A164119 -id:A164119 - cp's OEIS frontend

A161460 Positive integers k such that there is no m different from k where both d(k) = d(m) and d(k+1) = d(m+1), where d(k) is the number of positive divisors of k.

1, 2, 3, 4, 8, 15, 16, 24, 35, 48, 63, 64, 80, 99, 288, 528, 575, 624, 728, 960, 1023, 1024, 1088, 1295, 2303, 2400, 4095, 4096, 5328, 6399, 6723, 9408, 9999, 14640, 15624, 28223, 36863, 38415, 46655, 50175, 50624, 57121, 59048, 59049, 65535, 65536, 83520

Offset: 1

Leroy Quet, Jun 10 2009

nonn

Are these values known to be correct, or are they just conjectures? - Leroy Quet, Jun 20 2009 [Answer: all the numbers listed in the Data are known to be correct with the exception of 50175 and 59049, which remain conjectural at this time; see the Mathar link. - Jon E. Schoenfield, Feb 08 2021]
Numbers k that are uniquely identified by the values of the ordered pair (d(k), d(k+1)). - Jon E. Schoenfield, Aug 11 2019
Conjecture: 2 is the only term that is neither a square nor 1 less than a square. - Jon E. Schoenfield, Aug 12 2019

			d(15) = 4, and d(15+1) = 5. Any positive integer m+1 with exactly 5 divisors must be of the form p^4, where p is prime. So m = p^4 - 1 = (p^2+1)*(p+1)*(p-1). Now, in order for d(m) to have exactly 4 divisors, m must either be of the form q^3 or q*r, where q and r are distinct primes. But no p is such that (p^2+1)*(p+1)*(p-1) = q^3. And the only p where (p^2+1)*(p+1)*(p-1) = q*r is p=2 (and so q=5, r=3). So there is only one m where both d(m) = 4 and d(m+1) = 5, which is m=15. Therefore 15 is in this sequence.

R. J. Mathar, Recurring pairs of consecutive entries in the number-of-divisors function, vixra:1911.0287 (2019).

Cf. A000005, A164119.

Extended with J. Brennen's values of Jun 11 2009 by R. J. Mathar, Jun 16 2009

a(47) from Jon E. Schoenfield, Feb 08 2021

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.

Original entry on oeis.org

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

Offset: 1

Jon E. Schoenfield, Feb 20 2021

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.

			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

Cf. A161460, A164119.

cp's OEIS Frontend

A161460 Positive integers k such that there is no m different from k where both d(k) = d(m) and d(k+1) = d(m+1), where d(k) is the number of positive divisors of k.

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