The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation. Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
 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. 0
 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, 0, 0, -81, 15, -49, 0, 0, 0, -30, 0, -20, 0, -27, 0, -23, 0, 0, -36, 24, 64, 0, 0, 0, -169, 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 (list; table; graph; refs; listen; history; text; internal format)
 OFFSET 1,5 COMMENTS Absolute values of the nonzero terms here are the terms of A164119 (but the terms there are in ascending order). The positive terms here are the terms of A161460 (although 50175 and 59049 there are conjectural). 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. 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). 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.) Conjecture: T(n,k) < 0 for all T(n,k) where both n and k are even, other than the case n=k=2. LINKS EXAMPLE 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. 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. 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. 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. 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. . Table begins:   n\k| 1    2  3    4  5     6  7            8    9        10   ---+-------------------------------------------------------    1 | 0    1  0    0  0     0  0            0    0         0    2 | 0    2  3   -5  0   -11  0          -23    0       -47    3 | 0    4  0   -9  0   -49  0         -169    0     57121    4 | 0   -6  8  -14 15   -27  0          -39   35      -111    5 | 0   16  0  -81  0     0  0      -130321    0         0    6 | 0  -12  0  -20  0   -44 63         -153   99      -175    7 | 0    0  0   64  0     0  0         -729    0         0    8 | 0  -30 24  -54  0  -152  0         -104 -195      -890    9 | 0  -36  0 -225  0 -1444  0         -441    0 -96393124   10 | 0 -112 48 -176 80  -368  0         -272 6723     -2511   11 | 0    0  0    0  0  1024  0 -25937424601    0         0   12 | 0  -60  0  -84  0  -260  0         -350 -224      -495 CROSSREFS Cf. A161460, A164119. Sequence in context: A230295 A147592 A108885 * A072740 A226288 A185146 Adjacent sequences:  A341651 A341652 A341653 * A341655 A341656 A341657 KEYWORD sign,tabl AUTHOR Jon E. Schoenfield, Feb 20 2021 STATUS approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.

Last modified April 19 10:30 EDT 2021. Contains 343112 sequences. (Running on oeis4.)