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

1,3

Links

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

Example

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.

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

Prog

(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

Crossrefs

Cf. A248034 (first term is 0).

Sequence in context: A318225 A360257 A335497 * A359031 A248034 A358967

Adjacent sequences: A309869 A309870 A309871 * A309873 A309874 A309875

Keyword

base,nonn,look

Author

Maxim Skorohodov, Aug 21 2019

Status

approved