A060798 Numbers k such that difference between the upper and lower central divisors of k is 1.

2, 4, 6, 12, 20, 30, 42, 56, 72, 90, 110, 132, 156, 182, 210, 240, 272, 306, 342, 380, 420, 462, 506, 552, 600, 650, 702, 756, 812, 870, 930, 992, 1056, 1122, 1190, 1260, 1332, 1406, 1482, 1560, 1640, 1722, 1806, 1892, 1980, 2070, 2162, 2256, 2352, 2450

Offset: 1

Author

Labos Elemer, Apr 27 2001

Keywords

• nonn
• easy

Comments

From David A. Corneth, Sep 02 2018: (Start)

Theorem: a(1) = 2, a(2) = 4; a(n) = n*(n-1) for n > 2.

Proof:

If a(n) is a square m^2 then the upper central divisor is m and by definition of the sequence the lower one is m-1. But m-1 and m are coprime, and (m-1)|m^2 implies m-1 = 1, i.e. a(n) = 4.

if a(n) is not a square then it has an even number of divisors with m and m-1 the central divisors, so it has the form m*(m-1), i.e. is oblong (see A002378). QED (End)

Examples

			The divisors of 2 are 1 and 2, so the upper central divisor is 2 and the lower central divisor is 1, so a(1)=2 is a member.
k = 4032 = 2*2*2*2*2*2*3*3*7 is here because its central divisors (the 21st and 22nd divisors) are {63,64} which differ by 1.
		

Links

• Harvey P. Dale, Table of n, a(n) for n = 1..2000 [First 1000 terms from Harry J. Smith]
• Index entries for linear recurrences with constant coefficients, signature (3, -3, 1).

Crossrefs

Cf. A000196, A002378, A033677, A060775.

Programs

Mathematica

dulcdQ[n_]:=Module[{d=Divisors[n],len},len=Floor[Length[d]/2];d[[len+1]] - d[[len]]==1]; Select[Range[2500],dulcdQ] (* or *) Join[{2,4},Table[ n(n-1),{n,3,60}]] (* after David A. Corneth's comment and formula *) (* Harvey P. Dale, Aug 28 2018 *)

PARI

{ n=-1; for (m=1, 999000, d=divisors(m); if (m==1 || (d[1 + length(d)\2] - d[length(d)\2]) == 1, write("b060798.txt", n++, " ", m)); ) } \\ Harry J. Smith, Jul 13 2009

PARI

first(n) = res = List([2, 4]); for(i = 3, n, listput(res, i*(i-1))); res \\ David A. Corneth, Sep 02 2018

Formula

Solutions to A033677(k) - A060775(k) = 1, where k = j*(j+1) and at least one of j and j+1 is composite.

Except at n < 5, this sequence satisfies a(n+1) = 3*a(n) - 3*a(n-1) + a(n-2). - Georgi Guninski, Jun 07 2010 [This follows from Corneth's theorem above. - N. J. A. Sloane, Sep 02 2018]

G.f.: (2*x^2-2*x+1)*(x^3-x^2-x-1) / (x-1)^3. - Colin Barker, Apr 16 2014 [This follows from Corneth's theorem above. - N. J. A. Sloane, Sep 02 2018]

Extensions

Start and offset changed by N. J. A. Sloane, Sep 02 2018 at the suggestion of Harvey P. Dale. Further edited by N. J. A. Sloane, Sep 02 2018