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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.

%I A123402 #11 Feb 09 2023 22:23:02
%S A123402 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,
%T A123402 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,
%U A123402 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
%N 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.
%H A123402 Reed Kelly, <a href="https://web.archive.org/web/20210802052221/http://www.keldesign.com/math/CollatzPascal.pdf">Collatz-Pascal Triangle</a>
%H A123402 Reed Kelly, <a href="/A123402/a123402.pdf">Collatz-Pascal Triangle</a>, Oct 12 2006 [Local copy]
%F A123402 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).
%e A123402 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...
%t A123402 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]]
%Y A123402 Cf. A007318, A069202.
%K A123402 easy,nonn,tabl
%O A123402 0,1
%A A123402 _Reed Kelly_, Oct 14 2006