0] &, DeleteCases[IntegerDigits[-1 + Range[200], 2], {_, 0, 0, _}]]
    A112310 = Map[Length, A112309]
    (* Peter J. C. Moses, Mar 03 2015 *)

Formula

a(n) = A007953(A104326(n)). - Amiram Eldar, Oct 10 2023

Extensions

Extended by Ray Chandler, Dec 01 2005
Merged with a sequence from Casey Mongoven, Mar 20 2006, by Franklin T. Adams-Watters, Dec 19 2006

A095791 Number of digits in lazy-Fibonacci-binary representation of n (A104326).

Original entry on oeis.org

1, 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9
Offset: 0

Views

Author

Clark Kimberling, Jun 05 2004

Keywords

Comments

The lazy Fibonacci representation of n >= 0 is obtained by replacing every string of 0's in the binary representation of n by a single 0, thus obtaining a finite zero-one sequence (d(2), d(3), d(4), ..., d(k)), and then forming d(2)*F(2) + d(3)*F(3) + ... + d(k)*F(k), as in the Mathematica program. The lazy Fibonacci representation is often called the maximal Fibonacci representation, in contrast to the Zeckendorf representation, also called the minimal Fibonacci representation. - Clark Kimberling, Mar 04 2015
Regarding the References, the lazy Fibonacci representation is sometimes attributed to Erdős and Joo, but it is also found in Brown and Ferns. - Clark Kimberling, Mar 04 2015

Examples

			The lazy Fibonacci representation of 14 is 8+3+2+1, which in binary notation is 10111, which consists of 5 digits.
		

Crossrefs

Programs

  • Mathematica
    t=DeleteCases[IntegerDigits[-1+Range[200],2],{_,0,0,_}];
    A181632=Flatten[t]
    A095791=Map[Length,t]
    A112309=Map[DeleteCases[Reverse[#] Fibonacci[Range[Length[#]]+1],0]&,t]
    A112310=Map[Length,A112309]
    (* Peter J. C. Moses, Mar 03 2015 *)
  • PARI
    a(n)=if(n<2,1,a(floor(n*(-1+sqrt(5))/2))+1) \\ Benoit Cloitre, Dec 17 2006
    
  • PARI
    a(n)=if(n<0,0,c=1;s=n;while(floor(s*2/(1+sqrt(5)))>0,c++;s=floor(s*2/(1+sqrt(5))));c) \\ Benoit Cloitre, May 24 2007

Formula

1, 1, then F(3) 2's, then F(4) 3's, then F(5) 4's, ..., then F(k+1) k's, ...
a(0)=a(1)=1 then a(n) = a(floor(n/tau))+1 where tau=(1+sqrt(5))/2. - Benoit Cloitre, Dec 17 2006
a(n) is the least k such that f^(k)(n)=0 where f^(k+1)(x)=f(f^(k)(x)) and f(x)=floor(x/phi) where phi=(1+sqrt(5))/2 (see PARI/GP program). - Benoit Cloitre, May 24 2007
a(n) = A070939(A104326(n)). - Amiram Eldar, Oct 10 2023

A356771 a(n) is the sum of the Fibonacci numbers in common in the Zeckendorf and dual Zeckendorf representations of n.

Original entry on oeis.org

0, 1, 2, 0, 4, 0, 1, 7, 0, 1, 2, 3, 12, 0, 1, 2, 0, 4, 5, 6, 20, 0, 1, 2, 3, 4, 0, 1, 7, 8, 9, 10, 11, 33, 0, 1, 2, 0, 4, 5, 6, 7, 0, 1, 2, 3, 12, 13, 14, 15, 13, 17, 18, 19, 54, 0, 1, 2, 3, 4, 0, 1, 7, 8, 9, 10, 11, 12, 0, 1, 2, 0, 4, 5, 6, 20, 21, 22, 23, 24
Offset: 0

Views

Author

Rémy Sigrist, Aug 27 2022

Keywords

Comments

The Zeckendorf and dual Zeckendorf representations both express a number n as a sum of distinct positive Fibonacci numbers; these distinct Fibonacci numbers can be encoded in binary (see A022290 for the decoding function):
- in the Zeckendorf representation (or greedy Fibonacci representation):
- Fibonacci numbers are as big as possible (see A035517),
- and the corresponding binary encoding, A003714(n),
cannot have two consecutive 1's;
- in the dual Zeckendorf representation (or lazy Fibonacci representation):
- Fibonacci numbers are as small as possible (see A112309),
- and the corresponding binary encoding, A003754(n+1),
cannot have two consecutive nonleading 0's.
See A356326 for a similar sequence.

Examples

			For n = 28:
- using F(k) = A000045(k),
- the Zeckendorf representation of 28 is F(8) + F(5) + F(3),
- the dual Zeckendorf representation of 28 is F(7) + F(6) + F(5) + F(3),
- F(5) and F(3) appear in both representations,
- so a(28) = F(5) + F(3) = 7.
		

Crossrefs

Programs

  • PARI
    \\ See Links section.

Formula

a(n) = A022290(A003714(n) AND A003754(n+1)) (where AND denotes the bitwise AND operator).
a(n) = 0 iff n belongs to A331467.
a(n) = n iff n belongs to A000071.

A375272 The number of factors of n of the form p^Fibonacci(k), where p is a prime and k >= 2, when the factorization is uniquely done using the dual Zeckendorf representation of the exponents in the prime factorization of n.

Original entry on oeis.org

0, 1, 1, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 2, 2, 1, 2, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 2, 2, 1, 2, 2, 3, 1, 3, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 3, 2, 3, 2, 2, 1, 3, 1, 2, 2, 3, 2, 3, 1, 2, 2, 3, 1, 3, 1, 2, 2, 2, 2, 3, 1, 3, 2, 2, 1, 3, 2, 2, 2
Offset: 1

Views

Author

Amiram Eldar, Aug 09 2024

Keywords

Comments

First differs from A086435 at n = 36. Differs from A266226 at n = 1, 36, ... .
The number of dual-Zeckendorf-infinitary divisors of n (defined in A331109) that are prime powers (A246655).
a(n) depends only on the prime signature of n.
Analogous to A064547 (binary representation) and A318464 (Zeckendorf representation).

Examples

			For n = 8 = 2^3, the dual Zeckendorf representation of 3 is 11, i.e., 3 = Fibonacci(2) + Fibonacci(3) = 1 + 2. Therefore 8 = 2^(1+2) = 2^1 * 2^2, and a(8) = 2.
For n = 256 = 2^8, the dual Zeckendorf representation of 8 is 1011, i.e., 8 = Fibonacci(2) + Fibonacci(3) + Fibonacci(5) = 1 + 2 + 5. Therefore 256 = 2^(1+2+5) = 2^1 * 2^2 * 2^5, and a(256) = 3.
		

Crossrefs

Programs

  • Mathematica
    toDualZeck[n_] := Module[{s = 0, v = 0, i = 0, f}, While[s < n, s += Fibonacci[i + 2]; v += 2^i; i++]; i--; While[i >= 0, f = Fibonacci[i + 2]; If[s - f >= n, s -= f; v -= 2^i]; i--]; v]; (* A003754, after Rémy Sigrist's PARI code in A112309 *)
    f[p_, e_] := DigitCount[toDualZeck[e], 2, 1]; a[1] = 0; a[n_] := Plus @@ f @@@ FactorInteger[n]; Array[a, 100]
  • PARI
    todualzeck(n) = {my (s=0, v=0); for (i=0, oo, if (s>=n, forstep (j=i-1, 0, -1, if (s-fibonacci(2+j)>=n, s-=fibonacci(2+j); v-=2^j;);); return (v);); s+=fibonacci(2+i); v+=2^i;);} \\ A003754, Rémy Sigrist's code in A112309
    a(n) = vecsum(apply(x -> hammingweight(todualzeck(x)), factor(n)[, 2]));

Formula

Additive with a(p^e) = A112310(e).
a(n) = log_2(A331109(n)).
Sum_{k=1..n} a(k) ~ n * (log(log(n)) + B + C), where B is Mertens's constant (A077761), C = Sum_{k>=2} (A112310(k)-A112310(k-1)) * P(k) = 0.18790467121403662496..., and P(s) is the prime zeta function.

A357121 Irregular triangle T(n, k), n > 0, k = 1..A352104(n); the n-th row contains, in ascending order, the terms in the lazy tribonacci representation of n.

Original entry on oeis.org

1, 2, 1, 2, 4, 1, 4, 2, 4, 1, 2, 4, 1, 7, 2, 7, 1, 2, 7, 4, 7, 1, 4, 7, 2, 4, 7, 1, 2, 4, 7, 2, 13, 1, 2, 13, 4, 13, 1, 4, 13, 2, 4, 13, 1, 2, 4, 13, 1, 7, 13, 2, 7, 13, 1, 2, 7, 13, 4, 7, 13, 1, 4, 7, 13, 2, 4, 7, 13, 1, 2, 4, 7, 13, 4, 24, 1, 4, 24, 2, 4, 24
Offset: 1

Views

Author

Rémy Sigrist, Sep 12 2022

Keywords

Comments

See A357120 for the sequence corresponding to greedy tribonacci representations.

Examples

			Triangle T(n, k) begins:
     1: [1]
     2: [2]
     3: [1, 2]
     4: [4]
     5: [1, 4]
     6: [2, 4]
     7: [1, 2, 4]
     8: [1, 7]
     9: [2, 7]
    10: [1, 2, 7]
    11: [4, 7]
    12: [1, 4, 7]
    13: [2, 4, 7]
    14: [1, 2, 4, 7]
    15: [2, 13]
		

Crossrefs

Programs

  • PARI
    See Links section.

Formula

T(n, 1) = 2^A080843(n-1).
Showing 1-5 of 5 results.