A176698 A symmetrical sum triangle sequence:a(n)=vector(a(n-1)).Reverse(vector(a(n-1));a(0)=1;a(1)=2;t(n,m)=2+a(n)-a(m)-a(n-m).
1, 1, 1, 1, 2, 1, 1, 8, 8, 1, 1, 28, 34, 28, 1, 1, 104, 130, 130, 104, 1, 1, 400, 502, 522, 502, 400, 1, 1, 1584, 1982, 2078, 2078, 1982, 1584, 1, 1, 6416, 7998, 8390, 8466, 8390, 7998, 6416, 1, 1, 26464, 32878, 34454, 34826, 34826, 34454, 32878, 26464, 1, 1
Offset: 0
Examples
{1}, {1, 1}, {1, 2, 1}, {1, 8, 8, 1}, {1, 28, 34, 28, 1}, {1, 104, 130, 130, 104, 1}, {1, 400, 502, 522, 502, 400, 1}, {1, 1584, 1982, 2078, 2078, 1982, 1584, 1}, {1, 6416, 7998, 8390, 8466, 8390, 7998, 6416, 1}, {1, 26464, 32878, 34454, 34826, 34826, 34454, 32878, 26464, 1}, {1, 110784, 137246, 143654, 145210, 145506, 145210, 143654, 137246, 110784, 1}
Crossrefs
Cf. A025227
Programs
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Mathematica
a[0] := 1; a[1] := 2; a[n_] := a[n] = Table[a[i], {i, 0, n - 1}].Table[a[n - 1 - i], {i, 0, n - 1}]; t[n_, m_] := 2 + (-a[m] - a[n - m] + a[n]); Table[Table[t[n, m], {m, 0, n}], {n, 0, 10}]
Formula
a(n)=vector(a(n-1)).Reverse(vector(a(n-1));
a(0)=1;a(1)=2;\q t(n,m)=2+a(n)-a(m)-a(n-m)
Comments