OFFSET
1,3
COMMENTS
In other words, a(n) is the length of the shortest subsequence ending at a(n-1) which has a unique rhyme scheme among all rhyme schemes of subsequences of the sequence thus far. Alternatively, this is (the length of the longest subsequence ending at a(n-1) whose rhyme scheme has occurred before as that of another subsequence) plus 1.
The rhyme scheme of a subsequence is found by assigning 1 to the first unique value, 2 to the second unique value, 3 to the third, and so on.
LINKS
Neal Gersh Tolunsky, Table of n, a(n) for n = 1..10000
Neal Gersh Tolunsky, Ordinal Transform of 500000 terms
EXAMPLE
a(6)=4 because the length-4 subsequence a(2..5) = 1,2,2,3 has the shortest unique rhyme scheme (1,2,2,3 or A,B,B,C,) which does not occur elsewhere as the rhyme scheme of any other subsequence in the sequence thus far. No shorter subsequence ending in a(5) with a unique rhyme scheme exists in the sequence thus far. For example, a(6) cannot be 3 because the length-3 subsequence a(3..5) = 2,2,3 has the same rhyme scheme (1,1,2 or A,A,B) as that of the subsequence a(1..3) = 1,1,2.
PROG
(Python)
from itertools import count, islice
def rs(t): # rhyme scheme of t
s, k, m = [], 1, dict()
for e in t:
if e not in m: m[e] = k; k += 1
s.append(m[e])
return tuple(s)
def agen(): # generator of terms
a, R, maxL = [1], set(), 0 # maxL = max length of rhyme schemes stored
for n in count(1):
yield a[-1]
for k in range(1, n+1):
if k > maxL: # must increase length of rhyme schemes stored
maxL += 1
R.update(rs(a[i:i+maxL]) for i in range(n-maxL))
if rs(a[-k:]) not in R:
break
an = k
R.update(rs(a[-i:]) for i in range(1, maxL+1))
a.append(an)
print(list(islice(agen(), 90))) # Michael S. Branicky, Oct 13 2024
CROSSREFS
KEYWORD
nonn
AUTHOR
Neal Gersh Tolunsky, Oct 11 2024
STATUS
approved