A125115 Differences between consecutive abundant numbers.

6, 2, 4, 6, 6, 4, 2, 6, 6, 2, 4, 6, 4, 2, 6, 2, 4, 4, 2, 6, 4, 2, 2, 4, 4, 2, 6, 6, 6, 6, 2, 4, 6, 6, 4, 2, 6, 6, 2, 4, 6, 6, 4, 2, 2, 4, 4, 2, 6, 4, 2, 2, 4, 6, 6, 6, 6, 6, 2, 4, 6, 2, 4, 4, 2, 6, 6, 6, 4, 2, 2, 4, 6, 2, 4, 6, 6, 4, 2, 6, 2

Offset: 1

Author

Jason G. Wurtzel, Nov 21 2006

Keywords

• easy
• nonn

Comments

One may think that a(n) is always even and greater than 1. This is not the case as can be seen with A096399 or A228382. - Michel Marcus, Aug 21 2013

Examples

			a(1) = 6 because 18 - 12 = 6; a(4) = 6 because 30 - 24 = 6.
		

Links

• Muniru A Asiru, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A005101, A096399, A228382, A302991.

Programs

GAP

A:=Filtered([1..350],n->Sigma(n)>2*n);;  a:=List([1..Length(A)-1],i->A[i+1]-A[i]); # Muniru A Asiru, Jun 09 2018

Mathematica

#[[2]] - #[[1]]&/@Partition[Select[Range[300], DivisorSigma[1, #] > 2# &], 2, 1] (* Harvey P. Dale, Dec 02 2006 *)
Differences[Select[Range[300], DivisorSigma[1, #] > 2# &]] (* Alonso del Arte, Apr 29 2019 *)

PARI

lista(nn) = {lastab = 0; for (i=1, nn, if (sigma(i) > 2*i,if (lastab, print1(i - lastab, ", ")); lastab = i;););} \\ Michel Marcus, Aug 21 2013

Formula

From Amiram Eldar, Oct 21 2020: (Start)

a(n) = A005101(n+1) - A005101(n).

Asymptotic mean: lim_{n->oo} (1/n) Sum_{k=1..n} a(k) = 1/A302991 = 4.0384... (End)

Extensions

More terms from Michel Marcus, Aug 21 2013