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A123403 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.

%I A123403 #6 Feb 09 2023 14:48:48
%S A123403 4,2,3,5,9,15,27,25,47,89,107,119,241,545,699,1471,3313,4288,15661,
%T A123403 31739,30813,35143,92101,123614,384815,788429,1532363,2995379,6281191,
%U A123403 13569969,16900339,26062940,28141406,57780803,122540851,263162577
%N A123403 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.
%H A123403 Reed Kelly, <a href="https://web.archive.org/web/20210802052221/http://www.keldesign.com/math/CollatzPascal.pdf">Collatz-Pascal Triangle</a>
%F A123403 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).
%t A123403 (*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]
%Y A123403 Cf. A123402.
%K A123403 easy,nonn,tabl
%O A123403 1,1
%A A123403 _Reed Kelly_, Oct 14 2006