A108571 Any digit d in the sequence says: "I am part of an integer in which you'll find d digits d".

1, 22, 122, 212, 221, 333, 1333, 3133, 3313, 3331, 4444, 14444, 22333, 23233, 23323, 23332, 32233, 32323, 32332, 33223, 33232, 33322, 41444, 44144, 44414, 44441, 55555, 122333, 123233, 123323, 123332, 132233, 132323, 132332, 133223, 133232, 133322, 155555

Offset: 1

Author

Eric Angelini, Jul 05 2005

Keywords

• base
• easy
• fini
• nonn

Comments

The sequence is finite. Last term: 999999999888888887777777666666555554444333221.

Number of terms is 66712890763701234740813164553708284. - Zak Seidov, Jan 02 2007

Fixed points of A139337. - Reinhard Zumkeller, Apr 14 2008

Sequence contains squares (A181392) and cubes (A225886^3) but no higher powers, see Comments in A181392. - Giovanni Resta, May 19 2013

Examples

			23323 is in the sequence because it has two 2's and three 3's.
23332 is in the sequence because it has two 2's and three 3's.
23333 is not in the sequence because it has only one 2 and four 3's.
		

Links

• T. D. Noe, Table of n, a(n) for n=1..21056 (terms < 10^10)
• Michael S. Branicky, Python program

Crossrefs

Cf. A127007, A139337, A078348 (subsequence of primes), A181392, A225886.

Programs

PARI

is(n)={ vecmin(n=vecsort(digits(n))) && #n==normlp(Set(n),1) && !for(i=1,#n, n[i+n[i]-1]==n[i] || return; i+n[i]>#n || n[i+n[i]]>n[i] || return; n[i]>1 && i+=n[i]-1)} \\ M. F. Hasler, Sep 22 2014

Python

# see link for a function that directly generates terms
def ok(n): s = str(n); return all(s.count(d) == int(d) for d in set(s))
def aupto(limit): return [m for m in range(1, limit+1) if ok(m)]
print(aupto(155555)) # Michael S. Branicky, Jan 22 2021