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Number of distinct differences between consecutive divisors of n (ordered by size).
15

%I #27 Nov 10 2023 09:09:48

%S 1,1,2,1,2,1,3,2,3,1,3,1,3,2,4,1,3,1,4,3,3,1,4,2,3,3,5,1,5,1,5,3,3,3,

%T 5,1,3,3,5,1,4,1,5,4,3,1,5,2,5,3,5,1,4,3,6,3,3,1,7,1,3,4,6,3,5,1,5,3,

%U 6,1,6,1,3,3,5,3,5,1,7,4,3,1,6,3,3,3,7,1,7,2,5,3,3,3,6,1,5,4,6,1,5,1,7,5,3

%N Number of distinct differences between consecutive divisors of n (ordered by size).

%C Number of all differences for n is d(n)-1 = A000005(n)-1. Increments are not necessarily different, so a(n)<=d(n)-1.

%H Reinhard Zumkeller, <a href="/A060682/b060682.txt">Table of n, a(n) for n = 2..10000</a>

%H A. Balog, P. Erdős and G. Tenenbaum, <a href="http://dx.doi.org/10.1007/978-1-4612-3464-7_6">On Arithmetic Functions Involving Consecutive Divisors</a>, In: Analytical Number Theory, pp. 77-90, Birkhäuser, Basel, 1990.

%H Jason Earls, <a href="https://pdfs.semanticscholar.org/4559/ac50797ddeda688576630c4d92229440a0a3.pdf#page=259">Smarandache iterations of the first kind on functions involving divisors and prime factors</a>, in Smarandache Notions Journal (2004), Vol. 14.1, page 264.

%e For n=70, divisors={1,2,5,7,10,14,35,70}; differences={1,3,2,3,4,21,35}; a(70) = number of distinct differences = 6.

%t a[n_ ] := Length[Union[Drop[d=Divisors[n], 1]-Drop[d, -1]]]

%o (Haskell)

%o import Data.List (nub, genericLength)

%o a060682 = genericLength . nub . a193829_row

%o -- _Reinhard Zumkeller_, Jun 25 2015

%o (PARI) a(n) = my(d=divisors(n)); #vecsort(vector(#d-1, k, d[k+1] - d[k]),,8); \\ _Michel Marcus_, Jul 04 2017

%Y Cf. A000005, A060680, A060681, A060683.

%Y Cf. A193829.

%K nonn

%O 2,3

%A _Labos Elemer_, Apr 19 2001

%E Edited by _Dean Hickerson_, Jan 22 2002