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%I #29 Nov 27 2020 04:54:51
%S 1,0,1,2,3,5,8,12,17,22,28,35,43,52,60,69,77,86,92,103,112,123,137,
%T 151,168,180,194,204,224,245,261,280,301,318,335,352,369,387,413,433,
%U 459,482,507,528,552,586,614,638,669,701,733,761,791,824,855,885,917,952,985,1020
%N a(0)=1, a(n) for n >= 1 is the number of distinct sums of two elements in [a(0), ..., a(n-1)], chosen without replacement.
%C a(n) <= A000217(n)-n for n >= 1.
%C Without replacement means a(i)+a(i) is not included. However, if a(i)=a(j), a(i)+a(j) still counts because they have two different indices. If you include a(i)+a(i), the sequence becomes A000012 (all ones).
%C If you include the distinct sums between 3 elements and more, you arrive at the sequence 1, 0, followed by A000079 (2^n).
%C Same rule as in A247184, but with a(0)=1.
%e a(1) gives the number of distinct sums between two elements of [1]. There aren't two elements so a(1)=0.
%e a(2) gives the number of distinct sums between two elements of [1,0]. The only sum are 1+0, so a(2) = 1.
%e a(3) gives the number of distinct sums between two elements of [1,0,1]. The two sums are 1+0 and 1+1 so a(3)=2.
%p s:= proc(n) option remember; `if`(n=0, {},
%p {s(n-1)[], seq(a(i)+a(n), i=0..n-1)})
%p end:
%p a:= proc(n) option remember;
%p `if`(n=0, 1, nops(s(n-1)))
%p end:
%p seq(a(n), n=0..60); # _Alois P. Heinz_, Nov 16 2020
%t a[0] = 1; a[1] = 0;
%t a[n_Integer?Positive] := a[n] = Length[Union[Total[Subsets[Array[a, n, 0], {2}], {2}]]];
%t Array[a, 61, 0] (* _Jan Mangaldan_, Nov 23 2020 *)
%o (PARI) my(v=[1], w=[], n=1); while(n<75, for(i=2, #v, w=concat(w,v[i-1]+v[#v])); w=vecsort(w,,8); v=concat(v, #w); n++); v
%Y Cf. A000217, A000012, A247184, A247185.
%K nonn
%O 0,4
%A _Derek Orr_, Nov 15 2020
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