cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

A262313 Decimal expansion of the limit of the probability that a random binary word is an instance of the Zimin pattern "abacaba" as word length approaches infinity.

Original entry on oeis.org

1, 1, 9, 4, 4, 3, 6, 9, 5, 2, 5, 2, 8, 6, 3, 3, 7, 3, 0, 0, 0, 1, 1, 8, 5, 8, 6, 1, 2, 6, 8, 8, 5, 1, 0, 4, 8, 1, 5, 9, 0, 7, 9, 8, 8, 8, 1, 6, 8, 0, 8, 3, 3, 0, 8, 6, 3, 0, 6, 5, 2, 2, 2, 0, 2, 8, 9, 1, 4, 4, 5, 5, 9, 4, 2, 1, 0, 7, 7, 6, 1, 0, 7, 2
Offset: 0

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Author

Danny Rorabaugh, Sep 17 2015

Keywords

Comments

Word W over alphabet L is an instance of "abacaba" provided there exists a nonerasing monoid homomorphism f:{a,b,c}*->L* such that f(W)=abacaba. For example "01011010001011010" is an instance of "abacaba" via the homomorphism defined by f(a)=010, f(b)=11, f(c)=0. For a proof of the formula or more information on Zimin words, see Rorabaugh (2015).

Examples

			The constant is 0.11944369525286337300011858612688510481590798881680833086306522202891445594210776107239...
		

Crossrefs

Cf. A003000, A123121, A262312 (aba).

Formula

The constant is Sum_{n=1..infinity} A003000(n)*(Sum_{i=0..infinity} G_n(i)+H_n(i)), with:
G_n(i) = (-1)^i * r_n((1/2)^(2*2^i)) * (Product_{j=0..i-1} s_n((1/2)^(2*2^j))) / (Product_{k=0..i} 1-2*(1/2)^(2*2^k)),
r_n(x) = 2*x^(2n+1) - x^(4n) + x^(5n) - 2*x^(5n+1) + x^(6n),
s_n(x) = 1 - 2*x^(1-n) + x^(-n);
H_n(i) = (-1)^i * u_n((1/2)^(2*2^i)) * (Product_{j=0..i-1} v_n((1/2)^(2*2^j))) / (Product_{k=0..i} 1-2*(1/2)^(2*2^k)),
u_n(x) = 2*x^(4n+1) - x^(5n) + 2*x^(5n+1) + x^(6n),
v_n(x) = 1 - 2*x^(1-n) + x^(-n) - 2*x^(1-2n) + x^(-2n).
The inside sum is an alternating series and the outside sum has positive terms and a simple tail bound. Consequentially, we have the following bounds with any positive integers N and K:
Lower bound, Sum_{n=1..N} A003000(n)*(Sum_{i=0..2K-1} G_n(i)+H_n(i));
Upper bound, (1/2)^N + Sum_{n=1..N} A003000(n)*(Sum_{i=0..2K} G_n(i)+H_n(i)).

A262500 Number of binary, minimal instances of Zimin word Z_n that begin with 0.

Original entry on oeis.org

1, 3, 1751
Offset: 1

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Author

Danny Rorabaugh, Sep 24 2015

Keywords

Comments

Zimin words are defined recursively by Z_1 = x_1, Z_{n+1} = Z_nx_{n+1}Z_n. Using a different alphabet: Z_1 = a, Z_2 = aba, Z_3 = abacaba, ... .
Word W over alphabet L is an instance of Z_n provided there exists a nonerasing monoid homomorphism f:{x_1,...,x_n}*->L* such that f(W)=Z_n. For example "abracadabra" is an instance of Z_2 via the homomorphism defined by f(x_1)=abra, f(x_2)=cad.
An instance W is minimal if no proper substring of W is also an instance.
The total number of minimal Z_n-instances over the alphabet {0,1} is 2*a(n).
The minimal, binary Z_3-instances have lengths ranging from 7 to 25. There exist minimal, binary Z_4-instances over 10000 letters long.

Examples

			The a(1)=1 instance of Z_1 is '0'.
The a(2)=3 instances of Z_2 are '000', '010', and '0110'. '01110' is not a minimal instance because it contains Z_2-instance '111' as a proper subword.
		

Crossrefs

A342378 List of numbers whose binary expansion is an instance of the Zimin pattern ABA.

Original entry on oeis.org

5, 7, 9, 11, 13, 15, 17, 18, 19, 21, 22, 23, 25, 27, 29, 31, 33, 34, 35, 37, 38, 39, 41, 42, 43, 45, 46, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 66, 67, 68, 69, 70, 71, 73, 74, 75, 76, 77, 78, 79, 81, 82, 83, 85, 86, 87, 89, 90, 91, 93, 94, 95, 97, 99, 101
Offset: 1

Views

Author

Peter Kagey, Mar 09 2021

Keywords

Comments

This is A091066 with some terms of A020330 removed.

Examples

			The binary expansion of 182 is 10110110, which is an instance of the pattern ABA with A=10 and B=1101.
		

Crossrefs

Programs

  • Python
    def ok(n):
      b = bin(n)[2:]
      for i in range(1, (len(b)+1)//2):
        if b[:i] == b[-i:]: return True
      return False
    def aupto(lim): return [m for m in range(lim+1) if ok(m)]
    print(aupto(101)) # Michael S. Branicky, Mar 09 2021
Showing 1-3 of 3 results.