A091975 a(1) = 1; for n>1, a(n) = largest integer k such that the word a(1)a(2)...a(n-1) is of is of the form xy^(k_1)Y^(k_2)y^(k_3)Y^(k_4)...y^(k_(m-1))Y^(k_m) where y has positive length and Y=reverse(y) and k_1+k_2+k_3+...+k_m = k.
1, 1, 2, 1, 1, 2, 2, 2, 3, 1, 1, 2, 1, 1, 2, 2, 2, 3, 2, 1, 1, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 3, 3, 2, 2, 2, 4, 1, 1, 2, 1, 1, 2, 2, 2, 3, 1, 1, 2, 1, 1, 2, 2, 2, 3, 2, 1, 1, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 3, 3, 2, 2, 2, 4, 2, 1, 1, 2, 2, 2, 3, 1, 1, 2, 1, 1, 2, 2, 2, 3, 1, 1, 2, 2, 2, 3, 2, 1, 1, 2, 2, 2, 3
Offset: 1
Keywords
Links
- Michael S. Branicky, Table of n, a(n) for n = 1..10000
- F. J. van de Bult, D. C. Gijswijt, J. P. Linderman, N. J. A. Sloane and Allan Wilks, A Slow-Growing Sequence Defined by an Unusual Recurrence, J. Integer Sequences, Vol. 10 (2007), #07.1.2.
- F. J. van de Bult, D. C. Gijswijt, J. P. Linderman, N. J. A. Sloane and Allan Wilks, A Slow-Growing Sequence Defined by an Unusual Recurrence [pdf, ps].
- Index entries for sequences related to Gijswijt's sequence
Programs
-
Python
def k(s): maxk = 1 for m in range(1, len(s)+1): i, y, kk = 1, s[-m:], len(s)//m if kk <= maxk: return maxk yY = [y, y[::-1]] while s[-(i+1)*m:-i*m] in yY: i += 1 maxk = max(maxk, i) def aupton(terms): alst = [1] for n in range(2, terms+1): alst.append(k(alst)) return alst print(aupton(105)) # Michael S. Branicky, Nov 05 2023
Comments