A383398 a(n) is the smallest number whose sum with any previous term is abundant.

1, 11, 19, 29, 59, 349, 521, 2071, 66949, 223231, 3660191, 4552181, 5500081, 10161979, 12235619, 47859629

Offset: 1

Author

Jakub Buczak, Apr 25 2025

Keywords

• nonn
• hard
• more

Comments

The terms are generally either prime or semiprime. This results in all known terms to be deficient (see A005100).

If a(1) is an even abundant number, then the set of all the terms is simply the set of all the even abundant numbers (see A173490).

I conjecture that all the terms are odd integers ending in 1 or 9. The odd nature of the terms seems particularly likely, as the sum of a(n) that's even with any previous term would need to be an odd abundant number (see A005231).

This is also equivalent to the sum of any 2 terms being an abundant number.

Examples

			29 is a member, because 29+19, 29+11 and 29+1 are all abundant numbers.
		

Crossrefs

Cf. A383399, A005101, A005231, A005100, A173490.

Cf. A000040, A001358.

Programs

Maple

q:= n-> is(numtheory[sigma](n)>2*n):
a:= proc(n) option remember; local k, l;
      l:= [seq(a(i), i=1..n-1)]:
      for k while not andmap(j-> q(k+j), l) do od; k
    end:
seq(a(n), n=1..10);  # Alois P. Heinz, Apr 25 2025

Mathematica

a[1] = 1; a[n_] := a[n] = Module[{k = a[n-1] + 1}, While[AnyTrue[Array[a, n-1], DivisorSigma[-1, #+k] <= 2 &], k++]; k]; Array[a, 10] (* Amiram Eldar, Apr 26 2025 *)

PARI

isabundant(n) = (sigma(n) > 2*n) ;
isok(k, n, va) = for (i=1, n-1, if (! isabundant(k+va[i]), return(0));); return(1);
lista(nn) = my(va=vector(nn)); for (n=1, nn, my(k=if (n==1, 1, 1+va[n-1])); while(! isok(k, n, va), k++); k; va[n] = k;); va; \\ Michel Marcus, Apr 26 2025

Extensions

a(16) from Amiram Eldar, Apr 26 2025