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.

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

Offset

0,1

Links

Table of n, a(n) for n=0..96.

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 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).

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...

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 A368203 A375735

Adjacent sequences: A123399 A123400 A123401 * A123403 A123404 A123405

Keyword

easy,nonn,tabl

Author

Reed Kelly, Oct 14 2006

Status

approved