A023758 -id:A023758 - cp's OEIS frontend

A077436 Let B(n) be the sum of binary digits of n. This sequence contains n such that B(n) = B(n^2).

0, 1, 2, 3, 4, 6, 7, 8, 12, 14, 15, 16, 24, 28, 30, 31, 32, 48, 56, 60, 62, 63, 64, 79, 91, 96, 112, 120, 124, 126, 127, 128, 157, 158, 159, 182, 183, 187, 192, 224, 240, 248, 252, 254, 255, 256, 279, 287, 314, 316, 317, 318, 319, 351, 364, 365, 366, 374, 375, 379, 384

Offset: 1

Giuseppe Melfi, Nov 30 2002

Superset of A023758.

Hare, Laishram, & Stoll show that this sequence contains infinitely many odd numbers. In particular for each k in {12, 13, 16, 17, 18, 19, 20, ...} there are infinitely many terms in this sequence with binary digit sum k. - Charles R Greathouse IV, Aug 25 2015

			The element 79 belongs to the sequence because 79=(1001111) and 79^2=(1100001100001), so B(79)=B(79^2)

Reinhard Zumkeller, Table of n, a(n) for n = 1..10476, all terms <= 2^20
Karam Aloui, Damien Jamet, Hajime Kaneko, Steffen Kopecki, Pierre Popoli, and Thomas Stoll, On the binary digits of n and n^2, arXiv:2203.05451 [math.NT], 2022.
K. G. Hare, S. Laishram, and T. Stoll, The sum of digits of n and n^2, International Journal of Number Theory 7:7 (2011), pp. 1737-1752.
Giuseppe Melfi, On simultaneous binary expansions of n and n^2, arXiv:math/0402458 [math.NT], 2004.
Giuseppe Melfi, Su alcune successioni di interi (English with an Italian title)

Cf. A058369, A000120, A000290, A083567.
Cf. A211676 (number of n-bit numbers in this sequence).
A261586 is a subsequence. Subsequence of A352084.

import Data.List (elemIndices)
import Data.Function (on)
a077436 n = a077436_list !! (n-1)
a077436_list = elemIndices 0
   $ zipWith ((-) `on` a000120) [0..] a000290_list
-- Reinhard Zumkeller, Apr 12 2011

[n: n in [0..400] | &+Intseq(n, 2) eq &+Intseq(n^2, 2)]; // Vincenzo Librandi, Aug 30 2015

select(t -> convert(convert(t,base,2),`+`) = convert(convert(t^2,base,2),`+`), [$0..1000]); # Robert Israel, Aug 27 2015

t={}; Do[If[DigitCount[n, 2, 1] == DigitCount[n^2, 2, 1], AppendTo[t, n]], {n, 0, 364}]; t
f[n_] := Total@ IntegerDigits[n, 2]; Select[Range[0, 384], f@ # == f[#^2] &] (* Michael De Vlieger, Aug 27 2015 *)

is(n)=hammingweight(n)==hammingweight(n^2) \\ Charles R Greathouse IV, Aug 25 2015

def ok(n): return bin(n).count('1') == bin(n**2).count('1')
print([m for m in range(400) if ok(m)]) # Michael S. Branicky, Mar 11 2022

A159918(a(n)) = A000120(a(n)). - Reinhard Zumkeller, Apr 25 2009

Initial 0 added by Reinhard Zumkeller, Apr 28 2012, Apr 12 2011

A384877 Irregular triangle read by rows where row k lists the lengths of maximal anti-runs (increasing by more than 1) in the binary indices of n.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 2, 1, 2, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 2, 2, 3, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 2, 2, 3, 1, 2, 1, 1, 2, 2, 3, 3, 1, 3, 1, 2, 2, 2

Offset: 0

Gus Wiseman, Jun 17 2025

A binary index of n is any position of a 1 in its reversed binary expansion. The binary indices of n are row n of A048793.

			The binary indices of 182 are {2,3,5,6,8}, with maximal anti-runs ((2),(3,5),(6,8)) so row 182 is (1,2,2).
Triangle begins:
   0: ()
   1: (1)
   2: (1)
   3: (1,1)
   4: (1)
   5: (2)
   6: (1,1)
   7: (1,1,1)
   8: (1)
   9: (2)
  10: (2)
  11: (1,2)
  12: (1,1)
  13: (2,1)
  14: (1,1,1)
  15: (1,1,1,1)

Row-sums are A000120.
Positions of rows of the form (1,1,...) are A023758.
Positions of first appearances of each distinct row appear to be A052499.
For runs instead of anti-runs we have A245563, reverse A245562.
Row-lengths are A384890.
A355394 counts partitions without a neighborless part, singleton case A355393.
A356606 counts strict partitions without a neighborless part, complement A356607.
A384175 counts subsets with all distinct lengths of maximal runs, complement A384176.
Cf. A044813, A048793, A069010, A164707, A243815, A246029, A328592, A384177, A384877, A384879, A384893.

bpe[n_]:=Join@@Position[Reverse[IntegerDigits[n,2]],1];
Table[Length/@Split[bpe[n],#2!=#1+1&],{n,0,100}]

A264977 a(0) = 0, a(1) = 1, a(2n) = 2a(n), a(2*n+1) = a(n) XOR a(n+1).

Original entry on oeis.org

0, 1, 2, 3, 4, 1, 6, 7, 8, 5, 2, 7, 12, 1, 14, 15, 16, 13, 10, 7, 4, 5, 14, 11, 24, 13, 2, 15, 28, 1, 30, 31, 32, 29, 26, 7, 20, 13, 14, 3, 8, 1, 10, 11, 28, 5, 22, 19, 48, 21, 26, 15, 4, 13, 30, 19, 56, 29, 2, 31, 60, 1, 62, 63, 64, 61, 58, 7, 52, 29, 14, 19, 40, 25, 26, 3, 28, 13, 6, 11, 16, 9, 2, 11, 20, 1, 22

Offset: 0

Antti Karttunen, Dec 10 2015

a(n) is the n-th Stern polynomial (cf. A260443, A125184) evaluated at X = 2 over the field GF(2).

For n >= 1, a(n) gives the index of the row where n occurs in array A277710.

			In this example, binary numbers are given zero-padded to four bits.
a(2) = 2a(1) = 2 * 1 = 2.
a(3) = a(1) XOR a(2) = 1 XOR 2 = 0001 XOR 0010 = 0011 = 3.
a(4) = 2a(2) = 2 * 2 = 4.
a(5) = a(2) XOR a(3) = 2 XOR 3 = 0010 XOR 0011 = 0001 = 1.
a(6) = 2a(3) = 2 * 3 = 6.
a(7) = a(3) XOR a(4) = 3 XOR 4 = 0011 XOR 0100 = 0111 = 7.

Antti Karttunen, Table of n, a(n) for n = 0..16384

Cf. A000225, A001511, A002487, A003987, A010060, A011655, A048675, A055396, A125184, A248663, A260443, A265397, A277330.
Cf. A023758 (the fixed points).
Cf. also A068156, A168081, A265407.
Cf. A277700 (binary weight of terms).
Cf. A277701, A277712, A277713 (positions of 1's, 2's and 3's in this sequence).
Cf. A277711 (position of the first occurrence of each n in this sequence).
Cf. also arrays A277710, A099884.

recurXOR[0] = 0; recurXOR[1] = 1; recurXOR[n_] := recurXOR[n] = If[EvenQ[n], 2recurXOR[n/2], BitXor[recurXOR[(n - 1)/2 + 1], recurXOR[(n - 1)/2]]]; Table[recurXOR[n], {n, 0, 100}] (* Jean-François Alcover, Oct 23 2016 *)

class Memoize:
    def _init_(self, func):
        self.func=func
        self.cache={}
    def _call_(self, arg):
        if arg not in self.cache:
            self.cache[arg] = self.func(arg)
        return self.cache[arg]
@Memoize
def a(n): return n if n<2 else 2*a(n//2) if n%2==0 else a((n - 1)//2)^a((n + 1)//2)
print([a(n) for n in range(51)]) # Indranil Ghosh, Jul 27 2017

a(0) = 0, a(1) = 1, a(2*n) = 2*a(n), a(2*n+1) = a(n) XOR a(n+1).
a(n) = A248663(A260443(n)).
a(n) = A048675(A277330(n)). - Antti Karttunen, Oct 27 2016
Other identities. For all n >= 0:
a(n) = n - A265397(n).
From Antti Karttunen, Oct 28 2016: (Start)
A000035(a(n)) = A000035(n). [Preserves the parity of n.]
A010873(a(n)) = A010873(n). [a(n) mod 4 = n mod 4.]
A001511(a(n)) = A001511(n) = A055396(A277330(n)). [In general, the 2-adic valuation of n is preserved.]
A010060(a(n)) = A011655(n).
a(n) <= n.
For n >= 2, a((2^n)+1) = (2^n) - 3.
The following two identities are so far unproved:
For n >= 2, a(3*A000225(n)) = a(A068156(n)) = 5.
For n >= 2, a(A068156(n)-2) = A062709(n) = 2^n + 3.
(End)

A368900 LCM-transform of Doudna sequence.

Original entry on oeis.org

1, 2, 3, 2, 5, 1, 3, 2, 7, 1, 1, 1, 5, 1, 3, 2, 11, 1, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 5, 1, 3, 2, 13, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 11, 1, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 5, 1, 3, 2, 17, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 13, 1, 1, 1, 1, 1, 1, 1, 1

Offset: 1

David James Sycamore and Antti Karttunen, Jan 10 2024

nonn

Let's define "property S" for sequences as follows: If s is any sequence of positive natural numbers, normalized to begin with offset 1, then it satisfies the S-property if LCM-transform(s) is equal to the sequence obtained by applying A014963 to sequence s, or in other words, when for all n >= 1, lcm {s(1)..s(n)} / lcm {s(1)..s(n-1)} = A014963(s(n)). This holds if and only if, for all n >= 1, when, either (case A): s(n) is of the form p^k, p prime, then gcd(s(n), lcm {s(1)..s(n-1)}) must be equal to p^(k-1), or (case B): when s(n) is not a prime power, then gcd(s(n), lcm {s(1)..s(n-1)}) must be equal to s(n). Together the cases (A) and (B) reduce to the condition that each prime power should appear in s before any of its multiples do.
Clearly the Doudna-sequence satisfies the property by the way of its construction, as do many of its variants like A356867 (see A369060).
Also, for any base-2 related permutation b that keeps all the numbers of range [2^k, 2^(1+k)[ in the same range, i.e., if for all n >= 1, A000523(b(n)) = A000523(n), then the above property is automatically satisfied.
Furthermore, because in Doudna-sequence no multiple of any term is located on the same row as the term itself (see the tree-illustration in A005940), it follows that any composition of A005940 with any such base-2 related permutation as mentioned above also automatically satisfies the S-property, for example, the permutations A163511, A243353, A253563, A253565, A366260, A366263 and A366275.
Note: Like A005940 itself, also this sequence might be more logical with the starting offset 0 instead of 1, to better align with the underlying mapping from the binary expansion of n to the prime factorization. - Antti Karttunen, Jan 24 2024

Antti Karttunen, Table of n, a(n) for n = 1..65537
A. Nowicki, Strong divisibility and LCM-sequences, arXiv:1310.2416 [math.NT], 2013.
A. Nowicki, Strong divisibility and LCM-sequences, Am. Math. Mnthly 122 (2015), 958-966. [Defines the LCM-transform operation]
Index entries for sequences related to binary expansion of n

Cf. A000040, A000225, A000265, A005940, A005941, A007814, A023758, A368901.
List of LCM-transforms of permutations (permutation given in parentheses):
Cf. A265576 (A064413; note that the EKG sequence permutation does not satisfy the S-property).
In all following cases, the permutation satisfies the S-property:
Cf. A014963 (A000027),
Cf. A369028 (A253563), A369029 (A253565), A369030 (A163511), A369053 (A243353).
Cf. A369031 (A122111), A369032 (A241909), A369033 (A241916).
Cf. A369041 (A003188), A369042 (A006068), A369043 (A193231), A369044 (A057889), A369041 (A054429). [Base-2 related permutations]
Cf. A369060 (A356867).
Other permutations that have the same property: A303767, (and when used as an offset=1 sequence): A052330.

nn = 120; Array[Set[{s[#], a[#]}, {#, #}] &, 2]; j = 2;
Do[If[EvenQ[n],
  Set[s[n], 2 s[n/2]],
  Set[s[n],
    Times @@ Power @@@ Map[{Prime[PrimePi[#1] + 1], #2} & @@ # &,
      FactorInteger[s[(n + 1)/2]]]]];
  k = LCM[j, s[n]]; a[n] = k/j; j = k, {n, 3, nn}];
Array[a, nn] (* Michael De Vlieger, Mar 24 2024 *)

up_to = 16384;
LCMtransform(v) = { my(len = length(v), b = vector(len), g = vector(len)); b[1] = g[1] = 1; for(n=2,len, g[n] = lcm(g[n-1],v[n]); b[n] = g[n]/g[n-1]); (b); };
A005940(n) = { my(p=2, t=1); n--; until(!n\=2, if((n%2), (t*=p), p=nextprime(p+1))); (t) };
v368900 = LCMtransform(vector(up_to,i,A005940(i)));
A368900(n) = v368900[n];

A000265(n) = (n>>valuation(n,2));
A209229(n) = (n && !bitand(n,n-1));
A368900(n)  = if(1==n, 1, my(x=A000265(n-1)); if(A209229(1+x), prime(1+valuation(n-1,2)), 1));

a(n) = A368901(n) / A368901(n-1) = lcm {1..A005940(n)} / lcm {1..A005940(n-1)}.
a(n) = A005940(n) / gcd(A005940(n), A368901(n-1)).
a(n) = A014963(A005940(n)). [Because A005940 satisfies the property given in the comments]
For n >= 1, Product_{d|n} a(A005941(d)) = n. [Implied by above]
For n >= 1, a(n) = A369030(1+A054429(n-1)).
For n > 1, if n-1 is a number of the form 2^i - 2^j with i >= j, then a(n) = prime(1+j), otherwise a(n) = 1.

A384878 Position of first appearance of n in the flattened version of the triangle A384877, whose m-th row lists the lengths of maximal anti-runs in the binary indices of m.

Original entry on oeis.org

1, 6, 34, 178, 882, 4210, 19570, 89202, 400498, 1776754

Offset: 1

Gus Wiseman, Jun 23 2025

A binary index of n is any position of a 1 in its reversed binary expansion. The binary indices of n are row n of A048793.

			The set of binary indices of each nonnegative integer and its partition into anti-runs begins:
  0: {}      {{}}
  1: {1}     {{1}}
  2: {2}     {{2}}
  3: {1,2}   {{1},{2}}
  4: {3}     {{3}}
  5: {1,3}   {{1,3}}
  6: {2,3}   {{2},{3}}
  7: {1,2,3} {{1},{2},{3}}
The flattened version begins: {}, {1}, {2}, {1}, {2}, {3}, {1,3}, {2}, {3}, {1}, {2}, {3}. Of these sets, the first of length 2 is the sixth (starting with 0), so we have a(2) = 6.

For runs instead of anti-runs we have A001792.
The unflattened version is A052499.
Positions of first appearances in A384877, see A000120, A245562, A245563, A384890.
A023758 lists differences of powers of 2.
A384175 counts subsets with all distinct lengths of maximal runs, complement A384176.
Cf. A044813, A048793, A069010, A384879, A384893.

bpe[n_]:=Join@@Position[Reverse[IntegerDigits[n,2]],1];
mnrm[s_]:=If[Min@@s==1,mnrm[DeleteCases[s-1,0]]+1,0];
q=Join@@Table[Length/@Split[bpe[n],#2!=#1+1&],{n,0,100}];
Table[Position[q,i][[1,1]],{i,mnrm[q]}]

A200649 Number of 1's in the Stolarsky representation of n.

Original entry on oeis.org

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

Offset: 1

Casey Mongoven, Nov 19 2011

For the Stolarsky representation of n, see the C. Mongoven link.

Conjecture: a(n) is the length of row n-1 of A385886. To obtain it, first take maximal anti-run lengths of binary indices of each nonnegative integer (giving A384877), then remove all duplicate rows (giving A385886), and finally take the length of each remaining row. For sum instead of length we appear to have A200648. For runs minus 1 instead of anti-runs see A200650. - Gus Wiseman, Jul 21 2025

			The Stolarsky representation of 19 is 11101. This has 4 1's. So a(19) = 4.

Amiram Eldar, Table of n, a(n) for n = 1..10000
Casey Mongoven, Description of Stolarsky Representations.

Cf. A072649, A130312, A135818, A200651.
For length instead of number of 1's we have A200648.
For 0's instead of 1's we have A200650.
Stolarsky representation is listed by A385888, ranks A200714.
A000120 counts 1's in binary expansion.
A384877 lists anti-run lengths of binary indices, duplicates removed A385886.
A384890 counts maximal anti-runs of binary indices, ranked by A385816.
Cf. A023758, A044813, A048793, A069010, A164707, A245562, A245563, A328592, A384893.

stol[n_] := stol[n] = If[n == 1, {}, If[n != Round[Round[n/GoldenRatio]*GoldenRatio], Join[stol[Floor[n/GoldenRatio^2] + 1], {0}], Join[stol[Round[n/GoldenRatio]], {1}]]];
a[n_] := Count[stol[n], 1]; Array[a, 100] (* Amiram Eldar, Jul 07 2023 *)

stol(n) = {my(phi=quadgen(5)); if(n==1, [], if(n != round(round(n/phi)*phi), concat(stol(floor(n/phi^2) + 1), [0]), concat(stol(round(n/phi)), [1])));}
a(n) = vecsum(stol(n)); \\ Amiram Eldar, Jul 07 2023

a(n) = a(n - A130312(n-1)) + (A072649(n-1) - A072649(n - A130312(n-1) - 1)) mod 2 for n > 2 with a(1) = 0, a(2) = 1. - Mikhail Kurkov, Oct 19 2021 [verification needed]

a(n) = A200648(n) - A200650(n). - Amiram Eldar, Jul 07 2023

More terms from Amiram Eldar, Jul 07 2023

A200648 Length of Stolarsky representation of n.

Original entry on oeis.org

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

Offset: 1

Casey Mongoven, Nov 19 2011

For the Stolarsky representation of n, see the C. Mongoven link.

Conjecture: a(n) is the sum of row n-1 of A385886. To obtain it, first take maximal anti-run lengths of binary indices of each nonnegative integer (giving A384877), then remove all duplicate rows (giving A385886), and finally take the sum of each remaining row. For length instead of sum we appear to have A200649. - Gus Wiseman, Jul 21 2025

			The Stolarsky representation of 19 is 11101. This is of length 5. So a(19) = 5.

Amiram Eldar, Table of n, a(n) for n = 1..10000
Casey Mongoven, Description of Stolarsky Representations.

Cf. A135817, A200651.
Counting just ones gives A200649.
Counting just zeros gives A200650.
Stolarsky representation is listed by A385888, ranks A200714.
A000120 counts 1's in binary expansion.
A384890 counts maximal anti-runs of binary indices, ranks A385816.
A385886 lists maximal anti-run lengths of binary indices.
Cf. A023758, A048793, A052499, A069010, A072649, A083368, A135818, A245562, A245563, A348366, A384877, A384893.

stol[n_] := stol[n] = If[n == 1, {}, If[n != Round[Round[n/GoldenRatio]*GoldenRatio], Join[stol[Floor[n/GoldenRatio^2] + 1], {0}], Join[stol[Round[n/GoldenRatio]], {1}]]];
a[n_] := If[n == 1, 1, Length[stol[n]]]; Array[a, 100] (* Amiram Eldar, Jul 07 2023 *)

stol(n) = {my(phi=quadgen(5)); if(n==1, [], if(n != round(round(n/phi)*phi), concat(stol(floor(n/phi^2) + 1), [0]), concat(stol(round(n/phi)), [1])));}
a(n) = if(n == 1, 1, #stol(n)); \\ Amiram Eldar, Jul 07 2023

a(n) = A200649(n) + A200650(n). - Michel Marcus, Mar 14 2023

More terms from Amiram Eldar, Jul 07 2023

A384879 Numbers whose binary indices have all distinct lengths of maximal anti-runs (increasing by more than 1).

Original entry on oeis.org

1, 2, 4, 5, 8, 9, 10, 11, 13, 16, 17, 18, 19, 20, 21, 22, 25, 26, 32, 33, 34, 35, 36, 37, 38, 40, 41, 42, 43, 44, 49, 50, 52, 53, 64, 65, 66, 67, 68, 69, 70, 72, 73, 74, 75, 76, 80, 81, 82, 83, 84, 85, 86, 88, 97, 98, 100, 101, 104, 105, 106, 128, 129, 130

Offset: 1

Gus Wiseman, Jun 17 2025

nonn

A binary index of n is any position of a 1 in its reversed binary expansion. The binary indices of n are row n of A048793.

			The binary indices of 813 are {1,3,4,6,9,10}, with maximal anti-runs ((1,3),(4,6,9),(10)), with lengths (2,3,1), so 813 is in the sequence.
The terms together with their binary expansions and binary indices begin:
    1:       1 ~ {1}
    2:      10 ~ {2}
    4:     100 ~ {3}
    5:     101 ~ {1,3}
    8:    1000 ~ {4}
    9:    1001 ~ {1,4}
   10:    1010 ~ {2,4}
   11:    1011 ~ {1,2,4}
   13:    1101 ~ {1,3,4}
   16:   10000 ~ {5}
   17:   10001 ~ {1,5}
   18:   10010 ~ {2,5}
   19:   10011 ~ {1,2,5}
   20:   10100 ~ {3,5}
   21:   10101 ~ {1,3,5}
   22:   10110 ~ {2,3,5}
   25:   11001 ~ {1,4,5}
   26:   11010 ~ {2,4,5}

Subsets of this type are counted by A384177, for runs A384175 (complement A384176).
These are the indices of strict rows in A384877, see A384878, A245563, A245562, A246029.
A000120 counts binary indices.
A098859 counts Wilf partitions (distinct multiplicities), complement A336866.
A356606 counts strict partitions without a neighborless part, complement A356607.
A384890 counts maximal anti-runs in binary indices, runs A069010.
Cf. A023758, A044813, A048793, A164707, A242882, A243815, A325325, A328592, A384879, A384884, A384886, A384893.

bpe[n_]:=Join@@Position[Reverse[IntegerDigits[n,2]],1];
Select[Range[100],UnsameQ@@Length/@Split[bpe[#],#2!=#1+1&]&]

A101082 Numbers n such that binary representation contains bit strings "10" and "01" (possibly overlapping).

Original entry on oeis.org

5, 9, 10, 11, 13, 17, 18, 19, 20, 21, 22, 23, 25, 26, 27, 29, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 49, 50, 51, 52, 53, 54, 55, 57, 58, 59, 61, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92

Offset: 1

Rick L. Shepherd, Nov 29 2004

Subsequence of A062289; set difference A062289 minus A043569.
Complement of A023758. Also numbers not the sum of consecutive powers of 2. - Omar E. Pol, Mar 04 2013
Equivalently, numbers not the difference of two powers of two. - Charles R Greathouse IV, Mar 07 2013
The terms >=9 are bases in which a power of 2 exists, which does not contain a digit that is a power of 2. In base 10, 2^16 = 65536 is such a number, as it does not contain any one-digit power of 2, which in base 10 are 1, 2, 4 and 8. - Patrick Wienhöft, Jul 28 2016

			29 = 11101_2 is a term, "10" and "01" are contained (here overlapping).

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Patrick Wienhöft, Python program
Index entries for 2-automatic sequences.

Complement: A023758.

Cf. A043569, A062289, A049502.

a101082 n = a101082_list !! (n-1)
a101082_list = filter ((> 0) . a049502) [0..]
-- Reinhard Zumkeller, Jun 17 2015

Select[Range@ 120, Function[d, Times @@ Total@ Map[Map[Function[k, Boole@ MatchQ[#, k]], {{1, 0}, {0, 1}}] &, Partition[d, 2, 1]] > 0]@ IntegerDigits[#, 2] &] (* Michael De Vlieger, Dec 23 2016 *)
Select[Range[100],With[{c=IntegerDigits[#,2]},SequenceCount[c,{1,0}]>0&&SequenceCount[c,{0,1}]>0]&] (* Harvey P. Dale, Jun 08 2025 *)

is(n)=n>>=valuation(n,2);n+1!=1<Charles R Greathouse IV, Mar 07 2013

def A101082(n):
    def f(x): return n+((k:=x.bit_length())*(k-1)>>1)+sum(1 for i in range(k) if (1<Chai Wah Wu, Feb 23 2025

a(n) ~ n. In particular a(n) = n + (log_2 n)^2/2 + O(log n). - Charles R Greathouse IV, Mar 07 2013

A049502(a(n)) > 0. - Reinhard Zumkeller, Jun 17 2015

A200650 Number of 0's in Stolarsky representation of n.

Original entry on oeis.org

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

Offset: 1

Casey Mongoven, Nov 19 2011

For the Stolarsky representation of n, see the C. Mongoven link.
a(n+1), n >= 1, gives the size of the n-th generation of each of the "[male-female] pair of Fibonacci rabbits" in the Fibonacci rabbits tree read right-to-left by row, the first pair (the root) being the 0th generation. (Cf. OEIS Wiki link below.) - Daniel Forgues, May 07 2015
From Daniel Forgues, May 07 2015: (Start)
Concatenation of:
  0:                      1,
  1:                      0,
  2:                      0,
  3:                   1, 0,
  4:                1, 1, 0,
  5:          2, 1, 1, 1, 0,
  6: 2, 2, 1, 2, 1, 1, 1, 0,
(...),
where row n, n >= 3, is row n-1 prepended by incremented row n-2. (End)
For n >= 3, this algorithm yields the next F_n terms of the sequence, where F_n is the n-th Fibonacci number (A000045). Since it is asymptotic to (phi^n)/sqrt(5), the number of terms thus obtained grows exponentially at each step! - Daniel Forgues, May 22 2015
Conjecture: a(n) is one less than the length of row n-1 of A385817. To obtain it, first take maximal run lengths of binary indices of each nonnegative integer (giving A245563), then remove all duplicate rows (giving A385817), and finally take the length of each remaining row and subtract 1. For sum instead of length we appear to have A200648. For anti-runs instead of runs we appear to have A341259 = A200649-1. - Gus Wiseman, Jul 21 2025
How is this related to A117479? - R. J. Mathar, Aug 10 2025

			The Stolarsky representation of 19 is 11101. This has one 0. So a(19) = 1.

Kenny Lau, Table of n, a(n) for n = 1..20000
Casey Mongoven, Description of Stolarsky Representations.
OEIS Wiki, Fibonacci rabbits per generation.

For length instead of number of 0's we have A200648.
For sum (or number of 1's) instead of number of 0's we have A200649.
Stolarsky representation is listed by A385888, ranks A200714.
A000120 counts 1's in binary expansion, 0's A023416.
A069010 counts maximal runs of binary indices, ranked by A385889.
A245563 lists maximal run lengths of binary indices, duplicates removed A385817.
Cf. A000045, A023758, A135818, A200651, A245562, A328592, A385886.

stol[n_] := stol[n] = If[n == 1, {}, If[n != Round[Round[n/GoldenRatio]*GoldenRatio], Join[stol[Floor[n/GoldenRatio^2] + 1], {0}], Join[stol[Round[n/GoldenRatio]], {1}]]];
a[n_] := If[n == 1, 1, Count[stol[n], 0]]; Array[a, 100] (* Amiram Eldar, Jul 07 2023 *)

stol(n) = {my(phi=quadgen(5)); if(n==1, [], if(n != round(round(n/phi)*phi), concat(stol(floor(n/phi^2) + 1), [0]), concat(stol(round(n/phi)), [1])));}
a(n) = if(n == 1, 1,  my(s = stol(n)); #s - vecsum(s)); \\ Amiram Eldar, Jul 07 2023

a(n) = A200648(n) - A200649(n). - Amiram Eldar, Jul 07 2023

Corrected and extended by Kenny Lau, Jul 04 2016

cp's OEIS Frontend

A077436 Let B(n) be the sum of binary digits of n. This sequence contains n such that B(n) = B(n^2).

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A384877 Irregular triangle read by rows where row k lists the lengths of maximal anti-runs (increasing by more than 1) in the binary indices of n.

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A264977 a(0) = 0, a(1) = 1, a(2*n) = 2*a(n), a(2*n+1) = a(n) XOR a(n+1).

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A368900 LCM-transform of Doudna sequence.

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A384878 Position of first appearance of n in the flattened version of the triangle A384877, whose m-th row lists the lengths of maximal anti-runs in the binary indices of m.

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A200649 Number of 1's in the Stolarsky representation of n.

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A200648 Length of Stolarsky representation of n.

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A264977 a(0) = 0, a(1) = 1, a(2n) = 2a(n), a(2*n+1) = a(n) XOR a(n+1).