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A277144
Lexicographically least sequence of nonnegative integers that avoids 5/4-powers.
1
0, 0, 0, 0, 1, 1, 1, 1, 0, 2, 0, 2, 1, 0, 1, 0, 0, 1, 0, 1, 1, 2, 1, 2, 0, 0, 0, 0, 1, 3, 1, 1, 0, 1, 0, 2, 1, 0, 1, 3, 0, 2, 0, 0, 1, 1, 1, 1, 0, 0, 0, 2, 1, 2, 1, 0, 0, 1, 0, 1, 1, 0, 1, 2, 0, 2, 0, 0, 1, 3, 1, 1, 0, 0, 0, 3, 1, 1, 1, 0, 0, 2, 0, 1, 1, 0, 1, 3, 0, 1, 0, 0, 1, 2, 1, 1, 0, 0, 0, 2, 1, 1, 1, 0, 0, 3, 0, 1, 1, 0, 1, 2, 0, 1, 0, 0, 1, 3, 1, 1
OFFSET
0,10
COMMENTS
This sequence is 6-regular.
From M. F. Hasler, Oct 03 2016: (Start)
Avoiding 5/4-powers means avoiding factors of the form xyx where 3|x| = |y| >= 1.
The index of the first occurrence of 0, 1, 2, 3,.... is 0, 4, 9, 29, 725, 7351, 12607, .... (not in the OEIS, neither if the indices are shifted by 1). (End)
LINKS
Lara Pudwell and Eric Rowland, Avoiding fractional powers over the natural numbers, arXiv:1510.02807 [math.CO] (2015).
Eric Rowland and Manon Stipulanti, Avoiding 5/4-powers on the alphabet of nonnegative integers, arXiv:2005.03158 [math.CO], 2020.
FORMULA
From Eric Rowland, May 08 2020: (Start)
a(6*n + 123061) = a(n + 5920) + 3 if n = 0,2 mod 8;
a(6*n + 123061) = a(n + 5920) + 1 if n = 4,6 mod 8;
a(6*n + 123061) = a(n + 5920) + 2 if n = 1 mod 2.
(End)
EXAMPLE
The sequence cannot start 0,0,0,0,0,... because this would be a 5/4-power (with y = (0,0,0)). Thus it must start 0,0,0,0,1,.... This cannot be followed by a 0 (y = (0,0,1)), and the same applies after the next two 1's (with y = (0,1,1), or y = (1,1,1)). Only after the fourth 1, a 0 is again possible. Thereafter, neither a 1 (x = 1, y = 110) nor a 0 (x = 00, y = 001111) is possible, so it must be followed by a 2. - M. F. Hasler, Oct 03 2016
PROG
(PARI) {a=vector(2000); for(n=5, #a, until(0, for(x=1, n\5, a[n-x+1..n]==a[n-5*x+1..n-4*x]&&a[n]++&&next(2)); break)); a[1..200]} \\ M. F. Hasler, Oct 03 2016
CROSSREFS
Sequence in context: A203399 A323068 A072627 * A069848 A194702 A118682
KEYWORD
nonn
AUTHOR
Eric Rowland, Oct 01 2016
STATUS
approved