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

0,1

Comments

Number of identical consecutive integers in A290205.

Links

Michael Turniansky, Table of n, a(n) for n = 0..455

Wikipedia, Scrabble

Prog

(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

Crossrefs

Cf. A113172, A290205.

Sequence in context: A229193 A010212 A341702 * A040151 A010213 A140378

Adjacent sequences: A270001 A270002 A270003 * A270005 A270006 A270007

Keyword

nonn,word

Author

Michael Turniansky, Jul 24 2017

Status

approved