A197911 -id:A197911 - cp's OEIS frontend

A050292 a(2n) = 2n - a(n), a(2n+1) = 2n + 1 - a(n) (for n >= 0).

0, 1, 1, 2, 3, 4, 4, 5, 5, 6, 6, 7, 8, 9, 9, 10, 11, 12, 12, 13, 14, 15, 15, 16, 16, 17, 17, 18, 19, 20, 20, 21, 21, 22, 22, 23, 24, 25, 25, 26, 26, 27, 27, 28, 29, 30, 30, 31, 32, 33, 33, 34, 35, 36, 36, 37, 37, 38, 38, 39, 40, 41, 41, 42, 43, 44, 44, 45, 46, 47, 47, 48, 48, 49, 49, 50, 51, 52, 52, 53, 54

Offset: 0

Eric W. Weisstein

Note that the first equation implies a(0)=0, so there is no need to specify an initial value.
Maximal cardinality of a double-free subset of {1, 2, ..., n}, or in other words, maximal size of a subset S of {1, 2, ..., n} with the property that if x is in S then 2x is not. a(0)=0 by convention.
Least k such that a(k)=n is equal to A003159(n).
To construct the sequence: let [a, b, c, a, a, a, b, c, a, b, c, ...] be the fixed point of the morphism a -> abc, b ->a, c -> a, starting from a(1) = a, then write the indices of a, b, c, that of a being written twice; see A092606. - Philippe Deléham, Apr 13 2004
Number of integers from {1,...,n} for which the subtraction of 1 changes the parity of the number of 1's in their binary expansion. - Vladimir Shevelev, Apr 15 2010
Number of integers from {1,...,n} the factorization of which over different terms of A050376 does not contain 2. - Vladimir Shevelev, Apr 16 2010
a(n) modulo 2 is the Prouhet-Thue-Morse sequence A010060. Each number n appears A026465(n+1) times. - Philippe Deléham, Oct 19 2011
Another way of stating the last two comments from Philippe Deléham: the sequence can be obtained by replacing each term of the Thue-Morse sequence A010060 by the run number that term is in. - N. J. A. Sloane, Dec 31 2013

			Examples for n = 1 through 8: {1}, {1}, {1,3}, {1,3,4}, {1,3,4,5}, {1,3,4,5}, {1,3,4,5,7}, {1,3,4,5,7}.
Binary expansion of 5 is 101, so Sum{i>=0} b_i*(-1)^i = 2. Therefore a(5) = 10/3 + 2/3 = 4. - _Vladimir Shevelev_, Apr 15 2010

S. R. Finch, Mathematical Constants, Cambridge, 2003, Section 2.26.
Wang, E. T. H. "On Double-Free Sets of Integers." Ars Combin. 28, 97-100, 1989.

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
Steven R. Finch, Triple-Free Sets of Integers [From Steven Finch, Apr 20 2019]
Ralf Stephan, Some divide-and-conquer sequences ...
Ralf Stephan, Table of generating functions
Eric Weisstein's World of Mathematics, Double-Free Set.

Bisection of A123087.

Cf. A001045, A050291-A050296, A050321, A035263, A121701, A030308, A080277, A120385, A197911.

a050292 n = a050292_list !! (n-1)
a050292_list = scanl (+) 0 a035263_list
-- Reinhard Zumkeller, Jan 21 2013

A050292:=n->add((-1)^k*floor(n/2^k), k=0..n); seq(A050292(n), n=0..100); # Wesley Ivan Hurt, Feb 14 2014

a[n_] := a[n] = If[n < 2, 1, n - a[Floor[n/2]]]; Table[ a[n], {n, 1, 75}]
Join[{0},Accumulate[Nest[Flatten[#/.{0->{1,1},1->{1,0}}]&,{0},7]]] (* Harvey P. Dale, Apr 29 2018 *)

PARI
```
a(n)=if(n<2,1,n-a(floor(n/2)))
```

from sympy.ntheory import digits
def A050292(n): return ((n<<1)+sum((0,1,-1,0)[i] for i in digits(n,4)[1:]))//3 # Chai Wah Wu, Jan 30 2025

Partial sums of A035263. Close to (2/3)*n.
a(n) = A123087(2*n) = n - A123087(n). - Max Alekseyev, Mar 05 2023
From Benoit Cloitre, Nov 24 2002: (Start)
a(1)=1, a(n) = n - a(floor(n/2));
a(n) = (2/3)*n + (1/3)*A065359(n);
more generally, for m>=0, a(2^m*n) - 2^m*a(n) = A001045(m)*A065359(n) where A001045(m) = (2^m - (-1)^m)/3 is the Jacobsthal sequence;
a(A039004(n)) = (2/3)*A039004(n);
a(2*A039004(n)) = 2*a(A039004(n));
a(A003159(n)) = n;
a(A003159(n)-1) = n-1;
a(n) mod 2 = A010060(n) the Thue-Morse sequence;
a(n+1) - a(n) = A035263(n+1);
a(n+2) - a(n) = abs(A029884(n)).
(End)
G.f.: (1/(x-1)) * Sum_{i>=0} (-1)^i*x^(2^i)/(x^(2^i)-1). - Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Feb 17 2003
a(n) = Sum_{k>=0} (-1)^k*floor(n/2^k). - Benoit Cloitre, Jun 03 2003
a(A091785(n)) = 2n; a(A091855(n)) = 2n-1. - Philippe Deléham, Mar 26 2004
a(2^n) = (2^(n+1) + (-1)^n)/3. - Vladimir Shevelev, Apr 15 2010
If n = Sum_{i>=0} b_i*2^i is the binary expansion of n, then a(n) = 2n/3 + (1/3)Sum_{i>=0} b_i*(-1)^i. Thus a(n) = 2n/3 + O(log(n)). - Vladimir Shevelev, Apr 15 2010
Moreover, the equation a(3m)=2m has infinitely many solutions, e.g., a(3*2^k)=2*2^k; on the other hand, a((4^k-1)/3)=(2*(4^k-1))/9+k/3, i.e., limsup |a(n)-2n/3| = infinity. - Vladimir Shevelev, Feb 23 2011
a(n) = Sum_{k>=0} A030308(n,k)*A001045(k+1). - Philippe Deléham, Oct 19 2011
From Peter Bala, Feb 02 2013: (Start)
Product_{n >= 1} (1 + x^((2^n - (-1)^n)/3 )) = (1 + x)^2(1 + x^3)(1 + x^5)(1 + x^11)(1 + x^21)... = 1 + sum {n >= 1} x^a(n) = 1 + 2x + x^2 + x^3 + 2x^4 + 2x^5 + .... Hence this sequence lists the numbers representable as a sum of distinct Jacobsthal numbers A001045 = [1, 1', 3, 5, 11, 21, ...], where we distinguish between the two occurrences of 1 by writing them as 1 and 1'. For example, 9 occurs twice in the present sequence because 9 = 5 + 3 + 1 and 9 = 5 + 3 + 1'. Cf. A197911 and A080277. See also A120385.
(End)

Extended with formula by Christian G. Bower, Sep 15 1999
Corrected and extended by Reinhard Zumkeller, Aug 16 2006
Extended with formula by Philippe Deléham, Oct 19 2011
Entry revised to give a simpler definition by N. J. A. Sloane, Jan 03 2014

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.]

A056832 All a(n) = 1 or 2; a(1) = 1; get next 2^k terms by repeating first 2^k terms and changing last element so sum of first 2^(k+1) terms is odd.

Original entry on oeis.org

1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 2, 1

Offset: 1

Jonas Wallgren, Aug 30 2000

Dekking (2016) calls this the Toeplitz sequence or period-doubling sequence. - N. J. A. Sloane, Nov 08 2016
Fixed point of the morphism 1->12 and 2->11 (1 -> 12 -> 1211 -> 12111212 -> ...). - Benoit Cloitre, May 31 2004
a(n) is multiplicative. - Christian G. Bower, Jun 03 2005
a(n) is the least k such that A010060(n-1+k) = 1 - A010060(n-1); the sequence {a(n+1)-1} is the characteristic sequence for A079523. - Vladimir Shevelev, Jun 22 2009
The squarefree part of the even part of n. - Peter Munn, Dec 03 2020

			1 -> 1,2 -> 1,2,1,1 -> 1,2,1,1,1,2,1,2 -> 1,2,1,1,1,2,1,2,1,2,1,1,1,2,1,1.
Here we have 1 element, then 2 elements, then 4, 8, 16, etc.

Manfred R. Schroeder, Fractals, Chaos, Power Laws, W. H. Freeman, NY, 1991; pp. 277-279.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
F. Michel Dekking, Morphisms, Symbolic Sequences, and Their Standard Forms, Journal of Integer Sequences, Vol. 19 (2016), Article 16.1.1.
A. Hof, O. Knill and B. Simon, Singular continuous spectrum for palindromic Schroedinger operators, Commun. Math. Phys. 174 (1995), 149-159.
Kostas Karamanos, From Symbolic Dynamics to a Digital Approach: Chaos and Transcendence, in: Michel Planat (ed.), Noise, Oscillators and Algebraic Randomness, Lecture Notes in Physics, Vol. 550, Springer, Berlin, Heidelberg, 2000. (Short version. See p. 359)
Kostas Karamanos, From symbolic dynamics to a digital approach, International Journal of Bifurcation and Chaos, Vol. 11, No. 6 (2001), pp. 1683-1694. (Full version. See p. 1685)
Manfred R. Schroeder, Letter to N. J. A. Sloane, May 05 1994.
Eric Weisstein's World of Mathematics, Even Part.
Eric Weisstein's World of Mathematics, Squarefree Part.
Index entries for sequences that are fixed points of mappings

Cf. A197911 (partial sums).
Essentially same as first differences of Thue-Morse, A010060. - N. J. A. Sloane, Jul 02 2015
See A035263 for an equivalent version.
Limit of A317956(n) for large n.
Row/column 2 of A059895.
Positions of 1s: A003159.
Positions of 2s: A036554.
A002425, A006519, A079523, A096268, A214682, A234957 are used in a formula defining this sequence.
A059897 is used to express relationship between terms of this sequence.

a056832 n = a056832_list !! (n-1)
a056832_list = 1 : f [1] where
   f xs = y : f (y : xs) where
          y = 1 + sum (zipWith (*) xs $ reverse xs) `mod` 2
-- Reinhard Zumkeller, Jul 29 2014

Nest[ Function[l, {Flatten[(l /. {1 -> {1, 2}, 2 -> {1, 1}})]}], {1}, 7] (* Robert G. Wilson v, Mar 03 2005 *)
Table[Mod[-(-1)^(n + 1) (-1)^n Numerator[EulerE[2 n + 1, 1]], 3] , {n, 0, 120}] (* Michael De Vlieger, Aug 15 2016, after Jean-François Alcover at A002425 *)

a(n)=numerator(2/n*(4^n-1)*bernfrac(2*n))%3

a(n)=if(n<1, 0, valuation(n,2)%2+1) /* Michael Somos, Jun 18 2005 */

def A056832(n): return 1+((~n&n-1).bit_length()&1) # Chai Wah Wu, Jan 09 2023

a(n) = ((-1)^(n+1)*A002425(n)) modulo 3. - Benoit Cloitre, Dec 30 2003
a(1)=1, a(n) = 1 + ((Sum_{i=1..n-1} a(i)*a(n-i)) mod 2). - Benoit Cloitre, Mar 16 2004
a(n) is multiplicative with a(2^e) = 1 + (1-(-1)^e)/2, a(p^e)=1 if p > 2. - Michael Somos, Jun 18 2005
[a(2^n+1) .. a(2^(n+1)-1)] = [a(1) .. a(2^n-1)]; a(2^(n+1)) = 3 - a(2^n).
For n > 0, a(n) = 2 - A035263(n). - Benoit Cloitre, Nov 24 2002
a(n)=2 if n-1 is in A079523; a(n)=1 otherwise. - Vladimir Shevelev, Jun 22 2009
a(n) = A096268(n-1) + 1. - Reinhard Zumkeller, Jul 29 2014
From Peter Munn, Dec 03 2020: (Start)
a(n) = A007913(A006519(n)) = A006519(n)/A234957(n).
a(n) = A059895(n, 2) = n/A214682(n).
a(n*k) = (a(n) * a(k)) mod 3.
a(A059897(n, k)) = A059897(a(n), a(k)).
(End)
Asymptotic mean: lim_{m->oo} (1/m) * Sum__{k=1..m} a(k) = 4/3. - Amiram Eldar, Mar 09 2021

A265747 Numbers written in Jacobsthal greedy base.

Original entry on oeis.org

0, 1, 2, 10, 11, 100, 101, 102, 110, 111, 200, 1000, 1001, 1002, 1010, 1011, 1100, 1101, 1102, 1110, 1111, 10000, 10001, 10002, 10010, 10011, 10100, 10101, 10102, 10110, 10111, 10200, 11000, 11001, 11002, 11010, 11011, 11100, 11101, 11102, 11110, 11111, 20000, 100000, 100001, 100002, 100010, 100011, 100100

Offset: 0

Antti Karttunen, Dec 17 2015

These are called "Jacobsthal Representation Numbers" in Horadam's 1996 paper.
Sum_{i=0..} digit(i)*A001045(2+digit(i)) recovers n from such representation a(n), where digit(0) stands for the least significant digit (at the right), and A001045(k) gives the k-th Jacobsthal number.
No larger digits than 2 will occur, which allows representing the same sequence in a more compact form by base-3 coding in A265746.
Sequence A197911 gives the terms with no digit "2" in their representation, while its complement A003158 gives the terms where "2" occurs at least once.
Numbers beginning with digit "2" in this representation are given by A020988(n) [= 2*A002450(n) = 2*A001045(2n)].

			For n=7, when selecting the terms of A001045 with the greedy algorithm, we need terms A001045(4) + A001045(2) + A001045(2) = 5 + 1 + 1, thus a(7) = "102".
For n=10, we need A001045(4) + A001045(4) = 5+5, thus a(10) = "200".

Antti Karttunen, Table of n, a(n) for n = 0..10923
A. F. Horadam, Jacobsthal Representation Numbers, Fib Quart. 34 (1996), 40-54. (See especially page 50, which is page 11 in PDF.)

Cf. A001045, A002450, A020988, A007089, A197911, A003158.
Cf. A265745 (sum of digits).
Cf. A265746 (same numbers interpreted in base-3, then shown in decimal).
Cf. A084639 (positions of repunits).
Cf. A007961, A014417, A014418, A244159 for analogous sequences.

jacob[n_] := (2^n - (-1)^n)/3; maxInd[n_] := Floor[Log2[3*n + 1]]; A265747[n_] := A265747[n] = 10^(maxInd[n] - 2) + A265747[n - jacob[maxInd[n]]]; A265747[0] = 0; Array[A265747, 100, 0] (* Amiram Eldar, Jul 21 2023 *)

A130249(n) = floor(log(3*n + 1) / log(2));
A001045(n) = (2^n - (-1)^n) / 3;
A265747(n) = {if(n==0, 0, my(d=n - A001045(A130249(n))); 10^(A130249(n)-2) + if(d == 0, 0, A265747(d)));} \\ Amiram Eldar, Jul 21 2023

def greedyJ(n): m = (3*n+1).bit_length() - 1; return (m, (2**m-(-1)**m)//3)
def a(n):
    if n == 0: return 0
    place, value = greedyJ(n)
    return 10**(place-2) + a(n - value)
print([a(n) for n in range(49)]) # Michael S. Branicky, Jul 11 2021

a(0) = 0; for n >= 1, a(n) = 10^(A130249(n)-2) + a(n-A001045(A130249(n))).

a(n) = A007089(A265746(n)).

cp's OEIS Frontend

A050292 a(2n) = 2n - a(n), a(2n+1) = 2n + 1 - a(n) (for n >= 0).

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

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A056832 All a(n) = 1 or 2; a(1) = 1; get next 2^k terms by repeating first 2^k terms and changing last element so sum of first 2^(k+1) terms is odd.

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A265747 Numbers written in Jacobsthal greedy base.

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A296371 Number of integer partitions of n using Jacobsthal numbers.

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