A382606 Natural numbers ordered by the probability (highest to lowest) to occur in the sum of repeated rolls of a fair 6-sided die.

6, 5, 11, 12, 10, 16, 17, 15, 21, 22, 27, 23, 26, 28, 32, 33, 38, 37, 39, 43, 44, 49, 48, 50, 54, 55, 60, 59, 53, 65, 61, 66, 64, 70, 71, 76, 75, 77, 81, 82, 87, 72, 86, 88, 92, 93, 98, 97, 103, 99, 104, 102, 108, 109, 114, 115, 113, 110, 119, 120, 125, 124, 126

Offset: 1

Author

Sergio Pimentel, Mar 31 2025

Keywords

• nonn

Comments

The asymptotic probability for large n is 2/7 since the average roll of a die is 7/2.

Only terms with probability > 2/7 occur. - Michael S. Branicky, Apr 01 2025

Of any six consecutive integers, at least one is present and gives a maximum in the sequence (i.e., all terms preceding it are smaller). - Javier Múgica, May 01 2025

Examples

			The probability of achieving a '6' in n>=6 rolls is 1/6 + 5/36 + 10/216 + 10/1296 + 5/7776 + 1/46656 which is about 36.02%.
The probability of achieving a '1' is just 1/6 (about 16.67%). 6 is the highest of all, so a(1) = 6.
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 1..10000

Crossrefs

Complement of A382607. Cf. A365443.

Programs

Python

from fractions import Fraction
from math import factorial, prod
from itertools import count, islice
from sympy.utilities.iterables import partitions
def prob(n): return sum(factorial(N:=sum(p.values()))//prod(factorial(v) for v in p.values())*Fraction(1, 6**N) for p in partitions(n, k=6))
def agen(): # generator of terms
    n, vdict = 1, dict()
    for k in count(1):
        vdict[prob(k)] = k
        if k%6 == 0:
            s = [vdict[v] for v in sorted(vdict, reverse=True) if v > Fraction(2, 7)]
            yield from (s[i-1] for i in range(n, len(s)-1))
            n = len(s) - 1
print(list(islice(agen(), 20))) # Michael S. Branicky, Apr 01 2025