Reminder: The OEIS is hiring a new managing editor, and the application deadline is January 26.
%I #7 Jan 30 2019 03:19:33
%S 6,10,21,28,104,115,136,329,496,2133,2171,6821,8128,24331,32896,50579,
%T 79421,103729,226859,357769,704791,1092521,1224829,2048129,2247829,
%U 2685341,5177371,6967489,9393509,11089121,12648871,13651441,16974079,25153171,30663671
%N Non-prime-powers k at which the variance of the first differences of the logarithms of the divisors of k, scaled by log(k), reaches a new minimum.
%C For any positive integer k, each divisor d of k and its complement q=k/d have as their geometric mean the value sqrt(d*q) = sqrt(d*k/d) = sqrt(k), so log(d), log(sqrt(k)), and log(q) form an arithmetic progression; consequently, the logarithms of the divisors of k are distributed symmetrically about log(sqrt(k)) = log(k)/2.
%C If k is a prime power p^m (where p is prime and m >= 1), the divisors of k form the geometric progression {1, p, p^2, ..., p^m}, so the real-valued sequence consisting of their logarithms is the arithmetic progression {0, log(p), 2*log(p), ..., m*log(p)}, with constant difference log(p); if we divide that real-valued sequence by log(k), we get the rational-valued sequence {0, 1/m, 2/m, ..., 1} (an arithmetic progression with constant difference 1/m). If k is not a prime power, the differences between the logarithms of successive divisors of k will vary.
%C Consider the question: For which non-prime-powers k are the divisors of k distributed most evenly (on a logarithmic scale)? One way to quantify the evenness of their distribution is to scale the logarithms of the divisors by dividing them by log(k) (to fit them to the interval [0,1]) and compute the population variance of their first differences. The non-prime-powers k at which this variance reaches a new minimum are the terms of this integer sequence.
%C All terms in the sequence appear to be of the form p^m * q where p and q are prime and q is close to p^(m+1).
%C Of the first 35 terms, all but 9 are odd semiprimes of the form k = p*(p^2 - 2); cf. A240436. Such numbers have 4 divisors -- in ascending order, d_1 = 1, d_2 = p, d_3 = p^2 - 2, and d_4 = p*(p^2 - 2) -- and as k increases, the values log(d_i)/log(k) approach {0, 1/3, 2/3, 1}.
%C Conjectures:
%C 1. Other than A240436(1)=4 (a prime power), A240436 is a subsequence of this sequence.
%C 2. Only nine terms in this sequence are not semiprimes of the form p*(p^2 - 2):
%C 10 = 2 * 5
%C 104 = 2^3 * 13
%C 136 = 2^3 * 17
%C 2133 = 3^3 * 79
%C 32896 = 2^7 * 257
%C and the first four perfect numbers (6, 28, 496, and 8128).
%e k = 115 = 5 * (5^2 - 2) = 5 * 23 has 4 divisors: 1, 5, 23, and 115. We have
%e | first |
%e | differences |
%e d | log(d)/log(k) | (mean = 1/3) | (diff - mean)^2
%e ----+---------------+---------------+------------------
%e 1 | 0.00000000000 | |
%e ----+---------------+ 0.33919092389 | 0.000034311367177
%e 5 | 0.33919092389 | |
%e ----+---------------+ 0.32161815221 | 0.000137245468708
%e 23 | 0.66080907611 | |
%e ----+---------------+ 0.33919092389 | 0.000034311367177
%e 115 | 1.00000000000 | |
%e ----+---------------+---------------+------------------
%e sum: 0.000205868203062
%e .
%e Population variance = 0.000205868203062 / 3
%e = 0.000068622734354
%e .
%e The population variance 0.000068622734354 is smaller than that for any smaller value of k, so k=115 in the sequence.
%e The first several terms and their population variances are
%e k population variance
%e --- -------------------
%e 6 0.00572866817332422
%e 10 0.00208701124916037
%e 21 0.00151420255078270
%e 28 0.00025693510366524
%e 104 0.00024474680031955
%e 115 0.00006862273435404
%e 136 0.00001864750547090
%e 329 0.00001148749359549
%e 496 0.00000258435797989
%e 2133 0.00000130263831477
%Y Cf. A000396, A240436.
%K nonn
%O 1,1
%A _Jon E. Schoenfield_, Jan 29 2019