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Search: id:A078122
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| A078122 |
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Infinite lower triangular matrix, M, that satisfies [M^3](i,j) = M(i+1,j+1) for all i,j>=0 where [M^n](i,j) denotes the element at row i, column j, of the n-th power of matrix M, with M(0,k)=1 and M(k,k)=1 for all k>=0. |
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19
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| 1, 1, 1, 1, 3, 1, 1, 12, 9, 1, 1, 93, 117, 27, 1, 1, 1632, 3033, 1080, 81, 1, 1, 68457, 177507, 86373, 9801, 243, 1, 1, 7112055, 24975171, 15562314, 2371761, 88452, 729, 1, 1, 1879090014, 8786827629, 6734916423, 1291958181, 64392813, 796797, 2187, 1
(list; table; graph; listen)
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OFFSET
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0,5
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COMMENT
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M also satisfies: [M^(3k)](i,j) = [M^k](i+1,j+1) for all i,j,k >=0; thus [M^(3^n)](i,j) = M(i+n,j+n) for all n >= 0.
Conjecture: the sum of the n-th row equals the number of partitions of 3^n into powers of 3 (A078125).
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FORMULA
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M(1, j)=A078124(j), M(j+1, j)=3^j, M(j+2, j)=A016142(j).
M(n, k) = the coefficient of x^(3^n - 3^(n-k)) in the power series expansion of 1/Product_{j=0..n-k}(1-x^(3^j)) whenever 0<=k<n for all n>0 (conjecture).
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EXAMPLE
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The cube of the matrix is the same matrix excluding the first row and column:
[1,_0,_0,0]^3=[_1,__0,_0,_0]
[1,_1,_0,0]___[_3,__1,_0,_0]
[1,_3,_1,0]___[12,__9,_1,_0]
[1,12,_9,1]___[93,117,27,_1]
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MATHEMATICA
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m[i_, j_] := m[i, j]=If[j==0||i==j, 1, m3[i-1, j-1]]; m2[i_, j_] := m2[i, j]=Sum[m[i, k]m[k, j], {k, j, i}]; m3[i_, j_] := m3[i, j]=Sum[m[i, k]m2[k, j], {k, j, i}]; Flatten[Table[m[i, j], {i, 0, 8}, {j, 0, i}]]
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CROSSREFS
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Cf. A078121, A078123, A078124, A078125, A016142.
Sequence in context: A113340 A134523 A098778 this_sequence A128592 A156584 A129619
Adjacent sequences: A078119 A078120 A078121 this_sequence A078123 A078124 A078125
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KEYWORD
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nonn,tabl
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AUTHOR
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Paul D. Hanna (pauldhanna(AT)juno.com), Nov 18 2002
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