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

%I A382606 #37 May 06 2025 16:14:48
%S A382606 6,5,11,12,10,16,17,15,21,22,27,23,26,28,32,33,38,37,39,43,44,49,48,
%T A382606 50,54,55,60,59,53,65,61,66,64,70,71,76,75,77,81,82,87,72,86,88,92,93,
%U A382606 98,97,103,99,104,102,108,109,114,115,113,110,119,120,125,124,126
%N A382606 Natural numbers ordered by the probability (highest to lowest) to occur in the sum of repeated rolls of a fair 6-sided die.
%C A382606 The asymptotic probability for large n is 2/7 since the average roll of a die is 7/2.
%C A382606 Only terms with probability > 2/7 occur. - _Michael S. Branicky_, Apr 01 2025
%C A382606 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
%H A382606 Alois P. Heinz, <a href="/A382606/b382606.txt">Table of n, a(n) for n = 1..10000</a>
%e A382606 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%.
%e A382606 The probability of achieving a '1' is just 1/6 (about 16.67%). 6 is the highest of all, so a(1) = 6.
%o A382606 (Python)
%o A382606 from fractions import Fraction
%o A382606 from math import factorial, prod
%o A382606 from itertools import count, islice
%o A382606 from sympy.utilities.iterables import partitions
%o A382606 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))
%o A382606 def agen(): # generator of terms
%o A382606     n, vdict = 1, dict()
%o A382606     for k in count(1):
%o A382606         vdict[prob(k)] = k
%o A382606         if k%6 == 0:
%o A382606             s = [vdict[v] for v in sorted(vdict, reverse=True) if v > Fraction(2, 7)]
%o A382606             yield from (s[i-1] for i in range(n, len(s)-1))
%o A382606             n = len(s) - 1
%o A382606 print(list(islice(agen(), 20))) # _Michael S. Branicky_, Apr 01 2025
%Y A382606 Complement of A382607. Cf. A365443.
%K A382606 nonn
%O A382606 1,1
%A A382606 _Sergio Pimentel_, Mar 31 2025