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%I #24 May 05 2019 11:51:49
%S 1,3,1,8,1,8,1,17,1
%N Length of longest run of consecutive odd numbers having exactly n divisors.
%C a(n)=1 for n odd, since every number with an odd number of divisors is a square, and no two squares are consecutive odd numbers.
%C The start of the first run of exactly k consecutive odd numbers having exactly n divisors is A319046(n,k).
%C From _David Wasserman_, May 04 2019: (Start)
%C 7 <= a(10) <= 8.
%C 14 <= a(12) <= 59. Dickson's conjecture implies a(12) >= 39. Schinzel's Hypothesis H implies a(12) >= 41. (End)
%e From _David Wasserman_, May 04 2019: (Start)
%e A run of 17 consecutive odd numbers with 8 divisors begins at 237805775327, so a(8) >= 17; a run of 18 or more consecutive odd numbers would include at least two that are multiples of 9, and every multiple of 9 having 8 divisors is also a multiple of 27, but the two multiples of 9 cannot both be multiples of 27, so a(8) = 17.
%e A run of 5 consecutive odd numbers with 14 divisors begins at 10943266106145622193005970311, so a(14) >= 5. A run of 6 consecutive odd numbers with 14 divisors would include at least two that are multiples of 3, and these two would differ by 6. These must be 3^13, 3^6*p for p prime > 3, or 3*p^6 for p prime > 3. But 3*p^6 = 3 (mod 27), while 3^13 and 3^6*p = 0 (mod 27), so no two of these can differ by 6. Therefore no such run exists, and a(14) = 5. (End)
%Y Cf. A119479 (analog for consecutive integers), A319046.
%K nonn,more,hard
%O 1,2
%A _Jon E. Schoenfield_, Dec 22 2018
%E a(6)-a(9) from _David Wasserman_, Feb 07 2019
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