cp's OEIS Frontend

A182289 Triangle read by rows. Let p be one of the parts of size A135010(n,k) in one of the partitions of n and S(n,k) = sum of all preceding parts to p in the mentioned partition of n. So T(n,k) = 2*S(n,k) + A135010(n,k).

%I A182289 #34 Aug 28 2012 15:55:59
%S A182289 1,3,2,5,5,3,7,7,7,6,2,4,9,9,9,9,9,8,3,5,11,11,11,11,11,11,11,10,6,2,
%T A182289 10,4,9,3,6,13,13,13,13,13,13,13,13,13,13,13,12,8,3,12,5,11,4,7,15,15,
%U A182289 15,15,15,15,15,15,15,15,15,15,15,15,15,14,10,6
%N A182289 Triangle read by rows. Let p be one of the parts of size A135010(n,k) in one of the partitions of n and S(n,k) = sum of all preceding parts to p in the mentioned partition of n. So T(n,k) = 2*S(n,k) + A135010(n,k).
%C A182289 Consider a physical model of the partitions of n in which each part p of size A135010(n,j) is represented by a right circular cylinder with radius j and height 2. T(n,k) is also the distance (or coordinate X) from the axis Y to the center of the base of cylinder of the part p in the structure of A135010.
%e A182289 Written as an irregular triangle the sequence begins:
%e A182289 1;
%e A182289 3,2;
%e A182289 5,5,3;
%e A182289 7,7,7,6,2,4;
%e A182289 9,9,9,9,9,8,3,5;
%e A182289 11,11,11,11,11,11,11,10,6,2,10,4,9,3,6;
%e A182289 13,13,13,13,13,13,13,13,13,13,13,12,8,3,12,5,11,4,7;
%e A182289 15,15,15,15,15,15,15,15,15,15,15,15,15,15,15,14,10,6,2,14,10,4,14,9,3,14,6,13,5,10,4,8;
%Y A182289 Row n starts with A000041(n-1) terms equal to A005408(n-1). Row n has length A138137(n). Right border gives A000027.
%Y A182289 Cf. A135010.
%K A182289 nonn,tabf
%O A182289 1,2
%A A182289 _Omar E. Pol_, Aug 14 2012