A124672 Oblong (promic) abundant numbers = abundant numbers of the form k(k+1).

12, 20, 30, 42, 56, 72, 90, 132, 156, 210, 240, 272, 306, 342, 380, 420, 462, 552, 600, 650, 702, 756, 812, 870, 930, 992, 1056, 1122, 1190, 1260, 1332, 1482, 1560, 1640, 1722, 1806, 1980, 2070, 2256, 2352, 2450, 2550, 2652, 2862, 2970, 3080, 3192, 3306

Offset: 1

Author

Tanya Khovanova, Dec 27 2006

Keywords

• nonn
• easy

Comments

Promic numbers are highly divisible, so most of them are abundant.

Examples

			56 is in the sequence because 56=7*8 and the sum of its divisors 1+2+4+7+8+14+28+56=120 > 2*56.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000

Crossrefs

Intersection of A002378 (oblong numbers) and A005101 (abundant numbers).

Cf. A077804 (deficient oblong numbers).

Programs

Maple

with(numtheory): a:=proc(k) if sigma(k*(k+1))>2*k*(k+1) then k*(k+1) else fi end: seq(a(k),k=1..75); # Emeric Deutsch, Jan 01 2007
isA005101 := proc(n) if numtheory[sigma](n) > 2*n then RETURN(true) ; else RETURN(false) ; fi ; end : for k from 1 to 80 do if isA005101(k*(k+1)) then printf("%d,",k*(k+1)) ; fi ; od ; # R. J. Mathar, Jan 07 2007

Mathematica

s = {}; Do[ob = n*(n + 1); If[DivisorSigma[1, ob] > 2*ob, AppendTo[s, ob]], {n, 1, 100}]; s (* Amiram Eldar, Jun 07 2019 *)

PARI

helper(n)=my(k=sqrtint(n)); if(k*(k+1)>n, k, k+1)
list(lim)=my(v=List(),last=4/3,cur); forfactored(n=4,helper(lim\1), cur=sigma(n,-1); if(cur*last>2, listput(v, (n[1]-1)*n[1])); last=cur); Vec(v) \\ Charles R Greathouse IV, Mar 16 2022

Formula

If k > 2 is 0 or 2 mod 3, then k*(k+1) is in this sequence; the bounds n^2 < a(n) < (9/4)*n^2 + 6n + 4 can be derived from this. Probably a(n) ~ kn^2 with k near 1.496. - Charles R Greathouse IV, Mar 16 2022

Extensions

More terms from Emeric Deutsch, Jan 01 2007

More terms from R. J. Mathar, Jan 07 2007