A380593 Starting position of the first occurrence of the longest monochromatic arithmetic progression of difference n in the Rudin-Shapiro sequence (A020987).
7, 14, 28, 28, 31, 43, 95, 56, 43, 62, 453, 86, 99, 190, 39, 112, 495, 86, 366, 124, 81, 321, 203, 172, 1006, 81, 233, 380, 2019, 78, 993, 224, 980, 990, 888, 172, 1084, 732, 4057, 248, 2007, 162, 164, 642, 1215, 406, 1729, 344, 1398, 2012, 1988, 162, 1765
Offset: 1
Keywords
Examples
For n = 3, the longest length of the monochromatic arithmetic progression in the Rudin-Shapiro sequence is given by A364995(3)=5. Let r(i) be the i-th term of the Rudin-Shapiro sequence (A020985). We have r(28) = r(31) = r(34) = r(37) = r(40), and no j<i exists such that r(j) = r(j+3) = r(j+2*3) = r(j+3*3) = r(j+4*3). So a(3)=28.
References
- B. Sobolewski, On monochromatic arithmetic progressions in binary words associated with pattern sequences, Theoretical Computer Science 1018 (2024), 114815.
Links
- Gandhar Joshi, Table of n, a(n) for n = 1..1022
- Bartosz Sobolewski, On monochromatic arithmetic progressions in binary words associated with pattern sequences, arXiv:2204.05287 [math.CO], 2023.
Programs
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Walnut
# Replace in the code below: every n with the desired constant difference, and every q with the longest MAP length for difference n given by A364995(n). def rs_n_map "Ak (k
RS[i]=RS[i+n*k] & Aj (j ~(Ak (k
RS[j]=RS[j+n*k])"; # While using Walnut to find values above and including a(1022), one has to run the code with constant value estimates and through trial and error hone in onto the smallest value that returns TRUE.
Comments