

A327283


Irregular triangle T(n,k) read by rows: "residual summands" in reduced Collatz sequences (see Comments for definition and explanation).


0



1, 1, 5, 1, 5, 19, 73, 347, 1, 7, 29, 103, 373, 1631, 1, 5, 23, 133, 1, 11, 1, 5, 19, 65, 451, 1, 7, 53, 1, 5, 31, 125, 503, 2533, 1, 1, 5, 19, 185, 1, 7, 29, 151, 581, 2255, 10861, 1, 5, 23, 85, 287, 925
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OFFSET

1,3


COMMENTS

Let R_s be the reduced Collatz sequence (cf. A259663) starting with s and let R_s(k), k >= 0 be the kth term in R_s. Then R_(2n1)(k) = (3^k*(2n1) + T(n,k))/2^j, where j is the total number of halving steps from R_(2n1)(0) to R_(2n1)(k). T(n,k) is defined here as the "residual summand".
The sequence without duplicates is a permutation of A116641.


LINKS

Table of n, a(n) for n=1..52.


FORMULA

T(n,k) = 2^j*R_(2n1)(k)  3^k*(2n1), as defined in Comments.
T(n,1) = 1; for k>1: T(n,k) = 3*T(n,k1) + 2^i, where i is the total number of halving steps from R_(2n1)(0) to R_(2n1)(k1).


EXAMPLE

Triangle starts:
1;
1, 5;
1;
1, 5, 19, 73, 347;
1, 7, 29, 103, 373, 1631;
1, 5, 23, 133;
1, 11;
1, 5, 19, 65, 451;
1, 7, 53;
1, 5, 31, 125, 503, 2533;
1;
1, 5, 19, 185;
1, 7, 29, 151, 581, 2255, 10861;
...
T(5,4)=103 because R_9(4) = 13; the number of halving steps from R_9(0) to R_9(4) is 6, and 13 = (81*9 + 103)/64.


CROSSREFS

Cf. A116623, A116641, A259663.
Sequence in context: A145825 A101692 A281105 * A105060 A229096 A290797
Adjacent sequences: A327280 A327281 A327282 * A327284 A327285 A327286


KEYWORD

nonn,tabf


AUTHOR

Bob Selcoe, Sep 15 2019


STATUS

approved



