cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-7 of 7 results.

A088172 First differences of A019300.

Original entry on oeis.org

1, 2, 3, 7, 13, 26, 53, 106, 211, 422, 845, 1689, 3379, 6758, 13515, 27031, 54061, 108122, 216245, 432489, 864979, 1729958, 3459915, 6919830, 13839661, 27679322, 55358643, 110717287, 221434573, 442869146, 885738293, 1771476586, 3542953171
Offset: 1

Views

Author

Gary W. Adamson, Sep 21 2003

Keywords

Comments

Partial sums give A019300, while partial sums mod 2 = the Thue-Morse sequence, A010060. a(n) mod 2 = A035263: 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, ...

Crossrefs

Cf. A019300, A010060, A035263.

Programs

  • Mathematica
    tm = Table[ If[ OddQ[ Count[ IntegerDigits[n, 2], 1]], 1, 0], {n, 0, 35}]; Table[ FromDigits[ Take[ tm, n + 1], 2] - FromDigits[ Take[ tm, n], 2], {n, 1, 33}] (* Robert G. Wilson v *)

Formula

a(n+1) = 2*a(n) + 0 or +-1. - Robert G. Wilson v, Sep 27 2003
a(n+1) = 2*a(n) + A029883(n). - Philippe Deléham, Mar 21 2004

Extensions

More terms from Robert G. Wilson v, Sep 27 2003

A035263 Trajectory of 1 under the morphism 0 -> 11, 1 -> 10; parity of 2-adic valuation of 2n: a(n) = A000035(A001511(n)).

Original entry on oeis.org

1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1
Offset: 1

Views

Author

Karamanos Konstantinos, N. J. A. Sloane

Keywords

Comments

First Feigenbaum symbolic (or period-doubling) sequence, corresponding to the accumulation point of the 2^{k} cycles through successive bifurcations.
To construct the sequence: start with 1 and concatenate: 1,1, then change the last term (1->0; 0->1) gives: 1,0. Concatenate those 2 terms: 1,0,1,0, change the last term: 1,0,1,1. Concatenate those 4 terms: 1,0,1,1,1,0,1,1 change the last term: 1,0,1,1,1,0,1,0, etc. - Benoit Cloitre, Dec 17 2002
Let T denote the present sequence. Here is another way to construct T. Start with the sequence S = 1,0,1,,1,0,1,,1,0,1,,1,0,1,,... and fill in the successive holes with the successive terms of the sequence T (from paper by Allouche et al.). - Emeric Deutsch, Jan 08 2003 [Note that if we fill in the holes with the terms of S itself, we get A141260. - N. J. A. Sloane, Jan 14 2009]
From N. J. A. Sloane, Feb 27 2009: (Start)
In more detail: define S to be 1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1,0,1___...
If we fill the holes with S we get A141260:
1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0,
........1.........0.........1.........1.........0.......1.........1.........0...
- the result is
1..0..1.1.1..0..1.0.1..0..1.1.1..0..1.1.1..0..1.0.1.... = A141260.
But instead, if we define T recursively by filling the holes in S with the terms of T itself, we get A035263:
1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0,
........1.........0.........1.........1.........1.......0.........1.........0...
- the result is
1..0..1.1.1..0..1.0.1..0..1.1.1..0..1.1.1..0..1.1.1.0.1.0.1..0..1.1.1..0..1.0.1.. = A035263. (End)
Characteristic function of A003159, i.e., A035263(n)=1 if n is in A003159 and A035263(n)=0 otherwise (from paper by Allouche et al.). - Emeric Deutsch, Jan 15 2003
This is the sequence of R (=1), L (=0) moves in the Towers of Hanoi puzzle: R, L, R, R, R, L, R, L, R, L, R, R, R, ... - Gary W. Adamson, Sep 21 2003
Manfred Schroeder, p. 279 states, "... the kneading sequences for unimodal maps in the binary notation, 0, 1, 0, 1, 1, 1, 0, 1..., are obtained from the Morse-Thue sequence by taking sums mod 2 of adjacent elements." On p. 278, in the chapter "Self-Similarity in the Logistic Parabola", he writes, "Is there a closer connection between the Morse-Thue sequence and the symbolic dynamics of the superstable orbits? There is indeed. To see this, let us replace R by 1 and C and L by 0." - Gary W. Adamson, Sep 21 2003
Partial sums modulo 2 of the sequence 1, a(1), a(1), a(2), a(2), a(3), a(3), a(4), a(4), a(5), a(5), a(6), a(6), ... . - Philippe Deléham, Jan 02 2004
Parity of A007913, A065882 and A065883. - Philippe Deléham, Mar 28 2004
The length of n-th run of 1's in this sequence is A080426(n). - Philippe Deléham, Apr 19 2004
Also parity of A005043, A005773, A026378, A104455, A117641. - Philippe Deléham, Apr 28 2007
Equals parity of the Towers of Hanoi, or ruler sequence (A001511), where the Towers of Hanoi sequence (1, 2, 1, 3, 1, 2, 1, 4, ...) denotes the disc moved, labeled (1, 2, 3, ...) starting from the top; and the parity of (1, 2, 1, 3, ...) denotes the direction of the move, CW or CCW. The frequency of CW moves converges to 2/3. - Gary W. Adamson, May 11 2007
A conjectured identity relating to the partition sequence, A000041: p(x) = A(x) * A(x^2) when A(x) = the Euler transform of A035263 = polcoeff A174065: (1 + x + x^2 + 2x^3 + 3x^4 + 4x^5 + ...). - Gary W. Adamson, Mar 21 2010
a(n) is 1 if the number of trailing zeros in the binary representation of n is even. - Ralf Stephan, Aug 22 2013
From Gary W. Adamson, Mar 25 2015: (Start)
A conjectured identity relating to the partition sequence, A000041 as polcoeff p(x); A003159, and its characteristic function A035263: (1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, ...); and A036554 indicating n-th terms with zeros in A035263: (2, 6, 8, 10, 14, 18, 22, ...).
The conjecture states that p(x) = A(x) = A(x^2) when A(x) = polcoeffA174065 = the Euler transform of A035263 = 1/(1-x)*(1-x^3)*(1-x^4)*(1-x^5)*... = (1 + x + x^2 + 2x^3 + 3x^4 + 4x^5 + ...) and the aerated variant = the Euler transform of the complement of A035263: 1/(1-x^2)*(1-x^6)*(1-x^8)*... = (1 + x^2 + x^4 + 2x^6 + 3x^8 + 4x^10 + ...).
(End)
The conjecture above was proved by Jean-Paul Allouche on Dec 21 2013.
Regarded as a column vector, this sequence is the product of A047999 (Sierpinski's gasket) regarded as an infinite lower triangular matrix and A036497 (the Fredholm-Rueppel sequence) where the 1's have alternating signs, 1, -1, 0, 1, 0, 0, 0, -1, .... - Gary W. Adamson, Jun 02 2021
The numbers of 1's through n (A050292) can be determined by starting with the binary (say for 19 = 1 0 0 1 1) and writing: next term is twice current term if 0, otherwise twice plus 1. The result is 1, 2, 4, 9, 19. Take the difference row, = 1, 1, 2, 5, 10; and add the odd-indexed terms from the right: 5, 4, 3, 2, 1 = 10 + 2 + 1 = 13. The algorithm is the basis for determining the disc configurations in the tower of Hanoi game, as shown in the Jul 24 2021 comment of A060572. - Gary W. Adamson, Jul 28 2021

References

  • Karamanos, Kostas. "From symbolic dynamics to a digital approach." International Journal of Bifurcation and Chaos 11.06 (2001): 1683-1694. (Full version. See p. 1685)
  • Karamanos, K. (2000). From symbolic dynamics to a digital approach: chaos and transcendence. In Michel Planat (Ed.), Noise, Oscillators and Algebraic Randomness (Lecture Notes in Physics, pp. 357-371). Springer, Berlin, Heidelberg. (Short version. See p. 359)
  • Manfred R. Schroeder, "Fractals, Chaos, Power Laws", W. H. Freeman, 1991
  • S. Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 892, column 2, Note on p. 84, part (a).

Links

Crossrefs

Parity of A001511. Anti-parity of A007814.
Absolute values of first differences of A010060. Apart from signs, same as A029883. Essentially the same as A056832.
Swapping 0 and 1 gives A096268.
Cf. A033485, A050292 (partial sums), A089608, A088172, A019300, A039982, A073675, A121701, A141260, A000041, A174065, A220466, A154269 (Mobius transform).
Limit of A317957(n) for large n.
Cf. A060572

Programs

  • Haskell
    import Data.Bits (xor)
a035263 n = a035263_list !! (n-1)
a035263_list = zipWith xor a010060_list $ tail a010060_list
-- Reinhard Zumkeller, Mar 01 2012
  • Maple
    nmax:=105: for p from 0 to ceil(simplify(log[2](nmax))) do for n from 1 to ceil(nmax/(p+2)) do a((2*n-1)*2^p) := (p+1) mod 2 od: od: seq(a(n), n=1..nmax); # Johannes W. Meijer, Feb 07 2013
A035263 := n -> 1 - padic[ordp](n, 2) mod 2:
seq(A035263(n), n=1..105); # Peter Luschny, Oct 02 2018
  • Mathematica
    a[n_] := a[n] = If[ EvenQ[n], 1 - a[n/2], 1]; Table[ a[n], {n, 1, 105}] (* Or *)
Rest[ CoefficientList[ Series[ Sum[ x^(2^k)/(1 + (-1)^k*x^(2^k)), {k, 0, 20}], {x, 0, 105}], x]]
f[1] := True; f[x_] := Xor[f[x - 1], f[Floor[x/2]]]; a[x_] := Boole[f[x]] (* Ben Branman, Oct 04 2010 *)
a[n_] := If[n == 0, 0, 1 - Mod[ IntegerExponent[n, 2], 2]]; (* Jean-François Alcover, Jul 19 2013, after Michael Somos *)
Nest[ Flatten[# /. {0 -> {1, 1}, 1 -> {1, 0}}] &, {0}, 7] (* Robert G. Wilson v, Jul 23 2014 *)
SubstitutionSystem[{0->{1,1},1->{1,0}},1,{7}][[1]] (* Harvey P. Dale, Jun 06 2022 *)
  • PARI
    {a(n) = if( n==0, 0, 1 - valuation(n, 2)%2)}; /* Michael Somos, Sep 04 2006 */
  • PARI
    {a(n) = if( n==0, 0, n = abs(n); subst( Pol(binary(n)) - Pol(binary(n-1)), x, 1)%2)}; /* Michael Somos, Sep 04 2006 */
  • PARI
    {a(n) = if( n==0, 0, n = abs(n); direuler(p=2, n, 1 / (1 - X^((p<3) + 1)))[n])}; /* Michael Somos, Sep 04 2006 */
  • Python
    def A035263(n): return (n&-n).bit_length()&1 # Chai Wah Wu, Jan 09 2023
  • Scheme
    (define (A035263 n) (let loop ((n n) (i 1)) (cond ((odd? n) (modulo i 2)) (else (loop (/ n 2) (+ 1 i)))))) ;; (Use mod instead of modulo in R6RS) Antti Karttunen, Sep 11 2017

Formula

Absolute values of first differences (A029883) of Thue-Morse sequence (A001285 or A010060). Self-similar under 10->1 and 11->0.
Series expansion: (1/x) * Sum_{i>=0} (-1)^(i+1)*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+1)/2^k)-floor(n/2^k)). - Benoit Cloitre, Jun 03 2003
Another g.f.: Sum_{k>=0} x^(2^k)/(1+(-1)^k*x^(2^k)). - Ralf Stephan, Jun 13 2003
a(2*n) = 1-a(n), a(2*n+1) = 1. - Ralf Stephan, Jun 13 2003
a(n) = parity of A033485(n). - Philippe Deléham, Aug 13 2003
Equals A088172 mod 2, where A088172 = 1, 2, 3, 7, 13, 26, 53, 106, 211, 422, 845, ... (first differences of A019300). - Gary W. Adamson, Sep 21 2003
a(n) = a(n-1) - (-1)^n*a(floor(n/2)). - Benoit Cloitre, Dec 02 2003
a(1) = 1 and a(n) = abs(a(n-1) - a(floor(n/2))). - Benoit Cloitre, Dec 02 2003
a(n) = 1 - A096268(n+1); A050292 gives partial sums. - Reinhard Zumkeller, Aug 16 2006
Multiplicative with a(2^k) = 1 - (k mod 2), a(p^k) = 1, p > 2. Dirichlet g.f.: Product_{n = 4 or an odd prime} (1/(1-1/n^s)). - Christian G. Bower, May 18 2005
a(-n) = a(n). a(0)=0. - Michael Somos, Sep 04 2006
Dirichlet g.f.: zeta(s)*2^s/(2^s+1). - Ralf Stephan, Jun 17 2007
a(n+1) = a(n) XOR a(ceiling(n/2)), a(1) = 1. - Reinhard Zumkeller, Jun 11 2009
Let D(x) be the generating function, then D(x) + D(x^2) == x/(1-x). - Joerg Arndt, May 11 2010
a(n) = A010060(n) XOR A010060(n+1); a(A079523(n)) = 0; a(A121539(n)) = 1. - Reinhard Zumkeller, Mar 01 2012
a((2*n-1)*2^p) = (p+1) mod 2, p >= 0 and n >= 1. - Johannes W. Meijer, Feb 07 2013
a(n) = A000035(A001511(n)). - Omar E. Pol, Oct 29 2013
a(n) = 2-A056832(n) = (5-A089608(n))/4. - Antti Karttunen, Sep 11 2017, after Benoit Cloitre
For n >= 0, a(n+1) = M(2n) mod 2 where M(n) is the Motzkin number A001006 (see Deutsch and Sagan 2006 link). - David Callan, Oct 02 2018
a(n) = A038712(n) mod 3. - Kevin Ryde, Jul 11 2019
Given any n in the form (k * 2^m, k odd), extract k and m. Categorize the results into two outcomes of (k, m, even or odd). If (k, m) is (odd, even) substitute 1. If (odd, odd), denote the result 0. Example: 5 = (5 * 2^0), (odd, even, = 1). (6 = 3 * 2^1), (odd, odd, = 0). - Gary W. Adamson, Jun 23 2021

Extensions

Alternative description added to the name by Antti Karttunen, Sep 11 2017

A039963 The period-doubling sequence A035263 repeated.

Original entry on oeis.org

1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

An example of a d-perfect sequence.
Motzkin numbers mod 2. - Benoit Cloitre, Mar 23 2004
Let {a, b, c, c, a, b, a, b, a, b, c, c, a, b, ...} be the fixed point of the morphism: a -> ab, b -> cc, c -> ab, starting from a; then the sequence is obtained by taking a = 1, b = 1, c = 0. - Philippe Deléham, Mar 28 2004
The asymptotic mean of this sequence is 2/3 (Rowland and Yassawi, 2015; Burns, 2016). - Amiram Eldar, Jan 30 2021
The Gilbreath transform of floor(log_2(n)) (A000523). - Thomas Scheuerle, Sep 02 2024

Links

Crossrefs

Cf. A000523, A005802, A081706.
Motzkin numbers A001006 read mod 2,3,4,5,6,7,8,11: A039963, A039964, A299919, A258712, A299920, A258711, A299918, A258710.

Programs

  • Mathematica
    Flatten[ Nest[ Function[l, {Flatten[(l /. {a -> {a, b}, b -> {c, c}, c -> {a, b}})]}], {a}, 7] /. {a -> {1}, b -> {1}, c -> {0}}] (* Robert G. Wilson v, Feb 26 2005 *)
  • PARI
    A039963(n) = 1 - valuation(n\2+1,2)%2; \\ Max Alekseyev, Oct 23 2021
  • Python
    def A039963(n): return ((m:=(n>>1)+1)&-m).bit_length()&1 # Chai Wah Wu, Jan 09 2023

Formula

a(n) = A035263(1+floor(n/2)). - Benoit Cloitre, Mar 23 2004
a(n) = A040039(n) mod 2 = A002212(n+1) mod 2. a(0) = a(1) = 1, for n>=2: a(n) = ( a(n) + Sum_{k=0..n-2} a(k)*a(n-2-k)) mod 2. - Philippe Deléham, Mar 26 2004
a(n) = (A(n+2) - A(n)) mod 2, for A = A019300, A001285, A010060, A010059, A000069, A001969. - Philippe Deléham, Mar 28 2004
a(n) = A001006(n) mod 2. - Christian G. Bower, Jun 12 2005
a(n) = (-1)^n*(A096268(n+1) - A096268(n)). - Johannes W. Meijer, Feb 02 2013
a(n) = 1 - A007814(floor(n/2)+1) mod 2 = A005802(n) mod 2. - Max Alekseyev, Oct 23 2021

Extensions

More terms from Christian G. Bower, Jun 12 2005
Edited by N. J. A. Sloane at the suggestion of Andrew S. Plewe and Ralf Stephan, Jul 13 2007

A048707 Numerators of ratios converging to Thue-Morse constant.

Original entry on oeis.org

0, 1, 6, 105, 27030, 1771476585, 7608434000728254870, 140350834813144189858090274002849666665, 47758914269546354982683078068829456704164423862093743397580034411621752859030
Offset: 0

Views

Author

Antti Karttunen, Mar 09 1999

Keywords

Comments

Also interpret each iteration of the construction of the Thue-Morse constant as a binary number converted to a decimal number. Thus (0_b, 01_b, 0110_b, 01101001_b ...) gives the present sequence in decimal. - Robert G. Wilson v, Sep 22 2006
a(n) corresponds to the binary value of the truth-table for the xor operator with n-arguments. - Joe Riel (joer(AT)san.rr.com), Jan 31 2010

Links

Crossrefs

The denominators are given by A001146. Consists of every 2^n-th term of A019300. Cf. A048708 (same sequence in hexadecimal) and A014571, A010060, A014572.
Cf. A080814, A080815, A133468.

Programs

  • Mathematica
    Table[ FromDigits[ Nest[ Flatten[ #1 /. {0 -> {0, 1}, 1 -> {1, 0}}] &, {0}, n], 2], {n, 0, 8}] (* Robert G. Wilson v, Sep 22 2006 *)
  • Scheme
    ;returns all but the last element of a list
(define rdc(lambda(x)(if(null? (cdr x))'()(cons (car x) (rdc (cdr x))))))
;gets the two's complement of a given bit
(define twosComplement (lambda (x)(if (eq? x #\0) "1" "0" )))
;gets the two's complement of a string
(define complementOfCurrent (lambda (x y z)(if (eq? (string-length y) z) y (complementOfCurrent (list->string (cdr (string->list x))) (string-append y (twosComplement (string-ref x 0))) z))))
;concatenates the two's complement of a string onto the current string, giving the next element in the TM sequence
(define concatenateComplement (lambda (x i)(if(zero? i) x (concatenateComplement(string-append x (complementOfCurrent x "" (string-length x)))(- i 1)))))
;generates the TM sequence of length 2^x
(define generateThue (lambda (x)(concatenateComplement "0" x)))
;if a bit is 1, get 2^i, where i is the index of that bit from right-left
(define F (lambda (c i)(if (eq? c #\1) (expt 2 i) 0)))
;gathers the sum of 2^index for all indices corresponding to a 1
(define fn (lambda (x sum i stop)(if (eq? i stop) sum (fn (list->string (rdc (string->list x))) (+ sum (F (string-ref x (-(string-length x) 1)) i)) (+ i 1)stop))))
(define f (lambda (x)(fn (generateThue x) 0 0 (string-length (generateThue x)))))
;format: (f x)
;example: (f 10)
;by Ariel S Koiman, Apr 23 2013

Formula

a(0) = 0, a(n) = (a(n-1)+1)*((2^(2^(n-1)))-1).

A320916 Consider A010060 as a 2-adic number ...100110010110, then a(n) is its approximation up to 2^n.

Original entry on oeis.org

0, 0, 2, 6, 6, 22, 22, 22, 150, 406, 406, 406, 2454, 2454, 10646, 27030, 27030, 92566, 92566, 92566, 616854, 616854, 2714006, 6908310, 6908310, 6908310, 40462742, 107571606, 107571606, 376007062, 376007062, 376007062, 2523490710, 6818458006, 6818458006, 6818458006
Offset: 0

Views

Author

Jianing Song, Oct 26 2018

Keywords

Comments

This is another interpretation of A010060 as a number, in a different way as considering it as a binary number.
Consider the g.f. of A010060. As a real-valued (or complex-valued) function it only converges for |x| < 1. In 2-adic field it only converges for |x|_2 < 1 as well, but here |x|_2 is a different metric. For a 2-adic number x, |x|_2 < 1 iff x is an even 2-adic integer.

Examples

			a(1) =     0_2 =  0.
a(2) =    10_2 =  2.
a(3) =   110_2 =  6.
a(4) =  0110_2 =  6.
a(5) = 10110_2 = 22.
...

Links

Crossrefs

Cf. A010060, A122570, A019300 (bit reversal).

Programs

  • Mathematica
    With[{nmax = 50}, Table[FromDigits[#[[-n;;]], 2], {n, 0, nmax}] & [ThueMorse[Range[nmax, 0, -1]]]] (* or *)
A320916[n_] := FromDigits[ThueMorse[Range[n-1, 0, -1]], 2]; Array[A320916, 51, 0] (* Paolo Xausa, Oct 18 2024 *)
  • PARI
    a(n) = sum(i=0, n-1, 2^i*(hammingweight(i)%2))

Formula

a(n) = Sum_{i=0..n-1} A010060(i)*2^i (empty sum yields 0 for n = 0).

A101556 A Thue-Morse convolution.

Original entry on oeis.org

0, 1, 11, 110, 1101, 11010, 110100, 1101001, 11010011, 110100110, 1101001100, 11010011001, 110100110010, 1101001100101, 11010011001011, 110100110010110, 1101001100101101, 11010011001011010, 110100110010110100
Offset: 0

Views

Author

Paul Barry, Dec 06 2004

Keywords

Comments

A019300 in binary.
Convolution of A010060 and 10^n.

Links

Crossrefs

Cf. A010060, A019300.

Programs

  • Mathematica
    Table[Sum[ThueMorse[n - k]*10^k, {k, 0, n}], {n, 0, 50}] (* G. C. Greubel, May 23 2018 *)

Formula

a(n) = Sum_{k=0..n} A010060(n-k) * 10^k.

A248875 First n elements of the Kolakoski sequence read as a ternary number.

Original entry on oeis.org

0, 1, 5, 17, 52, 157, 473, 1420, 4262, 12788, 38365, 115097, 345293, 1035880, 3107641, 9322925, 27968776, 83906329, 251718989, 755156969, 2265470908, 6796412726, 20389238179, 61167714538, 183503143616, 550509430849, 1651528292549, 4954584877649, 14863754632948
Offset: 0

Views

Author

Danny Rorabaugh, Mar 04 2015

Keywords

Links

Crossrefs

Cf. A000002, A019300.

Formula

a(n+1) = 3*a(n) + A000002(n+1).
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