A338551 Number of ways to make a checkout score of n in darts.

0, 1, 1, 4, 7, 14, 20, 31, 39, 55, 65, 86, 96, 126, 133, 171, 179, 223, 228, 286, 283, 352, 348, 422, 408, 497, 467, 569, 534, 642, 594, 720, 654, 791, 719, 863, 775, 942, 831, 1012, 894, 1082, 945, 1159, 991, 1216, 1037, 1263, 1062, 1311, 1081, 1340, 1110, 1366

Offset

1,4

Comments

In other words, the number of ways to achieve a score of n using at most 3 darts and finishing on a double. The maximum checkout score is 170, so this is a finite sequence.

Links

Carmen Bruni, Table of n, a(n) for n = 1..170

Wikipedia, Darts

Prog

(Python)
def darts(n):
  if n > 170 or n <= 1:
    return 0
  ans = 0
  singles = list(range(1, 21)) + [25]
  doubles = list(map(lambda x: 2*x, singles))
  triples = list(map(lambda x: 3*x, singles[:-1]))
  throws = singles+doubles+triples
  for i in range(len(throws)):
    for j in range(len(throws)):
      for k in range(len(doubles)):
        dart1 = throws[i]
        dart2 = throws[j]
        dart3 = doubles[k]
        if dart1 + dart2 + dart3 == n:
          ans += 1
    for j in range(len(doubles)):
      dart1 = throws[i]
      dart2 = doubles[j]
      if dart1 + dart2 == n:
        ans += 1
  return ans + (n in doubles)
for i in range(1, 171):
  print(darts(i))
(PARI) seq()={my(s=x*(1-x^20)/(1-x)+x^25, d=subst(s, x, x^2), g=s+d+subst(s-x^25, x, x^3)); Vecrev((1+g+g^2)*d/x)} \\ Andrew Howroyd, Nov 04 2020

Crossrefs

Cf. A008575, A242718, A242678, A242681, A167213, A076119, A244512, A117883.

Sequence in context: A161426 A115759 A157615 * A188319 A171378 A147478

Adjacent sequences: A338548 A338549 A338550 * A338552 A338553 A338554

Keyword

nonn,fini,full

Author

Carmen Bruni, Nov 02 2020

Status

approved