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A199337 Number of highly composite numbers not divisible by n. 8
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, 14, 107, 8, 122, 51, 26, 53, 14, 12, 145, 67, 37, 11, 163, 14, 180, 26, 13, 86, 202, 18, 107, 54, 53, 37, 222, 40, 26, 15, 67, 107, 252, 8, 275, 122, 16, 79, 37 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,3

COMMENTS

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

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

LINKS

David A. Corneth, Table of n, a(n) for n = 1..600 (using b-file of A002182)

S. Ramanujan, Highly composite numbers, 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 alternative link)

EXAMPLE

a(6) = 3 because among highly composite numbers, only 1, 2, and 4 are not divisible by 6.

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

MATHEMATICA

(* let t be terms of b002182 *) Table[Length[Select[t, Mod[#, n] > 0 &]], {n, 100}] (* T. D. Noe, Mar 18 2012 *)

CROSSREFS

Cf. A002182, A106037.

Cf. A329570, A329571 (bounds from the Ramanujan formula).

Sequence in context: A294643 A029614 A143615 * A279789 A249389 A016606

Adjacent sequences:  A199334 A199335 A199336 * A199338 A199339 A199340

KEYWORD

nonn

AUTHOR

J. Lowell, Nov 05 2011

EXTENSIONS

Extended by T. D. Noe, Mar 18 2012

STATUS

approved

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Last modified February 24 09:21 EST 2020. Contains 332209 sequences. (Running on oeis4.)