A187187 -id:A187187 - cp's OEIS frontend

A187188 Parse the infinite string 0123456789012345678901234567890... into distinct phrases 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 01, 23, 45, 67, 89, 012, 34, 56, 78, 90, 12, 345, ...; a(n) = length of n-th phrase.

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 4, 4, 4, 4, 4, 5, 6, 5, 5, 6, 5, 5, 6, 5, 5, 6, 5, 5, 6, 7, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 9, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 9, 9, 9, 10, 11, 10, 11, 10, 11, 10, 11, 10, 11, 10, 11, 10, 11, 10, 11, 10, 11, 10, 11, 12, 12, 12, 12, 12, 13, 12, 12, 12, 12

Offset: 1

N. J. A. Sloane, Mar 06 2011

nonn

See A187180-A187187 for further details.

Answers a question raised by Sergio Verdu (personal communication, Mar 05 2011).

			The sequence begins
1   1   1   1   1   1   1   1   1   1
2   2   2   2   2   3   2   2   2   2   2   3
3   3   3   3   3   3   3   3
4   4   4   4   4   5   4   4   4   4   4   5
6   5   5   6   5   5   6   5   5   6   5   5
6   7
6   6   6   6   6
7   7   7   7   7   7   7   7   7
8   8   8   8   8   9   8   8   8   8   8   9
9   9   9   9   9   9   9   9
10  11  10  11  10  11  10  11  10  11 10  11  10  11  10  11  10  11  10  11
12  12  12  12  12  13  12  12  12  12  12  13
...

N. J. A. Sloane, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, -1).

See A187180-A187188 for alphabets of size 2 through 10.

See also A109337, A187199, A187200.

After the initial block of 10 1's, the sequence is quasi-periodic with period 100, increasing by 10 after each block. In more detail:
a(n) = 1 for 1 <= n <= 10.
For n >= 10, write n = 11 + 100i + j with i >= 0, 0 <= j <= 99.
Then for 0 <= j <= 79, a(n) = 10i + f(j),
where f(0) ... f(79) is the following 80-term sequence:
[2   2   2   2   2   3   2   2   2   2   2   3
3   3   3   3   3   3   3   3
4   4   4   4   4   5   4   4   4   4   4   5
6   5   5   6   5   5   6   5   5   6   5   5
6   7
6   6   6   6   6
7   7   7   7   7   7   7   7   7
8   8   8   8   8   9   8   8   8   8   8   9
9   9   9   9   9   9   9   9]
(this has been broken into blocks to make it easier to see),
and for 80 <= j <= 99, a(n) = 10i+10 if j is even, a(n) = 10i+11 if j is odd.
Examples:
n=120 = 11 + 100*1 + 9, i=1, j=9, a(120)=10+f(9) = 10+2 = 12
n=292 = 11 + 100*2 + 81, i=2, j=81. a(292)=20+11=31

A288533 Parse A004736 into distinct phrases [1], [2], [1,3], [2,1], [4], [3], [2,1,5], [4,3], [2,1,6], ...; a(n) is the length of the n-th phrase.

Original entry on oeis.org

1, 1, 2, 2, 1, 1, 3, 2, 3, 1, 3, 2, 1, 2, 2, 2, 1, 2, 4, 1, 1, 2, 3, 3, 2, 3, 5, 1, 3, 3, 3, 1, 1, 2, 2, 4, 3, 2, 3, 4, 4, 1, 3, 4, 4, 2, 1, 2, 2, 5, 5, 1, 2, 4, 3, 5, 1, 1, 2, 3, 4, 5, 2, 2, 3, 5, 5, 3, 1, 3, 3, 3, 4, 5, 1, 2, 2, 4, 5, 6, 1, 2, 4, 4, 6, 4, 1, 2, 3, 4, 4, 6, 2, 1, 2, 3, 3, 5, 5, 4, 1, 2, 3, 5, 6, 6, 1, 1, 2, 3, 4, 5, 7, 3, 2, 3, 4, 4, 7, 6, 1, 3, 3, 4, 5, 6, 5, 1, 2, 2

Offset: 1

Lewis Chen, Jun 11 2017

nonn

The phrases are formed by the Ziv-Lempel encoding described in A106182. - Neal Gersh Tolunsky, Nov 30 2023

			Consider the infinite sequence [1,2,1,3,2,1,4,3,2,1,5,4,3,2,1,...], i.e., A004736. We can first take [1] since we've never used it before. Then [2]. For the third term, we've already used [1], so we must instead take [1,3].

Neal Gersh Tolunsky, Table of n, a(n) for n = 1..10000

Cf. A109337, A106182, A187180, A187181, A187182, A187183, A187184, A187185, A187186, A187187, A187188, A187199, A187200.

# you should use program from internal format
a = set()
i = 2
s = "1"
seq = ""
while i < 100:
    j = i
    while j > 0:
        if s not in a:
            seq = seq + "," + str(len(s)-len(s.replace(",",""))+1)
            a.add(s)
            s = str(j)
        else:
            s = s + "," + str(j)
        j -= 1
    i += 1
print(seq[1:])

cp's OEIS Frontend

A187188 Parse the infinite string 0123456789012345678901234567890... into distinct phrases 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 01, 23, 45, 67, 89, 012, 34, 56, 78, 90, 12, 345, ...; a(n) = length of n-th phrase.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Formula