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%I #37 Jan 17 2020 10:43:38
%S 0,1,3,3,8,3,14,9,12,8,26,4,37,14,8,18,53,12,67,8,14,26,86,9,54,37,40,
%T 14,107,8,122,51,26,53,14,12,145,67,37,11,163,14,180,26,13,86,202,18,
%U 107,54,53,37,222,40,26,15,67,107,252,8,275,122,16,79,37
%N Number of highly composite numbers not divisible by n.
%C The sequence is well defined since for any n there is m such that n | A002182(k) for all k >= m. This follows from eq.(54) in Ramanujan (1915): [log_p P] <= e_p <= 2*[log_p P'], where for any N = A002182(k), P = gpf(N) is the greatest prime factor, e_p = valuation(N, p) is the exponent of any p in the prime factorization, P' = nextprime(P+1) and [.] = floor: The right inequality gives N <= Product_{prime p} p^(2*[log_p P']) = A003418(P')^2, so P -> oo as N -> oo. Then the left inequality implies e_p -> oo for any p, as N -> oo. - _M. F. Hasler_, Jan 03 2020
%C Sequences A329570 and A329571 give the gpf P as above and L = A003418(P) such that all A002182(k) >= L^2 are divisible by n. - _M. F. Hasler_, Jan 07 2020
%H David A. Corneth, <a href="/A199337/b199337.txt">Table of n, a(n) for n = 1..600</a> (using b-file of A002182)
%H S. Ramanujan, <a href="https://doi.org/10.1112/plms/s2_14.1.347">Highly composite numbers</a>, Proceedings of the London Mathematical Society ser. 2, vol. XIV, no. 1 (1915): 347-409. (DOI: 10.1112/plms/s2_14.1.347, a variant of better quality with an additional footnote is available at this <a href="http://ramanujan.sirinudi.org/Volumes/published/ram15.html">alternative link</a>)
%e a(6) = 3 because among highly composite numbers, only 1, 2, and 4 are not divisible by 6.
%e To illustrate the comment, we prove that n = 12 = A002182(5), respectively n = 60 = A002182(9), divide all A002182(k) >= n (whence a(12) = 5 - 1 = 4, a(30) = 9 - 1 = 8): From eq.(54) we have e_2 >= 2 and e_3 >= 1 when [log_2 P] >= 2, [log_3 P] >= 1, which is the case for P >= 5. To get gpf(N) >= 5, use the other side of the inequality, e_p <= 2*[log_p P'] with P = 3, P' = 5: This gives e_2 <= 4; e_3, e_5 <= 2; e_p = 0 for p > 5. Thus all N = a(n) > 2^4*3^2*5^2 = 3600 must have gpf(N) > 3, i.e., gpf(N) >= 5. This implies e_2 >= 2 and e_3 >= 1 and also e_5 >= 1, so we have 12 | N and 60 | N for all N = A002182(k) > 3600. The terms between 12 (resp. 60) and 3600 are also multiples of 12 (resp. 60), which completes the proof. - _M. F. Hasler_, Jan 03 2020
%t (* let t be terms of b002182 *) Table[Length[Select[t, Mod[#, n] > 0 &]], {n, 100}] (* _T. D. Noe_, Mar 18 2012 *)
%Y Cf. A002182, A106037.
%Y Cf. A329570, A329571 (bounds from the Ramanujan formula).
%K nonn
%O 1,3
%A _J. Lowell_, Nov 05 2011
%E Extended by _T. D. Noe_, Mar 18 2012
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