

A123403


Combining the conditional dividebytwo 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 nth term from the (2n)th row.


0



4, 2, 3, 5, 9, 15, 27, 25, 47, 89, 107, 119, 241, 545, 699, 1471, 3313, 4288, 15661, 31739, 30813, 35143, 92101, 123614, 384815, 788429, 1532363, 2995379, 6281191, 13569969, 16900339, 26062940, 28141406, 57780803, 122540851, 263162577
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OFFSET

1,1


LINKS

Table of n, a(n) for n=1..36.
Reed Kelly, CollatzPascal Triangle


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, m1), 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).


MATHEMATICA

(*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]


CROSSREFS

Cf. A123402.
Sequence in context: A284307 A160079 A226939 * A276957 A275847 A243961
Adjacent sequences: A123400 A123401 A123402 * A123404 A123405 A123406


KEYWORD

easy,nonn,tabl


AUTHOR

Reed Kelly, Oct 14 2006


STATUS

approved



