A156044 - OEIS

%I #2 Mar 30 2012 17:34:33

%S 1,1,1,1,2,1,1,2,2,1,1,3,3,3,1,1,3,4,4,3,1,1,5,6,7,6,5,1,1,5,8,9,9,8,

%T 5,1,1,8,11,14,14,14,11,8,1,1,9,15,18,20,20,18,15,9,1,1,13,20,26,28,

%U 30,28,26,20,13,1

%N A triangle sequence made symmetrical by reverse coefficients: t0(n,m)=2 + PartitionsP[n] - PartitionsP[m] - PartitionsP[n - m]; t(n,m)=(t0(n,m)+Reverse[t0(n,m)])/2

%C Row sums are:

%C {1, 2, 4, 6, 11, 16, 31, 46, 82, 126, 206,...}.

%C The idea here was tom design a symmetrical analog of a Pascal's triangle using p

%C partitions and addition and subtraction instead of multiplication.

%F t0(n,m)=2 + PartitionsP[n] - PartitionsP[m] - PartitionsP[n - m];

%F t(n,m)=(t0(n,m)+Reverse[t0(n,m)])/2

%e {1},

%e {1, 1},

%e {1, 2, 1},

%e {1, 2, 2, 1},

%e {1, 3, 3, 3, 1},

%e {1, 3, 4, 4, 3, 1},

%e {1, 5, 6, 7, 6, 5, 1},

%e {1, 5, 8, 9, 9, 8, 5, 1},

%e {1, 8, 11, 14, 14, 14, 11, 8, 1},

%e {1, 9, 15, 18, 20, 20, 18, 15, 9, 1},

%e {1, 13, 20, 26, 28, 30, 28, 26, 20, 13, 1}

%t Clear[t];

%t t[n_, m_] = 2 + PartitionsP[n] - PartitionsP[m] - PartitionsP[n - m];

%t Table[(Table[t[n, m], {m, 0, n}] + Reverse[Table[t[n, m], {m, 0, n}]])/2, {n, 0, 10}];

%t Flatten[%]

%K nonn,tabl,uned

%O 0,5

%A _Roger L. Bagula_, Feb 02 2009