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%I #25 Sep 23 2019 13:57:06
%S 1,1,5,1,5,19,73,347,1,7,29,103,373,1631,1,5,23,133,1,11,1,5,19,65,
%T 451,1,7,53,1,5,31,125,503,2533,1,1,5,19,185,1,7,29,151,581,2255,
%U 10861,1,5,23,85,287,925
%N Irregular triangle T(n,k) read by rows: "residual summands" in reduced Collatz sequences (see Comments for definition and explanation).
%C Let R_s be the reduced Collatz sequence (cf. A259663) starting with s and let R_s(k), k >= 0 be the k-th term in R_s. Then R_(2n-1)(k) = (3^k*(2n-1) + T(n,k))/2^j, where j is the total number of halving steps from R_(2n-1)(0) to R_(2n-1)(k). T(n,k) is defined here as the "residual summand".
%C The sequence without duplicates is a permutation of A116641.
%F T(n,k) = 2^j*R_(2n-1)(k) - 3^k*(2n-1), as defined in Comments.
%F T(n,1) = 1; for k>1: T(n,k) = 3*T(n,k-1) + 2^i, where i is the total number of halving steps from R_(2n-1)(0) to R_(2n-1)(k-1).
%e Triangle starts:
%e 1;
%e 1, 5;
%e 1;
%e 1, 5, 19, 73, 347;
%e 1, 7, 29, 103, 373, 1631;
%e 1, 5, 23, 133;
%e 1, 11;
%e 1, 5, 19, 65, 451;
%e 1, 7, 53;
%e 1, 5, 31, 125, 503, 2533;
%e 1;
%e 1, 5, 19, 185;
%e 1, 7, 29, 151, 581, 2255, 10861;
%e ...
%e 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.
%Y Cf. A116623, A116641, A259663.
%K nonn,tabf
%O 1,3
%A _Bob Selcoe_, Sep 15 2019
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