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 1-10 of 60 results. Next

A272882 Discriminator of the Cantor numbers (A005823).

Original entry on oeis.org

1, 3, 5, 5, 7, 11, 13, 15, 19, 21, 31, 33, 37, 39, 41, 41, 43, 47, 49, 59, 61, 65, 67, 95, 97, 101, 103, 113, 115, 119, 121, 123, 127, 129, 139, 141, 145, 147, 175, 177, 181, 183, 193, 195, 199, 201, 283, 285, 289, 291, 301, 303, 307, 309, 337, 339, 343, 345, 355, 357, 361, 363, 365, 365, 367, 371, 373, 383, 385, 389, 391, 419
Offset: 1

Views

Author

Jeffrey Shallit, May 08 2016

Keywords

Comments

The discriminator of the sequence is the least positive integer k such that the first n terms of the sequence are distinct modulo k.

Crossrefs

Cf. A005823.

Formula

The linked paper by Haque and Shallit gives a conjectured recursive description of the sequence.

A339636 Counterexamples to a conjecture about integers representable as the quotient of two Cantor numbers (A005823).

Original entry on oeis.org

529, 592, 601, 616, 5368, 50281, 4072741, 4074361, 4088941, 4245688, 37884151, 316980400, 329892001, 329893621, 330023221, 330024841, 330039421, 331204201, 331205821, 331220401, 331958485, 344321272
Offset: 1

Views

Author

Jeffrey Shallit, Dec 11 2020

Keywords

Comments

Let C be the Cantor numbers (A005823), and let A be the set of integers congruent to 1 (mod 3) representable as the quotient of two nonzero elements of C (A339637). It is easy to see that if (3/2)*3^i < n < 2*3^i for some i, then n cannot be in A. Initial empirical data suggested that these are the only integers congruent to 1 (mod 3) not in A. However, there are "sporadic" counterexamples enumerated by this sequence entry, whose structure is not well understood.
A simple automaton-based (or breadth-first search) algorithm can establish in O(n) time whether n is in A or not.
Conjecture: every number of the form 23*3^(4k+3) - 20 is not representable.

Crossrefs

Extensions

a(11)-a(22) computed by Robert Dougherty-Bliss added by Jeffrey Shallit, Mar 11 2025

A339637 Numbers congruent to 1 (mod 3) that are the quotient of two Cantor numbers (A005823).

Original entry on oeis.org

1, 4, 7, 10, 13, 19, 22, 25, 28, 31, 34, 37, 40, 55, 58, 61, 64, 67, 70, 73, 76, 79, 82, 85, 88, 91, 94, 97, 100, 103, 106, 109, 112, 115, 118, 121, 163, 166, 169, 172, 175, 178, 181, 184, 187, 190, 193, 196, 199, 202, 205, 208, 211, 214, 217, 220, 223, 226, 229, 232
Offset: 1

Views

Author

Jeffrey Shallit, Dec 11 2020

Keywords

Comments

Let C be the Cantor numbers (A005823), and let A be the set of integers congruent to 1 (mod 3) representable as the quotient of two nonzero elements of C. It is easy to see that if (3/2)*3^i < n < 2*3^i for some i, then n cannot be in A. Initial empirical data suggested that these are the only integers congruent to 1 (mod 3) not in A. However, there are additional "sporadic" counterexamples enumerated by A339636, whose structure is not well understood.
A simple automaton-based (or breadth-first search) algorithm can establish in O(n) time whether n is in A or not.

Examples

			106 is in the sequence, because 106=1462376/13796, and 1462376 in base 3 is 2202022000002, and 13796 in base 3 is 200220222.
		

Crossrefs

A117482 Numbers with no 1's in their base-3, base-4, and base-5 expansions. Intersection of A005823, A023709, and A023725.

Original entry on oeis.org

0, 2, 60, 62, 512, 560, 2178, 15072, 15074, 15090, 40992, 40994, 41004, 41010, 43760, 43818, 43820, 45260, 45704, 45710, 45738, 45740, 45744, 53192, 53198, 57512, 57518, 57570, 163842, 163854, 164000, 164018, 164024
Offset: 1

Views

Author

Zak Seidov, Apr 26 2006

Keywords

Comments

No more terms <= 1000000. - Emeric Deutsch, Apr 30 2006
No more terms <= 5000000. - Harvey P. Dale, Jan 17 2014

Examples

			560 is in the sequence because in base 3 it is 202202, in base 4 it is 20300 and in base 5 it is 4220.
		

Crossrefs

Programs

  • Maple
    a:=proc(n) if member(1,convert(convert(n,base,3),set) union convert(convert(n,base,4),set) union convert(convert(n,base,5),set))=false then n else fi end: seq(a(n),n=0..170000); # Emeric Deutsch, Apr 30 2006
  • Mathematica
    Select[Range[0,200000],Table[DigitCount[#,n,1],{n,3,5}]=={0,0,0}&] (* Harvey P. Dale, Jan 17 2014 *)

A005836 Numbers whose base-3 representation contains no 2.

Original entry on oeis.org

0, 1, 3, 4, 9, 10, 12, 13, 27, 28, 30, 31, 36, 37, 39, 40, 81, 82, 84, 85, 90, 91, 93, 94, 108, 109, 111, 112, 117, 118, 120, 121, 243, 244, 246, 247, 252, 253, 255, 256, 270, 271, 273, 274, 279, 280, 282, 283, 324, 325, 327, 328, 333, 334, 336, 337, 351, 352
Offset: 1

Views

Author

Keywords

Comments

3 does not divide binomial(2s, s) if and only if s is a member of this sequence, where binomial(2s, s) = A000984(s) are the central binomial coefficients.
This is the lexicographically earliest increasing sequence of nonnegative numbers that contains no arithmetic progression of length 3. - Robert Craigen (craigenr(AT)cc.umanitoba.ca), Jan 29 2001
In the notation of A185256 this is the Stanley Sequence S(0,1). - N. J. A. Sloane, Mar 19 2010
Complement of A074940. - Reinhard Zumkeller, Mar 23 2003
Sums of distinct powers of 3. - Ralf Stephan, Apr 27 2003
Numbers n such that central trinomial coefficient A002426(n) == 1 (mod 3). - Emeric Deutsch and Bruce E. Sagan, Dec 04 2003
A039966(a(n)+1) = 1; A104406(n) = number of terms <= n.
Subsequence of A125292; A125291(a(n)) = 1 for n>1. - Reinhard Zumkeller, Nov 26 2006
Also final value of n - 1 written in base 2 and then read in base 3 and with finally the result translated in base 10. - Philippe LALLOUET (philip.lallouet(AT)wanadoo.fr), Jun 23 2007
a(n) modulo 2 is the Thue-Morse sequence A010060. - Dennis Tseng, Jul 16 2009
Also numbers such that the balanced ternary representation is the same as the base 3 representation. - Alonso del Arte, Feb 25 2011
Fixed point of the morphism: 0 -> 01; 1 -> 34; 2 -> 67; ...; n -> (3n)(3n+1), starting from a(1) = 0. - Philippe Deléham, Oct 22 2011
It appears that this sequence lists the values of n which satisfy the condition sum(binomial(n, k)^(2*j), k = 0..n) mod 3 <> 0, for any j, with offset 0. See Maple code. - Gary Detlefs, Nov 28 2011
Also, it follows from the above comment by Philippe Lallouet that the sequence must be generated by the rules: a(1) = 0, and if m is in the sequence then so are 3*m and 3*m + 1. - L. Edson Jeffery, Nov 20 2015
Add 1 to each term and we get A003278. - N. J. A. Sloane, Dec 01 2019

Examples

			12 is a term because 12 = 110_3.
This sequence regarded as a triangle with rows of lengths 1, 1, 2, 4, 8, 16, ...:
   0
   1
   3,  4
   9, 10, 12, 13
  27, 28, 30, 31, 36, 37, 39, 40
  81, 82, 84, 85, 90, 91, 93, 94, 108, 109, 111, 112, 117, 118, 120, 121
... - _Philippe Deléham_, Jun 06 2015
		

References

  • Richard K. Guy, Unsolved Problems in Number Theory, 3rd Edition, Springer, 2004, Section E10, pp. 317-323.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A039966 (characteristic function).
For generating functions Product_{k>=0} (1+a*x^(b^k)) for the following values of (a,b) see: (1,2) A000012 and A000027, (1,3) A039966 and A005836, (1,4) A151666 and A000695, (1,5) A151667 and A033042, (2,2) A001316, (2,3) A151668, (2,4) A151669, (2,5) A151670, (3,2) A048883, (3,3) A117940, (3,4) A151665, (3,5) A151671, (4,2) A102376, (4,3) A151672, (4,4) A151673, (4,5) A151674.
Row 3 of array A104257.
Summary of increasing sequences avoiding arithmetic progressions of specified lengths (the second of each pair is obtained by adding 1 to the first):
3-term AP: A005836 (>=0), A003278 (>0);
4-term AP: A005839 (>=0), A005837 (>0);
5-term AP: A020654 (>=0), A020655 (>0);
6-term AP: A020656 (>=0), A005838 (>0);
7-term AP: A020657 (>=0), A020658 (>0);
8-term AP: A020659 (>=0), A020660 (>0);
9-term AP: A020661 (>=0), A020662 (>0);
10-term AP: A020663 (>=0), A020664 (>0).
See also A000452.

Programs

  • Haskell
    a005836 n = a005836_list !! (n-1)
    a005836_list = filter ((== 1) . a039966) [0..]
    -- Reinhard Zumkeller, Jun 09 2012, Sep 29 2011
    
  • Julia
    function a(n)
        m, r, b = n, 0, 1
        while m > 0
            m, q = divrem(m, 2)
            r += b * q
            b *= 3
        end
    r end; [a(n) for n in 0:57] |> println # Peter Luschny, Jan 03 2021
  • Maple
    t := (j, n) -> add(binomial(n,k)^j, k=0..n):
    for i from 1 to 400 do
        if(t(4,i) mod 3 <>0) then print(i) fi
    od; # Gary Detlefs, Nov 28 2011
    # alternative Maple program:
    a:= proc(n) option remember: local k, m:
    if n=1 then 0 elif n=2 then 1 elif n>2 then k:=floor(log[2](n-1)): m:=n-2^k: procname(m)+3^k: fi: end proc:
    seq(a(n), n=1.. 20); # Paul Weisenhorn, Mar 22 2020
    # third Maple program:
    a:= n-> `if`(n=1, 0, irem(n-1, 2, 'q')+3*a(q+1)):
    seq(a(n), n=1..100);  # Alois P. Heinz, Jan 26 2022
  • Mathematica
    Table[FromDigits[IntegerDigits[k, 2], 3], {k, 60}]
    Select[Range[0, 400], DigitCount[#, 3, 2] == 0 &] (* Harvey P. Dale, Jan 04 2012 *)
    Join[{0}, Accumulate[Table[(3^IntegerExponent[n, 2] + 1)/2, {n, 57}]]] (* IWABUCHI Yu(u)ki, Aug 01 2012 *)
    FromDigits[#,3]&/@Tuples[{0,1},7] (* Harvey P. Dale, May 10 2019 *)
  • PARI
    A=vector(100);for(n=2,#A,A[n]=if(n%2,3*A[n\2+1],A[n-1]+1));A \\ Charles R Greathouse IV, Jul 24 2012
    
  • PARI
    is(n)=while(n,if(n%3>1,return(0));n\=3);1 \\ Charles R Greathouse IV, Mar 07 2013
    
  • PARI
    a(n) = fromdigits(binary(n-1),3);  \\ Gheorghe Coserea, Jun 15 2018
    
  • Python
    def A005836(n):
        return int(format(n-1,'b'),3) # Chai Wah Wu, Jan 04 2015
    

Formula

a(n) = A005823(n)/2 = A003278(n)-1 = A033159(n)-2 = A033162(n)-3.
Numbers n such that the coefficient of x^n is > 0 in prod (k >= 0, 1 + x^(3^k)). - Benoit Cloitre, Jul 29 2003
a(n+1) = Sum_{k=0..m} b(k)* 3^k and n = Sum( b(k)* 2^k ).
a(2n+1) = 3a(n+1), a(2n+2) = a(2n+1) + 1, a(0) = 0.
a(n+1) = 3*a(floor(n/2)) + n - 2*floor(n/2). - Ralf Stephan, Apr 27 2003
G.f.: (x/(1-x)) * Sum_{k>=0} 3^k*x^2^k/(1+x^2^k). - Ralf Stephan, Apr 27 2003
a(n) = Sum_{k = 1..n-1} (1 + 3^A007814(k)) / 2. - Philippe Deléham, Jul 09 2005
From Reinhard Zumkeller, Mar 02 2008: (Start)
A081603(a(n)) = 0.
If the offset were changed to zero, then: a(0) = 0, a(n+1) = f(a(n)+1, a(n)+1) where f(x, y) = if x < 3 and x <> 2 then y else if x mod 3 = 2 then f(y+1, y+1) else f(floor(x/3), y). (End)
With offset a(0) = 0: a(n) = Sum_{k>=0} A030308(n,k)*3^k. - Philippe Deléham, Oct 15 2011
a(2^n) = A003462(n). - Philippe Deléham, Jun 06 2015
We have liminf_{n->infinity} a(n)/n^(log(3)/log(2)) = 1/2 and limsup_{n->infinity} a(n)/n^(log(3)/log(2)) = 1. - Gheorghe Coserea, Sep 13 2015
a(2^k+m) = a(m) + 3^k with 1 <= m <= 2^k and 1 <= k, a(1)=0, a(2)=1. - Paul Weisenhorn, Mar 22 2020
Sum_{n>=2} 1/a(n) = 2.682853110966175430853916904584699374821677091415714815171756609672281184705... (calculated using Baillie and Schmelzer's kempnerSums.nb, see Links). - Amiram Eldar, Feb 12 2022
A065361(a(n)) = n-1. - Rémy Sigrist, Feb 06 2023
a(n) ≍ n^k, where k = log 3/log 2 = 1.5849625007. (I believe the constant varies from 1/2 to 1.) - Charles R Greathouse IV, Mar 29 2024

Extensions

Offset corrected by N. J. A. Sloane, Mar 02 2008
Edited by the Associate Editors of the OEIS, Apr 07 2009

A062756 Number of 1's in ternary (base-3) expansion of n.

Original entry on oeis.org

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

Views

Author

Ahmed Fares (ahmedfares(AT)my-deja.com), Jul 16 2001

Keywords

Comments

Fixed point of the morphism: 0 ->010; 1 ->121; 2 ->232; ...; n -> n(n+1)n, starting from a(0)=0. - Philippe Deléham, Oct 25 2011

Crossrefs

Cf. A080846, A343785 (first differences).
Cf. A081606 (indices of !=0).
Indices of terms 0..6: A005823, A023692, A023693, A023694, A023695, A023696, A023697.
Numbers of: A077267 (0's), A081603 (2's), A160384 (1's+2's).
Other bases: A000120, A160381, A268643.

Programs

  • Haskell
    a062756 0 = 0
    a062756 n = a062756 n' + m `mod` 2 where (n',m) = divMod n 3
    -- Reinhard Zumkeller, Feb 21 2013
    
  • Mathematica
    Table[Count[IntegerDigits[i, 3], 1], {i, 0, 200}]
    Nest[Join[#, # + 1, #] &, {0}, 5] (* IWABUCHI Yu(u)ki, Sep 08 2012 *)
  • PARI
    a(n)=if(n<1,0,a(n\3)+(n%3)%2) \\ Paul D. Hanna, Feb 24 2006
    
  • PARI
    a(n)=hammingweight(digits(n,3)%2); \\ Ruud H.G. van Tol, Dec 10 2023
    
  • Python
    from sympy.ntheory import digits
    def A062756(n): return digits(n,3)[1:].count(1) # Chai Wah Wu, Dec 23 2022

Formula

a(0) = 0, a(3n) = a(n), a(3n+1) = a(n)+1, a(3n+2) = a(n). - Vladeta Jovovic, Jul 18 2001
G.f.: (Sum_{k>=0} x^(3^k)/(1+x^(3^k)+x^(2*3^k)))/(1-x). In general, the generating function for the number of digits equal to d in the base b representation of n (0 < d < b) is (Sum_{k>=0} x^(d*b^k)/(Sum_{i=0..b-1} x^(i*b^k)))/(1-x). - Franklin T. Adams-Watters, Nov 03 2005 [For d=0, use the above formula with d=b: (Sum_{k>=0} x^(b^(k+1))/(Sum_{i=0..b-1} x^(i*b^k)))/(1-x), adding 1 if you consider the representation of 0 to have one zero digit.]
a(n) = a(floor(n/3)) + (n mod 3) mod 2. - Paul D. Hanna, Feb 24 2006

Extensions

More terms from Vladeta Jovovic, Jul 18 2001

A001196 Double-bitters: only even length runs in binary expansion.

Original entry on oeis.org

0, 3, 12, 15, 48, 51, 60, 63, 192, 195, 204, 207, 240, 243, 252, 255, 768, 771, 780, 783, 816, 819, 828, 831, 960, 963, 972, 975, 1008, 1011, 1020, 1023, 3072, 3075, 3084, 3087, 3120, 3123, 3132, 3135, 3264, 3267, 3276, 3279, 3312, 3315, 3324, 3327, 3840, 3843
Offset: 0

Views

Author

N. J. A. Sloane, based on an email from Bart la Bastide (bart(AT)xs4all.nl)

Keywords

Comments

Numbers whose set of base 4 digits is {0,3}. - Ray Chandler, Aug 03 2004
n such that there exists a permutation p_1, ..., p_n of 1, ..., n such that i + p_i is a power of 4 for every i. - Ray Chandler, Aug 03 2004
The first 2^n terms of the sequence could be obtained using the Cantor-like process for the segment [0, 4^n-1]. E.g., for n=1 we have [0, {1, 2}, 3] such that numbers outside of braces are the first 2 terms of the sequence; for n=2 we have [0, {1, 2}, 3, {4, 5, 6, 7, 8, 9, 10, 11}, 12, {13, 14}, 15] such that the numbers outside of braces are the first 4 terms of the sequence, etc. - Vladimir Shevelev, Dec 17 2012
From Emeric Deutsch, Jan 26 2018: (Start)
Also, the indices of the compositions having only even parts. For the definition of the index of a composition, see A298644. For example, 195 is in the sequence since its binary form is 11000011 and the composition [2,4,2] has only even parts. 132 is not in the sequence since its binary form is 10000100 and the composition [1,4,1,2] also has odd parts.
The command c(n) from the Maple program yields the composition having index n. (End)
After the k-th step of generating the Koch snowflake curve, label the edges of the curve consecutively 0..3*4^k-1 starting from a vertex of the originating triangle. a(0), a(1), ... a(2^k-1) are the labels of the edges contained in one edge of the originating triangle. Add 4^k to each label to get the labels for the next edge of the triangle. Compare with A191108 in respect of the Sierpinski arrowhead curve. - Peter Munn, Aug 18 2019

Crossrefs

3 times the Moser-de Bruijn sequence A000695.
Two digits in other bases: A005823, A097252-A097262.
Digit duplication in other bases: A338086, A338754.
Main diagonal of A054238.
Cf. A191108.

Programs

  • C
    int a_next(int a_n) { int t = a_n << 1; return a_n ^ t ^ (t + 3); } /* Falk Hüffner, Jan 24 2022 */
  • Haskell
    a001196 n = if n == 0 then 0 else 4 * a001196 n' + 3 * b
                where (n',b) = divMod n 2
    -- Reinhard Zumkeller, Feb 21 2014
    
  • Maple
    Runs := proc (L) local j, r, i, k: j := 1: r[j] := L[1]: for i from 2 to nops(L) do if L[i] = L[i-1] then r[j] := r[j], L[i] else j := j+1: r[j] := L[i] end if end do: [seq([r[k]], k = 1 .. j)] end proc: RunLengths := proc (L) map(nops, Runs(L)) end proc: c := proc (n) ListTools:-Reverse(convert(n, base, 2)): RunLengths(%) end proc: A := {}: for n to 3350 do if type(product(1+c(n)[j], j = 1 .. nops(c(n))), odd) = true then A := `union`(A, {n}) else  end if end do: A; # most of the Maple  program is due to W. Edwin Clark. - Emeric Deutsch, Jan 26 2018
    # second Maple program:
    a:= proc(n) option remember;
         `if`(n<2, 3*n, 4*a(iquo(n, 2, 'r'))+3*r)
        end:
    seq(a(n), n=0..100);  # Alois P. Heinz, Jan 24 2022
  • Mathematica
    fQ[n_] := Union@ Mod[Length@# & /@ Split@ IntegerDigits[n, 2], 2] == {0}; Select[ Range@ 10000, fQ] (* Or *)
    fQ[n_] := Union@ Join[IntegerDigits[n, 4], {0, 3}] == {0, 3}; Select[ Range@ 10000, fQ] (* Robert G. Wilson v, Dec 24 2012 *)
  • PARI
    a(n) = 3*fromdigits(binary(n),4); \\ Kevin Ryde, Nov 07 2020
    
  • Python
    def inA001196(n):
        while n != 0:
            if n%4 == 1 or n%4 == 2:
                return 0
            n = n//4
        return 1
    n, a = 0, 0
    while n < 20:
        if inA001196(a):
            print(n,a)
            n = n+1
        a = a+1 # A.H.M. Smeets, Aug 19 2019
    
  • Python
    from itertools import groupby
    def ok2lb(n):
      if n == 0: return True  # by convention
      return all(len(list(g))%2 == 0 for k, g in groupby(bin(n)[2:]))
    print([i for i in range(3313) if ok2lb(i)]) # Michael S. Branicky, Jan 04 2021
    
  • Python
    def A001196(n): return 3*int(bin(n)[2:],4) # Chai Wah Wu, Aug 21 2023
    

Formula

a(2n) = 4*a(n), a(2n+1) = 4*a(n) + 3.
a(n) = 3 * A000695(n).
Sum_{n>=1} 1/a(n) = 0.628725478158702414849086504025451177643560169366348272891020450593453403709... (calculated using Baillie and Schmelzer's kempnerSums.nb, see Links). - Amiram Eldar, Feb 12 2022

A032924 Numbers whose ternary expansion contains no 0.

Original entry on oeis.org

1, 2, 4, 5, 7, 8, 13, 14, 16, 17, 22, 23, 25, 26, 40, 41, 43, 44, 49, 50, 52, 53, 67, 68, 70, 71, 76, 77, 79, 80, 121, 122, 124, 125, 130, 131, 133, 134, 148, 149, 151, 152, 157, 158, 160, 161, 202, 203, 205, 206, 211, 212, 214, 215, 229, 230, 232, 233, 238, 239
Offset: 1

Views

Author

Keywords

Comments

Complement of A081605. - Reinhard Zumkeller, Mar 23 2003
Subsequence of A154314. - Reinhard Zumkeller, Jan 07 2009
The first 28 terms are the range of A059852 (Morse codes for letters, when written in base 3) union {44, 50} (which correspond to Morse codes of Ü and Ä). Subsequent terms represent the Morse code of other symbols in the same coding. - M. F. Hasler, Jun 22 2020

Crossrefs

Zeroless numbers in some other bases <= 10: A000042 (base 2), A023705 (base 4), A248910 (base 6), A255805 (base 8), A255808 (base 9), A052382 (base 10).

Programs

  • Haskell
    a032924 n = a032924_list !! (n-1)
    a032924_list = iterate f 1 where
       f x = 1 + if r < 2 then x else 3 * f x'  where (x', r) = divMod x 3
    -- Reinhard Zumkeller, Mar 07 2015, May 04 2012
    
  • Maple
    f:= proc(n) local L,i,m;
       L:= convert(n,base,2);
       m:= nops(L);
       add((1+L[i])*3^(i-1),i=1..m-1);
    end proc:
    map(f, [$2..101]); # Robert Israel, Aug 04 2015
  • Mathematica
    Select[Range@ 240, Last@ DigitCount[#, 3] == 0 &] (* Michael De Vlieger, Aug 05 2015 *)
    Flatten[Table[FromDigits[#,3]&/@Tuples[{1,2},n],{n,5}]] (* Harvey P. Dale, May 28 2016 *)
  • PARI
    apply( {A032924(n)=if(n<3,n,3*self()((n-1)\2)+2-n%2)}, [1..99]) \\ M. F. Hasler, Jun 22 2020
    
  • PARI
    a(n) = fromdigits(apply(d->d+1,binary(n+1)[^1]), 3); \\ Kevin Ryde, Jun 23 2020
    
  • Python
    def a(n): return sum(3**i*(int(b)+1) for i, b in enumerate(bin(n+1)[:2:-1]))
    print([a(n) for n in range(1, 61)]) # Michael S. Branicky, Aug 15 2022
    
  • Python
    def is_A032924(n):
        while n > 2:
           n,r = divmod(n,3)
           if r==0: return False
        return n > 0
    print([n for n in range(250) if is_A032924(n)]) # M. F. Hasler, Feb 15 2023
    
  • Python
    def A032924(n): return int(bin(m:=n+1)[3:],3) + (3**(m.bit_length()-1)-1>>1) # Chai Wah Wu, Oct 13 2023

Formula

a(n) = A107680(n) + A107681(n). - Reinhard Zumkeller, May 20 2005
A081604(A107681(n)) <= A081604(A107680(n)) = A081604(a(n)) = A000523(n+1). - Reinhard Zumkeller, May 20 2005
A077267(a(n)) = 0. - Reinhard Zumkeller, Mar 02 2008
a(1)=1, a(n+1) = f(a(n)+1,a(n)+1) where f(x,y) = if x<3 and x<>0 then y, else if x mod 3 = 0 then f(y+1,y+1), else f(floor(x/3),y). - Reinhard Zumkeller, Mar 02 2008
a(2*n) = a(2*n-1)+1, n>0. - Zak Seidov, Jul 27 2009
A212193(a(n)) = 0. - Reinhard Zumkeller, May 04 2012
a(2*n+1) = 3*a(n)+1. - Robert Israel, Aug 05 2015
G.f.: x/(1-x)^2 + Sum_{m >= 1} 3^(m-1)*x^(2^(m+1)-1)/((1-x^(2^m))*(1-x)). - Robert Israel, Aug 04 2015
A065361(a(n)) = n. - Rémy Sigrist, Feb 06 2023
Sum_{n>=1} 1/a(n) = 3.4977362637842652509313189236131190039368413460747606236619907531632476445332666030262441154353753276457... (calculated using Baillie and Schmelzer's kempnerSums.nb, see Links). - Amiram Eldar, Apr 14 2025

A097262 Numbers whose set of base 16 digits is {0,F}, where F base 16 = 15 base 10.

Original entry on oeis.org

0, 15, 240, 255, 3840, 3855, 4080, 4095, 61440, 61455, 61680, 61695, 65280, 65295, 65520, 65535, 983040, 983055, 983280, 983295, 986880, 986895, 987120, 987135, 1044480, 1044495, 1044720, 1044735, 1048320, 1048335, 1048560, 1048575
Offset: 0

Views

Author

Ray Chandler, Aug 03 2004

Keywords

Comments

n such that there exists a permutation p_1, ..., p_n of 1, ..., n such that i + p_i is a power of 16 for every i.

Crossrefs

Programs

  • Magma
    [n: n in [0..1110000] | Set(IntegerToSequence(n, 16)) subset {0, 15}]; // Vincenzo Librandi, Jun 05 2012
  • Mathematica
    f[n_] := FromDigits[ IntegerDigits[n, 2] /. {1 -> 15}, 16]; Array[f, 32, 0] (* or *)
    FromDigits[#, 16] & /@ Tuples[{0, 15}, 6] (* Harvey P. Dale, Sep 22 2011 *) (* or much slower *)
    fQ[n_] := Union@ Join[{0, 15}, IntegerDigits[n, 16]] == {0, 15}; Select[ Range[0, 11000000 ], fQ] (* Robert G. Wilson v, May 12 2012 *)

Formula

a(n) = 15*A033052(n).
a(2n) = 16*a(n), a(2n+1) = a(2n)+15.

A191106 Increasing sequence generated by these rules: a(1)=1, and if x is in a then 3x-2 and 3x are in a.

Original entry on oeis.org

1, 3, 7, 9, 19, 21, 25, 27, 55, 57, 61, 63, 73, 75, 79, 81, 163, 165, 169, 171, 181, 183, 187, 189, 217, 219, 223, 225, 235, 237, 241, 243, 487, 489, 493, 495, 505, 507, 511, 513, 541, 543, 547, 549, 559, 561, 565, 567, 649, 651, 655, 657, 667, 669, 673, 675, 703, 705, 709, 711, 721, 723, 727, 729, 1459, 1461, 1465, 1467, 1477
Offset: 1

Views

Author

Clark Kimberling, May 26 2011

Keywords

Comments

Related sequences for various choices of i and k as defined in A190803:
A003278: (i,k) = (-2,-1)
A191106: (i,k) = (-2, 0)
A191107: (i,k) = (-2, 1)
A191108: (i,k) = (-2, 2)
A153775: (i,k) = (-1, 0)
A147991: (i,k) = (-1, 1)
A191109: (i,k) = (-1, 2)
A005836: (i,k) = ( 0, 1)
A191110: (i,k) = ( 0, 2)
A132140: (i,k) = ( 1, 2)
For a=A191106, we have closure properties: the integers in (2+a)/3 comprise a; the integers in a/3 comprise a.
For k >= 1, m = a(i), 1 <= i <= 2^k seems to be m such that m/(3^k+1) is in the Cantor set (except that m = 0 and m = 3^k+1 do not appear). For k >= 2, m = (a(i)-1)/2, 1 <= i <= 2^k seems to be m such that m/((3^k-1)/2) is in the Cantor set. - Peter Munn, Jul 06 2019
Every even number is the sum of two (possibly equal) terms. More specifically: terms a(1) through a(2^n) = 3^n sum to even numbers 2 times 1 through 3^n. Every even number is infinitely often the difference of two terms. Since the sequence is equal to 2*A005836(n) + 1, these properties follow immediately from similar properties of A005836 for every number. - Aad Thoen, Feb 17 2022
if A_n=(a(1),a(2),...,a(2^n)), then A_(n+1)=(A_n,A_n+2*3^n), similar to A003278. - Arie Bos, Jul 26 2022

Examples

			1 -> 3 -> 7,9 -> 19,21,25,27 -> ...
		

Crossrefs

Cf. A005823, A005836, A054591, A088917 (characteristic function), A173934, A190803, A191108.
Partial sums of A061393.
Similar formula as A003278, A_(n+1)=(A_n,A_n+2*3^n).

Programs

  • Mathematica
    h = 3; i = -2; j = 3; k = 0; f = 1; g = 9;
    a = Union[Flatten[NestList[{h # + i, j # + k} &, f, g]]]  (* A191106; regarding g, see note at A190803 *)
    b = (a + 2)/3; c = a/3; r = Range[1, 900];
    d = Intersection[b, r](* illustrates closure property *)
    e = Intersection[c, r](* illustrates closure property *)
    2 FromDigits[#, 3]&/@Tuples[{0, 1}, 7] + 1 (* Vincenzo Librandi, Jul 10 2019 *)

Formula

a(n) = 2*A005836(n) + 1. - Charles R Greathouse IV, Sep 06 2011
a(n) = A005823(n) + 1. - Vladimir Shevelev, Dec 17 2012
a(n) = (A191108(n) + 1)/2. - Peter Munn, Jul 09 2019
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