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A123402
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Combining the conditional divide-by-two concept from Collatz sequences with Pascal's triangle, one can construct 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 new term is divided by two if it is even.
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1
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4, 2, 2, 1, 2, 1, 1, 3, 3, 1, 1, 2, 3, 2, 1, 1, 3, 5, 5, 3, 1, 1, 2, 4, 5, 4, 2, 1, 1, 3, 3, 9, 9, 3, 3, 1, 1, 2, 3, 6, 9, 6, 3, 2, 1, 1, 3, 5, 9, 15, 15, 9, 5, 3, 1, 1, 2, 4, 7, 12, 15, 12, 7, 4, 2, 1, 1, 3, 3, 11, 19, 27, 27, 19, 11, 3, 3, 1, 1, 2, 3, 7, 15, 23, 27, 23, 15, 7, 3, 2, 1, 1, 3, 5, 5, 11, 19
(list; table; graph; refs; listen; history; internal format)
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OFFSET
| 0,1
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LINKS
| R. Kelly, Collatz-Pascal Triangle
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FORMULA
| 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).
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EXAMPLE
| For the row starting with (1,2,4,5,8,...) the subsequent row is computed as follows: 0+1->1, 1+2->3, (2+4)/2->3, 4+5->9, 5+8->13...
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MATHEMATICA
| CollatzPascalTriangle[init_, n_] := Module[{CPT, ROWA, ROWB, a, i, j}, If[ListQ[init], ROWA = init, ROWA = {4}]; CPT = {ROWA}; ROWA = Flatten[{0, ROWA, 0}]; For[i = 1, i < n, i++, ROWB = {0}; For[j = 1, j < Length[ROWA], j++, a = ROWA[[j]] + ROWA[[j + 1]]; a = a/(2 - Mod[a, 2]); ROWB = Append[ROWB, a]; ]; CPT = Append[CPT, Rest[ROWB]]; ROWA = Append[ROWB, 0]]; CPT] Flatten[ CollatzPascalTriangle[{4}, 20]]
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CROSSREFS
| Cf. A007318, A069202.
Sequence in context: A136737 A004551 A016511 * A205032 A088570 A102888
Adjacent sequences: A123399 A123400 A123401 * A123403 A123404 A123405
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KEYWORD
| easy,nonn,tabl
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AUTHOR
| Reed Kelly (math(AT)keldesign.com), Oct 14 2006
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