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Search: id:A123403
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| A123403 |
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Combining the conditional divide-by-two concept from Collatz sequences with Pascal's triangle, we can arrive at a new kind of triangle. Start with an initial row of just 4. To compute subsequent rows, start by appending a zero to the beginning and end of the previous row. Like Pascal's triangle, add adjacent terms of the previous row to create each of the subsequent terms. The only change is that each term is divided by two if it is even. Then take the center of this triangle. In other words, take the n-th term from the (2n)th row. |
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| 4, 2, 3, 5, 9, 15, 27, 25, 47, 89, 107, 119, 241, 545, 699, 1471, 3313, 4288, 15661, 31739, 30813, 35143, 92101, 123614, 384815, 788429, 1532363, 2995379, 6281191, 13569969, 16900339, 26062940, 28141406, 57780803, 122540851, 263162577
(list; table; graph; listen)
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OFFSET
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1,1
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LINKS
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R. Kelly, Collatz-Pascal Triangle
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FORMULA
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Define a(n, m) for integers m, n: a(0, 0)=4, a(n, m) := 0 for m<0 and n<m, set x(n+1, m) = a(n, m)+a(n, m-1), if ( x(n+1, m) is even ), then a(n+1, m) = x(n+1, m)/2, otherwise a(n+1, m) = x(n+1, m). Now consider the terms a(2n, n).
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MATHEMATICA
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(*Returns the center row of the CPT*) CollatzPascalCenter[init_, n_] := Module[{CPT, CENTER, ROWA, ROWB, a, i, j}, If[ListQ[init], CPT = {init}, CPT = {{0, 4, 0}}]; CENTER = {4}; For[i = 1, i < n, i++, ROWA = CPT[[i]]; ROWB = {0}; For[j = 1, j < Length[ROWA], j++, a = ROWA[[j]] + ROWA[[j + 1]]; a = a/(2 - Mod[a, 2]); If[And[EvenQ[Length[ROWA]], (j == Length[ROWA]/2)], CENTER = Append[CENTER, a], ]; ROWB = Append[ROWB, a]; ]; ROWB = Append[ROWB, 0]; CPT = Append[CPT, ROWB]; ]; CENTER] CollatzPascalCenter[, 200]
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CROSSREFS
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Cf. A123402.
Sequence in context: A166016 A010646 A160079 this_sequence A098317 A095185 A128009
Adjacent sequences: A123400 A123401 A123402 this_sequence A123404 A123405 A123406
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KEYWORD
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easy,nonn,tabl
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AUTHOR
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Reed Kelly (math(AT)keldesign.com), Oct 14 2006
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