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

%I A383398 #27 May 02 2025 19:40:05
%S A383398 1,11,19,29,59,349,521,2071,66949,223231,3660191,4552181,5500081,
%T A383398 10161979,12235619,47859629
%N A383398 a(n) is the smallest number whose sum with any previous term is abundant.
%C A383398 The terms are generally either prime or semiprime. This results in all known terms to be deficient (see A005100).
%C A383398 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 A383398 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 A383398 This is also equivalent to the sum of any 2 terms being an abundant number.
%e A383398 29 is a member, because 29+19, 29+11 and 29+1 are all abundant numbers.
%p A383398 q:= n-> is(numtheory[sigma](n)>2*n):
%p A383398 a:= proc(n) option remember; local k, l;
%p A383398       l:= [seq(a(i), i=1..n-1)]:
%p A383398       for k while not andmap(j-> q(k+j), l) do od; k
%p A383398     end:
%p A383398 seq(a(n), n=1..10);  # _Alois P. Heinz_, Apr 25 2025
%t A383398 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 A383398 (PARI) isabundant(n) = (sigma(n) > 2*n) ;
%o A383398 isok(k, n, va) = for (i=1, n-1, if (! isabundant(k+va[i]), return(0));); return(1);
%o A383398 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 A383398 Cf. A383399, A005101, A005231, A005100, A173490.
%Y A383398 Cf. A000040, A001358.
%K A383398 nonn,hard,more
%O A383398 1,2
%A A383398 _Jakub Buczak_, Apr 25 2025
%E A383398 a(16) from _Amiram Eldar_, Apr 26 2025