%I #8 Sep 17 2023 02:02:48
%S 1,1,4,8,18,30,58,90,153,233,365,533,806,1142,1652,2308,3243,4431,
%T 6103,8203,11080,14710,19540,25612,33612,43570,56476,72548,93080,
%U 118490,150699,190315,240046,301042,376887,469515,583993,723073,893815,1100615,1352888
%N a(n) = Sum_{k=0..n} p(k) where the p(k) are the partial sums of row n of A365676.
%C If one reverses row n in the definition formula before accumulating, one gets A000070.
%o (Python) # Using function A365676Row from A365676.
%o from itertools import accumulate
%o def A365675List(size: int) -> list[int]:
%o return [sum(accumulate(A365676Row(n))) for n in range(size)]
%o print(A365675List(41))
%Y Cf. A365676, A000070.
%K nonn
%O 0,3
%A _Peter Luschny_, Sep 16 2023