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.

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A269160 Formula for Wolfram's Rule 30 cellular automaton: a(n) = n XOR (2n OR 4n).

Original entry on oeis.org

0, 7, 14, 13, 28, 27, 26, 25, 56, 63, 54, 53, 52, 51, 50, 49, 112, 119, 126, 125, 108, 107, 106, 105, 104, 111, 102, 101, 100, 99, 98, 97, 224, 231, 238, 237, 252, 251, 250, 249, 216, 223, 214, 213, 212, 211, 210, 209, 208, 215, 222, 221, 204, 203, 202, 201, 200, 207, 198, 197, 196, 195, 194, 193, 448, 455, 462
Offset: 0

Views

Author

Antti Karttunen, Feb 20 2016

Keywords

Comments

Take n, write it in binary, see what Rule 30 would do to that state, convert it to decimal: that is a(n). For example, we can see in A110240 that 7 = 111_2 becomes 25 = 11001_2 under Rule 30, which is shown here by a(7) = 25. - N. J. A. Sloane, Nov 25 2016
The sequence is injective: no value occurs more than once.
Fibbinary numbers (A003714) give all integers n>=0 for which a(n) = A048727(n) and for which a(n) = A269161(n).

Crossrefs

Cf. A110240 (iterates starting from 1).
Cf. A269162 (left inverse).
Cf. A269163 (same sequence sorted into ascending order).
Cf. A269164 (values missing from this sequence).
Cf. also A048727, A269161.

Programs

Formula

a(n) = n XOR (2n OR 4n) = A003987(n, A003986(2*n, 4*n)).
Other identities. For all n >= 0:
a(2*n) = 2*a(n).
a(n) = A057889(A269161(A057889(n))). [Rule 30 is the mirror image of rule 86.]
A269162(a(n)) = n.
For all n >= 1:
A070939(a(n)) - A070939(n) = 2. [The binary length of a(n) is two bits longer than that of n for all nonzero values.]
G.f.: (3*x + 2*x^2 +x^3)/(1 - x^4) + Sum_{k>=1}(2^(k + 1)*x^(2^(k - 1))/((1 + x^(2^(k + 1)))*(1 - x))). - Miles Wilson, Jan 24 2025
a(n) = A387312(2*n). - Mia Boudreau, Sep 04 2025

A057890 In base 2, either a palindrome or becomes a palindrome if trailing 0's are omitted.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 15, 16, 17, 18, 20, 21, 24, 27, 28, 30, 31, 32, 33, 34, 36, 40, 42, 45, 48, 51, 54, 56, 60, 62, 63, 64, 65, 66, 68, 72, 73, 80, 84, 85, 90, 93, 96, 99, 102, 107, 108, 112, 119, 120, 124, 126, 127, 128, 129, 130, 132, 136, 144, 146
Offset: 1

Views

Author

Marc LeBrun, Sep 25 2000

Keywords

Comments

Symmetric bit strings (bit-reverse palindromes), including as many leading as trailing zeros.
Fixed points of A057889, complement of A057891
n such that A000265(n) is in A006995. - Robert Israel, Jun 07 2016

Examples

			10 is included, since 01010 is a palindrome, but 11 is not because 1011 is not.
		

Crossrefs

Programs

  • Haskell
    a057890 n = a057890_list !! (n-1)
    a057890_list = 0 : filter ((== 1) . a178225 . a000265) [1..]
    -- Reinhard Zumkeller, Oct 21 2011
    
  • Maple
    dmax:= 10: # to get all terms < 2^dmax
    revdigs:= proc(n)
      local L, Ln, i;
      L:= convert(n, base, 2);
      Ln:= nops(L);
      add(L[i]*2^(Ln-i), i=1..Ln);
    end proc;
    P[0]:= {0}:
    P[1]:= {1}:
    for d from 2 to dmax do
      if d::even then
        P[d]:= { seq(2^(d/2)*x + revdigs(x), x=2^(d/2-1)..2^(d/2)-1)}
      else
        m:= (d-1)/2;
        B:={seq(2^(m+1)*x + revdigs(x), x=2^(m-1)..2^m-1)};
        P[d]:= B union map(`+`, B, 2^m)
      fi
    od:
    A:= `union`(seq(seq(map(`*`,P[d],2^k),k=0..dmax-d),d=0..dmax)):
    sort(convert(A,list)); # Robert Israel, Jun 07 2016
  • Mathematica
    PaleQ[n_Integer, base_Integer] := Module[{idn, trim = n/base^IntegerExponent[n, base]}, idn = IntegerDigits[trim, base]; idn == Reverse[idn]]; Select[Range[0, 150], PaleQ[#, 2] &] (* Lei Zhou, Dec 13 2013 *)
    pal2Q[n_]:=Module[{id=Drop[IntegerDigits[n,2],-IntegerExponent[n,2]]},id==Reverse[id]]; Join[{0},Select[Range[200],pal2Q]] (* Harvey P. Dale, Feb 26 2015 *)
    A057890Q = If[# > 0 && EvenQ@#, #0[#/2], # == #~IntegerReverse~2] &; Select[0~Range~146, A057890Q] (* JungHwan Min, Mar 29 2017 *)
    Select[Range[0, 200], PalindromeQ[IntegerDigits[#, 2] /. {b__, 0..} -> {b} ]&] (* Jean-François Alcover, Sep 18 2018 *)
  • PARI
    bitrev(n) = subst(Pol(Vecrev(binary(n>>valuation(n,2))), 'x), 'x, 2);
    is(n) = my(x = n >> valuation(n,2)); x == bitrev(x);
    concat(0, select(is,vector(147,n,n)))  \\ Gheorghe Coserea, Jun 07 2016
    
  • PARI
    is(n)=n==0 || Vecrev(n=binary(n>>valuation(n,2)))==n \\ Charles R Greathouse IV, Aug 25 2016
  • Python
    A057890 = [n for n in range(10**6) if bin(n)[2:].rstrip('0') == bin(n)[2:].rstrip('0')[::-1]] # Chai Wah Wu, Aug 12 2014
    

Formula

A030101(A030101(n)) = A030101(n). - David W. Wilson, Jun 09 2009, Jun 18 2009
A178225(A000265(a(n))) = 1. - Reinhard Zumkeller, Oct 21 2011
a(7*2^n-4*n-4) = 4^n + 1, a(10*2^n-4*n-6) = 2*4^n + 1. - Gheorghe Coserea, Apr 05 2017

A278233 Filter-sequence for GF(2)[X]-factorization: sequence that gives the least natural number with the same prime signature that (0, 1)-polynomial encoded in the binary expansion of n has when it is factored over GF(2).

Original entry on oeis.org

1, 2, 2, 4, 4, 6, 2, 8, 6, 12, 2, 12, 2, 6, 8, 16, 16, 30, 2, 36, 4, 6, 6, 24, 2, 6, 12, 12, 6, 24, 2, 32, 6, 48, 6, 60, 2, 6, 12, 72, 2, 12, 6, 12, 24, 30, 2, 48, 6, 6, 32, 12, 6, 60, 2, 24, 12, 30, 2, 72, 2, 6, 12, 64, 36, 30, 2, 144, 4, 30, 6, 120, 2, 6, 24, 12, 6, 60, 6, 144, 4, 6, 30, 36, 64, 30, 2, 24, 6, 120, 2, 60, 6, 6, 12, 96, 2, 30, 12, 12, 30, 96, 2
Offset: 1

Views

Author

Antti Karttunen, Nov 16 2016

Keywords

Comments

a(n) = the least number with the same prime signature as A091203(n).
This sequence works as an A046523-analog in the polynomial ring GF(2)[X] and can be used as a filter which matches with (and thus detects) any sequence in the database where a(n) depends only on the exponents of irreducible factors when the polynomial corresponding to n (via base-2 encoding) is factored over GF(2). These sequences are listed in the Crossrefs section, "Sequences that partition N into ...".
Matching in this context means that the sequence a matches with the sequence b iff for all i, j: a(i) = a(j) => b(i) = b(j). In other words, iff the sequence b partitions the natural numbers to the same or coarser equivalence classes (as/than the sequence a) by the distinct values it obtains.

Examples

			3 is "11" in binary, encodes polynomial x + 1, and 7 is "111" in binary, encodes polynomial x^2 + x + 1, both which are irreducible over GF(2). We can multiply their codes with carryless multiplication A048720 as A048720(3,7) = 9, A048720(9,3) = 27, A048720(9,7) = 63. Now a(27) = a(63) because the exponents occurring in both codes 27 and 63 are one 1 and two 2's, and their order is not significant when computing prime signature. Moreover a(27) = a(63) = 12 because that is the least number with a prime signature (1,2) in the more familiar domain of natural numbers.
a(25) = 2, because 25 is "11001" in binary, encoding polynomial x^4 + x^3 + 1, which is irreducible in the ring GF(2)[X], i.e., 25 is in A014580, whose initial term is 2.
		

Crossrefs

Cf. A014580 (gives the positions of 2's), A048720, A057889, A091203, A091205, A193231, A235042, A278231, A278238, A278239.
Similar filtering sequences: A046523, A278222, A278226, A278236, A278243.
Sequences that partition N into same or coarser equivalence classes: A091220, A091221, A091222, A106493, A106494.
Cf. also A304529, A304751, A305788 (rgs-transform), A305789.

Programs

Formula

a(n) = A046523(A091203(n)) = A046523(A091205(n)) = A046523(A235042(n)). [Because of the "sorting" essentially performed by A046523, any map from GF(2)[X] to Z can be used, as long as it is fully (cross-)multiplicative and preserves also the exponents intact.]
Other identities. For all n >= 1:
a(A014580(n)) = 2.
a(n) = a(A057889(n)) = a(A193231(n)).
a(A000695(n)) = A278238(n).
a(A277699(n)) = A278239(n).

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

Views

Author

Keywords

Comments

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

Crossrefs

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. A369041 (A003188), A369042 (A006068), A369043 (A193231), A369044 (A057889), A369041 (A054429). [Base-2 related permutations]
Other permutations that have the same property: A303767, (and when used as an offset=1 sequence): A052330.

Programs

  • Mathematica
    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 *)
  • PARI
    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];
    
  • PARI
    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));

Formula

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.

A268389 a(n) = greatest k such that polynomial (X+1)^k divides the polynomial (in polynomial ring GF(2)[X]) that is encoded in the binary expansion of n. (See the comments for details).

Original entry on oeis.org

0, 0, 1, 0, 2, 1, 0, 0, 1, 2, 0, 1, 0, 0, 3, 0, 4, 1, 0, 2, 0, 0, 1, 1, 0, 0, 2, 0, 1, 3, 0, 0, 1, 4, 0, 1, 0, 0, 2, 2, 0, 0, 1, 0, 3, 1, 0, 1, 0, 0, 5, 0, 1, 2, 0, 0, 2, 1, 0, 3, 0, 0, 1, 0, 2, 1, 0, 4, 0, 0, 1, 1, 0, 0, 3, 0, 1, 2, 0, 2, 0, 0, 1, 0, 6, 1, 0, 0, 1, 3, 0, 1, 0, 0, 2, 1, 0, 0, 2, 0, 1, 5, 0, 0, 3, 1, 0, 2, 0, 0, 1, 0, 1, 2, 0, 1, 0, 0, 4, 3
Offset: 1

Views

Author

Antti Karttunen, Feb 10 2016

Keywords

Comments

a(n) gives the number of iterations of map k -> A006068(k)/2 that are required (when starting from k = n) until k is an odd number.
A001317 gives the record positions and particularly, A001317(n) gives the first occurrence of n in this sequence.
When polynomials over GF(2) are encoded in the binary representation of n in a natural way where each polynomial b(n)*X^n+...+b(0)*X^0 over GF(2) is represented by the binary number b(n)*2^n+...+b(0)*2^0 in N (each coefficient b(k) is either 0 or 1), then a(n) = the number of times polynomial X+1 (encoded by 3, "11" in binary) divides the polynomial encoded by n.

Examples

			For n = 5 ("101" in binary) which encodes polynomial x^2 + 1, we see that it can be factored over GF(2) as (X+1)(X+1), and thus a(5) = 2.
For n = 8 ("1000" in binary) which encodes polynomial x^3, we see that it is not divisible in ring GF(2)[X] by polynomial X+1, thus a(8) = 0.
For n = 9 ("1001" in binary) which encodes polynomial x^3 + 1, we see that it can be factored over GF(2) as (X+1)(X^2 + X + 1), and thus a(9) = 1.
		

Crossrefs

Cf. A268669 (quotient left after (X+1)^a(n) has been divided out).
Cf. A268395 (partial sums).
Cf. A000069 (positions of zeros), A268679 (this sequence without zeros).
Cf. also A037096, A037097, A136386.

Programs

  • Mathematica
    f[n_] := Which[n == 1, 0, OddQ@ #, 0, EvenQ@ #, 1 + f[#/2]] &@ Fold[BitXor, n, Quotient[n, 2^Range[BitLength@ n - 1]]]; Array[f, {120}] (* Michael De Vlieger, Feb 12 2016, after Jan Mangaldan at A006068 *)
  • PARI
    a(n) = {my(b = binary(n), p = Pol(binary(n))*Mod(1,2), k = poldegree(p)); while (type(p/(x+1)^k*Mod(1,2)) != "t_POL", k--); k;} \\ Michel Marcus, Feb 12 2016
    
  • Scheme
    ;; This employs the given recurrence and uses memoization-macro definec:
    (definec (A268389 n) (if (odd? (A006068 n)) 0 (+ 1 (A268389 (/ (A006068 n) 2)))))
    (define (A268389 n) (let loop ((n n) (s 0)) (let ((k (A006068 n))) (if (odd? k) s (loop (/ k 2) (+ 1 s)))))) ;; Computed in a loop, no memoization.

Formula

If A006068(n) is odd, then a(n) = 0, otherwise a(n) = 1 + a(A006068(n)/2).
Other identities. For all n >= 0:
a(A001317(n)) = n. [The sequence works as a left inverse of A001317.]
A048720(A268669(n),A048723(3,a(n))) = A048720(A268669(n),A001317(a(n))) = n.
A048724^a(n) (A268669(n)) = n. [The a(n)-th fold application (power) of A048724 when applied to A268669(n) gives n back.]
a(n) = A007949(A235042(n)).
a(A057889(n)) = a(n).

A235027 Reverse the bits of prime divisors of n (with 2 -> 2), and multiply together: a(0)=0, a(1)=1, a(2)=2, a(p) = revbits(p) for odd primes p, a(u*v) = a(u) * a(v) for composites.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 13, 12, 11, 14, 15, 16, 17, 18, 25, 20, 21, 26, 29, 24, 25, 22, 27, 28, 23, 30, 31, 32, 39, 34, 35, 36, 41, 50, 33, 40, 37, 42, 53, 52, 45, 58, 61, 48, 49, 50, 51, 44, 43, 54, 65, 56, 75, 46, 55, 60, 47, 62, 63, 64, 55, 78, 97
Offset: 0

Views

Author

Antti Karttunen, Jan 02 2014

Keywords

Comments

This is not a permutation of integers: a(25) = 25 = 5*5 = a(19) is the first case which breaks the injectivity. However, the first 24 terms are equal with A057889, which is a GF(2)[X]-analog of this sequence and which in contrary to this, is bijective. This stems from the fact that the set of irreducible GF(2)[X] polynomials (A014580) is closed under bit-reversal (A056539), while primes (A000040) are not.
Sequence A290078 gives the positions n where the ratio a(n)/n obtains new record values.
Note, instead of A056539 we could as well use A057889 to reverse the bits of n, and also A030101 when restricted to odd primes.

Examples

			a(33) = a(3*11) = a(3) * a(11) = 3 * 13 = 39 (because 3, in binary '11', stays same when reversed, while 11 (eleven), in binary '1011', changes to '1101' = 13).
		

Crossrefs

A235028 gives the fixed points. A235030 numbers such that n <> a(a(n)), or equally A001222(a(n)) > A001222(n). A235145 the number of iterations needed to reach a fixed point or cycle of 2, A235146 its records.

Programs

  • Mathematica
    f[p_, e_] := IntegerReverse[p, 2]^e; f[2, e_] := 2^e; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100, 0] (* Amiram Eldar, Sep 03 2023 *)
  • PARI
    revbits(n) = fromdigits(Vecrev(binary(n)), 2);
    a(n) = {my(f = factor(n)); for (k=1, #f~, if (f[k,1] != 2, f[k,1] = revbits(f[k,1]););); factorback(f);} \\ Michel Marcus, Aug 05 2017

Formula

Completely multiplicative with a(0)=0, a(1)=1, a(p) = A056539(p) for primes p (which maps 2 to 2, and reverses the binary representation of odd primes), and a(u*v) = a(u) * a(v) for composites.
Equally, after a(0)=0, a(p * q * ... * r) = A056539(p) * A056539(q) * ... * A056539(r), for primes p, q, etc., not necessarily distinct.
a(0)=0, a(1)=1, a(n) = A056539(A020639(n)) * a(n/A020639(n)).

A269174 Formula for Wolfram's Rule 124 cellular automaton: a(n) = (n OR 2n) AND ((n XOR 2n) OR (n XOR 4n)).

Original entry on oeis.org

0, 3, 6, 7, 12, 15, 14, 11, 24, 27, 30, 31, 28, 31, 22, 19, 48, 51, 54, 55, 60, 63, 62, 59, 56, 59, 62, 63, 44, 47, 38, 35, 96, 99, 102, 103, 108, 111, 110, 107, 120, 123, 126, 127, 124, 127, 118, 115, 112, 115, 118, 119, 124, 127, 126, 123, 88, 91, 94, 95, 76, 79, 70, 67, 192, 195, 198, 199, 204, 207, 206, 203, 216
Offset: 0

Views

Author

Antti Karttunen, Feb 22 2016

Keywords

Crossrefs

Cf. A269175.
Cf. A269176 (numbers not present in this sequence).
Cf. A269177 (same sequence sorted into ascending order, duplicates removed).
Cf. A269178 (numbers that occur only once).
Cf. A267357 (iterates from 1 onward).

Programs

Formula

a(n) = A163617(n) AND A269173(n).
a(n) = A163617(n) AND (A048724(n) OR A048725(n)).
a(n) = (n OR 2n) AND ((n XOR 2n) OR (n XOR 4n)).
Other identities. For all n >= 0:
a(2*n) = 2*a(n).
a(n) = A057889(A161903(A057889(n))). [Rule 124 is the mirror image of rule 110.]
G.f.: (-3*x^3 - 2*x^2 - 3*x)/(x^4 - 1) + Sum_{k>=1}((2^(k + 1)*x^(2^k) - 2^(k + 1)*x^(14*2^(k - 2)))/((x^(2^(k + 2)) - 1)*(x - 1))). - Miles Wilson, Jan 25 2025

A269161 Formula for Wolfram's Rule 86 cellular automaton: a(n) = 4n XOR (2n OR n).

Original entry on oeis.org

0, 7, 14, 11, 28, 27, 22, 19, 56, 63, 54, 51, 44, 43, 38, 35, 112, 119, 126, 123, 108, 107, 102, 99, 88, 95, 86, 83, 76, 75, 70, 67, 224, 231, 238, 235, 252, 251, 246, 243, 216, 223, 214, 211, 204, 203, 198, 195, 176, 183, 190, 187, 172, 171, 166, 163, 152, 159, 150, 147, 140, 139, 134, 131, 448, 455, 462, 459
Offset: 0

Views

Author

Antti Karttunen, Feb 20 2016

Keywords

Comments

The sequence is injective: no value occurs more than once.
Fibbinary numbers (A003714) give all integers n>=0 for which a(n) = A048727(n) and for which a(n) = A269160(n).

Crossrefs

Cf. A265281 (iterates starting from 1).
Cf. also A048727, A269160.

Programs

Formula

a(n) = 4n XOR (2n OR n) = A003987(4*n, A003986(2*n, n)).
a(n) = 4*n XOR A163617(n).
Other identities. For all n >= 0:
a(2*n) = 2*a(n).
a(n) = A057889(A269160(A057889(n))). [Rule 86 is the mirror image of rule 30.]

A366283 a(n) = gcd(n, A366275(n)), where A366275 is the Cat's tongue permutation.

Original entry on oeis.org

1, 1, 2, 3, 4, 1, 6, 1, 8, 9, 2, 1, 12, 1, 2, 1, 16, 1, 18, 1, 4, 3, 2, 1, 24, 25, 2, 1, 4, 1, 2, 1, 32, 3, 2, 5, 36, 1, 2, 3, 8, 1, 6, 1, 4, 3, 2, 1, 48, 1, 50, 1, 4, 1, 2, 55, 8, 1, 2, 1, 4, 1, 2, 1, 64, 1, 6, 1, 4, 3, 10, 1, 72, 1, 2, 15, 4, 7, 6, 1, 16, 3, 2, 1, 12, 5, 2, 3, 8, 1, 6, 7, 4, 3, 2, 1, 96, 1, 2, 1, 100
Offset: 0

Views

Author

Antti Karttunen, Oct 07 2023

Keywords

Crossrefs

Differs from related A364255 for the first time at n=25, where a(25) = 25, while A364255(25) = 5.

Programs

Formula

a(n) = gcd(n,A366282(n)) = gcd(A366275(n),A366282(n)).
a(n) = n / A366284(n) = A366275(n) / A366285(n).

A057891 In base 2, neither a palindrome nor becomes a palindrome if trailing 0's are omitted.

Original entry on oeis.org

11, 13, 19, 22, 23, 25, 26, 29, 35, 37, 38, 39, 41, 43, 44, 46, 47, 49, 50, 52, 53, 55, 57, 58, 59, 61, 67, 69, 70, 71, 74, 75, 76, 77, 78, 79, 81, 82, 83, 86, 87, 88, 89, 91, 92, 94, 95, 97, 98, 100, 101, 103, 104, 105, 106, 109, 110, 111, 113, 114, 115, 116, 117, 118
Offset: 1

Views

Author

Marc LeBrun, Sep 25 2000

Keywords

Comments

These could be called "asymmetric bit strings".
Fixed pairs of A057889, complement of A057890.
If these numbers are converted to their binary polynomial, one of the roots of that polynomial will have absolute values other than 1 or 0. For example 11 = 2^3 + 2^1 + 2^0, the absolute values of the roots of x^3 + x + 1 are 0.682328... and 1.21061... which are not 1 or 0, so 11 is in the sequence. The first number with this property which is not a term is A057890(53) = 107. - Benedict W. J. Irwin, Sep 07 2017 and Andrey Zabolotskiy, Oct 13 2017

Examples

			11 is included because 1011 is asymmetrical, but 12 is not because 001100 is a palindrome.
		

Crossrefs

Cf. A061917, A006995. Complement of A057890.

Programs

  • Haskell
    a057891 n = a057891_list !! (n-1)
    a057891_list = filter ((== 0) . a178225 . a000265) [1..]
    -- Reinhard Zumkeller, Oct 21 2011

Formula

A030101(A030101(n)) != A030101(n). - David Wilson, Jun 09 2009, Jun 18 2009
A178225(A000265(a(n))) = 0. - Reinhard Zumkeller, Oct 21 2011

Extensions

Edited by N. J. A. Sloane, Jun 09 2009 at the suggestion of Ray Chandler
A-numbers in formula corrected by R. J. Mathar, Jun 18 2009
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