1,5

Row sums = A005993, Paraffin numbers: 1, 2, 6, 10, 19, 28, 44, 60...

Row sums are; {1, 2, 6, 10, 19, 28, 44, 60, 85, 110, 146,...}

Given the triangle of A003983, replace each of the terms by its square.

p(x,n)=Sum[x^i*If[i == Floor[n/2] && Mod[n, 2] == 0, 0, If[i <= ( then than equal) Floor[n/2], 2*i + 1, -(2*(n - i) + 1)]], {i, 0, n}]/(1 - x);

t(n,m)=coefficients(p(x,n),x)

The triangle of A003983 is:

1;

1, 1;

1, 2, 1;

1, 2, 2, 1;

1, 2, 3, 2, 1;

...

Replacing each term by its square, we get:

1;

1, 1;

1, 4, 1;

1, 4, 4, 1;

1, 4, 9, 4, 1;

...

{1},

{1, 1},

{1, 4, 1},

{1, 4, 4, 1},

{1, 4, 9, 4, 1},

{1, 4, 9, 9, 4, 1},

{1, 4, 9, 16, 9, 4, 1},

{1, 4, 9, 16, 16, 9, 4, 1},

{1, 4, 9, 16, 25, 16, 9, 4, 1},

{1, 4, 9, 16, 25, 25, 16, 9, 4, 1},

{1, 4, 9, 16, 25, 36, 25, 16, 9, 4, 1} (End)

Clear[p, n, i];

p[x_, n_] = Sum[x^i*If[i ==Floor[n/2] && Mod[n, 2] == 0, 0, If[i <= Floor[n/2], 2*i + 1, -(2*(n - i) + 1)]], {i, 0, n}]/(1 - x);

Table[CoefficientList[FullSimplify[p[x, n]], x], {n, 1, 11}];

Flatten[%]

Cf. A003983, A106314, A005993.

Sequence in context: A053239 A046569 A046596 this_sequence A152716 A110812 A151904

Adjacent sequences: A106311 A106312 A106313 this_sequence A106315 A106316 A106317

nonn,tabl

Gary W. Adamson (qntmpkt(AT)yahoo.com), Apr 28 2005

Additional comments from Roger L. Bagula and Gary W. Adamson (rlbagulatftn(AT)yahoo.com), Apr 02 2009