A357930 a(0) = 0; for n > 0, let S = concatenation of a(0)..a(n-1); a(n) is the number of times the digit at a(n-1) digits back from the end of S appears in S.

0, 1, 1, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 10, 7, 7, 8, 8, 7, 7, 8, 8, 7, 13, 7, 14, 7, 13, 13, 15, 15, 13, 15, 15, 6, 17, 16, 19, 8, 10, 10, 22, 10, 23

Offset

0,4

Links

Michael De Vlieger, Table of n, a(n) for n = 0..9999

Michael De Vlieger, Table of n, a(n) for n = 0..10^5

Michael De Vlieger, Scalar scatterplot of a(n), n = 0..2^16, with points assigned a color function as to digit d that is a(n-1) digits back. Black indicates d=0, red d=1, ..., magenta d=9.

Michael De Vlieger, Log-log scatterplot of a(n), n = 0..2^20, with points assigned a color function as to digit d that is a(n-1) digits back. Black indicates d=0, red d=1, ..., magenta d=9.

Scott R. Shannon, Image of the first 5000000 terms.

Example

a(7) = 3 as a(6) = 3 and the string concatenation of a(0)..a(6) = "0112223", and the digit 3 digits back from the end of the string concatenation is 2, and 2 has appeared three times in the string.

Prog

(MATLAB)
function a = A357930( max_n )
    a = 0; s = '0'; c = zeros(1, 10); c(1) = 1;
    for n = 2:max_n
        k = c(s(end-a(n-1))-47); sk = num2str(k);
        c(sk-47) = c(sk-47)+1; s = [s sk]; a(n) = k;
    end
end % Thomas Scheuerle, Oct 21 2022

Crossrefs

Cf. A117707, A356348, A000217, A351753.

Sequence in context: A225687 A083291 A169894 * A003056 A117707 A163352

Adjacent sequences: A357927 A357928 A357929 * A357931 A357932 A357933

Keyword

nonn,base,look

Author

Scott R. Shannon, Oct 21 2022

Status

approved