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A338381 Smallest number m such that tau(k) * k = m has exactly n solutions when tau(k) is the number of divisors of k.

%I A338381 #48 Mar 31 2023 09:18:42
%S A338381 1,108,2688,21000000,8400000
%N A338381 Smallest number m such that tau(k) * k = m has exactly n solutions when tau(k) is the number of divisors of k.
%C A338381 The map k -> k*tau(k) = m is not injective (A038040), this sequence lists the smallest integers m that have exactly n preimages.
%C A338381 This sequence is not increasing with a(5) < a(4).
%C A338381 a(6) <= 4124832465600000, a(7) <= 33195080318400000. - _David A. Corneth_, Oct 28 2020
%C A338381 a(6) <= 1508867287200000, a(8) <= 2544150895374925200000, a(9) <= 55487699012097891000000. - _Daniel Suteu_, Oct 28 2020
%D A338381 Richard K. Guy, Unsolved Problems in Number Theory, 3rd Edition, Springer, 2004, Section B12, p. 102-103.
%D A338381 D. Wells, The Penguin Dictionary of Curious and Interesting Numbers, Revised Edition, Penguin Books, London, England, 1997, entry 168, page 127.
%e A338381 a(1) = 1 because 1 * tau(1) = 1.
%e A338381 a(2) = 108 because 18 * tau(18) = 27 * tau(27) = 108 and 108 is the smallest number with 2 preimages.
%e A338381 a(3) = 2688 because 168 * tau(168) = 192 * tau(192) = 224 * tau(224) = 2688 and 2688 is the smallest number with 3 preimages.
%e A338381 a(4) = 21000000 and the corresponding 4 values of k are: 210000, 350000, 375000, 500000.
%e A338381 a(5) = 8400000 and the corresponding 5 values of k are: 105000, 120000, 140000, 175000, 200000. Thanks to _Amiram Eldar_ for these values of k for a(4) and a(5).
%o A338381 (PARI) isok(k, n) = sumdiv(k, d, d*numdiv(d) == k) == n;
%o A338381 a(n) = my(k=1); while (! isok(k,n), k++); k; \\ _Michel Marcus_, Oct 28 2020
%Y A338381 Cf. A000005, A038040, A327166.
%Y A338381 Cf. A212490 (similar for k*sigma(k)).
%K A338381 nonn,more
%O A338381 1,2
%A A338381 _Bernard Schott_, Oct 23 2020
%E A338381 a(4)-a(5) from _Amiram Eldar_, Oct 23 2020