A309872 For each term, take the last digit of the previous term and count all the appearances of that digit up to and including the previous term; the first term is 1.

1, 1, 2, 1, 3, 1, 4, 1, 5, 1, 6, 1, 7, 1, 8, 1, 9, 1, 10, 1, 12, 2, 3, 2, 4, 2, 5, 2, 6, 2, 7, 2, 8, 2, 9, 2, 10, 2, 11, 16, 3, 3, 4, 3, 5, 3, 6, 4, 4, 5, 4, 6, 5, 5, 6, 6, 7, 3, 7, 4, 7, 5, 7, 6, 8, 3, 8, 4, 8, 5, 8, 6, 9, 3, 9, 4, 9, 5, 9, 6, 10, 3, 10, 4, 10, 5, 10, 6, 11, 23, 11

Offset: 1

Author

Maxim Skorohodov, Aug 21 2019

Keywords

• base
• nonn
• look

Examples

			To get the eleventh term, you need to get the last digit of the tenth term, which is 1, and then count all the 1's already in the sequence: 1, 1, 2, 1, 3, 1, 4, 1, 5, 1; there are six 1's, so the eleventh term is 6.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A248034 (first term is 0).

Programs

Maple

Q:= Array(0..9):
A:= Vector(100):
Q[1]:= 1:
A[1]:= 1:
for n from 2 to 100 do
  d:= A[n-1] mod 10;
  A[n]:= Q[d];
  L:= convert(%,base,10);
  for i in L do Q[i]:= Q[i]+1 od
od:
convert(A,list); # Robert Israel, Feb 18 2020

PARI

f = vector(base=10); for (n=1, 91, v = if (n==1, 1, f[1+(v%base)]); apply (d -> f[1+d]++, if (v, digits(v, base), [0])); print1 (v ", ")) \\ Rémy Sigrist, Aug 21 2019

Python

s, a, n = "1", [1], 1
while n < 100:
    n = n+1
    d = s[len(s)-1]
    i, aa = 0, 0
    while i < len(s):
        if s[i] == d:
            aa = aa+1
        i = i+1
    s, a = s+str(aa), a+[aa]
for n in range(1, 92): print(a[n-1], end=', ') # A.H.M. Smeets, Aug 22 2019