%I
%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
