A309261 The sum of the number of times each unique digit in the previous number occurred in the numbers before that with a(1) = 0.
0, 0, 1, 0, 2, 0, 3, 0, 4, 0, 5, 0, 6, 0, 7, 0, 8, 0, 9, 0, 10, 12, 3, 1, 3, 2, 2, 3, 3, 4, 1, 4, 2, 4, 3, 5, 1, 5, 2, 5, 3, 6, 1, 6, 2, 6, 3, 7, 1, 7, 2, 7, 3, 8, 1, 8, 2, 8, 3, 9, 1, 9, 2, 9, 3, 10, 22, 10, 24, 16, 16, 18, 18, 20, 27, 18, 22, 15, 21, 35, 16, 25, 24, 24
Offset: 1
Examples
For n=2, the previous term a(1) is 0 by definition and the digit 0 has never been seen before so a(2) is 0. For n=3, a(2) is 0, the digit 0 has been seen once before so a(3) is 1. For n=22, a(21) is 10, digit 1 has been seen once before a(21) and digit 0 has been seen eleven times, so a(22) = 12. For n=68, a(67) is 22, digit 2 has been seen ten times before, we only consider unique digits of a(n-1), so a(68) = 10.
Links
- Rémy Sigrist, Table of n, a(n) for n = 1..20000
- Rémy Sigrist, Density plot of the first million terms
Crossrefs
Cf. A004207.
Programs
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PARI
{ f=vector(base=10); for(n=1, 84, if(n==1, v=0, d=if(v, digits(v, base), [0]); s=Set(d); v=sum(i=1, #s, f[1+s[i]]); apply(t -> f[1+t]++, d)); print1(v ", ")) } \\ Rémy Sigrist, Jul 24 2019 -
Python
from collections import Counter from itertools import count, islice def agen(): # generator of terms an, c = 0, Counter() while True: yield an s = str(an) an = sum(c[d] for d in set(s)) c.update(s) print(list(islice(agen(), 84))) # Michael S. Branicky, Mar 24 2025