A383398 - OEIS

%I #27 May 02 2025 19:40:05

%S 1,11,19,29,59,349,521,2071,66949,223231,3660191,4552181,5500081,

%T 10161979,12235619,47859629

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

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

%C 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).

%C 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).

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

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

%p q:= n-> is(numtheory[sigma](n)>2*n):

%p a:= proc(n) option remember; local k, l;

%p       l:= [seq(a(i), i=1..n-1)]:

%p       for k while not andmap(j-> q(k+j), l) do od; k

%p     end:

%p seq(a(n), n=1..10);  # _Alois P. Heinz_, Apr 25 2025

%t 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 *)

%o (PARI) isabundant(n) = (sigma(n) > 2*n) ;

%o isok(k, n, va) = for (i=1, n-1, if (! isabundant(k+va[i]), return(0));); return(1);

%o 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

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

%Y Cf. A000040, A001358.

%K nonn,hard,more

%O 1,2

%A _Jakub Buczak_, Apr 25 2025

%E a(16) from _Amiram Eldar_, Apr 26 2025