A338264 Triangle read by rows: T(n,k) is the frequency of occurrence of binary words, under iterations of making the n-bit binary representation of k palindromic, by binary-subtracting its bit reversal, 0 <= k < 2^n.

0, 0, 0, 1, 0, 2, 0, 0, 0, 2, 0, 4, 0, 0, 0, 0, 2, 0, 0, 1, 0, 9, 0, 12, 0, 2, 0, 0, 6, 0, 0, 0, 0, 0, 0, 0, 4, 0, 0, 2, 0, 2, 0, 0, 0, 16, 0, 22, 0, 0, 0, 2, 0, 4, 0, 0, 12, 0, 0, 0, 0, 0, 0, 0

Offset

1,6

Comments

This sequence is a side result of the same process used for generating A338203.

All operations are performed with n bits. In particular, the bit-reversal of k considers k to be an n-bit binary word and k is a palindrome if it is equal to its bit reversal in this sense. The subtraction is performed modulo 2^n.

The sequence is divided into blocks of length 2^n, n = 1, 2, 3, ... .

Sequence looks random if only a small amount of data is seen, but shows self-similarity and scale invariance if seen from a broader perspective.

An example is provided in "links" section.

T(n,0) is set to zero by definition.

If for any number the operation of subtracting its binary reversed bitsequence is carried out, the result is either 0 or one of the numbers k in the set T(n,k) > 0. Numbers k with bigger values T(n,k) are more likely the result of such operation.

Links

Thomas Scheuerle, Table of n, a(n) for n = 1..16382. Max: T(13,2^13-1)

Thomas Scheuerle, See here selfsimilarity in graph of first 131070 elements

Thomas Scheuerle, Fast Walsh-Hadamard transform (MATLAB)plot(fwht(a(1:2^16),2^16,'hadamard'))

Example

Triangle begins:
  0, 0;   Data starts with n=1
  0, 1, 0, 2;
  0, 0, 0, 2, 0, 4, 0, 0;
  0, 0, 2, 0, 0, 1, 0, 9, 0, 12, 0, 2, 0, 0, 6, 0;
  ...
n=4 Four bit wordlength.
For each row of n we let m iterate from 0 to 2^n-1.
For this example let's take a snapshot at m=5 in binary : 0101
0101 - 1010 = 1011 (2^4 + 5 - 10)=11 -> T((2^n-1)+11):=T((2^n-1)+11)+1
(we increment counts for frequency)
1011 - 1101 = 1110 (2^4 + 11 - 13)=14 -> T((2^n-1)+14):=T((2^n-1)+14)+1
1110 - 0111 = 0111 (14 - 7)=7         -> T((2^n-1)+7):=T((2^n-1)+7)+1
0111 - 1110 = 1001 (2^4 + 7 -14)=9    -> T((2^n-1)+9):=T((2^n-1)+9)+1
Now 1001 is palindromic, we stop m=5 and proceed with m=6.
If we did not stop we would reach zero. T(n,0) is zero by definition.

Prog

(MATLAB)
sequence = [];
for numberofBits = 1:maxNumberofBits
    numbersPerWordlength = 2^numberofBits;
    frequency_list = zeros(1, numbersPerWordlength);
    for n = 1:numbersPerWordlength
        word = n-1;
        while word > 0
            word_reversed = 0;
            % reverse bit order
            for i = 1:numberofBits
                word_reversed = bitset(word_reversed, numberofBits-i+1, bitget(word, i));
            end
            % binary subtraction with worlength of numbersPerWordlength
            if word >= word_reversed
              word = word-word_reversed;
            else
              word = numbersPerWordlength + word - word_reversed;
            end
            if word > 0 % if == 0 it was already a palindrome
              frequency_list(word+1) = frequency_list(word+1)+1;
            end
        end
    end
    sequence = [sequence frequency_list];
end

Crossrefs

Cf. A338203.

Sequence in context: A368118 A328590 A219204 * A353509 A326067 A318879

Adjacent sequences: A338261 A338262 A338263 * A338265 A338266 A338267

Keyword

nonn,base,tabf

Author

Thomas Scheuerle, Oct 19 2020

Status

approved