A270004 Run-length encoding of iterated Scrabble function.
12, 1, 4, 2, 1, 1, 1, 2, 1, 3, 2, 1, 1, 2, 1, 3, 4, 1, 1, 1, 3, 1, 2, 3, 3, 3, 1, 1, 1, 5, 6, 3, 1, 2, 4, 3, 1, 2, 1, 1, 1, 1, 3, 2, 2, 1, 1, 5, 2, 1, 1, 3, 2, 7, 1, 4, 1, 4, 1, 4, 1, 1, 1, 1, 1, 3, 1, 2, 1, 4, 2, 1, 1, 4, 1, 1, 8, 2, 2, 3, 3, 2, 2, 3, 1, 3, 3, 2, 1, 1, 2, 2, 2, 3, 1, 2, 3, 7, 1, 1, 5, 3, 1, 1, 6
Offset: 0
Links
- Michael Turniansky, Table of n, a(n) for n = 0..455
- Wikipedia, Scrabble
Programs
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Python
from num2words import num2words tp = {"aeilnorstu": 1, "dg": 2, "bcmp":3, "fhvwy":4, "k":5, "jx":8, "qz":10} def pts(c): return ([tp[s] for s in tp if c in s]+[0])[0] def A113172(n): return sum(map(pts, num2words(n).replace(" and", ""))) def A290205(n): while n not in {12, 4, 7, 8, 9}: n = A113172(n) return 12 if n == 12 else 4 def aupton(terms): alst, prev, k, rl = [], A290205(0), 1, 1 while len(alst) < terms: while A290205(k) == prev: k += 1; rl += 1 alst.append(rl); rl = 0; prev = 12 if prev == 4 else 4 return alst print(aupton(105)) # Michael S. Branicky, Dec 01 2021
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