A007895 -id:A007895 - cp's OEIS frontend

A010049 Second-order Fibonacci numbers.

0, 1, 1, 3, 5, 10, 18, 33, 59, 105, 185, 324, 564, 977, 1685, 2895, 4957, 8462, 14406, 24465, 41455, 70101, 118321, 199368, 335400, 563425, 945193, 1583643, 2650229, 4430290, 7398330, 12342849, 20573219, 34262337, 57013865, 94800780, 157517532, 261545777

Offset: 0

N. J. A. Sloane

Number of parts in all compositions of n+1 with no 1's. E.g. a(5)=10 because in the compositions of 6 with no part equal to 1, namely 6,4+2,3+3,2+4,2+2+2, the total number of parts is 10. - Emeric Deutsch, Dec 10 2003

D. E. Knuth, The Art of Computer Programming. Addison-Wesley, Reading, MA, Vol. 1, p. 83.
Cornelius Gerrit Lekkerkerker, Voorstelling van natuurlijke getallen door een som van getallen van Fibonacci, Simon Stevin, Vol. 29 (1952), pp. 190-195.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Carlos A. Rico A. and Ana Paula Chaves, Double-Recurrence Fibonacci Numbers and Generalizations, arXiv:1903.07490 [math.NT], 2019.
T. Amdeberhan and M. B. Can, M. Jensen, Divisors and specializations of Lucas polynomials, arXiv preprint arXiv:1406.0432 [math.CO], 2014.
Kálmán Liptai, László Németh, Tamás Szakács, and László Szalay, On certain Fibonacci representations, arXiv:2403.15053 [math.NT], 2024. See p. 2.
Mengmeng Liu and Andrew Yezhou Wang, The Number of Designated Parts in Compositions with Restricted Parts, J. Int. Seq., Vol. 23 (2020), Article 20.1.8.
Jia Huang, Compositions with restricted parts, arXiv:1812.11010 [math.CO], 2018.
Tamás Szakács, Linear recursive sequences and factorials, Ph. D. Thesis, Univ. Debrecen (Hungary, 2024). See pp. 2, 50, 57.
Loïc Turban, Lattice animals on a staircase and Fibonacci numbers, arXiv:cond-mat/0011038 [cond-mat.stat-mech], 2000; J. Phys. A 33 (2000) 2587-2595.
Index entries for linear recurrences with constant coefficients, signature (2,1,-2,-1).

Partial sums of A006367.
Cf. A000032, A000045, A001629, A007895, A023610.
Cf. A029907, A094440, A134410, A135830.

a:=List([0..40],n->Sum([0..n-1],k->(k+1)*Binomial(n-k-1,k)));; Print(a); # Muniru A Asiru, Dec 31 2018

a010049 n = a010049_list !! n
a010049_list = uncurry c $ splitAt 1 a000045_list where
   c us (v:vs) = (sum $ zipWith (*) us (1 : reverse us)) : c (v:us) vs
-- Reinhard Zumkeller, Nov 01 2013

[((2*n+3)*Fibonacci(n)-n*Fibonacci(n-1))/5: n in [0..40]]; // Vincenzo Librandi, Dec 31 2018

with(combinat): A010049 := proc(n) options remember; if n <= 1 then n else A010049(n-1)+A010049(n-2)+fibonacci(n-2); fi; end;

CoefficientList[Series[(z - z^2)/(z^2 + z - 1)^2, {z, 0, 100}], z] (* Vladimir Joseph Stephan Orlovsky, Jul 01 2011 *)
CoefficientList[Series[x (1 - x) / (1 - x - x^2)^2, {x, 0, 60}], x] (* Vincenzo Librandi, Jun 11 2013 *)
LinearRecurrence[{2, 1, -2, -1}, {0, 1, 1, 3}, 38] (* Amiram Eldar, Jan 11 2020 *)

a(n)=([0,1,0,0; 0,0,1,0; 0,0,0,1; -1,-2,1,2]^n*[0;1;1;3])[1,1] \\ Charles R Greathouse IV, Jul 20 2016

def A010049():
    a, b, c, d = 0, 1, 1, 3
    while True:
        yield a
        a, b, c, d = b, c, d, 2*(d-b)+c-a
a = A010049(); [next(a) for i in range(38)]  # Peter Luschny, Nov 20 2013

def A010049(n): return (1/5)*(n*lucas_number2(n-1, 1, -1) + 3*fibonacci(n))
[A010049(n) for n in (0..40)] # G. C. Greubel, Apr 06 2022

First differences of A001629.
From Wolfdieter Lang, May 03 2000: (Start)
a(n) = ((2*n+3)*F(n) - n*F(n-1))/5, F(n)=A000045(n) (Fibonacci numbers) (Turban reference eq.(2.12)).
G.f.: x*(1-x)/(1-x-x^2)^2. (Turban reference eq.(2.10)). (End)
Recurrence: a(0)=0, a(1)=1, a(2)=1, a(n+2) = a(n+1) + a(n) + F(n). - Benoit Cloitre, Sep 02 2002
Set A(n) = a(n+1) + a(n-1), B(n) = a(n+1) - a(n-1). Then A(n+2) = A(n+1) + A(n) + Lucas(n) and B(n+2) = B(n+1) + B(n) + Fibonacci(n). The polynomials F_2(n,-x) = Sum_{k=0..n} C(n,k)*a(n-k)*(-x)^k appear to satisfy a Riemann hypothesis; their zeros appear to lie on the vertical line Re x = 1/2 in the complex plane. Compare with the polynomials F(n,-x) defined in A094440. For a similar conjecture for polynomials involving the second-order Lucas numbers see A134410. - Peter Bala, Oct 24 2007
a(n) = -A001629(n+2) + 2*A001629(n+1) + A000045(n+1). - R. J. Mathar, Nov 16 2007
Starting (1, 1, 3, 5, 10, ...), = row sums of triangle A135830. - Gary W. Adamson, Nov 30 2007
a(n) = F(n) + Sum_{k=0..n-1} F(k)*F(n-1-k), where F = A000045. - Reinhard Zumkeller, Nov 01 2013
a(n) = Sum_{k=0..n-1} (k+1)*binomial(n-k-1, k). - Peter Luschny, Nov 20 2013
a(n) = Sum_{i=0..n-1} Sum_{j=0..i} F(j-1)*F(i-j-1), where F = A000045. - Carlos A. Rico A., Jul 14 2016
a(n) = Sum_{k = F(n+1)..F(n+2)-1} A007895(k), where F(n) is the n-th Fibonacci number (Lekkerkerker, 1952). - Amiram Eldar, Jan 11 2020
a(n) = (1/5)*(n*A000032(n-1) + 3*A000045(n)). - G. C. Greubel, Apr 06 2022
E.g.f.: 2*exp(x/2)*(5*x*cosh(sqrt(5)*x/2) + 3*sqrt(5)*sinh(sqrt(5)*x/2))/25. - Stefano Spezia, Dec 04 2023

More terms from Emeric Deutsch, Dec 10 2003

A060502 a(n) = number of occupied digit slopes in the factorial base representation of n (see comments for the definition); number of drops in the n-th permutation of list A060117.

Original entry on oeis.org

0, 1, 1, 2, 1, 1, 1, 2, 2, 3, 2, 2, 1, 2, 1, 2, 2, 2, 1, 1, 2, 2, 1, 1, 1, 2, 2, 3, 2, 2, 2, 3, 3, 4, 3, 3, 2, 3, 2, 3, 3, 3, 2, 2, 3, 3, 2, 2, 1, 2, 2, 3, 2, 2, 1, 2, 2, 3, 2, 2, 2, 3, 2, 3, 3, 3, 2, 2, 3, 3, 2, 2, 1, 2, 1, 2, 2, 2, 2, 3, 2, 3, 3, 3, 1, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 2, 2, 1, 1, 2, 2, 3, 3, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 2, 2, 1, 1, 1

Offset: 0

Antti Karttunen, Mar 22 2001

From Antti Karttunen, Aug 11-24 2016: (Start)
a(n) gives the number of occupied "digit slopes" in the factorial base representation of n, or more formally, the number of distinct elements in a multiset [(i_x - d_x) | where d_x ranges over each nonzero digit present in factorial base representation of n and i_x is that digit's position from the right]. Here one-based indexing is used, thus the least significant digit is in position 1. Each value {digit's position} - {digit's value} determines on which slope that particular nonzero digit is. The nonzero digits for which (position - digit) = 0, are said to be on the "maximal slope" (see A260736), those with value 1 on "sub-maximal", etc.
The number of occupied digit slopes translates directly to the number of drops in the n-th permutation as given in the list A060117 because only the largest (and thus leftmost) of all nonzero digits on any particular slope adds a (single) drop to the permutation, when constructed by the unranking algorithm employed in A060117.
The original definition of this sequence is (essentially):
  a(n) = the average of digits (where "digits" may eventually obtain also any values > 9) in each siteswap pattern A060498(n) constructed from each permutation in list A060117, which is equal to number of balls used in that pattern.
The equivalence of the old and the new definitions is seen from the following (as kindly pointed by Olivier Gérard in personal mail): For any permutation p of [1..n], Sum(i=1..n) p(i)-i = 0 (whether taken modulo n or not), thus Sum(i=1..n) (p(i)-i modulo n) = Sum(i={set of nondrops}) (p(i)-i) + Sum(i={set of drops}) (n + (p(i)-i)) = 0 + n * #{set of drops}, where drops is the set of those i where p[i] < i and nondrops are those i for which p[i] >= 1.
Involution A225901 maps this metric to another metric A275806 which gives the number of distinct nonzero digits in factorial base representation of n. See also A275811.
A007489 (repunits in this context) gives the positions where a(n) = A084558(n) (the length of factorial base representation of n). These are also the positions of records.
(End)

			For n=23 ("321" in factorial base representation, A007623), all the digits are maximal for their positions (they occur on the "maximal slope"), thus there is only one distinct digit slope present and a(23)=1. Also, for the 23rd permutation in the ordering A060117, [2341], there is just one drop, as p[4] = 1 < 4.
For n=29 ("1021"), there are three nonzero digits, where both 2 and the rightmost 1 are on the maximal slope, while the most significant 1 is on the "sub-sub-sub-maximal", thus there are two occupied slopes in total, and a(29) = 2. In the 29th permutation of A060117, [23154], there are two drops as p[3] = 1 < 3 and p[5] = 4 < 5.
For n=37 ("1201"), there are three nonzero digits, where the rightmost 1 is on the maximal slope, 2 is on the submaximal, and the most significant 1 is on the "sub-sub-sub-maximal", thus there are three occupied slopes in total, and a(37) = 3. In the 37th permutation of A060117, [51324], there are three drops at indices 2, 4 and 5.

Antti Karttunen, Table of n, a(n) for n = 0..40320
Index entries for sequences related to factorial base representation

Cf. A000120, A001221, A007489, A007623, A033312, A060117, A060118, A060130, A060498, A225901, A275734, A275806.
Cf. A007489 (positions of records, the first occurrence of each n).
Cf. A276001, A276002, A276003 (positions where a(n) obtains values 1, 2, 3).
Cf. also A060500, A060501, A060125, A084558, A153880, A255411, A260736, A273670, A275804, A275811, A275946, A275947, A275849, A275956, A276004,
A276005, A276010.

# The following program follows the original 2001 interpretation of this sequence:
A060502 := n -> avg(Perm2SiteSwap3(PermUnrank3R(n)));
with(group);
permul := (a, b) -> mulperms(b, a);
# factorial_base(n) gives the digits of A007623(n) as a list, uncorrupted even when there are digits > 9:
factorial_base := proc(nn) local n, a, d, j, f; n := nn; if(0 = n) then RETURN([0]); fi; a := []; f := 1; j := 2; while(n > 0) do d := floor(`mod`(n, (j*f))/f); a := [d, op(a)]; n := n - (d*f); f := j*f; j := j+1; od; RETURN(a); end;
# PermUnrank3R(r) gives the permutation with rank r in list A060117:
PermUnrank3R := proc(r) local n; n := nops(factorial_base(r)); convert(PermUnrank3Raux(n+1, r, []), 'permlist', 1+(((r+2) mod (r+1))*n)); end;
PermUnrank3Raux := proc(n, r, p) local s; if(0 = r) then RETURN(p); else s := floor(r/((n-1)!)); RETURN(PermUnrank3Raux(n-1, r-(s*((n-1)!)), permul(p, [[n, n-s]]))); fi; end;
Perm2SiteSwap3 := proc(p) local ip,n,i,a; n := nops(p); ip := convert(invperm(convert(p,'disjcyc')),'permlist',n); a := []; for i from 1 to n do if(0 = ((ip[i]-i) mod n)) then a := [op(a),0]; else a := [op(a), n-((ip[i]-i) mod n)]; fi; od; RETURN(a); end;
avg := a -> (convert(a, `+`)/nops(a));

From Antti Karttunen, Aug 11-21 2016: (Start)
The following formula reflects the original definition of computing the average, with a few unnecessary steps eliminated:
a(n) = 1/s * Sum_{i=1..s} ((p[i]-i) modulo s), where p is the permutation of rank n as ordered in the list A060117, and s is its size (the number of its elements) computed as s = 1+A084558(n).
a(n) = Sum_{i=1..s} [p[i]a(n) = 1/s * Sum_{i=1..s} ((i-p[i]) modulo s). [If inverse permutations from list A060118 are used, then we just flip the order of difference that is used in the first formula].
Following formulas do not need intermediate construction of permutation lists:
a(n) = A001221(A275734(n)).
a(n) = A275806(A225901(n)).
a(n) = A000120(A276010(n)).
Other identities and observations. For all n >= 0:
a(n) = A275946(n) + A275947(n).
a(n) = A060500(A060125(n)).
a(n) = A060128(n) + A276004(n).
a(n) = A060129(n) - A060500(n).
a(n) = A084558(n) - A275849(n) = 1 + A084558(n) - A060501(n).
a(A007489(n)) = n. [Particularly, A007489(n) gives the position of the first occurrence of each n.]
A060128(n) <= a(n) <= A060129(n).
a(n!) = 1.
a(A033312(n)) = 1 for all n > 1.
a(A059590(n)) = A000120(n).
a(A060112(n)) = A007895(n).
a(n) = a(A153880(n)) = a(A255411(n)). [The shift-operations do not change the number of distinct slopes.]
a(A275804(n)) = A060130(A275804(n)). [A275804 gives all the positions where this coincides with A060130.]
(End)

Entry revised, with a new interpretation and formulas. Maple-code cleaned up. - Antti Karttunen, Aug 11 2016

Another new interpretation added and the original definition moved to the comments - Antti Karttunen, Aug 24 2016

A055778 Number of 1's in the base-phi representation of n.

Original entry on oeis.org

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

Offset: 0

Robert Lozyniak (11(AT)onna.com), Jul 12 2000

Uses greedy algorithm (start with largest possible power of phi, then work downward) - see pseudo-code below.
Conjecture: For all n, A007895(n) <= a(n). There is equality at 1, 7, 18, 19, 47, 48, 54, 123, 124, 130, 141, 142, 322, 323, 329, 340, 341, 369, 370, 376, 843, 844, 850, 861, 862, 890, 891, 897, 966, 967, 973, 984, 985, 2207, 2208, 2214, 2225, 2226, 2254, 2255, 2261, 2330, 2331, 2337, 2348, 2349, 2529, 2530, 2536, 2547, 2548, 2576, 2577, 2583, ... - Dale Gerdemann, Apr 01 2012
From Michel Dekking, Feb 06 2021: (Start)
Here is a proof that there are infinitely n many such that A007895(n) = a(n).
Let F(n) = A000045(n) be the n-th Fibonacci number, and let L(n) = A000032(n) be the n-th Lucas number.
Then a(L(2k)) = 2, since L(2k) = phi^(2k) + phi^(-2k), where phi is the golden ratio.
On the other hand, A007895(L(2k)) = 2, since L(n) = F(n+1) + F(n-1) for all n > 1. So for k>1 one has A007895(L(2k)) = a(L(2k)) = 2.
(End)
Gerdemann's conjecture above is true:  we can run the Bellman-Ford algorithm to determine the lowest-weight path from the initial state to a final state in the weighted directed graph derived from the automaton "saka" in my paper cited below in the Links section, and verify that they all have nonnegative weight. - Jeffrey Shallit, May 07 2023
Furthermore, the set of n, in Zeckendorf representation, for which A007895(n) = a(n), is accepted by an 8-state finite automaton. - Jeffrey Shallit, May 08 2023

			The phi-expansions for n<=15 are:
   n   phi-rep(n)     a(n)
   0       0.           0
   1       1.           1
   2      10.01         2
   3     100.01         2
   4     101.01         3
   5    1000.1001       3
   6    1010.0001       3
   7   10000.0001       2
   8   10001.0001       3
   9   10010.0101       4
  10   10100.0101       4
  11   10101.0101       5
  12  100000.101001     4
  13  100010.001001     4
  14  100100.001001     4
  15  100101.001001     5
- _Joerg Arndt_, Jan 30 2012

Carmine Suriano, Table of n, a(n) for n = 0..5000
Michel Dekking, Points of increase of the sum of digits function of the base phi expansion, arXiv:2003.14125 [math.CO], 2020.
F. Michel Dekking, The sum of digits functions of the Zeckendorf and the base phi expansions, Theoretical Computer Science, 2021.
Ron Knott, Using Powers of Phi to represent Integers (Base Phi) (inspiration for this sequence).
Jeffrey Shallit, Proving Properties of phi-Representations with the Walnut Theorem-Prover, arXiv:2305.02672 [math.NT], 2023.
Eric Weisstein's World of Mathematics, Phi Number System

A179242 Numbers that have two terms in their Zeckendorf representation.

Original entry on oeis.org

4, 6, 7, 9, 10, 11, 14, 15, 16, 18, 22, 23, 24, 26, 29, 35, 36, 37, 39, 42, 47, 56, 57, 58, 60, 63, 68, 76, 90, 91, 92, 94, 97, 102, 110, 123, 145, 146, 147, 149, 152, 157, 165, 178, 199, 234, 235, 236, 238, 241, 246, 254, 267, 288, 322, 378, 379, 380, 382, 385, 390

Offset: 1

Emeric Deutsch, Jul 05 2010

nonn

A007895(a(n)) = 2. - Reinhard Zumkeller, Mar 10 2013

			4 = 1+3;
6 = 1+5;
7 = 2+5;
9 = 1+8;
10 = 2+8;

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Cf. A035517, A007895, A179243, A179244, A179245, A179246, A179247, A179248, A179249, A179250, A179251, A179252, A179253.

Cf. A000045.

import Data.List (inits)
a179242 n = a179242_list !! (n-1)
a179242_list = concatMap h $ drop 3 $ inits $ drop 2 a000045_list where
   h is = reverse $ map (+ f) fs where
          (f:_:fs) = reverse is
-- Reinhard Zumkeller, Mar 10 2013

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: Q := {}: for i from fibonacci(5)-1 to 400 do if B(i) = 2 then Q := `union`(Q, {i}) else end if end do: Q;

f[n_] := (k = 1; ff = {}; While[(fi = Fibonacci[k]) <= n, AppendTo[ff, fi]; k++]; Drop[ff, 1]); z[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]; Select[ Range[400], Length[z[#]] == 2 &] (* Jean-François Alcover, Sep 27 2011 *)
zeck = DigitCount[Select[Range[5000], BitAnd[#, 2*#] == 0&], 2, 1];
Position[zeck, 2] // Flatten (* Jean-François Alcover, Jan 25 2018 *)

from math import comb, isqrt
from sympy import fibonacci
def A179242(n): return fibonacci((m:=isqrt(n<<3)+1>>1)+3)+fibonacci(n+1-comb(m, 2)) # Chai Wah Wu, Apr 09 2025

A179253 Numbers k that have 13 terms in their Zeckendorf representation.

Original entry on oeis.org

196417, 271442, 300099, 311045, 315226, 316823, 317433, 317666, 317755, 317789, 317802, 317807, 317809, 317810, 392835, 421492, 432438, 436619, 438216, 438826, 439059, 439148, 439182, 439195, 439200, 439202, 439203, 467860, 478806, 482987

Offset: 1

Emeric Deutsch, Jul 05 2010

nonn

A007895(a(n)) = 13. - Reinhard Zumkeller, Mar 10 2013

			196417 = 1 + 3 + 8 + 21 + 55 + 144 + 377 + 987 + 2584 + 6765 + 17711 + 46368 + 121393;
271442 = 1 + 3 + 8 + 21 + 55 + 144 + 377 + 987 + 2584 + 6765 + 17711 + 46368 + 196418;

Reinhard Zumkeller, Table of n, a(n) for n = 1..1000

Cf. A035517, A007895, A179242, A179243, A179244, A179245, A179246, A179247, A179248, A179249, A179250, A179251, A179252.

a179253 n = a179253_list !! (n-1)
a179253_list = filter ((== 13) . a007895) [1..]
-- Reinhard Zumkeller, Mar 10 2013

with(combinat): seq(add(fibonacci(2*k), k = 1 .. 13-m)+add(fibonacci(27-2*k+2), k = 1 .. m), m = 0 .. 13); # this program yields only the first 14 terms of the sequence
From R. J. Mathar, Jul 23 2010: (Start)
Lzto10 := proc(L) local i ; add( op(i,L)*combinat[fibonacci](i+1),i=1..nops(L) ) ; end proc:
zbits := proc(numbits,toset,upbits) local L,hibi ; if 2*toset-1 > numbits then return ; end if; if toset = 0 then L := [(seq(0,i=1..numbits)),op(upbits)] ; Lzto10(L); print(%) ; else for hibi from toset-1 to numbits -1 do if toset = 1 then procname(hibi,toset-1,[1,seq(0,i=1..numbits-hibi-1),op(upbits)]) ; else procname(hibi-1,toset-1,[0,1,seq(0,i=1..numbits-hibi-1),op(upbits)]) ; end if; end do; end if; return ; end proc:
ztot := 13 : for numbits from 2*ztot -1 to 50 do zbits(numbits-2,ztot-1,[0,1]) ; end do: (End)

Reap[For[m = 0; k = 1, k <= 10^8, k++, If[BitAnd[k, 2 k] == 0, m++; If[DigitCount[k, 2, 1] == 13, Print[m]; Sow[m]]]]][[2, 1]] (* Jean-François Alcover, Aug 20 2023 *)

More terms from R. J. Mathar, Jul 23 2010

A116543 Number of terms in greedy representation of n in terms of the Lucas numbers.

Original entry on oeis.org

0, 1, 1, 1, 1, 2, 2, 1, 2, 2, 2, 1, 2, 2, 2, 2, 3, 3, 1, 2, 2, 2, 2, 3, 3, 2, 3, 3, 3, 1, 2, 2, 2, 2, 3, 3, 2, 3, 3, 3, 2, 3, 3, 3, 3, 4, 4, 1, 2, 2, 2, 2, 3, 3, 2, 3, 3, 3, 2, 3, 3, 3, 3, 4, 4, 2, 3, 3, 3, 3, 4, 4, 3, 4, 4, 4, 1, 2, 2, 2, 2, 3, 3, 2, 3, 3, 3, 2, 3, 3, 3, 3, 4, 4, 2, 3, 3, 3, 3, 4, 4, 3, 4, 4, 4, 2

Offset: 0

James E Davis, Mar 28 2006, Jun 07 2006

I have been studying A007895 and similar sequences and created this sequence as an analog of A007895 for the Lucas sequence (A000032).

			a(12)=2 because 12=11+1.

Clark Kimberling, Table of n, a(n) for n = 0..10000
Ron Knott, Using the Fibonacci numbers to represent whole numbers.

Cf. A000032, A007895, A007953, A130310, A131343.

s = Reverse[Sort[Table[LucasL[n - 1], {n, 1, 22}]]];
t = Map[Length[Select[Reap[FoldList[(Sow[Quotient[#1, #2]]; Mod[#1, #2]) &, #, s]][[2,1]], # > 0 &]] &, Range[1000]] (* Peter J. C. Moses, Oct 18 2012 *)

Let L(n) = max(Lucas numbers < n). Then a(0) = 0, a(n) = 1 + a(n-L(n)).

a(n) = A007953(A130310(n)). - Amiram Eldar, Feb 17 2022

Edited by N. J. A. Sloane, Aug 10 2007

a(0) added by Amiram Eldar, Feb 17 2022

A135817 Length of Wythoff representation of n.

Original entry on oeis.org

1, 1, 2, 3, 2, 4, 3, 3, 5, 4, 4, 4, 3, 6, 5, 5, 5, 4, 5, 4, 4, 7, 6, 6, 6, 5, 6, 5, 5, 6, 5, 5, 5, 4, 8, 7, 7, 7, 6, 7, 6, 6, 7, 6, 6, 6, 5, 7, 6, 6, 6, 5, 6, 5, 5, 9, 8, 8, 8, 7, 8, 7, 7, 8, 7, 7, 7, 6, 8, 7, 7, 7, 6, 7, 6, 6, 8, 7, 7, 7, 6, 7, 6, 6, 7, 6, 6, 6, 5, 10, 9, 9, 9, 8, 9, 8, 8, 9, 8, 8, 8, 7, 9, 8, 8

Offset: 1

Wolfdieter Lang, Jan 21 2008

For the Wythoff representation of n see the W. Lang reference and A189921.
The Wythoff complementary sequences are A(n):=A000201(n) and B(n)=A001950(n), n>=1. The Wythoff representation of n=1 is A(1) and for n>=2 there is a unique representation as composition of A- or B-sequence applied to B(1)=2. E.g., n=4 is A(A(B(1))), written as AAB or as `110`, i.e., 1 for A and 0 for B.
The Wythoff orbit of 1 (starting always with B(1), applying any number of A- or B-sequences) produces every number n>1 just once. This produces a binary Wythoff code for n>1, ending always in 0 (for B(1)). See the W. Lang link for this code.

			W(4) = `110`, i.e., 4 = A(A(B(1))) with Wythoff's A and B sequences.

Wolfdieter Lang, The Wythoff and the Zeckendorf representations of numbers are equivalent, in G. E. Bergum et al. (editors), Application of Fibonacci numbers, vol. 6, Kluwer, Dordrecht, 1996, pp. 319-337. [See A317208 for a link.]

Amiram Eldar, Table of n, a(n) for n = 1..10000
Aviezri S. Fraenkel, From Enmity to Amity, American Mathematical Monthly 117 (2010) 646-648.
Clark Kimberling, The Zeckendorf array equals the Wythoff array, Fibonacci Quarterly 33 (1995) 3-8.
Wolfdieter Lang, Wythoff representations for n=1...150.

Cf. A135818 (number of 1's or A's in Wythoff representation of n).
Cf. A007895 (number of 0's or B's in Wythoff representation of n).
Row lengths of A189921.

z[n_] := Floor[(n + 1)*GoldenRatio] - n - 1; h[n_] := z[n] - z[n - 1]; w[n_] := Module[{m = n, zm = 0, hm, s = {}}, While[zm != 1, hm = h[m]; AppendTo[s, hm]; If[hm == 1, zm = z[m], zm = z[z[m]]]; m = zm]; s]; w[0] = 0; a[n_] := Length[w[n]]; Array[a, 100] (* Amiram Eldar, Jul 01 2023 *)

a(n) = number of digits in Wythoff representation of n>=1.
a(n) = length of Wythoff code for n>=1.
a(n) = number of applications of Wythoff sequences A or B on 1 in the Wythoff representation for n >=1.

A265745 a(n) is the number of Jacobsthal numbers (A001045) needed to sum to n using the greedy algorithm.

Original entry on oeis.org

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

Offset: 0

Antti Karttunen, Dec 17 2015

Sum of digits in "Jacobsthal greedy base", A265747.
It would be nice to know for sure whether this sequence gives also the least number of Jacobsthal numbers that add to n, i.e., that there cannot be even better nongreedy solutions.
The integer 63=21+21+21 has 3 for its 'non-greedy' solution, and a(63) = 5 for its greedy solution 63=43+11+5+3+1. - Yuriko Suwa, Jul 11 2021
Positions where a(n) is different from A372555(n) are n=63, 84, 148, 169, 191, 212, 234, 255, etc. See A372557. - Antti Karttunen, May 07 2024

			a(0) = 0, because no numbers are needed to form an empty sum, which is zero.
For n=1 we need just A001045(2) = 1, thus a(1) = 1.
For n=2 we need A001045(2) + A001045(2) = 1 + 1, thus a(2) = 2.
For n=4 we need A001045(3) + A001045(2) = 3 + 1, thus a(4) = 2.
For n=6 we form the greedy sum as A001045(4) + A001045(2) = 5 + 1, thus a(6) = 2. Alternatively, we could form the sum as A001045(3) + A001045(3) = 3 + 3, but the number of summands in that case is no less.
For n=7 we need A001045(4) + A001045(2) + A001045(2) = 5 + 1 + 1, thus a(7) = 3.
For n=8 we need A001045(4) + A001045(3) = 5 + 3, thus a(8) = 2.
For n=10 we need A001045(4) + A001045(4) = 5 + 5, thus a(10) = 2.

Antti Karttunen, Table of n, a(n) for n = 0..10923

Cf. A001045, A130249, A197911, A003158, A265746, A265747, A364377, A364379, A364380, A372288, A372555, A372557.
Cf. A054111 (apparently the positions of the first occurrence of each n > 0).
Cf. also A007895, A014420, A053610, A265404, A265743, A265744.

jacob[n_] := (2^n - (-1)^n)/3; maxInd[n_] := Floor[Log2[3*n + 1]]; A265745[n_] := A265745[n] = 1 + A265745[n - jacob[maxInd[n]]]; A265745[0] = 0; Array[A265745, 100, 0] (* Amiram Eldar, Jul 21 2023 *)

A130249(n) = floor(log(3*n + 1)/log(2));
A001045(n) = (2^n - (-1)^n) / 3;
A265745(n) = {if(n == 0, 0, my(d = n - A001045(A130249(n))); if(d == 0, 1, 1 + A265745(d)));} \\ Amiram Eldar, Jul 21 2023

def greedyJ(n): n1 = (3*n+1).bit_length() - 1; return (2**n1 - (-1)**n1)//3
def a(n): return 0 if n == 0 else 1 + a(n - greedyJ(n))
print([a(n) for n in range(107)]) # Michael S. Branicky, Jul 11 2021

a(0) = 0; for n >= 1, a(n) = 1 + a(n - A001045(A130249(n))). [This formula uses a simple greedy algorithm.]

A328211 Starts of runs of 4 consecutive Zeckendorf-Niven numbers (A328208).

Original entry on oeis.org

1, 2, 3, 123543, 124242, 545502, 1367583, 1856349, 2431230, 2465110, 2593590, 2783709, 3247389, 3479229, 3917823, 3942909, 4174749, 4303428, 4494390, 4920640, 5143830, 5710383, 6261309, 6493149, 6552903, 6956829, 7420509, 7470880, 8970948, 9107790, 9507069, 10952928

Offset: 1

Amiram Eldar, Oct 07 2019

nonn

Grundman proved that this sequence is infinite by showing the F(120k-6) + F(8) + F(6) + F(4) is a term for all k >= 1, where F(k) is the k-th Fibonacci number.

She also proved that the only starts of runs of 5 consecutive Zeckendorf-Niven numbers are 1 and 2.

			1 is in the sequence since 1, 2, 3 and 4 are in A328208: A007895(1) = 1 is a divisor of 1, A007895(2) = 1 is a divisor of 2, A007895(3) = 1 is a divisor of 3, and A007895(4) = 2 is a divisor of 4.

Amiram Eldar, Table of n, a(n) for n = 1..216
Helen G. Grundman, Consecutive Zeckendorf-Niven and lazy-Fibonacci-Niven numbers, Fibonacci Quarterly, Vol. 45, No. 3 (2007), pp. 272-276.

Cf. A005349, A007895, A141769, A328208.

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, {4}]; While[c < 32, If[aQ[k], v = Join[Rest[v], {k}]; If[AllTrue[Differences[v], # == 1 &], c++; AppendTo[s, k - 3]]]; k++]; s (* after Alonso del Arte at A007895 *)

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A010049 Second-order Fibonacci numbers.

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A328210 Starts of runs of 3 consecutive Zeckendorf-Niven numbers (A328208).

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A060502 a(n) = number of occupied digit slopes in the factorial base representation of n (see comments for the definition); number of drops in the n-th permutation of list A060117.

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A055778 Number of 1's in the base-phi representation of n.

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A179242 Numbers that have two terms in their Zeckendorf representation.

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A179253 Numbers k that have 13 terms in their Zeckendorf representation.

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