Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).
%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