OFFSET
1,1
COMMENTS
De Koninck and Ivić found that the least number k such that k, k+1, and k+2 are 3 consecutive integers that are S-abundant numbers is 171078830 (which is also the first term of A096536).
LINKS
Amiram Eldar, Table of n, a(n) for n = 1..10000
Jean-Marie De Koninck and Aleksandar Ivić, On a sum of divisors problem, Publications de l'Institut Mathématique (Beograd), New Series, Vol. 64 (78) (1998), pp. 9-20.
Wikipedia, Granville number.
MATHEMATICA
seq[kmax_] := Module[{s = {1}, a = {}, sum, q1 = False, q2}, Do[sum = Total[Select[Divisors[k], MemberQ[s, #] &]]; q2 = sum > k; If[!q2, AppendTo[s, k]]; If[q1 && q2, AppendTo[a, k-1]]; q1 = q2, {k, 2, kmax}]; a]; seq[40000]
PROG
(PARI) lista(nmax) = {my(c = 0, s, q1 = 0, q2); for(n=2, nmax, if(sumdiv(n, d, !bittest(c, d)*d) > 2*n, c+=1<<n; q2 = 1, q2 = 0); if(q1 && q2, print1(n-1, ", ")); q1 = q2) } \\ after M. F. Hasler at A181487
CROSSREFS
KEYWORD
nonn
AUTHOR
Amiram Eldar, Aug 11 2023
STATUS
approved