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 A123402 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. 2
 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; text; internal format)
 OFFSET 0,1 LINKS Reed Kelly, Collatz-Pascal Triangle Reed Kelly, Collatz-Pascal Triangle, Oct 12 2006 [Local copy] FORMULA Define a(n, m) for integers m, n: a(0, 0)=4, a(n, m) := 0 for m<0 and n1, 1+2->3, (2+4)/2->3, 4+5->9, 5+8->13... 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]] CROSSREFS Cf. A007318, A069202. Sequence in context: A285001 A016511 A250623 * A205032 A264752 A325527 Adjacent sequences: A123399 A123400 A123401 * A123403 A123404 A123405 KEYWORD easy,nonn,tabl AUTHOR Reed Kelly, Oct 14 2006 STATUS approved

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Last modified March 26 14:32 EDT 2023. Contains 361549 sequences. (Running on oeis4.)