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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A164862 a(n) is the smallest positive integer that, when written in binary, contains the binary representations of both A164861(n) and the reversal of the order of the digits of A164861(n) as (overlapping) substrings.

Original entry on oeis.org

27, 27, 51, 93, 51, 93, 99, 165, 231, 165, 107, 189, 99, 107, 119, 231, 119, 189, 195, 325, 455, 843, 717, 633, 325, 1619, 471, 717, 219, 381, 195, 1619, 231, 843, 219, 495, 455, 231, 471, 633, 495, 381, 387, 645, 903, 1161, 1675, 1421, 1935, 1161, 403, 1193
Offset: 1

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Author

Leroy Quet, Aug 28 2009

Keywords

Comments

Every integer that occurs in this sequence occurs at least twice.

Examples

			The second odd non-binary-palindrome is 13, which is 1101 in binary. The smallest positive integer that, when written in binary, contains both 1101 and its reverse (1011) is 27, which is 11011 in binary.

Crossrefs

Cf. A164861.

Extensions

Extended by Ray Chandler, Mar 14 2010

A154809 Numbers whose binary expansion is not palindromic.

Original entry on oeis.org

2, 4, 6, 8, 10, 11, 12, 13, 14, 16, 18, 19, 20, 22, 23, 24, 25, 26, 28, 29, 30, 32, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 46, 47, 48, 49, 50, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 64, 66, 67, 68, 69, 70, 71, 72, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 86, 87, 88
Offset: 1

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Author

Omar E. Pol, Jan 24 2009

Keywords

Comments

Complement of A006995.
The (a(n)-n+1)-th binary palindrome equals the greatest binary palindrome <= a(n). The corresponding formula identity is: A006995(a(n)-n+1)=A206913(a(n)). - Hieronymus Fischer, Mar 18 2012
A145799(a(n)) < a(n). - Reinhard Zumkeller, Sep 24 2015

Examples

			a(1)=2, since 2 = 10_2 is not binary palindromic.

Links

Crossrefs

Cf. A006995, A154810, A164861, A206913, A206915.
Cf. A145799.

Programs

  • Haskell
    a154809 n = a154809_list !! (n-1)
a154809_list = filter ((== 0) . a178225) [0..]
  • Magma
    [n: n in [0..600] | not (Intseq(n, 2) eq Reverse(Intseq(n, 2)))]; // Vincenzo Librandi, Jul 05 2015
  • Maple
    ispali:= proc(n) local L;
L:= convert(n,base,2);
ListTools:-Reverse(L)=L
end proc:
remove(ispali, [$1..1000]); # Robert Israel, Jul 05 2015
  • Mathematica
    palQ[n_Integer, base_Integer]:=Module[{idn=IntegerDigits[n, base]}, idn==Reverse[idn]]; Select[Range[1000], ! palQ[#, 2] &] (* Vincenzo Librandi, Jul 05 2015 *)
  • PARI
    isok(n) = binary(n) != Vecrev(binary(n)); \\ Michel Marcus, Jul 05 2015
  • Python
    def A154809(n):
    def f(x): return n+(x>>(l:=x.bit_length())-(k:=l+1>>1))-(int(bin(x)[k+1:1:-1],2)>(x&(1<Chai Wah Wu, Jul 24 2024

Formula

A030101(n) != n. - David W. Wilson, Jun 09 2009
A178225(a(n)) = 0. - Reinhard Zumkeller, Oct 21 2011
From Hieronymus Fischer, Feb 19 2012 and Mar 18 2012: (Start)
Inversion formula: If d is any number from this sequence, then the index number n for which a(n)=d can be calculated by n=d+1-A206915(A206913(d)).
Explicitly: Let p=A206913(d), m=floor(log_2(p)) and p>2, then: a(n)=d+1+(((5-(-1)^m)/2) + sum(k=1...floor(m/2)|(floor(p/2^k) mod 2)/2^k))*2^floor(m/2).
Example 1: d=1000, A206913(d)=975, A206915(975)=62, hence n=1001-62=939.
Example 2: d=10^6, A206913(d)=999471, A206915(999471)=2000, hence n=1000001-2000=998001.
Recursion formulas:
a(n+1)=a(n)+1+A178225(a(n)+1)
Also:
Case 1: a(n+1)=a(n)+2, if A206914(a(n))=a(n)+1;
Case 2: a(n+1)=a(n)+1, else.
Also:
Case 1: a(n+1)=a(n)+1, if A206914(a(n))>a(n)+1;
Case 2: a(n+1)=a(n)+2, else.
Iterative calculation formula:
Let f(0):=n+1, f(j):=n-1+A206915(A206913(f(j-1)) for j>0; then a(n)=f(j), if f(j)=f(j-1). The number of necessary steps is typically <4 and is limited by O(log_2(n)).
Example 3: n=1000, f(0)=1001, f(1)=1061, f(2)=1064=f(3), hence a(1000)=1064.
Example 4: n=10^6, f(0)=10^6+1, f(1)=1001999, f(2)=1002001=f(3), hence a(10^6)=1002001.
Formula identity:
a(n) = n-1 + A206915(A206913(a(n))).
(End)

Extensions

Extended by Ray Chandler, Mar 14 2010

A280506 Nonpalindromic part of n in base 2 (with carryless GF(2)[X] factorization): a(n) = A280500(n,A280505(n)).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 11, 1, 13, 1, 1, 1, 1, 1, 19, 1, 1, 11, 13, 1, 25, 13, 1, 1, 11, 1, 1, 1, 1, 1, 13, 1, 37, 19, 11, 1, 41, 1, 25, 11, 1, 13, 47, 1, 11, 25, 1, 13, 19, 1, 55, 1, 13, 11, 59, 1, 61, 1, 1, 1, 1, 1, 67, 1, 69, 13, 61, 1, 1, 37, 13, 19, 59, 11, 25, 1, 81, 41, 11, 1, 1, 25, 87, 11, 55, 1, 91, 13, 1, 47, 19, 1, 97, 11, 1, 25, 13, 1, 103
Offset: 1

Views

Author

Antti Karttunen, Jan 09 2017

Keywords

Comments

a(n) = number obtained when the maximal base-2 palindromic divisor of n, A280505(n), is divided out of n with carryless GF(2)[X] factorization (see examples of A280500 for the explanation).
Apart from 1, all terms are present in A164861 (form their proper subset).

Links

Crossrefs

Cf. A000265, A048720, A057889, A057891, A164861, A280500, A280502, A280504, A280508, A280509.
Cf. A057890 (positions of ones).

Programs

Formula

a(n) = A280500(n,A280505(n)).
Other identities. For all n >= 1:
a(2n) = a(A000265(n)) = a(n).
A048720(a(n), A280505(n)) = n.

Extensions

Erroneous claim removed from comments by Antti Karttunen, May 13 2018
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