A206923 Number of bisections of the n-th binary palindrome bit pattern until the result is not palindromic

1, 1, 2, 1, 3, 1, 3, 1, 2, 1, 4, 1, 2, 1, 4, 1, 2, 1, 1, 1, 1, 1, 4, 1, 2, 1, 1, 1, 1, 1, 4, 1, 2, 1, 1, 1, 3, 1, 1, 1, 1, 1, 2, 1, 1, 1, 5, 1, 2, 1, 1, 1, 3, 1, 1, 1, 1, 1, 2, 1, 1, 1, 5, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 5, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 2, 1

Offset

1,3

Comments

Let k=1, p(1)=A006995(n) and m(1)=number of bits in p(1); if p(k) is a binary palindrome > 1 then iterate k=k+1, m(k)=floor((m(k-1)+1)/2), p(k)=leftmost m(k) bits of p(k-1); else set a(n)=k endif.

Links

Table of n, a(n) for n=1..116.

Formula

Recursion: define f(x)=floor(A006995(x)/2^floor(floor(log_2(A006995(x))+1)/2)), for x=1,2,3,...

Case 1: a(n)=1+a(A206915(f(n))), if f(n) is a binary palindrome;

Case 2: a(n)=1, else.

Formally: a(n)=if (A178225(f(n))==1) then a(A206915(f(n)))+1 else 1.

Example

a(1)=a(2)=1, since A006995(1)=0 and A006995(2)=1;
a(5)=3, since A006995(5)=7=111_2 and so the iteration is 11==>11==>1;
a(9)=2, since A006995(9)=21=10101_2 and so the iteration is 10101==>101;
a(13)=2, since A006995(13)=45=101101_2 and so the iteration is 101101==>101;
a(15)=4, since A006995(15)=63=111111_2 and so the iteration is 111111==>111==>11==>1;
a(37)=3, since A006995(37)=341=101010101_2 and so the iteration is 101010101==>10101==>101;

Prog

(C)
/* quasi-C program fragment, omitting formal details, n>1 */
p=n;
p1=n+1;
k=0;
while (A178225(p)==1) && (p != p1)
{
  p1=p;
  k++;
  m=int(log(p)/log(2));
  p=int(p/2^int((m+1)/2));
}
return k;

Crossrefs

A006995, A206915, A178225, A154809, A206924, A206925.

Sequence in context: A278572 A136644 A111963 * A328570 A078079 A328049

Adjacent sequences: A206920 A206921 A206922 * A206924 A206925 A206926

Keyword

nonn,base

Author

Hieronymus Fischer, Mar 12 2012

Status

approved