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 A156044 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 0
 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, 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, 30, 28, 26, 20, 13, 1 (list; table; graph; refs; listen; history; text; internal format)
 OFFSET 0,5 COMMENTS Row sums are: {1, 2, 4, 6, 11, 16, 31, 46, 82, 126, 206,...}. The idea here was tom design a symmetrical analog of a Pascal's triangle using p partitions and addition and subtraction instead of multiplication. LINKS FORMULA t0(n,m)=2 + PartitionsP[n] - PartitionsP[m] - PartitionsP[n - m]; t(n,m)=(t0(n,m)+Reverse[t0(n,m)])/2 EXAMPLE {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, 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, 30, 28, 26, 20, 13, 1} MATHEMATICA Clear[t]; t[n_, m_] = 2 + PartitionsP[n] - PartitionsP[m] - PartitionsP[n - m]; Table[(Table[t[n, m], {m, 0, n}] + Reverse[Table[t[n, m], {m, 0, n}]])/2, {n, 0, 10}]; Flatten[%] CROSSREFS Sequence in context: A103691 A103441 A081206 * A180980 A048570 A090806 Adjacent sequences:  A156041 A156042 A156043 * A156045 A156046 A156047 KEYWORD nonn,tabl,uned AUTHOR Roger L. Bagula, Feb 02 2009 STATUS approved

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