A123403 - OEIS

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.

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, Collatz-Pascal 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, 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).

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