%I #8 Mar 02 2016 11:18:15
%S 2,2,2,2,2,2,3,2,2,2,3,3,2,2,2,4,3,3,2,2,2,5,4,3,3,2,2,2,6,5,4,3,3,2,
%T 2,2,7,6,5,4,3,3,2,2,2,9,7,6,5,4,3,3,2,2,2,11,9,7,6,5,4,3,3,2,2,2,13,
%U 11,9,7,6,5,4,3,3,2,2,2,16,13,11,9,7,6,5,4,3,3,2,2,2,19,16,13,11,9,7,6,5
%N The first term greater than one on each row of A068009. a(n) = A068009[n, A002024[n]].
%C In row 1 of A068009 the first term > 1 is found at position 1, for rows 2 & 3 at position 2, for rows 4,5,6 at position 3, for rows 7,8,9,10 at position 4 etc., thus it is natural to view this also as a triangular table.
%H Alois P. Heinz, <a href="/A068049/b068049.txt">Table of n, a(n) for n = 1..1000</a>
%p [seq(A000009(A025581(j-1))+1,j=1..99)];
%p A025581 := n-> binomial(1+floor(1/2+sqrt(2+2*n)),2)-(n+1);
%p N := 100; t1 := series(mul(1+x^k,k=1..N),x,N); A000009 := proc(n) coeff(t1,x,n); end;
%t a[n_] := PartitionsQ[(1/2)(Floor[Sqrt[2n]+1/2]^2 + Floor[Sqrt[2n]+1/2] - 2n)] + 1; Array[a, 100] (* _Jean-François Alcover_, Mar 02 2016 *)
%Y a(n) = A000009(A025581(n-1))+1. Specifically, the left edge is equal to A000009[n]+1 (i.e. apart from the first term = A052839) and the right edge is all-2 sequence A007395.
%K nonn,tabl
%O 1,1
%A _Antti Karttunen_, Feb 11 2002