A022342 -id:A022342 - cp's OEIS frontend

A088462 a(1)=1, a(n) = ceiling((n - a(a(n-1)))/2).

1, 1, 1, 2, 2, 3, 3, 4, 4, 4, 5, 5, 6, 6, 6, 7, 7, 8, 8, 8, 9, 9, 10, 10, 11, 11, 11, 12, 12, 13, 13, 13, 14, 14, 15, 15, 16, 16, 16, 17, 17, 18, 18, 18, 19, 19, 20, 20, 21, 21, 21, 22, 22, 23, 23, 23, 24, 24, 25, 25, 25, 26, 26, 27, 27, 28, 28, 28, 29, 29, 30, 30, 30, 31, 31, 32, 32

Offset: 1

Benoit Cloitre, Nov 12 2003

nonn

Partial sums of A004641. - Reinhard Zumkeller, Dec 05 2009

This sequence generates A004641; see comment at A004641. - Clark Kimberling, May 25 2011

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
N. J. A. Sloane, Families of Essentially Identical Sequences, Mar 24 2021 (Includes this sequence)

Cf. A005206.

The following sequences are all essentially the same, in the sense that they are simple transformations of each other, with A000201 as the parent: A000201, A001030, A001468, A001950, A003622, A003842, A003849, A004641, A005614, A014675, A022342, A088462, A096270, A114986, A124841. - N. J. A. Sloane, Mar 11 2021

[Floor((Sqrt(2)-1)*n+1/Sqrt(2)): n in [1..100]]; // Vincenzo Librandi, Jun 26 2017

Table[Floor[(Sqrt[2] - 1) n + 1 / Sqrt[2]], {n, 100}] (* Vincenzo Librandi, Jun 26 2017 *)

l=[0, 1, 1]
for n in range(3, 101): l.append(n - l[n - 1] - l[l[n - 2]])
print(l[1:]) # Indranil Ghosh, Jun 24 2017, after Altug Alkan

a(n) = floor((sqrt(2)-1)*n + 1/sqrt(2)).

a(1) = a(2) = 1; a(n) = n - a(n-1) - a(a(n-2)) for n > 2. - Altug Alkan, Jun 24 2017

A114986 Characteristic function of (A000201 prefixed with 0).

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0

Offset: 0

N. J. A. Sloane, Feb 28 2006

Lord, E. A, Ranganathan, S. and Subramaniam, A., Stacking sequences and symmetry properties of trigonal vacancy-ordered phases, Phil. Mag. A 82 (2002) 255-268.
N. J. A. Sloane, Families of Essentially Identical Sequences, Mar 24 2021 (Includes this sequence)
Index entries for characteristic functions

Essentially the same as A005614. Cf. A096270, A189479.

The following sequences are all essentially the same, in the sense that they are simple transformations of each other, with A000201 as the parent: A000201, A001030, A001468, A001950, A003622, A003842, A003849, A004641, A005614, A014675, A022342, A088462, A096270, A114986, A124841. - N. J. A. Sloane, Mar 11 2021

A054770 Numbers that are not the sum of distinct Lucas numbers 1,3,4,7,11, ... (A000204).

Original entry on oeis.org

2, 6, 9, 13, 17, 20, 24, 27, 31, 35, 38, 42, 46, 49, 53, 56, 60, 64, 67, 71, 74, 78, 82, 85, 89, 93, 96, 100, 103, 107, 111, 114, 118, 122, 125, 129, 132, 136, 140, 143, 147, 150, 154, 158, 161, 165, 169, 172, 176, 179, 183, 187, 190, 194, 197, 201, 205, 208, 212

Offset: 1

Antreas P. Hatzipolakis (xpolakis(AT)otenet.gr), May 28 2000

Alternatively, Lucas representation of n includes L_0 = 2. - Fred Lunnon, Aug 25 2001
Conjecture: this is the sequence of numbers for which the base phi representation includes phi itself, where phi = (1 + sqrt(5))/2 = the golden ratio. Example: let r = phi; then 6 = r^3 + r + r^(-4). - Clark Kimberling, Oct 17 2012
This conjecture is proved in my paper 'Base phi representations and golden mean beta-expansions', using the formula by Wilson/Agol/Carlitz et al. - Michel Dekking, Jun 25 2019
Numbers whose minimal Lucas representation (A130310) ends with 1. - Amiram Eldar, Jan 21 2023

G. C. Greubel, Table of n, a(n) for n = 1..5000
L. Carlitz, R. Scoville, and V. E. Hoggatt, Jr., Lucas representations, Fibonacci Quart. 10 (1972), 29-42, 70, 112.
Weiru Chen and Jared Krandel, Interpolating Classical Partitions of the Set of Positive Integers, arXiv:1810.11938 [math.NT], 2018. See sequence D2 p. 4.
Michel Dekking, Base phi representations and golden mean beta-expansions, arXiv:1906.08437 [math.NT], 2019.
Jared Krandel and Weiru Chen, Interpolating classical partitions of the set of positive integers, The Ramanujan Journal (2020).

Complement of A063732.

Cf. A003263, A003622, A022342, A130310.

[Floor(n*(Sqrt(5)+5)/2)-1: n in [1..60]]; // Vincenzo Librandi, Oct 30 2018

A054770 := n -> floor(n*(sqrt(5)+5)/2)-1;

Complement[Range[220],Total/@Subsets[LucasL[Range[25]],5]] (* Harvey P. Dale, Feb 27 2012 *)
Table[Floor[n (Sqrt[5] + 5) / 2] - 1, {n, 60}] (* Vincenzo Librandi, Oct 30 2018 *)

PARI
```
a(n)=floor(n*(sqrt(5)+5)/2)-1
```

from math import isqrt
def A054770(n): return (n+isqrt(5*n**2)>>1)+(n<<1)-1 # Chai Wah Wu, Aug 17 2022

a(n) = floor(((5+sqrt(5))/2)*n)-1 (conjectured by David W. Wilson; proved by Ian Agol (iagol(AT)math.ucdavis.edu), Jun 08 2000)
a(n) = A000201(n) + 2*n - 1. - Michel Dekking, Sep 07 2017
G.f.: x*(x+1)/(1-x)^2 + Sum_{i>=1} (floor(i*phi)*x^i), where phi = (1 + sqrt(5))/2. - Iain Fox, Dec 19 2017
Ian Agol tells me that David W. Wilson's formula is proved in the Carlitz, Scoville, Hoggatt paper 'Lucas representations'. See Equation (1.12), and use A(A(n))+n = B(n)+n-1 = A(n)+2*n-1, the well known formulas for the lower Wythoff sequence A = A000201, and the upper Wythoff sequence B = A001950. - Michel Dekking, Jan 04 2018

More terms from James Sellers, May 28 2000

A060143 a(n) = floor(n/tau), where tau = (1 + sqrt(5))/2.

Original entry on oeis.org

0, 0, 1, 1, 2, 3, 3, 4, 4, 5, 6, 6, 7, 8, 8, 9, 9, 10, 11, 11, 12, 12, 13, 14, 14, 15, 16, 16, 17, 17, 18, 19, 19, 20, 21, 21, 22, 22, 23, 24, 24, 25, 25, 26, 27, 27, 28, 29, 29, 30, 30, 31, 32, 32, 33, 33, 34, 35, 35, 36, 37, 37, 38, 38, 39, 40, 40, 41, 42, 42, 43, 43, 44, 45, 45

Offset: 0

Clark Kimberling, Mar 05 2001

Fibonacci base shift right: for n >= 0, a(n+1) = Sum_{k in A_n} F_{k-1}, where n = Sum_{k in A_n} F_k (unique) expression of n as a sum of "noncontiguous" Fibonacci numbers (with index >=2). - Michele Dondi (bik.mido(AT)tiscalenet.it), Dec 30 2001 [corrected, and aligned with sequence offset by Peter Munn, Jan 10 2018]
Numerators a(n) of fractions slowly converging to phi, the golden ratio: let a(1) = 0, b(n) = n - a(n); if (a(n) + 1) / b(n) < (1 + sqrt(5))/2, then a(n+1) = a(n) + 1, else a(n+1) = a(n). a(n) + b(n) = n and as n -> +infinity, a(n) / b(n) converges to (1 + sqrt(5))/2. For all n, a(n) / b(n) < (1 + sqrt(5))/2. - Robert A. Stump (bee_ess107(AT)msn.com), Sep 22 2002
a(10^n) gives the first few digits of phi=(sqrt(5)-1)/2.
Comment corrected, two alternative ways, by Peter Munn, Jan 10 2018: (Start)
(a(n) = a(n+1) or a(n) = a(n-1)) if and only if a(n) is in A066096.
a(n+1) = a(n+2) if and only if n is in A003622.
(End)
From Wolfdieter Lang, Jun 28 2011: (Start)
a(n+1) counts for n >= 1 the number of Wythoff A-numbers not exceeding n.
a(n+1) counts also the number of Wythoff B-numbers smaller than A(n+2), with the Wythoff A- and B-sequences A000201 and A001950, respectively.
a(n+1) = Sum_{j=1..n} A005614(j-1) for n >= 1 (no rounding problems like in the above definition, because the rabbit sequence A005614(n-1) for n >= 1, can be defined by a substitution rule).
a(n+1) = A(n+1)-(n+1) (serving, together with the last equation, as definition for A(n+1), given the rabbit sequence).
a(n+1) = A005206(n), n >= 0.
(End)
Let b(n) = floor((n+1)/phi). Then b(n) + b(b(n-1)) = n [Granville and Rasson]. - N. J. A. Sloane, Jun 13 2014

			a(6)= 3 so b(6) = 6 - 3 = 3. a(7) = 4 because (a(6) + 1) / b(6) = 4/3 which is < (1 + sqrt(5))/2. So b(7) = 7 - 4 = 3. a(8) = 4 because (a(7) + 1) / b(7) = 5/3 which is > (1 + sqrt(5))/2. - Robert A. Stump (bee_ess107(AT)msn.com), Sep 22 2002
From _Wolfdieter Lang_, Jun 28 2011: (Start)
There are a(4) = 2 (positive) Wythoff A-numbers <= 3, namely 1 and 3.
There are a(4) = 2 (positive) Wythoff B-numbers < A(4) = 6, namely 2 and 5.
a(4) = 2 = A(4) - 4 = 6 - 4.
(End)

William A. Tedeschi, Table of n, a(n) for n = 0..10000 [This replaces an earlier b-file computed by Harry J. Smith]
M. Celaya and F. Ruskey (Proposers), Another property of only the golden ratio, Problem 11651, Amer. Math. Monthly, 121 (2014), 550-551.
F. Michel Dekking, Morphisms, Symbolic Sequences, and Their Standard Forms, Journal of Integer Sequences, Vol. 19 (2016), Article 16.1.1.
Vincent Granville and Jean-Paul Rasson, A strange recursive relation, J. Number Theory 30 (1988), no. 2, 238--241. MR0961919(89j:11014). - From _N. J. A. Sloane_, Jun 13 2014
Jeffrey Shallit, The Hurt-Sada Array and Zeckendorf Representations, arXiv:2501.08823 [math.NT], 2025. See p. 2.

Cf. A000045 (Fibonacci numbers), A003622, A022342, A035336.
Terms that occur only once: A001950.
Terms that occur twice: A066096 (a version of A000201).
Numerator sequences for other values, as described in Robert A. Stump's 2002 comment: A074065 (sqrt(3)), A074840 (sqrt(2)).
Apart from initial terms, same as A005206.
First differences: A096270 (a version of A005614).
Partial sums: A183136.

[Floor(2*n/(1+Sqrt(5))): n in [0..80]]; // Vincenzo Librandi, Mar 29 2015

Floor[Range[0,80]/GoldenRatio] (* Harvey P. Dale, May 09 2013 *)

{ default(realprecision, 10); p=(sqrt(5) - 1)/2; for (n=0, 1000, write("b060143.txt", n, " ", floor(n*p)); ) } \\ Harry J. Smith, Jul 02 2009

from math import isqrt
def A060143(n): return (n+isqrt(5*n**2)>>1)-n # Chai Wah Wu, Aug 10 2022

a(n) = floor(phi(n)), where phi=(sqrt(5)-1)/2. [corrected by Casey Mongoven, Jul 18 2008]
a(F_n + 1) = F_{n-1} if F_n is the n-th Fibonacci number. [aligned with sequence offset by Peter Munn, Jan 10 2018]
a(1) = 0. b(n) = n - a(n). If (a(n) + 1) / b(n) < (1 + sqrt(5))/2, then a(n+1) = a(n) + 1, else a(n+1) = a(n). - Robert A. Stump (bee_ess107(AT)msn.com), Sep 22 2002 [corrected by Peter Munn, Jan 07 2018]
A006336(n) = A006336(n-1) + A006336(a(n)) for n>1. - Reinhard Zumkeller, Oct 24 2007
a(n) = floor(n*phi) - n, where phi = (1+sqrt(5))/2. - William A. Tedeschi, Mar 06 2008
Celaya and Ruskey give an interesting formula for a(n). - N. J. A. Sloane, Jun 13 2014

I merged three identical sequences to create this entry. Some of the formulas may need their initial terms adjusting now. - N. J. A. Sloane, Mar 05 2003

More terms from William A. Tedeschi, Mar 06 2008

A124841 Inverse binomial transform of A005614, the rabbit sequence: (1, 0, 1, 1, 0, ...).

Original entry on oeis.org

1, -1, 2, -3, 3, 0, -10, 35, -90, 200, -400, 726, -1188, 1716, -2080, 1820, -312, -2704, 5408, 455, -39195, 170313, -523029, 1352078, -3114774, 6548074, -12668578, 22492886, -36020998, 49549110, -49549110, 0, 182029056, -670853984, 1809734560, -4242470755

Offset: 0

Gary W. Adamson, Nov 10 2006

sign

As with every inverse binomial transform, the numbers are given by starting from the sequence (A005614) and reading the leftmost values of the array of repeated differences.

			Given 1, 0, 1, 1, 0, ..., take finite difference rows:
1, 0, 1, 1, 0, ...
_-1, 1, 0, -1, ...
___ 2, -1, -1, ...
_____ -3, 0, ...
________ 3, ...
Left border becomes the sequence.

Alois P. Heinz, Table of n, a(n) for n = 0..1000
N. J. A. Sloane, Families of Essentially Identical Sequences, Mar 24 2021 (Includes this sequence)

Cf. A124842.

The following sequences are all essentially the same, in the sense that they are simple transformations of each other, with A000201 as the parent: A000201, A001030, A001468, A001950, A003622, A003842, A003849, A004641, A005614, A014675, A022342, A088462, A096270, A114986, A124841. - N. J. A. Sloane, Mar 11 2021

A005614 = SubstitutionSystem[{0 -> {1}, 1 -> {1, 0}}, {1, 0}, 7] // Last;
Table[Differences[A005614, n], {n, 0, 35}][[All, 1]] (* Jean-François Alcover, Feb 06 2020 *)

Corrected and extended by R. J. Mathar, Nov 28 2011

A035612 Horizontal para-Fibonacci sequence: says which column of Wythoff array (starting column count at 1) contains n.

Original entry on oeis.org

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

Offset: 1

J. H. Conway and N. J. A. Sloane

Ordinal transform of A003603. Removing all 1's from this sequence and decrementing the remaining numbers generates the original sequence. - Franklin T. Adams-Watters, Aug 10 2012
It can be shown that a(n) is the index of the smallest Fibonacci number used in the Zeckendorf representation of n, where f(0)=f(1)=1. - Rachel Chaiser, Aug 18 2017
The asymptotic density of the occurrences of k = 1, 2, ..., is (2-phi)/phi^(k-1), where phi is the golden ratio (A001622). The asymptotic mean of this sequence is 1 + phi (A104457). - Amiram Eldar, Nov 02 2023

			After the first 6 we see "1 2 3 1 4 1 2" then 7.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Paul Curtz, Comments on A035612, Jan 25 2016.
N. J. A. Sloane, My favorite integer sequences, in Sequences and their Applications (Proceedings of SETA '98).
N. J. A. Sloane, Classic Sequences.

Cf. A019586, A035513, A035614.
Cf. A000045, A003603.
Cf. A003714, A007814, A022340, A022342, A026274.
Cf. A000012, A000027, A001045, A001610, A003622, A023548, A255671, A268034.
Cf. A001622, A104457, A132338.

a035612 = a007814 . a022340
-- Reinhard Zumkeller, Jul 20 2015, Mar 10 2013

f[1] = {1}; f[2] = {1, 2}; f[n_] := f[n] = Join[f[n-1], Most[f[n-2]], {n}]; f[11] (* Jean-François Alcover, Feb 22 2012 *)

The segment between the first M and the first M+1 is given by the segment before the first M-1.
a(A022342(n)) > 1; a(A026274(n) + 1) = 1. - Reinhard Zumkeller, Jul 20 2015
a(n) = v2(A022340(n)), where v2(n) = A007814(n), the dyadic valuation of n. - Ralf Stephan, Jun 20 2004. In other words, a(n) = A007814(A003714(n)) + 1, which is certainly true. - Don Reble, Nov 12 2005
From Rachel Chaiser, Aug 18 2017: (Start)
a(n) = a(p(n))+1 if n = b(p(n)) where p(n) = floor((n+2)/phi)-1 and b(n) = floor((n+1)*phi)-1 where phi=(1+sqrt(5))/2; a(n)=1 otherwise.
a(n) = 3 - n_1 + s_z(n-1) - s_z(n) + s_z(p(n-1)) - s_z(p(n)), where s_z(n) is the Zeckendorf sum of digits of n (A007895), and n_1 is the least significant digit in the Zeckendorf representation of n. (End)

Formula corrected by Tom Edgar, Jul 09 2018

A219641 a(n) = n minus (number of 1's in Zeckendorf expansion of n).

Original entry on oeis.org

0, 0, 1, 2, 2, 4, 4, 5, 7, 7, 8, 9, 9, 12, 12, 13, 14, 14, 16, 16, 17, 20, 20, 21, 22, 22, 24, 24, 25, 27, 27, 28, 29, 29, 33, 33, 34, 35, 35, 37, 37, 38, 40, 40, 41, 42, 42, 45, 45, 46, 47, 47, 49, 49, 50, 54, 54, 55, 56, 56, 58, 58, 59, 61, 61, 62, 63, 63, 66

Offset: 0

Antti Karttunen, Nov 24 2012

nonn

See A014417 for the Fibonacci number system representation, also known as Zeckendorf expansion.

Antti Karttunen, Table of n, a(n) for n = 0..10000
Paul Baird-Smith, Alyssa Epstein, Kristen Flint, and Steven J. Miller, The Zeckendorf Game, arXiv:1809.04881 [math.NT], 2018.

Cf. A007895, A014417. A022342 gives the positions of records, resulting the same sequence with duplicates removed: A219640. A035336 gives the positions of values that occur only once: A219639. Cf. also A219637, A219642. Analogous sequence for binary system: A011371, for factorial number system: A219651.

zeck = DigitCount[Select[Range[0, 500], BitAnd[#, 2*#] == 0&], 2, 1];
Range[0, Length[zeck]-1] - zeck (* Jean-François Alcover, Jan 25 2018 *)

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 str(x).count("1")
print([n - a(n) for n in range(101)]) # Indranil Ghosh, Jun 09 2017

Scheme
```
(define (A219641 n) (- n (A007895 n)))
```

a(n) = n - A007895(n).

A105774 A "fractal" transform of the Fibonacci numbers: a(1)=1; then if F(n) < k <= F(n+1), a(k) = F(n+1) - a(k - F(n)) where F(n) = A000045(n).

Original entry on oeis.org

1, 1, 2, 4, 4, 7, 7, 6, 12, 12, 11, 9, 9, 20, 20, 19, 17, 17, 14, 14, 15, 33, 33, 32, 30, 30, 27, 27, 28, 22, 22, 23, 25, 25, 54, 54, 53, 51, 51, 48, 48, 49, 43, 43, 44, 46, 46, 35, 35, 36, 38, 38, 41, 41, 40, 88, 88, 87, 85, 85, 82, 82, 83, 77, 77, 78, 80, 80, 69, 69, 70, 72, 72

Offset: 1

Benoit Cloitre, May 04 2005

nonn

Let tau = (1+sqrt(5))/2; then the missing numbers 3,5,8,10,13,16,18,21,... are given by round(tau^2*k) for k > 0 (A004937).
Indices n such that a(n) = a(n+1) are given by floor(tau^2*k) - 1 for k > 0 (A003622).
Numbers n such that a(n) differs from a(n+1) are given by floor(tau*k+1/tau) for k > 0 (A022342).
Indices n giving isolated terms (a(n) differs from a(n-1) and a(n+1)) are given by floor(tau*floor(tau^2*k)) for k > 0 (A003623).
Remove 0's from the first differences of sorted values; then you get a version of the infinite Fibonacci word (A001468). I.e., sorted values are 1,1,2,4,4,6,7,7,9,9,11,12,12,..., first differences are 0,1,2,0,2,1,0,2,0,2,1,0,2,0,1,...; removing 0's gives 1,2,2,1,2,2,1,2,1,2,2,1,2,1,2,2,1,2,... #{ k : a(k)=k}=infty.

			For 1 = F(2) < k <= F(3) = 2 the rule gives a(2) = 2 - a(1) = 1 ... if 5 = F(5) < k <= F(6) = 8 the rule forces a(6) = 8 - a(6-5) = 8 - a(1) = 7; a(7) = 8 - a(2) = 7; a(8) = 8 - a(3) = 6.

Antti Karttunen, Table of n, a(n) for n = 1..10946
Benoit Cloitre and Jeffrey Shallit, Some Fibonacci-Related Sequences, arXiv:2312.11706 [math.CO], 2023.
Jeffrey Shallit, Using automata to prove theorems about sequences, One FLAT World Seminar, 2024.

Cf. A000045, A001468, A003622, A003623, A004937, A006498, A022342, A072649, A085002, A105669, A105670, A105672, A093347, A093348, A283766.

a(A000045(n)) = A006498(n-1) for n >= 1. - Typo corrected by Antti Karttunen, Mar 17 2017
limsup a(n)/n = tau and liminf a(n)/n = (tau+2)/5 where tau = (1+sqrt(5))/2. - Corrected by Jeffrey Shallit, Dec 17 2023
a(n) mod 2 = A085002(n) - Benoit Cloitre, May 10 2005
a(1) = 1; for n > 1, a(n) = A000045(2+A072649(n-1)) - a(n-A000045(1 + A072649(n-1))). - Antti Karttunen, Mar 17 2017

A219640 Numbers m for which there exists k such that m = k - (number of 1's in Zeckendorf expansion of k); distinct values in A219641.

Original entry on oeis.org

0, 1, 2, 4, 5, 7, 8, 9, 12, 13, 14, 16, 17, 20, 21, 22, 24, 25, 27, 28, 29, 33, 34, 35, 37, 38, 40, 41, 42, 45, 46, 47, 49, 50, 54, 55, 56, 58, 59, 61, 62, 63, 66, 67, 68, 70, 71, 74, 75, 76, 78, 79, 81, 82, 83, 88, 89, 90, 92, 93, 95, 96, 97, 100, 101, 102

Offset: 0

Antti Karttunen, Nov 24 2012

nonn

These are the positive integers i for which there exists k such that A007895(i+k)=k.

Starting offset is zero, because a(0) = 0 is a special case. Start indexing from 1 when you want only nonzero natural numbers satisfying the same condition.

Antti Karttunen, Table of n, a(n) for n = 0..10947

Cf. A007895, A022342, A219641. Complement: A219638. Union of A219639 and A219637.
First differences: A261095.
Characteristic function: A261092.
Left inverses: A261093, A261094.
Analogous sequences for other bases: A005187, A219650.

(define (A219640 n) (A219641 (A022342 (+ 1 n))))

a(n) = A219641(A022342(n+1)).
Other identities. For all n >= 0:
A261093(a(n)) = n.
A261094(a(n)) = n.

Starting offset changed to 0 by Antti Karttunen, Aug 08 2015

A356874 Write n as Sum_{i in S} 2^(i-1), where S is a set of positive integers, then a(n) = Sum_{i in S} F_i, where F_i is the i-th Fibonacci number, A000045(i).

Original entry on oeis.org

0, 1, 1, 2, 2, 3, 3, 4, 3, 4, 4, 5, 5, 6, 6, 7, 5, 6, 6, 7, 7, 8, 8, 9, 8, 9, 9, 10, 10, 11, 11, 12, 8, 9, 9, 10, 10, 11, 11, 12, 11, 12, 12, 13, 13, 14, 14, 15, 13, 14, 14, 15, 15, 16, 16, 17, 16, 17, 17, 18, 18, 19, 19, 20, 13, 14, 14, 15, 15, 16, 16, 17, 16, 17, 17, 18, 18, 19, 19, 20

Offset: 0

Peter Munn, Sep 02 2022

This sequence looks on the Fibonacci sequence terms (from F_1 onwards) as an ordered set, considering F_1 and F_2 as distinct members mapped to the same integer value, namely 1. We run through all finite subsets, adding up the integer values from each subset (see table in the examples). In consequence, a number, k, occurs exactly A000121(k) times.
The definition is effectively an amendment of the definition of A022290 so that a(n) is a sum of Fibonacci terms starting with F_1 rather than F_2. If we took this a further step so that a(n) was a sum of Fibonacci terms starting with F_0, we would get this sequence with each term duplicated, that is 0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, ... .
Doubling n increments the indices of the Fibonacci terms that sum to a(n), so a(2n) is in this sense a Fibonacci successor to a(n). When n is even, this sense matches the meaning of successor as used in A022342; however, for odd n, the generated successor of a(n) is A000201(a(n)) -- note, in this respect, that A000201 is a column in the extended Wythoff array (A287870).

			n = 13: 13 = 8 + 4 + 1 = 2^(4-1) + 2^(3-1) + 2^(1-1); so a(n) = F_4 + F_3 + F_1 = 3 + 2 + 1 = 6.
Table showing initial terms with Fibonacci subsets:
  n   Fibonacci subset                           a(n)
  0  { }     (empty set)     ->  0              =  0
  1  { F_1 }                 ->  1              =  1
  2  {      F_2 }            ->  0 + 1          =  1
  3  { F_1, F_2 }            ->  1 + 1          =  2
  4  {           F_3 }       ->  0 + 0 + 2      =  2
  5  { F_1,      F_3 }       ->  1 + 0 + 2      =  3
  6  {      F_2, F_3 }       ->  0 + 1 + 2      =  3
  7  { F_1, F_2, F_3 }       ->  1 + 1 + 2      =  4
  8  {                F_4 }  ->  0 + 0 + 0 + 3  =  3
  9  { F_1,           F_4 }  ->  1 + 0 + 0 + 3  =  4

Amiram Eldar, Table of n, a(n) for n = 0..10000

Cf. A000045, A000121, A000201, A022342, A287870.

Even bisection: A022290.

a[n_] := Module[{d = IntegerDigits[n, 2], nd}, nd = Length[d]; Total[d * Fibonacci[Range[nd, 1, -1]]]]; Array[a, 100, 0] (* Amiram Eldar, Aug 08 2023 *)

a(n) = if(n==0,0,if(n%2==1,a(n-1)+1,a(n/2)+a(n\4))); \\ Peter Munn, Sep 06 2022

a(2n) = a(n) + a(floor(n/2)); a(2n+1) = a(2n) + 1.
a(2^k) = A000045(k+1).
For k >= 1, 2^k < n < 2^(k+1), a(n) = a(n - 2^k) + a(2^k).
a(2n) = A022290(n).
a(4n) = A022342(a(2n)+1).
a(4n+2) = A000201(a(2n+1)).

cp's OEIS Frontend

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