A236322 Number of (potentially overlapping) occurrences of n in the decimal representation of n^n.

1, 0, 0, 0, 1, 3, 0, 0, 1, 1, 2, 0, 0, 0, 0, 2, 1, 0, 2, 0, 2, 0, 0, 1, 2, 0, 0, 1, 0, 0, 2, 2, 2, 0, 3, 1, 1, 0, 1, 0, 1, 1, 1, 0, 3, 1, 0, 1, 1, 1, 1, 2, 2, 1, 0, 1, 1, 0, 1, 3, 2, 0, 1, 1, 0, 2, 0, 0, 0, 0, 1, 0, 1, 1, 2, 5, 2, 1, 2, 0, 3, 3, 2, 1, 0, 1, 0, 0, 0, 0, 5, 1, 3, 4, 2, 2, 1, 1, 10

Offset: 1

Author

Christian Perfect, Jan 22 2014

Keywords

• base
• nonn

Comments

Differs from A236314 at n=99.

Links

• Scott R. Shannon, Table of n, a(n) for n = 1..10000 (terms 1..999 from Christian Perfect).

Crossrefs

A049329 lists n where a(n) is nonzero.

Non-overlapping occurrences are counted by A236314.

Programs

Mathematica

a[n_] := Length[StringPosition @@ ToString /@ {n^n, n}]; Array[a, 99] (* Giovanni Resta, Jan 22 2014 *)

PARI

a(n) = my(m=Mod(n,10^#Str(n)));(m==n=n^n)+sum(i=0,1+log(n)/log(10),m==n\=10) \\ - M. F. Hasler, Jan 23 2014

Python

from itertools import count
def occurrences(string, sub):
    count = start = 0
    while True:
        start = string.find(sub, start) + 1
        if start > 0:
            count+=1
        else:
            return count
def a(n):
    return occurrences(str(n**n), str(n))