A382607 Natural numbers ordered by the probability (lowest to highest) to occur in the sum of repeated rolls of a fair 6-sided die.

1, 2, 3, 7, 4, 8, 13, 9, 14, 19, 18, 24, 25, 20, 30, 29, 31, 35, 36, 41, 40, 34, 46, 42, 47, 45, 51, 52, 57, 56, 58, 62, 63, 68, 67, 69, 73, 74, 79, 78, 84, 80, 85, 83, 89, 90, 95, 94, 96, 100, 101, 91, 106, 105, 107, 111, 112, 117, 116, 122, 118, 123, 128, 127

Offset

1,2

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

Links

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

Example

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%). 1 is the lowest of all, so a(1)=1.

Prog

(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) 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

Crossrefs

Complement of A382606. Cf. A365443.

Sequence in context: A118810 A357482 A026259 * A168087 A378704 A168085

Adjacent sequences: A382604 A382605 A382606 * A382608 A382609 A382610

Keyword

nonn

Author

Sergio Pimentel, Mar 31 2025

Status

approved