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A341654 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.

%I A341654 #12 Mar 07 2021 20:57:15
%S A341654 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 A341654 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 A341654 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 A341654 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 A341654 Absolute values of the nonzero terms here are the terms of A164119 (but the terms there are in ascending order).
%C A341654 The positive terms here are the terms of A161460 (although 50175 and 59049 there are conjectural).
%C A341654 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 A341654 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 A341654 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 A341654 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 A341654 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 A341654 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 A341654 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 A341654 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 A341654 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 A341654 .
%e A341654 Table begins:
%e A341654   n\k| 1    2  3    4  5     6  7            8    9        10
%e A341654   ---+-------------------------------------------------------
%e A341654    1 | 0    1  0    0  0     0  0            0    0         0
%e A341654    2 | 0    2  3   -5  0   -11  0          -23    0       -47
%e A341654    3 | 0    4  0   -9  0   -49  0         -169    0     57121
%e A341654    4 | 0   -6  8  -14 15   -27  0          -39   35      -111
%e A341654    5 | 0   16  0  -81  0     0  0      -130321    0         0
%e A341654    6 | 0  -12  0  -20  0   -44 63         -153   99      -175
%e A341654    7 | 0    0  0   64  0     0  0         -729    0         0
%e A341654    8 | 0  -30 24  -54  0  -152  0         -104 -195      -890
%e A341654    9 | 0  -36  0 -225  0 -1444  0         -441    0 -96393124
%e A341654   10 | 0 -112 48 -176 80  -368  0         -272 6723     -2511
%e A341654   11 | 0    0  0    0  0  1024  0 -25937424601    0         0
%e A341654   12 | 0  -60  0  -84  0  -260  0         -350 -224      -495
%Y A341654 Cf. A161460, A164119.
%K A341654 sign,tabl
%O A341654 1,5
%A A341654 _Jon E. Schoenfield_, Feb 20 2021