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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. 4
 1, 108, 2688, 21000000, 8400000 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS The application k -> k*tau(k) = m is not injective (A038040), this sequence proposes the smallest integers m that have exactly n preimages. This sequence is not increasing with a(5) < a(4). a(6) <= 4124832465600000, a(7) <= 33195080318400000. - David A. Corneth, Oct 28 2020 a(6) <= 1508867287200000, a(8) <= 2544150895374925200000, a(9) <= 55487699012097891000000. - Daniel Suteu, Oct 28 2020 REFERENCES Richard K. Guy, Unsolved Problems in Number Theory, 3rd Edition, Springer, 2004, Section B12, p. 102-103. D. Wells, The Penguin Dictionary of Curious and Interesting Numbers, Revised Edition, Penguin Books, London, England, 1997, entry 168, page 127. LINKS EXAMPLE a(1) = 1 because 1 * tau(1) = 1. a(2) = 108 because 18 * tau(18) = 27 * tau(27) = 108 and 108 is the smallest number with 2 preimages. a(3) = 2688 because 168 * tau(168) = 192 * tau(192) = 224 * tau(224) = 2688 and 2688 is the smallest number with 3 preimages. a(4) = 21000000 and the corresponding 4 values of k are: 210000, 350000, 375000, 500000. 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). PROG (PARI) isok(k, n) = sumdiv(k, d, d*numdiv(d) == k) == n; a(n) = my(k=1); while (! isok(k, n), k++); k; \\ Michel Marcus, Oct 28 2020 CROSSREFS Cf. A000005, A038040, A327166. Cf. A212490 (similar for k*sigma(k)). Sequence in context: A183357 A264681 A221018 * A269182 A244877 A223153 Adjacent sequences:  A338378 A338379 A338380 * A338382 A338383 A338384 KEYWORD nonn,more AUTHOR Bernard Schott, Oct 23 2020 EXTENSIONS a(4)-a(5) from Amiram Eldar, Oct 23 2020 STATUS approved

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Last modified October 17 17:21 EDT 2021. Contains 348065 sequences. (Running on oeis4.)