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 A176228 A symmetrical triangle sequence:t(n,m)=Binomial[n, m] + Fibonacci[n] + 1 0
 2, 3, 3, 3, 4, 3, 4, 6, 6, 4, 5, 8, 10, 8, 5, 7, 11, 16, 16, 11, 7, 10, 15, 24, 29, 24, 15, 10, 15, 21, 35, 49, 49, 35, 21, 15, 23, 30, 50, 78, 92, 78, 50, 30, 23, 36, 44, 71, 119, 161, 161, 119, 71, 44, 36, 57, 66, 101, 176, 266, 308, 266, 176, 101, 66, 57 (list; table; graph; refs; listen; history; text; internal format)
 OFFSET 0,1 COMMENTS Row sums are: {2, 6, 10, 20, 36, 68, 127, 240, 454, 862, 1640,...}. The sequence is designed as an leading ones "adjustable" sequence that will give Pascal's triangle. Replacing Fibonacci[n] with any a(n) will still adjust back to the original symmetrical triangle sequence. LINKS FORMULA t(n,m)=Binomial[n, m] + Fibonacci[n] + 1 EXAMPLE {2}, {3, 3}, {3, 4, 3}, {4, 6, 6, 4}, {5, 8, 10, 8, 5}, {7, 11, 16, 16, 11, 7}, {10, 15, 24, 29, 24, 15, 10}, {15, 21, 35, 49, 49, 35, 21, 15}, {23, 30, 50, 78, 92, 78, 50, 30, 23}, {36, 44, 71, 119, 161, 161, 119, 71, 44, 36}, {57, 66, 101, 176, 266, 308, 266, 176, 101, 66, 57} MATHEMATICA t[n_, m_] = Binomial[n, m] + Fibonacci[n] + 1; Table[Table[t[n, m], {m, 0, n}], {n, 0, 10}]; Flatten[%] CROSSREFS Sequence in context: A049837 A098201 A175239 * A129574 A130193 A084516 Adjacent sequences:  A176225 A176226 A176227 * A176229 A176230 A176231 KEYWORD nonn,tabl,uned AUTHOR Roger L. Bagula, Apr 12 2010 STATUS approved

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