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%I #12 Mar 07 2021 20:57:15
%S 0,0,1,0,2,0,0,4,3,0,0,-6,0,-5,0,0,16,8,-9,0,0,0,-12,0,-14,0,-11,0,0,
%T 0,0,-81,15,-49,0,0,0,-30,0,-20,0,-27,0,-23,0,0,-36,24,64,0,0,0,-169,
%U 0,0,0,-112,0,-54,0,-44,0,-39,0,-47,0,0,0,48,-225,0,0,63,-130321,35,57121,0,0
%N Table read by antidiagonals upward: |T(n,k)| is the smallest number j such that j and j+1 have n and k divisors, respectively, or 0 if no such number exists, and the sign of T(n,k) is positive iff there exists only one such number j.
%C Absolute values of the nonzero terms here are the terms of A164119 (but the terms there are in ascending order).
%C The positive terms here are the terms of A161460 (although 50175 and 59049 there are conjectural).
%C The smallest number j such that j and j+1 have 11 and 12 divisors, respectively, is 59049, and (per the link at A161460) it isn't yet known whether that's the only such number. So T(11,12) is either 59049 or -59049.
%C T(n,k)=0 whenever both n and k are odd (since every number with an odd number of divisors is a square, and no two squares are consecutive integers).
%C Terms where both n and k are even (so neither j nor j+1 is a square) tend to be relatively small negative numbers. (T(2,2)=+2 is a special case; see Example section.)
%C Conjecture: T(n,k) < 0 for all T(n,k) where both n and k are even, other than the case n=k=2.
%e The only number j with 1 divisor is 1, so 1 is the only nonzero number in row n=1; its successor j+1 is 2 (a prime, so it has 2 divisors), so T(1,2)=1, and T(1,k)=0 for all k != 2.
%e The only two consecutive integers that are primes are 2 and 3, so 2 is the only number j such that both j and j+1 have 2 divisors, thus T(2,2)=2.
%e For j and j+1 to have 2 and 3 divisors, respectively, j must be a prime p, and j+1 must be the square of a prime q, and the only solution to p + 1 = q^2 is at p=3, so T(2,3)=3.
%e Numbers j such that j and j+1 have 2 and 4 divisors, respectively, begin 5, 7, 13, 37, 61, ...; since 5 is the smallest such number, T(2,4)=-5.
%e Every number with 11 divisors is of the form p^10 (p prime), and primes p such that p^10 + 1 has 8 divisors begin 11, 19, 101, 139, ..., so T(11,8) = -(11^10) = -25937424601.
%e .
%e Table begins:
%e n\k| 1 2 3 4 5 6 7 8 9 10
%e ---+-------------------------------------------------------
%e 1 | 0 1 0 0 0 0 0 0 0 0
%e 2 | 0 2 3 -5 0 -11 0 -23 0 -47
%e 3 | 0 4 0 -9 0 -49 0 -169 0 57121
%e 4 | 0 -6 8 -14 15 -27 0 -39 35 -111
%e 5 | 0 16 0 -81 0 0 0 -130321 0 0
%e 6 | 0 -12 0 -20 0 -44 63 -153 99 -175
%e 7 | 0 0 0 64 0 0 0 -729 0 0
%e 8 | 0 -30 24 -54 0 -152 0 -104 -195 -890
%e 9 | 0 -36 0 -225 0 -1444 0 -441 0 -96393124
%e 10 | 0 -112 48 -176 80 -368 0 -272 6723 -2511
%e 11 | 0 0 0 0 0 1024 0 -25937424601 0 0
%e 12 | 0 -60 0 -84 0 -260 0 -350 -224 -495
%Y Cf. A161460, A164119.
%K sign,tabl
%O 1,5
%A _Jon E. Schoenfield_, Feb 20 2021
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