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.

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A000120 1's-counting sequence: number of 1's in binary expansion of n (or the binary weight of n).

Original entry on oeis.org

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

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The binary weight of n is also called Hamming weight of n. [The term "Hamming weight" was named after the American mathematician Richard Wesley Hamming (1915-1998). - Amiram Eldar, Jun 16 2021]
a(n) is also the largest integer such that 2^a(n) divides binomial(2n, n) = A000984(n). - Benoit Cloitre, Mar 27 2002
To construct the sequence, start with 0 and use the rule: If k >= 0 and a(0), a(1), ..., a(2^k-1) are the first 2^k terms, then the next 2^k terms are a(0) + 1, a(1) + 1, ..., a(2^k-1) + 1. - Benoit Cloitre, Jan 30 2003
An example of a fractal sequence. That is, if you omit every other number in the sequence, you get the original sequence. And of course this can be repeated. So if you form the sequence a(0 * 2^n), a(1 * 2^n), a(2 * 2^n), a(3 * 2^n), ... (for any integer n > 0), you get the original sequence. - Christopher.Hills(AT)sepura.co.uk, May 14 2003
The n-th row of Pascal's triangle has 2^k distinct odd binomial coefficients where k = a(n) - 1. - Lekraj Beedassy, May 15 2003
Fixed point of the morphism 0 -> 01, 1 -> 12, 2 -> 23, 3 -> 34, 4 -> 45, etc., starting from a(0) = 0. - Robert G. Wilson v, Jan 24 2006
a(n) is the number of times n appears among the mystery calculator sequences: A005408, A042964, A047566, A115419, A115420, A115421. - Jeremy Gardiner, Jan 25 2006
a(n) is the number of solutions of the Diophantine equation 2^m*k + 2^(m-1) + i = n, where m >= 1, k >= 0, 0 <= i < 2^(m-1); a(5) = 2 because only (m, k, i) = (1, 2, 0) [2^1*2 + 2^0 + 0 = 5] and (m, k, i) = (3, 0, 1) [2^3*0 + 2^2 + 1 = 5] are solutions. - Hieronymus Fischer, Jan 31 2006
The first appearance of k, k >= 0, is at a(2^k-1). - Robert G. Wilson v, Jul 27 2006
Sequence is given by T^(infinity)(0) where T is the operator transforming any word w = w(1)w(2)...w(m) into T(w) = w(1)(w(1)+1)w(2)(w(2)+1)...w(m)(w(m)+1). I.e., T(0) = 01, T(01) = 0112, T(0112) = 01121223. - Benoit Cloitre, Mar 04 2009
For n >= 2, the minimal k for which a(k(2^n-1)) is not multiple of n is 2^n + 3. - Vladimir Shevelev, Jun 05 2009
Triangle inequality: a(k+m) <= a(k) + a(m). Equality holds if and only if C(k+m, m) is odd. - Vladimir Shevelev, Jul 19 2009
a(k*m) <= a(k) * a(m). - Robert Israel, Sep 03 2023
The number of occurrences of value k in the first 2^n terms of the sequence is equal to binomial(n, k), and also equal to the sum of the first n - k + 1 terms of column k in the array A071919. Example with k = 2, n = 7: there are 21 = binomial(7,2) = 1 + 2 + 3 + 4 + 5 + 6 2's in a(0) to a(2^7-1). - Brent Spillner (spillner(AT)acm.org), Sep 01 2010, simplified by R. J. Mathar, Jan 13 2017
Let m be the number of parts in the listing of the compositions of n as lists of parts in lexicographic order, a(k) = n - length(composition(k)) for all k < 2^n and all n (see example); A007895 gives the equivalent for compositions into odd parts. - Joerg Arndt, Nov 09 2012
From Daniel Forgues, Mar 13 2015: (Start)
Just tally up row k (binary weight equal k) from 0 to 2^n - 1 to get the binomial coefficient C(n,k). (See A007318.)
0 1 3 7 15
0: O | . | . . | . . . . | . . . . . . . . |
1: | O | O . | O . . . | O . . . . . . . |
2: | | O | O O . | O O . O . . . |
3: | | | O | O O O . |
4: | | | | O |
Due to its fractal nature, the sequence is quite interesting to listen to.
(End)
The binary weight of n is a particular case of the digit sum (base b) of n. - Daniel Forgues, Mar 13 2015
The mean of the first n terms is 1 less than the mean of [a(n+1),...,a(2n)], which is also the mean of [a(n+2),...,a(2n+1)]. - Christian Perfect, Apr 02 2015
a(n) is also the largest part of the integer partition having viabin number n. The viabin number of an integer partition is defined in the following way. Consider the southeast border of the Ferrers board of the integer partition and consider the binary number obtained by replacing each east step with 1 and each north step, except the last one, with 0. The corresponding decimal form is, by definition, the viabin number of the given integer partition. "Viabin" is coined from "via binary". For example, consider the integer partition [2, 2, 2, 1]. The southeast border of its Ferrers board yields 10100, leading to the viabin number 20. - Emeric Deutsch, Jul 20 2017
a(n) is also known as the population count of the binary representation of n. - Chai Wah Wu, May 19 2020

Examples

			Using the formula a(n) = a(floor(n / floor_pow4(n))) + a(n mod floor_pow4(n)):
  a(4) = a(1) + a(0) = 1,
  a(8) = a(2) + a(0) = 1,
  a(13) = a(3) + a(1) = 2 + 1 = 3,
  a(23) = a(1) + a(7) = 1 + a(1) + a(3) = 1 + 1 + 2 = 4.
_Gary W. Adamson_ points out (Jun 03 2009) that this can be written as a triangle:
  0,
  1,
  1,2,
  1,2,2,3,
  1,2,2,3,2,3,3,4,
  1,2,2,3,2,3,3,4,2,3,3,4,3,4,4,5,
  1,2,2,3,2,3,3,4,2,3,3,4,3,4,4,5,2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,
  1,2,2,3,2,3,...
where the rows converge to A063787.
From _Joerg Arndt_, Nov 09 2012: (Start)
Connection to the compositions of n as lists of parts (see comment):
[ #]:   a(n)  composition
[ 0]:   [0]   1 1 1 1 1
[ 1]:   [1]   1 1 1 2
[ 2]:   [1]   1 1 2 1
[ 3]:   [2]   1 1 3
[ 4]:   [1]   1 2 1 1
[ 5]:   [2]   1 2 2
[ 6]:   [2]   1 3 1
[ 7]:   [3]   1 4
[ 8]:   [1]   2 1 1 1
[ 9]:   [2]   2 1 2
[10]:   [2]   2 2 1
[11]:   [3]   2 3
[12]:   [2]   3 1 1
[13]:   [3]   3 2
[14]:   [3]   4 1
[15]:   [4]   5
(End)
		

References

  • Jean-Paul Allouche and Jeffrey Shallit, Automatic Sequences, Cambridge Univ. Press, 2003, p. 119.
  • Donald E. Knuth, The Art of Computer Programming, vol. 4A, Combinatorial Algorithms, Section 7.1.3, Problem 41, p. 589. - N. J. A. Sloane, Aug 03 2012
  • Manfred R. Schroeder, Fractals, Chaos, Power Laws. W.H. Freeman, 1991, p. 383.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

The basic sequences concerning the binary expansion of n are this one, A000788, A000069, A001969, A023416, A059015, A007088.
Partial sums see A000788. For run lengths see A131534. See also A001792, A010062.
Number of 0's in n: A023416 and A080791.
a(n) = n - A011371(n).
Sum of digits of n written in bases 2-16: this sequence, A053735, A053737, A053824, A053827, A053828, A053829, A053830, A007953, A053831, A053832, A053833, A053834, A053835, A053836.
This is Guy Steele's sequence GS(3, 4) (see A135416).
Cf. A230952 (boustrophedon transform).
Cf. A070939 (length of binary representation of n).

Programs

  • Fortran
    c See link in A139351
    
  • Haskell
    import Data.Bits (Bits, popCount)
    a000120 :: (Integral t, Bits t) => t -> Int
    a000120 = popCount
    a000120_list = 0 : c [1] where c (x:xs) = x : c (xs ++ [x,x+1])
    -- Reinhard Zumkeller, Aug 26 2013, Feb 19 2012, Jun 16 2011, Mar 07 2011
    
  • Haskell
    a000120 = concat r
        where r = [0] : (map.map) (+1) (scanl1 (++) r)
    -- Luke Palmer, Feb 16 2014
    
  • Magma
    [Multiplicity(Intseq(n, 2), 1): n in [0..104]]; // Marius A. Burtea, Jan 22 2020
    
  • Magma
    [&+Intseq(n, 2):n in [0..104]]; // Marius A. Burtea, Jan 22 2020
  • Maple
    A000120 := proc(n) local w,m,i; w := 0; m := n; while m > 0 do i := m mod 2; w := w+i; m := (m-i)/2; od; w; end: wt := A000120;
    A000120 := n -> add(i, i=convert(n,base,2)): # Peter Luschny, Feb 03 2011
    with(Bits): p:=n->ilog2(n-And(n,n-1)): seq(p(binomial(2*n,n)),n=0..200) # Gary Detlefs, Jan 27 2019
  • Mathematica
    Table[DigitCount[n, 2, 1], {n, 0, 105}]
    Nest[Flatten[# /. # -> {#, # + 1}] &, {0}, 7] (* Robert G. Wilson v, Sep 27 2011 *)
    Table[Plus @@ IntegerDigits[n, 2], {n, 0, 104}]
    Nest[Join[#, # + 1] &, {0}, 7] (* IWABUCHI Yu(u)ki, Jul 19 2012 *)
    Log[2, Nest[Join[#, 2#] &, {1}, 14]] (* gives 2^14 term, Carlos Alves, Mar 30 2014 *)
  • PARI
    {a(n) = if( n<0, 0, 2*n - valuation((2*n)!, 2))};
    
  • PARI
    {a(n) = if( n<0, 0, subst(Pol(binary(n)), x ,1))};
    
  • PARI
    {a(n) = if( n<1, 0, a(n\2) + n%2)}; /* Michael Somos, Mar 06 2004 */
    
  • PARI
    a(n)=my(v=binary(n));sum(i=1,#v,v[i]) \\ Charles R Greathouse IV, Jun 24 2011
    
  • PARI
    a(n)=norml2(binary(n)) \\ better use {A000120=hammingweight}. - M. F. Hasler, Oct 09 2012, edited Feb 27 2020
    
  • PARI
    a(n)=hammingweight(n) \\ Michel Marcus, Oct 19 2013
    (Common Lisp) (defun floor-to-power (n pow) (declare (fixnum pow)) (expt pow (floor (log n pow)))) (defun enabled-bits (n) (if (< n 4) (n-th n (list 0 1 1 2)) (+ (enabled-bits (floor (/ n (floor-to-power n 4)))) (enabled-bits (mod n (floor-to-power n 4)))))) ; Stephen K. Touset (stephen(AT)touset.org), Apr 04 2007
    
  • Python
    def A000120(n): return bin(n).count('1') # Chai Wah Wu, Sep 03 2014
    
  • Python
    import numpy as np
    A000120 = np.array([0], dtype="uint8")
    for bitrange in range(25): A000120 = np.append(A000120, np.add(A000120, 1))
    print([A000120[n] for n in range(0, 105)]) # Karl-Heinz Hofmann, Nov 07 2022
    
  • Python
    def A000120(n): return n.bit_count() # Requires Python 3.10 or higher. - Pontus von Brömssen, Nov 08 2022
    
  • Python
    # Also see links.
    
  • SageMath
    def A000120(n):
        if n <= 1: return Integer(n)
        return A000120(n//2) + n%2
    [A000120(n) for n in range(105)]  # Peter Luschny, Nov 19 2012
    
  • SageMath
    def A000120(n) : return sum(n.digits(2)) # Eric M. Schmidt, Apr 26 2013
    
  • Scala
    (0 to 127).map(Integer.bitCount()) // _Alonso del Arte, Mar 05 2019
    

Formula

a(0) = 0, a(2*n) = a(n), a(2*n+1) = a(n) + 1.
a(0) = 0, a(2^i) = 1; otherwise if n = 2^i + j with 0 < j < 2^i, a(n) = a(j) + 1.
G.f.: Product_{k >= 0} (1 + y*x^(2^k)) = Sum_{n >= 0} y^a(n)*x^n. - N. J. A. Sloane, Jun 04 2009
a(n) = a(n-1) + 1 - A007814(n) = log_2(A001316(n)) = 2n - A005187(n) = A070939(n) - A023416(n). - Henry Bottomley, Apr 04 2001; corrected by Ralf Stephan, Apr 15 2002
a(n) = log_2(A000984(n)/A001790(n)). - Benoit Cloitre, Oct 02 2002
For n > 0, a(n) = n - Sum_{k=1..n} A007814(k). - Benoit Cloitre, Oct 19 2002
a(n) = n - Sum_{k>=1} floor(n/2^k) = n - A011371(n). - Benoit Cloitre, Dec 19 2002
G.f.: (1/(1-x)) * Sum_{k>=0} x^(2^k)/(1+x^(2^k)). - Ralf Stephan, Apr 19 2003
a(0) = 0, a(n) = a(n - 2^floor(log_2(n))) + 1. Examples: a(6) = a(6 - 2^2) + 1 = a(2) + 1 = a(2 - 2^1) + 1 + 1 = a(0) + 2 = 2; a(101) = a(101 - 2^6) + 1 = a(37) + 1 = a(37 - 2^5) + 2 = a(5 - 2^2) + 3 = a(1 - 2^0) + 4 = a(0) + 4 = 4; a(6275) = a(6275 - 2^12) + 1 = a(2179 - 2^11) + 2 = a(131 - 2^7) + 3 = a(3 - 2^1) + 4 = a(1 - 2^0) + 5 = 5; a(4129) = a(4129 - 2^12) + 1 = a(33 - 2^5) + 2 = a(1 - 2^0) + 3 = 3. - Hieronymus Fischer, Jan 22 2006
A fixed point of the mapping 0 -> 01, 1 -> 12, 2 -> 23, 3 -> 34, 4 -> 45, ... With f(i) = floor(n/2^i), a(n) is the number of odd numbers in the sequence f(0), f(1), f(2), f(3), f(4), f(5), ... - Philippe Deléham, Jan 04 2004
When read mod 2 gives the Morse-Thue sequence A010060.
Let floor_pow4(n) denote n rounded down to the next power of four, floor_pow4(n) = 4 ^ floor(log4 n). Then a(0) = 0, a(1) = 1, a(2) = 1, a(3) = 2, a(n) = a(floor(n / floor_pow4(n))) + a(n % floor_pow4(n)). - Stephen K. Touset (stephen(AT)touset.org), Apr 04 2007
a(n) = n - Sum_{k=2..n} Sum_{j|n, j >= 2} (floor(log_2(j)) - floor(log_2(j-1))). - Hieronymus Fischer, Jun 18 2007
a(n) = A138530(n, 2) for n > 1. - Reinhard Zumkeller, Mar 26 2008
a(A077436(n)) = A159918(A077436(n)); a(A000290(n)) = A159918(n). - Reinhard Zumkeller, Apr 25 2009
a(n) = A063787(n) - A007814(n). - Gary W. Adamson, Jun 04 2009
a(n) = A007814(C(2n, n)) = 1 + A007814(C(2n-1, n)). - Vladimir Shevelev, Jul 20 2009
For odd m >= 1, a((4^m-1)/3) = a((2^m+1)/3) + (m-1)/2 (mod 2). - Vladimir Shevelev, Sep 03 2010
a(n) - a(n-1) = { 1 - a(n-1) if and only if A007814(n) = a(n-1), 1 if and only if A007814(n) = 0, -1 for all other A007814(n) }. - Brent Spillner (spillner(AT)acm.org), Sep 01 2010
a(A001317(n)) = 2^a(n). - Vladimir Shevelev, Oct 25 2010
a(n) = A139351(n) + A139352(n) = Sum_k {A030308(n, k)}. - Philippe Deléham, Oct 14 2011
From Hieronymus Fischer, Jun 10 2012: (Start)
a(n) = Sum_{j = 1..m+1} (floor(n/2^j + 1/2) - floor(n/2^j)), where m = floor(log_2(n)).
General formulas for the number of digits >= d in the base p representation of n, where 1 <= d < p: a(n) = Sum_{j = 1..m+1} (floor(n/p^j + (p-d)/p) - floor(n/p^j)), where m=floor(log_p(n)); g.f.: g(x) = (1/(1-x))*Sum_{j>=0} (x^(d*p^j) - x^(p*p^j))/(1-x^(p*p^j)). (End)
a(n) = A213629(n, 1) for n > 0. - Reinhard Zumkeller, Jul 04 2012
a(n) = A240857(n,n). - Reinhard Zumkeller, Apr 14 2014
a(n) = log_2(C(2*n,n) - (C(2*n,n) AND C(2*n,n)-1)). - Gary Detlefs, Jul 10 2014
Sum_{n >= 1} a(n)/2n(2n+1) = (gamma + log(4/Pi))/2 = A344716, where gamma is Euler's constant A001620; see Sondow 2005, 2010 and Allouche, Shallit, Sondow 2007. - Jonathan Sondow, Mar 21 2015
For any integer base b >= 2, the sum of digits s_b(n) of expansion base b of n is the solution of this recurrence relation: s_b(n) = 0 if n = 0 and s_b(n) = s_b(floor(n/b)) + (n mod b). Thus, a(n) satisfies: a(n) = 0 if n = 0 and a(n) = a(floor(n/2)) + (n mod 2). This easily yields a(n) = Sum_{i = 0..floor(log_2(n))} (floor(n/2^i) mod 2). From that one can compute a(n) = n - Sum_{i = 1..floor(log_2(n))} floor(n/2^i). - Marek A. Suchenek, Mar 31 2016
Sum_{k>=1} a(k)/2^k = 2 * Sum_{k >= 0} 1/(2^(2^k)+1) = 2 * A051158. - Amiram Eldar, May 15 2020
Sum_{k>=1} a(k)/(k*(k+1)) = A016627 = log(4). - Bernard Schott, Sep 16 2020
a(m*(2^n-1)) >= n. Equality holds when 2^n-1 >= A000265(m), but also in some other cases, e.g., a(11*(2^2-1)) = 2 and a(19*(2^3-1)) = 3. - Pontus von Brömssen, Dec 13 2020
G.f.: A(x) satisfies A(x) = (1+x)*A(x^2) + x/(1-x^2). - Akshat Kumar, Nov 04 2023

A003714 Fibbinary numbers: if n = F(i1) + F(i2) + ... + F(ik) is the Zeckendorf representation of n (i.e., write n in Fibonacci number system) then a(n) = 2^(i1 - 2) + 2^(i2 - 2) + ... + 2^(ik - 2). Also numbers whose binary representation contains no two adjacent 1's.

Original entry on oeis.org

0, 1, 2, 4, 5, 8, 9, 10, 16, 17, 18, 20, 21, 32, 33, 34, 36, 37, 40, 41, 42, 64, 65, 66, 68, 69, 72, 73, 74, 80, 81, 82, 84, 85, 128, 129, 130, 132, 133, 136, 137, 138, 144, 145, 146, 148, 149, 160, 161, 162, 164, 165, 168, 169, 170, 256, 257, 258, 260, 261, 264
Offset: 0

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Author

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Comments

The name "Fibbinary" is due to Marc LeBrun.
"... integers whose binary representation contains no consecutive ones and noticed that the number of such numbers with n bits was fibonacci(n)". [posting to sci.math by Bob Jenkins (bob_jenkins(AT)burtleburtle.net), Jul 17 2002]
From Benoit Cloitre, Mar 08 2003: (Start)
A number m is in the sequence if and only if C(3m, m) (or equally, C(3m, 2m)) is odd.
a(n) == A003849(n) (mod 2). (End)
Numbers m such that m XOR 2*m = 3*m. - Reinhard Zumkeller, May 03 2005. [This implies that A003188(2*a(n)) = 3*a(n) holds for all n.]
Numbers whose base-2 representation contains no two adjacent ones. For example, m = 17 = 10001_2 belongs to the sequence, but m = 19 = 10011_2 does not. - Ctibor O. Zizka, May 13 2008
m is in the sequence if and only if the central Stirling number of the second kind S(2*m, m) = A007820(m) is odd. - O-Yeat Chan (math(AT)oyeat.com), Sep 03 2009
A000120(3*a(n)) = 2*A000120(a(n)); A002450 is a subsequence.
Every nonnegative integer can be expressed as the sum of two terms of this sequence. - Franklin T. Adams-Watters, Jun 11 2011
Subsequence of A213526. - Arkadiusz Wesolowski, Jun 20 2012
This is also the union of A215024 and A215025 - see the Comment in A014417. - N. J. A. Sloane, Aug 10 2012
The binary representation of each term m contains no two adjacent 1's, so we have (m XOR 2m XOR 3m) = 0, and thus a two-player Nim game with three heaps of (m, 2m, 3m) stones is a losing configuration for the first player. - V. Raman, Sep 17 2012
Positions of zeros in A014081. - John Keith, Mar 07 2022
These numbers are similar to Fibternary numbers A003726, Tribbinary numbers A060140 and Tribternary numbers. This sequence is a subsequence of Fibternary numbers A003726. The number of Fibbinary numbers less than any power of two is a Fibonacci number. We can generate this sequence recursively: start with 0 and 1; then, if x is in the sequence add 2x and 4x+1 to the sequence. The Fibbinary numbers have the property that the n-th Fibbinary number is even if the n-th term of the Fibonacci word is a. Respectively, the n-th Fibbinary number is odd (of the form 4x+1) if the n-th term of the Fibonacci word is b. Every number has a Fibbinary multiple. - Tanya Khovanova and PRIMES STEP Senior, Aug 30 2022
This is the ordered set S of numbers defined recursively by: 0 is in S; if x is in S, then 2*x and 4*x + 1 are in S. See Kimberling (2006) Example 3, in references below. - Harry Richman, Jan 31 2024

Examples

			From _Joerg Arndt_, Jun 11 2011: (Start)
In the following, dots are used for zeros in the binary representation:
  a(n)  binary(a(n))  n
    0:    .......     0
    1:    ......1     1
    2:    .....1.     2
    4:    ....1..     3
    5:    ....1.1     4
    8:    ...1...     5
    9:    ...1..1     6
   10:    ...1.1.     7
   16:    ..1....     8
   17:    ..1...1     9
   18:    ..1..1.    10
   20:    ..1.1..    11
   21:    ..1.1.1    12
   32:    .1.....    13
   33:    .1....1    14
   34:    .1...1.    15
   36:    .1..1..    16
   37:    .1..1.1    17
   40:    .1.1...    18
   41:    .1.1..1    19
   42:    .1.1.1.    20
   64:    1......    21
   65:    1.....1    22
(End)
		

References

  • Donald E. Knuth, The Art of Computer Programming: Fundamental Algorithms, Vol. 1, 2nd ed., Addison-Wesley, 1973, pp. 85, 493.

Crossrefs

A007088(a(n)) = A014417(n) (same sequence in binary). Complement: A004780. Char. function: A085357. Even terms: A022340, odd terms: A022341. First difference: A129761.
Other sequences based on similar restrictions on binary expansion: A003726 & A278038, A003754, A048715, A048718, A107907, A107909.
3*a(n) is in A001969.
Cf. A014081 (count 11 bits).

Programs

  • Haskell
    import Data.Set (Set, singleton, insert, deleteFindMin)
    a003714 n = a003714_list !! n
    a003714_list = 0 : f (singleton 1) where
       f :: Set Integer -> [Integer]
       f s = m : (f $ insert (4*m + 1) $ insert (2*m) s')
             where (m, s') = deleteFindMin s
    -- Reinhard Zumkeller, Jun 03 2012, Feb 07 2012
    
  • Maple
    A003714 := proc(n)
        option remember;
        if n < 3 then
            n ;
        else
            2^(A072649(n)-1) + procname(n-combinat[fibonacci](1+A072649(n))) ;
        end if;
    end proc:
    seq(A003714(n),n=0..10) ;
    # To produce a table giving n, a(n) (base 10), a(n) (base 2) - from N. J. A. Sloane, Sep 30 2018
    # binary: binary representation of n, in human order
    binary:=proc(n) local t1,L;
    if n<0 then ERROR("n must be nonnegative"); fi;
    if n=0 then return([0]); fi;
    t1:=convert(n,base,2); L:=nops(t1);
    [seq(t1[L+1-i],i=1..L)];
    end;
    for n from 0 to 100 do t1:=A003714(n); lprint(n, t1, binary(t1)); od:
  • Mathematica
    fibBin[n_Integer] := Block[{k = Ceiling[Log[GoldenRatio, n Sqrt[5]]], t = n, fr = {}}, While[k > 1, If[t >= Fibonacci[k], AppendTo[fr, 1]; t = t - Fibonacci[k], AppendTo[fr, 0]]; k--]; FromDigits[fr, 2]]; Table[fibBin[n], {n, 0, 61}] (* Robert G. Wilson v, Sep 18 2004 *)
    Select[Range[0, 270], ! MemberQ[Partition[IntegerDigits[#, 2], 2, 1], {1, 1}] &] (* Harvey P. Dale, Jul 17 2011 *)
    Select[Range[256], BitAnd[#, 2 #] == 0 &] (* Alonso del Arte, Jun 18 2012 *)
    With[{r = Range[10^5]}, Pick[r, BitAnd[r, 2 r], 0]] (* Eric W. Weisstein, Aug 18 2017 *)
    Select[Range[0, 299], SequenceCount[IntegerDigits[#, 2], {1, 1}] == 0 &] (* Requires Mathematica version 10 or later. -- Harvey P. Dale, Dec 06 2018 *)
  • PARI
    msb(n)=my(k=1); while(k<=n, k<<=1); k>>1
    for(n=1,1e4,k=bitand(n,n<<1);if(k,n=bitor(n,msb(k)-1),print1(n", "))) \\ Charles R Greathouse IV, Jun 15 2011
    
  • PARI
    select( is_A003714(n)=!bitand(n,n>>1), [0..266])
    {(next_A003714(n,t)=while(t=bitand(n+=1,n<<1), n=bitor(n,1<A003714(t)) \\ M. F. Hasler, Nov 30 2021
    
  • Python
    for n in range(300):
        if 2*n & n == 0:
            print(n, end=",") # Alex Ratushnyak, Jun 21 2012
    
  • Python
    def A003714(n):
        tlist, s = [1,2], 0
        while tlist[-1]+tlist[-2] <= n:
            tlist.append(tlist[-1]+tlist[-2])
        for d in tlist[::-1]:
            s *= 2
            if d <= n:
                s += 1
                n -= d
        return s # Chai Wah Wu, Jun 14 2018
    
  • Python
    def fibbinary():
        x = 0
        while True:
            yield x
            y = ~(x >> 1)
            x = (x - y) & y # Falk Hüffner, Oct 23 2021
    (C++)
    /* start with x=0, then repeatedly call x=next_fibrep(x): */
    ulong next_fibrep(ulong x)
    {
        // 2 examples:         //  ex. 1             //  ex.2
        //                     // x == [*]0 010101   // x == [*]0 01010
        ulong y = x | (x>>1);  // y == [*]? 011111   // y == [*]? 01111
        ulong z = y + 1;       // z == [*]? 100000   // z == [*]? 10000
        z = z & -z;            // z == [0]0 100000   // z == [0]0 10000
        x ^= z;                // x == [*]0 110101   // x == [*]0 11010
        x &= ~(z-1);           // x == [*]0 100000   // x == [*]0 10000
        return x;
    }
    /* Joerg Arndt, Jun 22 2012 */
    
  • Scala
    (0 to 255).filter(n => (n & 2 * n) == 0) // Alonso del Arte, Apr 12 2020
    (C#)
    public static bool IsFibbinaryNum(this int n) => ((n & (n >> 1)) == 0) ? true : false; // Frank Hollstein, Jul 07 2021

Formula

No two adjacent 1's in binary expansion.
Let f(x) := Sum_{n >= 0} x^Fibbinary(n). (This is the generating function of the characteristic function of this sequence.) Then f satisfies the functional equation f(x) = x*f(x^4) + f(x^2).
a(0) = 0, a(1) = 1, a(2) = 2, a(n) = 2^(A072649(n) - 1) + a(n - A000045(1 + A072649(n))). - Antti Karttunen
It appears that this sequence gives m such that A082759(3*m) is odd; or, probably equivalently, m such that A037011(3*m) = 1. - Benoit Cloitre, Jun 20 2003
If m is in the sequence then so are 2*m and 4*m + 1. - Henry Bottomley, Jan 11 2005
A116361(a(n)) <= 1. - Reinhard Zumkeller, Feb 04 2006
A085357(a(n)) = 1; A179821(a(n)) = a(n). - Reinhard Zumkeller, Jul 31 2010
a(n)/n^k is bounded (but does not tend to a limit), where k = 1.44... = A104287. - Charles R Greathouse IV, Sep 19 2012
a(n) = a(A193564(n+1))*2^(A003849(n) + 1) + A003849(n) for n > 0. - Daniel Starodubtsev, Aug 05 2021
There are Fibonacci(n+1) terms with up to n bits in this sequence. - Charles R Greathouse IV, Oct 22 2021
Sum_{n>=1} 1/a(n) = 3.704711752910469457886531055976801955909489488376627037756627135425780134020... (calculated using Baillie and Schmelzer's kempnerSums.nb, see Links). - Amiram Eldar, Feb 12 2022

Extensions

Edited by Antti Karttunen, Feb 21 2006
Cross reference to A007820 added (into O-Y.C. comment) by Jason Kimberley, Sep 14 2009
Typo corrected by Jeffrey Shallit, Sep 26 2014

A014417 Representation of n in base of Fibonacci numbers (the Zeckendorf representation of n). Also, binary words starting with 1 not containing 11, with the word 0 added.

Original entry on oeis.org

0, 1, 10, 100, 101, 1000, 1001, 1010, 10000, 10001, 10010, 10100, 10101, 100000, 100001, 100010, 100100, 100101, 101000, 101001, 101010, 1000000, 1000001, 1000010, 1000100, 1000101, 1001000, 1001001, 1001010, 1010000, 1010001, 1010010, 1010100, 1010101
Offset: 0

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Comments

Old name was: Representation of n in base of Fibonacci numbers (the Zeckendorf representation of n). Also, binary vectors not containing 11.
For n > 0, write n = Sum_{i >= 2} eps(i) Fib_i where eps(i) = 0 or 1 and no 2 consecutive eps(i) can be 1 (see A035517); then a(n) is obtained by writing the eps(i) in reverse order.
"One of the most important properties of the Fibonacci numbers is the special way in which they can be used to represent integers. Let's write j >> k <==> j >= k+2. Then every positive integer has a unique representation of the form n = F_k1 + F_k2 + ... + F_kr, where k1 >> k2 >> ... >> kr >> 0. (This is 'Zeckendorf's theorem.') ... We can always find such a representation by using a "greedy" approach, choosing F_k1 to be the largest Fibonacci number =< n, then choosing F_k2 to be the largest that is =< n - F_k1 and so on. Fibonacci representation needs a few more bits because adjacent 1's are not permitted; but the two representations are analogous." [Concrete Math.]
Since the binary representation of n in base of Fibonacci numbers allows only the successive bit pairs 00, 01, 10 and leaves 11 unused, we can use a ternary representation using all trits 0, 1, 2 where 00 --> 0, 01 --> 1 and 10 --> 2 (e.g. binary 1001010 as ternary 1022). - Daniel Forgues, Nov 30 2009
The same sequence also arises when considering the NegaFibonacci representations of the integers, as follows. Take the NegaFibonacci representations of n = 0, 1, 2, ... (A215022) and of n = -1, -2, -3, ... (A215023), sort the union of these two lists into increasing binary order, and we get A014417. Likewise the corresponding list of decimal representations, A003714, is the union of A215024 and A215025 sorted into increasing order. - N. J. A. Sloane, Aug 10 2012
Also, numbers, written in binary, such that no adjacent bits are equal to 1: A one-dimensional analog of the matrices considered in A228277/A228285, A228390, A228476, A228506 etc. - M. F. Hasler, Apr 27 2014
The sequence of final bits, starting with a(1), is the complement of the Fibonacci word A005614. - N. J. A. Sloane, Oct 03 2018
This representation is named after the Belgian Army doctor and mathematician Edouard Zeckendorf (1901-1983). - Amiram Eldar, Jun 11 2021

Examples

			The Zeckendorf expansions of 1, 2, ... are 1 = 1 = Fib_2 -> 1, 2 = 2 = Fib_3 -> 10, 3 = Fib_4 -> 100, 4 = 3+1 = Fib_4 + Fib_2 -> 101, 5 = 5 = Fib_5 -> 1000, 6 = 1+5 = Fib_2 + Fib_5 -> 1001, etc.
		

References

  • Ronald L. Graham, Donald E. Knuth and Oren Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990.
  • Donald E. Knuth, The Art of Computer Programming, Vol. 4A, Section 7.1.3, p. 169.
  • Edouard Zeckendorf, Représentation des nombres naturels par une somme des nombres de Fibonacci ou de nombres de Lucas, Bull. Soc. Roy. Sci. Liège 41, 179-182, 1972.

Crossrefs

a(n) = A003714(n) converted to binary.
See A104326 for dual Zeckendorf representation of n.

Programs

  • Haskell
    a014417 0 = 0
    a014417 n = foldl (\v z -> v * 10 + z) 0 $ a189920_row n
    -- Reinhard Zumkeller, Mar 10 2013
    
  • Maple
    A014417 := proc(n)
        local nshi,Z,i ;
        if n <= 1 then
            return n;
        end if;
        nshi := n ;
        Z := [] ;
        for i from A130234(n) to 2 by -1 do
            if nshi >= A000045(i) and nshi > 0 then
                Z := [1,op(Z)] ;
                nshi := nshi-A000045(i) ;
            else
                Z := [0,op(Z)] ;
            end if;
        end do:
        add( op(i,Z)*10^(i-1),i=1..nops(Z)) ;
    end proc: # R. J. Mathar, Jan 31 2015
  • Mathematica
    fb[n_Integer] := Block[{k = Ceiling[Log[GoldenRatio, n * Sqrt[5]]], t = n, fr = {}}, While[k > 1, If[t >= Fibonacci[k], AppendTo[fr, 1]; t = t - Fibonacci[k], AppendTo[fr, 0]]; k-- ]; FromDigits[fr]]; Table[ fb[n], {n, 0, 30}] (* Robert G. Wilson v, May 15 2004 *)
    r = Map[Fibonacci, Range[2, 12]]; Table[Total[FromDigits@ PadRight[{1}, Flatten@ #] &@ Reverse@ Position[r, #] & /@ Abs@ Differences@ NestWhileList[Function[k, k - SelectFirst[Reverse@ r, # < k &]], n + 1, # > 1 &]], {n, 0, 33}] (* Michael De Vlieger, Mar 27 2016, Version 10 *)
    FromDigits/@Select[Tuples[{0,1},7],SequenceCount[#,{1,1}]==0&] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Aug 14 2019 *)
  • PARI
    Zeckendorf(n)=my(k=0,v,m); while(fibonacci(k)<=n,k=k+1); m=k-1; v=vector(m-1); v[1]=1; n=n-fibonacci(k-1); while(n>0,k=0; while(fibonacci(k)<=n,k=k+1); v[m-k+2]=1; n=n-fibonacci(k-1)); v \\ Ralf Stephan
    
  • PARI
    Zeckendorf(n)= { local(k); a=0; while(n>0, k=0; while(fibonacci(k)<=n, k=k+1); a=a+10^(k-3); n=n-fibonacci(k-1); ); a }
    { for (n=0, 10000, Zeckendorf(n); print(n," ",a); write("b014417.txt", n, " ", a) ) } \\ Harry J. Smith, Jan 17 2009
    
  • Python
    from sympy import fibonacci
    def a(n):
        k=0
        x=0
        while n>0:
            k=0
            while fibonacci(k)<=n: k+=1
            x+=10**(k - 3)
            n-=fibonacci(k - 1)
        return x
    print([a(n) for n in range(101)]) # Indranil Ghosh, Jun 07 2017, after PARI code by Harry J. Smith

Extensions

Comment layout fixed by Daniel Forgues, Dec 07 2009
Typo corrected by Daniel Forgues, Mar 25 2010
Definition expanded and Duchene et al. reference added by N. J. A. Sloane, Aug 07 2018
Name corrected by Michel Dekking, Nov 30 2020

A001629 Self-convolution of Fibonacci numbers.

Original entry on oeis.org

0, 0, 1, 2, 5, 10, 20, 38, 71, 130, 235, 420, 744, 1308, 2285, 3970, 6865, 11822, 20284, 34690, 59155, 100610, 170711, 289032, 488400, 823800, 1387225, 2332418, 3916061, 6566290, 10996580, 18394910, 30737759, 51310978, 85573315, 142587180, 237387960, 394905492
Offset: 0

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Number of elements in all subsets of {1,2,...,n-1} with no consecutive integers. Example: a(5)=10 because the subsets of {1,2,3,4} that have no consecutive elements, i.e., {}, {1}, {2}, {3}, {4}, {1,3}, {1,4}, {2,4}, the total number of elements is 10. - Emeric Deutsch, Dec 10 2003
If g is either of the real solutions to x^2-x-1=0, g'=1-g is the other one and phi is any 2 X 2-matricial solution to the same equation, not of the form gI or g'I, then Sum'_{i+j=n-1} g^i phi^j = F_n + (A001629(n) - A001629(n-1)g')*(phi-g'I), where i,j >= 0, F_n is the n-th Fibonacci number and I is the 2 X 2 identity matrix... - Michele Dondi (blazar(AT)lcm.mi.infn.it), Apr 06 2004
Number of 3412-avoiding involutions containing exactly one subsequence of type 321.
Number of binary sequences of length n with exactly one pair of consecutive 1's. - George J. Schaeffer (gschaeff(AT)andrew.cmu.edu), Sep 02 2004
For this sequence the n-th term is given by (nF(n+1)-F(n)+nF(n-1))/5 where F(n) is the n-th Fibonacci number. - Mrs. J. P. Shiwalkar and M. N. Deshpande (dpratap_ngp(AT)sancharnet.in), Apr 20 2005
If an unbiased coin is tossed n times then there are 2^n possible strings of H and T. Out of these, number of strings with exactly one 'HH' is given by a(n) where a(n) denotes n-th term of this sequence. - Mrs. J. P. Shiwalkar and M. N. Deshpande (dpratap_ngp(AT)sancharnet.in), May 04 2005
a(n) is half the number of horizontal dominoes in all domino tilings of a horizontally aligned 2 X n rectangle; a(n+1) = the number of vertical dominoes in all domino tilings of a horizontally aligned 2 X n rectangle; thus 2*a(n)+a(n+1)=n*F(n+1) = the number of dominoes in all domino tilings of a 2 X n rectangle, where F=A000045, the Fibonacci sequence. - Roberto Tauraso, May 02 2005; Graeme McRae, Jun 02 2006
a(n+1) = (-i)^(n-1)*(d/dx)S(n,x)|A049310%20for%20the%20S-polynomials.%20-%20_Wolfdieter%20Lang">{x=i}, where i is the imaginary unit, n >= 1. First derivative of Chebyshev S-polynomials evaluated at x=i multiplied by (-i)^(n-1). See A049310 for the S-polynomials. - _Wolfdieter Lang, Apr 04 2007
For n >= 4, a(n) is the number of weak compositions of n-2 in which exactly one part is 0 and all other parts are either 1 or 2. - Milan Janjic, Jun 28 2010
For n greater than 1, a(n) equals the absolute value of (1 - (1/2 - i/2)*(1 + (-1)^(n + 1))) times the x-coefficient of the characteristic polynomial of the (n-1) X (n-1) tridiagonal matrix with i's along the main diagonal (i is the imaginary unit), 1's along the superdiagonal and the subdiagonal and 0's everywhere else (see Mathematica code below). - John M. Campbell, Jun 23 2011
For n > 0: a(n) = Sum_{k=1..n-1} (A039913(n-1,k)) / 2. - Reinhard Zumkeller, Oct 07 2012
The right-hand side of a binomial-coefficient identity [Gauthier]. - N. J. A. Sloane, Apr 09 2013
a(n) is the number of edges in the Fibonacci cube Gamma(n-1) (see the Klavzar 2005 reference, p. 149). Example: a(3)=2; indeed, the Fibonacci cube Gamma(2) is the path P(3) having 2 edges. - Emeric Deutsch, Aug 10 2014
a(n) is the number of c(i)'s, including repetitions, in p(n), where p(n)/q(n) is the n-th convergent p(n)/q(n) of the formal infinite continued fraction [c(0), c(1), ...]; e.g., the number of c(i)'s in p(3) = c(0)*c(1)*c(2)*c(3) + c(0)*c(1) + c(0)*c(3) + c(2)*c(3) + 1 is a(5) = 10. - Clark Kimberling, Dec 23 2015
Also the number of maximal and maximum cliques in the (n-1)-Fibonacci cube graph. - Eric W. Weisstein, Sep 07 2017
a(n+1) is the total number of fixed points in all permutations p on 1, 2, ..., n such that |k-p(k)| <= 1 for 1 <= k <= n. - Katharine Ahrens, Sep 03 2019
From Steven Finch, Mar 22 2020: (Start)
a(n+1) is the total binary weight (cf. A000120) of all A000045(n+2) binary sequences of length n not containing any adjacent 1's.
The only three 2-bitstrings without adjacent 1's are 00, 01 and 10. The bitsums of these are 0, 1 and 1. Adding these give a(3)=2.
The only five 3-bitstrings without adjacent 1's are 000, 001, 010, 100 and 101. The bitsums of these are 0, 1, 1, 1 and 2. Adding these give a(4)=5.
The only eight 4-bitstrings without adjacent 1's are 0000, 0001, 0010, 0100, 1000, 0101, 1010 and 1001. The bitsums of these are 0, 1, 1, 1, 1, 2, 2, and 2. Adding these give a(5)=10. (End)
Number of tilings of a 1 X n strip with monominoes (1 X 1 squares) and at least one domino (1 X 2 rectangles), where exactly one of the dominoes is colored gold. - Greg Dresden and Jiachen Weng, Jul 31 2025

Examples

			G.f. = x^2 + 2*x^3 + 5*x^4 + 10*x^5 + 20*x^6 + 38*x^7 + 71*x^8 + 130*x^9 + ... - _Michael Somos_, Jun 24 2018
		

References

  • Donald E. Knuth, Fundamental Algorithms, Addison-Wesley, 1968, p. 83, Eq. 1.2.8--(17). - Don Knuth, Feb 26 2019
  • Thomas Koshy, Fibonacci and Lucas Numbers with Applications, 2001, Chapter 15, page 187, "Hosoya's Triangle", and p. 375, eq. (32.13).
  • J. Riordan, Combinatorial Identities, Wiley, 1968, p. 101.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • S. Vajda, Fibonacci and Lucas numbers and the Golden Section, Ellis Horwood Ltd., Chichester, 1989, p. 183, Nr.(98).

Crossrefs

Row sums of triangles A058071, A134510, A134836.
First differences of A006478.

Programs

  • GAP
    List([0..40],n->Sum([0..n],k->Fibonacci(k)*Fibonacci(n-k))); # Muniru A Asiru, Jun 24 2018
    
  • Haskell
    a001629 n = a001629_list !! (n-1)
    a001629_list = f [] $ tail a000045_list where
       f us (v:vs) = (sum $ zipWith (*) us a000045_list) : f (v:us) vs
    -- Reinhard Zumkeller, Jan 18 2014, Oct 16 2011
    
  • Magma
    I:=[0,0,1,2]; [n le 4 select I[n] else 2*Self(n-1)+Self(n-2)-2*Self(n-3)-Self(n-4): n in [1..40]]; // Vincenzo Librandi, Nov 19 2014
    
  • Maple
    a:= n-> (<<2|1|0|0>, <1|0|1|0>, <-2|0|0|1>, <-1|0|0|0>>^n)[1,3]:
    seq(a(n), n=0..40); # Alois P. Heinz, Aug 01 2008
    # Alternative:
    A001629 := n -> `if`(n<2, 0, (n-1)*hypergeom([1-n/2, (3-n)/2], [1-n], -4)):
    seq(simplify(A001629(n)), n=0..37); # Peter Luschny, Apr 10 2018
  • Mathematica
    Table[Sum[Binomial[n-i, i] i, {i, 0, n}], {n, 0, 34}] (* Geoffrey Critzer, May 04 2009 *)
    Table[Abs[(1 -(1/2 -I/2)(1 - (-1)^n))*Coefficient[CharacteristicPolynomial[ Array[KroneckerDelta[#1, #2] I + KroneckerDelta[#1 + 1, #2] + KroneckerDelta[#1 -1, #2] &, {n-1, n-1}], x], x]], {n,2,50}] (* John M. Campbell, Jun 23 2011 *)
    LinearRecurrence[{2,1,-2,-1}, {0,0,1,2}, 40] (* Harvey P. Dale, Aug 26 2013 *)
    CoefficientList[Series[x^2/(1-x-x^2)^2, {x, 0, 40}], x] (* Vincenzo Librandi, Nov 19 2014 *)
    Table[(2nFibonacci[n-1] + (n-1)Fibonacci[n])/5, {n, 0, 40}] (* Vladimir Reshetnikov, May 08 2016 *)
    Table[With[{fibs=Fibonacci[Range[n]]},ListConvolve[fibs,fibs]],{n,-1,40}]//Flatten (* Harvey P. Dale, Aug 19 2018 *)
  • PARI
    Vec(1/(1-x-x^2)^2+O(x^99)) \\ Charles R Greathouse IV, Feb 03 2014
    
  • PARI
    a(n)=([0,1,0,0; 0,0,1,0; 0,0,0,1; -1,-2,1,2]^n)[2,4] \\ Charles R Greathouse IV, Jul 20 2016
    
  • SageMath
    def A001629(n): return (1/5)*(n*lucas_number2(n, 1, -1) - fibonacci(n))
    [A001629(n) for n in (0..40)] # G. C. Greubel, Apr 06 2022

Formula

G.f.: x^2/(1 - x - x^2)^2. - Simon Plouffe in his 1992 dissertation
a(n) = A037027(n-1, 1), n >= 1 (Fibonacci convolution triangle).
a(n) = 2*a(n-1) + a(n-2) - 2*a(n-3) - a(n-4), n > 3.
a(n) = Sum_{k=0..n} A000045(k)*A000045(n-k).
a(n+1) = Sum_{i=0..F(n)} A007895(i), where F = A000045, the Fibonacci sequence. - Claude Lenormand (claude.lenormand(AT)free.fr), Feb 04 2001
a(n) = Sum_{k=0..floor(n/2)-1} (k+1)*binomial(n-k-1, k+1). - Emeric Deutsch, Nov 15 2001
a(n) = floor( (1/5)*(n - 1/sqrt(5))*phi^n + 1/2 ) where phi=(1+sqrt(5))/2 is the golden ratio. - Benoit Cloitre, Jan 05 2003
a(n) = a(n-1) + A010049(n-1) for n > 0. - Emeric Deutsch, Dec 10 2003
a(n) = Sum_{k=0..floor((n-2)/2)} (n-k-1)*binomial(n-k-2, k). - Paul Barry, Jan 25 2005
a(n) = ((n-1)*F(n) + 2*n*F(n-1))/5, F(n)=A000045(n) (see Vajda and Koshy reference).
F'(n, 1), the first derivative of the n-th Fibonacci polynomial evaluated at 1. - T. D. Noe, Jan 18 2006
a(n) = a(n-1) + a(n-2) + F(n-1), where F=A000045, the Fibonacci sequence. - Graeme McRae, Jun 02 2006
a(n) = (1/5)*(n-1/sqrt(5))*((1+sqrt(5))/2)^n + (1/5)*(n+1/sqrt(5))*((1-sqrt(5))/2)^n. - Graeme McRae, Jun 02 2006
a(n) = A055244(n-1) - F(n-2). Example: a(6) = 20 = A055244(5) - F(3) = (23 - 3). - Gary W. Adamson, Jul 27 2007
a(n) = term (1,3) in the 4 X 4 matrix [2,1,0,0; 1,0,1,0; -2,0,0,1; -1,0,0,0]^n. - Alois P. Heinz, Aug 01 2008
a(n) = A214178(n,1) for n > 0. - Reinhard Zumkeller, Jul 08 2012
a(n) = ((n+1)*F(n-1) + (n-1)*F(n+1))/5. - Richard R. Forberg, Aug 04 2014
(n-2)*a(n) - (n-1)*a(n-1) - n*a(n-2) = 0, n > 1. - Michael D. Weiner, Nov 18 2014
a(n) = Sum_{i=0..n-1} Sum_{j=0..i} F(j-1)*F(i-j), where F(n) = A000045 Fibonacci Numbers. - Carlos A. Rico A., Jul 14 2016
a(n) = (n*Lucas(n) - Fibonacci(n))/5, where Lucas = A000032, Fibonacci = A000045. - Vladimir Reshetnikov, Sep 27 2016
a(n) = (n-1)*hypergeom([1-n/2, (3-n)/2], [1-n], -4) for n >= 2. - Peter Luschny, Apr 10 2018
a(n) = -(-1)^n a(-n) for all n in Z. - Michael Somos, Jun 24 2018
E.g.f.: (1/50)*exp(-2*x/(1+sqrt(5)))*(2*sqrt(5)-5*(-1+sqrt(5))*x+exp(sqrt(5)*x)*(-2*sqrt(5)+5*(1+sqrt(5))*x)). - Stefano Spezia, Sep 03 2019
From Peter Bala, Jan 14 2025: (Start)
a(2*n+1) is even and a(2*n) has the same parity as Fibonacci(n).
For n >= 1, a(n) = (2/n)*Sum_{k = 0..n} k*Fibonacci(k)*Fibonacci(n-k). (End)

A027941 a(n) = Fibonacci(2*n + 1) - 1.

Original entry on oeis.org

0, 1, 4, 12, 33, 88, 232, 609, 1596, 4180, 10945, 28656, 75024, 196417, 514228, 1346268, 3524577, 9227464, 24157816, 63245985, 165580140, 433494436, 1134903169, 2971215072, 7778742048, 20365011073, 53316291172, 139583862444, 365435296161, 956722026040
Offset: 0

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Comments

Also T(2n+1,n+1), T given by A027935. Also first row of Inverse Stolarsky array.
Third diagonal of array defined by T(i, 1)=T(1, j)=1, T(i, j)=Max(T(i-1, j)+T(i-1, j-1); T(i-1, j-1)+T(i, j-1)). - Benoit Cloitre, Aug 05 2003
Number of Schroeder paths of length 2(n+1) having exactly one up step starting at an even height (a Schroeder path is a lattice path starting from (0,0), ending at a point on the x-axis, consisting only of steps U=(1,1) (up steps), D=(1,-1) (down steps) and H=(2,0) (level steps) and never going below the x-axis). Schroeder paths are counted by the large Schroeder numbers (A006318). Example: a(1)=4 because among the six Schroeder paths of length 4 only the paths (U)HD, (U)UDD, H(U)D, (U)DH have exactly one U step that starts at an even height (shown between parentheses). - Emeric Deutsch, Dec 19 2004
Also: smallest number not writeable as the sum of fewer than n positive Fibonacci numbers. E.g., a(5)=88 because it is the smallest number that needs at least 5 Fibonacci numbers: 88 = 55 + 21 + 8 + 3 + 1. - Johan Claes, Apr 19 2005 [corrected for offset and clarification by Mike Speciner, Sep 19 2023] In general, a(n) is the sum of n positive Fibonacci numbers as a(n) = Sum_{i=1..n} A000045(2*i). See A001076 when negative Fibonacci numbers can be included in the sum. - Mike Speciner, Sep 24 2023
Except for first term, numbers a(n) that set a new record in the number of Fibonacci numbers needed to sum up to n. Position of records in sequence A007895. - Ralf Stephan, May 15 2005
Successive extremal petal bends beta(n) = a(n-2). See the Ring Lemma of Rodin and Sullivan in K. Stephenson, Introduction to Circle Packing (Cambridge U. P., 2005), pp. 73-74 and 318-321. - David W. Cantrell (DWCantrell(AT)sigmaxi.net)
a(n+1)= AAB^(n)(1), n>=1, with compositions of Wythoff's complementary A(n):=A000201(n) and B(n)=A001950(n) sequences. See the W. Lang link under A135817 for the Wythoff representation of numbers (with A as 1 and B as 0 and the argument 1 omitted). E.g., 4=`110`, 12=`1100`, 33=`11000`, 88=`110000`, ..., in Wythoff code. AA(1)=1=a(1) but for uniqueness reason 1=A(1) in Wythoff code. - N. J. A. Sloane, Jun 29 2008
Start with n. Each n generates a sublist {n-1,n-1,n-2,..,1}. Each element of each sublist also generates a sublist. Add numbers in all terms. For example, 3->{2,2,1} and both 2->{1,1}, so a(3) = 3 + 2 + 2 + 1 + 1 + 1 + 1 + 1 = 12. - Jon Perry, Sep 01 2012
For n>0: smallest number such that the inner product of Zeckendorf binary representation and its reverse equals n: A216176(a(n)) = n, see also A189920. - Reinhard Zumkeller, Mar 10 2013
Also, numbers m such that 5*m*(m+2)+1 is a square. - Bruno Berselli, May 19 2014
Also, number of nonempty submultisets of multisets of weight n that span an initial interval of integers (see 2nd example). - Gus Wiseman, Feb 10 2015
From Robert K. Moniot, Oct 04 2020: (Start)
Including a(-1):=0, consecutive terms (a(n-1),a(n))=(u,v) or (v,u) give all points on the hyperbola u^2-u+v^2-v-4*u*v=0 with both coordinates nonnegative integers. Note that this follows from identifying (1,u+1,v+1) with the Markov triple (1,Fibonacci(2n-1),Fibonacci(2n+1)). See A001519 (comments by Robert G. Wilson, Oct 05 2005, and Wolfdieter Lang, Jan 30 2015).
Let T(n) denote the n-th triangular number. If i, j are any two successive elements of the above sequence then (T(i-1) + T(j-1))/T(i+j-1) = 3/5. (End)

Examples

			a(5) = 88 = 2*33 + 12 + 4 + 1 + 5. a(6) = 232 = 2*88 + 33 + 12 + 4 + 1 + 6. - _Jon Perry_, Sep 01 2012
a(4) = 33 counts all nonempty submultisets of the last row: [1][2][3][4], [11][12][13][14][22][23][24][33][34], [111][112][113][122][123][124][133][134][222][223][233][234], [1111][1112][1122][1123][1222][1223][1233][1234]. - _Gus Wiseman_, Feb 10 2015
		

References

  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 12.

Crossrefs

Related to partial sums of Fibonacci(k*n) over n: A000071, A099919, A058038, A138134, A053606; this sequence is the case k=2.
Cf. A212336 for more sequences with g.f. of the type 1/(1 - k*x + k*x^2 - x^3).
Cf. A000225 (sublist connection).
Cf. A258993 (row sums, n > 0), A000967.

Programs

Formula

a(n) = Sum_{i=1..n} binomial(n+i, n-i). - Benoit Cloitre, Oct 15 2002
G.f.: Sum_{k>=1} x^k/(1-x)^(2*k+1). - Benoit Cloitre, Apr 21 2003
a(n) = Sum_{k=1..n} F(2*k), i.e., partial sums of A001906. - Benoit Cloitre, Oct 27 2003
a(n) = Sum_{k=0..n-1} U(k, 3/2) = Sum_{k=0..n-1} S(k, 3), with S(k, 3) = A001906(k+1). - Paul Barry, Nov 14 2003
G.f.: x/((1-x)*(1-3*x+x^2)) = x/(1-4*x+4*x^2-x^3).
a(n) = 4*a(n-1) - 4*a(n-2) + a(n-3) with n>=2, a(-1)=0, a(0)=0, a(1)=1.
a(n) = 3*a(n-1) - a(n-2) + 1 with n>=1, a(-1)=0, a(0)=0.
a(n) = Sum_{k=1..n} F(k)*L(k), where L(k) = Lucas(k) = A000032(k) = F(k-1) + F(k+1). - Alexander Adamchuk, May 18 2007
a(n) = 2*a(n-1) + (Sum_{k=1..n-2} a(k)) + n. - Jon Perry, Sep 01 2012
Sum {n >= 1} 1/a(n) = 3 - phi, where phi = 1/2*(1 + sqrt(5)) is the golden ratio. The ratio of adjacent terms r(n) := a(n)/a(n-1) satisfies the recurrence r(n+1) = (4*r(n) - 1)/(r(n) + 1) for n >= 2. - Peter Bala, Dec 05 2013
a(n) = S(n, 3) - S(n-1, 3) - 1, n >= 0, with Chebyshev's S-polynomials (see A049310), where S(-1, x) = 0. - Wolfdieter Lang, Aug 28 2014
a(n) = -1 + (2^(-1-n)*((3-sqrt(5))^n*(-1+sqrt(5)) + (1+sqrt(5))*(3+sqrt(5))^n)) / sqrt(5). - Colin Barker, Jun 03 2016
E.g.f.: (sqrt(5)*sinh(sqrt(5)*x/2) + 5*cosh(sqrt(5)*x/2))*exp(3*x/2)/5 - exp(x). - Ilya Gutkovskiy, Jun 03 2016
a(n) = Sum_{k=0..n} binomial(n+1,k+1)*Fibonacci(k). - Vladimir Kruchinin, Oct 14 2016
a(n) = Sum_{k=0..n-1} Sum_{i=0..n-1} C(k+i+1,k-i). - Wesley Ivan Hurt, Sep 21 2017
a(n)*a(n-2) = a(n-1)*(a(n-1) - 1) for n>1. - Robert K. Moniot, Aug 23 2020
a(n) = Sum_{k=1..n} C(2*n-k,k). - Wesley Ivan Hurt, Dec 22 2020
a(n) = Sum_{k = 1..2*n+2} (-1)^k*Fibonacci(k). - Peter Bala, Nov 14 2021
a(n) = (2*cosh((1 + 2*n)*arccsch(2)))/sqrt(5) - 1. - Peter Luschny, Nov 21 2021
a(n) = F(n + (n mod 2)) * L(n+1 - (n mod 2)), where L(n) = A000032(n) and F(n) = A000045(n) (Euler and Sadek, 2001). - Amiram Eldar, Jan 13 2022

Extensions

More terms from James Sellers, Sep 08 2000
Paul Barry's Nov 14 2003 formula, recurrences and g.f. corrected for offset 0 and index link for Chebyshev polynomials added by Wolfdieter Lang, Aug 28 2014

A035517 Triangular array read by rows, formed from Zeckendorf expansion of integers: repeatedly subtract the largest Fibonacci number you can until nothing remains. Row n give Z. expansion of n.

Original entry on oeis.org

0, 1, 2, 3, 1, 3, 5, 1, 5, 2, 5, 8, 1, 8, 2, 8, 3, 8, 1, 3, 8, 13, 1, 13, 2, 13, 3, 13, 1, 3, 13, 5, 13, 1, 5, 13, 2, 5, 13, 21, 1, 21, 2, 21, 3, 21, 1, 3, 21, 5, 21, 1, 5, 21, 2, 5, 21, 8, 21, 1, 8, 21, 2, 8, 21, 3, 8, 21, 1, 3, 8, 21, 34, 1, 34, 2, 34, 3, 34, 1, 3, 34, 5, 34, 1, 5, 34, 2, 5, 34
Offset: 0

Views

Author

Keywords

Comments

Row n has A007895(n) terms.
With the 2nd Maple program, B(n) yields the number of terms in the Zeckendorf expansion of n, while Z(n) yields the expansion itself. For example, B(100)=3 and Z(100)=3, 8, 89. [Emeric Deutsch, Jul 05 2010]

Examples

			0=0; 1=1; 2=2; 3=3; 4=1+3; 5=5; 6=1+5; 7=2+5; 8=8; 9=1+8; 10=2+8; ... so triangle begins
  0;
  1;
  2;
  3;
  1, 3;
  5;
  1, 5;
  2, 5;
  8;
  1, 8;
  2, 8;
  3, 8;
  1, 3, 8;
		

References

  • Zeckendorf, E., Représentation des nombres naturels par une somme des nombres de Fibonacci ou de nombres de Lucas, Bull. Soc. Roy. Sci. Liège 41, 179-182, 1972.

Crossrefs

Programs

  • Haskell
    a035517 n k = a035517_tabf !! n !! k
    a035517_row n = a035517_tabf !! n
    a035517_tabf = map reverse a035516_tabf
    -- Reinhard Zumkeller, Mar 10 2013
    
  • Maple
    with(combinat): B := proc (n) local A, ct, m, j: A := proc (n) local i: for i while fibonacci(i) <= n do n-fibonacci(i) end do end proc: ct := 0: m := n: for j while 0 < A(m) do ct := ct+1: m := A(m) end do: ct+1 end proc: F := proc (n) local i: for i while fibonacci(i) <= n do fibonacci(i) end do end proc: Z := proc (n) local j, z: for j to B(n) do z[j] := F(n-add(z[i], i = 1 .. j-1)) end do: seq(z[B(n)+1-k], k = 1 .. B(n)) end proc: for n to 25 do Z(n) end do;
    # Emeric Deutsch, Jul 05 2010
    # yields sequence in triangular form; end of this Maple program
  • Mathematica
    f[n_] := (k=1; ff={}; While[(fi = Fibonacci[k]) <= n, AppendTo[ff, fi]; k++]; Drop[ff,1]); ro[n_] := If[n == 0, 0, r = n; s = {}; fr = f[n];
    While[r > 0, lf = Last[fr]; If[lf <= r, r = r - lf; PrependTo[s, lf]]; fr = Drop[fr,-1]]; s]; Flatten[ro /@ Range[0, 42]] (* Jean-François Alcover, Jul 23 2011 *)
  • Python
    zeck, fib = [], [0, 1]
    from itertools import count, islice
    def agen(): # generator of terms
        for r in count(0):
            while fib[-1] < r:
                fib.append(fib[-2] + fib[-1])
            i = 1
            while fib[-i] > r: i += 1
            bigfib = fib[-i]
            zeck.append( ([] if r == bigfib else zeck[r-bigfib]) + [bigfib] )
            yield from zeck[r]  # row r of the triangle
    print(list(islice(agen(), 90))) # Michael S. Branicky, Apr 04 2022

Extensions

More terms from James Sellers, Dec 13 1999

A328845 The first Fibonacci based variant of arithmetic derivative: a(p) = A000045(p) for prime p, a(u*v) = a(u)*v + u*a(v), with a(0) = a(1) = 0.

Original entry on oeis.org

0, 0, 1, 2, 4, 5, 7, 13, 12, 12, 15, 89, 20, 233, 33, 25, 32, 1597, 33, 4181, 40, 53, 189, 28657, 52, 50, 479, 54, 80, 514229, 65, 1346269, 80, 289, 3211, 100, 84, 24157817, 8381, 725, 100, 165580141, 127, 433494437, 400, 105, 57337, 2971215073, 128, 182, 125, 4825, 984, 53316291173, 135, 500, 188, 12581, 1028487, 956722026041, 160
Offset: 0

Views

Author

Antti Karttunen, Oct 28 2019

Keywords

Crossrefs

Cf. A374046 (indices of even terms), A374047 (of odd terms), A374122 (of multiples of 3), A374202 (2-adic valuation), A374203 (3-adic valuation), A374205 (5-adic valuation), A374125 [a(n) mod 360].
Cf. A374106 [gcd(a(n), A113177(n))], A374035 [gcd(a(n), A328846(n))], A374116 [gcd(a(n), A328768(n))].
For variants of the same formula, see A003415, A258851, A328768, A328769, A328846, A371192.

Programs

  • Mathematica
    A328845[n_] := If[n <= 1, 0, n*Total[MapApply[#2*Fibonacci[#]/# &, FactorInteger[n]]]];
    Array[A328845, 100, 0] (* Paolo Xausa, Dec 16 2024 *)
  • PARI
    A328845(n) = if(n<=1, 0, my(f=factor(n)); n*sum(i=1, #f~, f[i, 2]*fibonacci(f[i,1])/f[i, 1]));

Formula

a(n) = n * Sum e_j * A000045(p_j)/p_j for n = Product p_j^e_j.
a(A000040(n)) = A030426(n).
A007895(a(n)) = A328847(n).

A113177 Fully additive with a(p) = Fibonacci(p); If, for p prime, p^(m_{n,p}) is the highest power of p dividing n with m>=0, then a(n) = Sum_{p prime} F(p)*(m_{n,p}), where F(p) = p-th Fibonacci number.

Original entry on oeis.org

0, 1, 2, 2, 5, 3, 13, 3, 4, 6, 89, 4, 233, 14, 7, 4, 1597, 5, 4181, 7, 15, 90, 28657, 5, 10, 234, 6, 15, 514229, 8, 1346269, 5, 91, 1598, 18, 6, 24157817, 4182, 235, 8, 165580141, 16, 433494437, 91, 9, 28658, 2971215073, 6, 26, 11, 1599, 235, 53316291173, 7, 94
Offset: 1

Views

Author

Leroy Quet, Oct 16 2005

Keywords

Examples

			12 = 2^2 * 3^1, so a(12) = F(2)*2 + F(3)*1 = 2 + 2 = 4.
		

Crossrefs

Cf. A373586 (indices of even terms), A373587 (of odd terms), A374052 (of multiples of 3), A374206 (2-adic valuation), A374207 (3-adic valuation), A374208 (5-adic valuation), A374209 [A007895(a(n))], A374124 [a(n) mod 360].
Cf. A374106 [gcd(a(n), A328845(n))], A374112 [gcd(a(n), A276085(n))].
For other completely additive sequences see the cross-references in A001414.

Programs

  • Mathematica
    b[t_]:=Fibonacci[First[t]]Last[t] a[n_]:=Apply[Plus, Map[b, FactorInteger[n]]] (* Esa Peuha, Oct 26 2005 *)
  • PARI
    { for(n=1,100,f=factor(n);s=0;for(i=1,matsize(f)[1],s+=fibonacci(f[i,1])*f[i,2]);print1(s,",")) } \\ Lambert Klasen, Oct 26 2005
    
  • Sage
    [0]+[sum([fibonacci(x[0])*x[1] for x in factor(n)]) for n in range(2,56)] # Danny Rorabaugh, Apr 03 2015

Formula

Totally additive with a(p) = A000045(p).

Extensions

More terms from Esa Peuha (esa.peuha(AT)helsinki.fi) and Lambert Klasen (lambert.klasen(AT)gmx.net), Oct 26 2005
Prefixed the name with a more succinct form of the definition given in comments. - Antti Karttunen, Jul 08 2024

A328209 Numbers m such that m and m+1 are consecutive Zeckendorf-Niven numbers (A328208).

Original entry on oeis.org

1, 2, 3, 4, 5, 12, 13, 21, 26, 55, 68, 80, 89, 92, 93, 110, 152, 183, 195, 207, 233, 236, 237, 254, 291, 304, 327, 364, 377, 380, 381, 398, 435, 471, 484, 555, 584, 605, 609, 639, 644, 759, 795, 834, 875, 894, 930, 987, 992, 1004, 1011, 1028, 1047, 1076, 1220
Offset: 1

Views

Author

Amiram Eldar, Oct 07 2019

Keywords

Examples

			12 is in the sequence since both 12 and 13 are in A328208: A007895(12) = 3 is a divisor of 12, and A007895(13) = 1 is a divisor of 13.
		

Crossrefs

Programs

  • Mathematica
    z[n_] := Length[DeleteCases[NestWhileList[# - Fibonacci[Floor[Log[Sqrt[5]*# + 3/2]/Log[GoldenRatio]]] &, n, # > 1 &], 0]]; aQ[n_] := Divisible[n, z[n]]; c = 0; k = 1; s = {}; v = Table[-1, {2}]; While[c < 60, If[aQ[k], v = Join[Rest[v], {k}]; If[AllTrue[Differences[v], # == 1 &], c++; AppendTo[s, k - 1]]]; k++]; s (* after Alonso del Arte at A007895 *)

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
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