A357067 Decimal expansion of the limit of A091411(k)/2^(k-1) as k goes to infinity.

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

Offset

1,1

Comments

In the article "The first occurrence of a number in Gijswijt's sequence", this constant is called epsilon_1. Its existence is proved in Theorem 7.2. The constant occurs in a direct formula (Theorem 7.11) for A091409(n), the first occurrence of the integer n in Gijswijt's sequence A090822.

Links

Table of n, a(n) for n=1..46.

Levi van de Pol, The first occurrence of a number in Gijswijt's sequence, arXiv:2209.04657 [math.CO], 2022.

Formula

Equal to 1 + Sum_{k>=1} A091579(k)/2^k. Proved in Corollary 7.3 of the article "The first occurrence of a number in Gijswijt's sequence".

Example

3.48669886438365597023...

Prog

(Python)
import math
from mpmath import *
# warning: 0.1 and mpf(1/10) are incorrect. Use mpf(1)/mpf(10)
mp.dps=60
def Cn(X):
    l=len(X)
    cn=1
    for i in range(1, int(l/2)+1):
        j=i
        while(X[l-j-1]==X[l-j-1+i]):
            j=j+1
            if j>=l:
                break
        candidate=int(j/i)
        if candidate>cn:
            cn=candidate
    return cn
def epsilon():
    A=[2] # level-2 Gijswijt sequence
    number=1 # number of S strings encountered
    position=0 # position of end of last S
    value=mpf(1) # approximation for epsilon1
    for i in range(1, 6000):
        k=Cn(A)
        A.append(max(2, k))
        if k<2:
            value=value+mpf(i-position)/mpf(2**number)
            position=mpf(i)
            number+=1
    return value
print("epsilon_1: ", epsilon())

Crossrefs

Cf. A090822, A091409, A091411, A091579, A357066, A357068.

Sequence in context: A050417 A327613 A213954 * A074212 A125715 A129283

Adjacent sequences: A357064 A357065 A357066 * A357068 A357069 A357070

Keyword

nonn,cons,more

Author

Levi van de Pol, Oct 22 2022

Status

approved