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 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 (list; graph; refs; listen; history; text; internal format)
 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 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". The sequence without duplicates is a permutation of A116641. LINKS FORMULA T(n,k) = 2^j*R_(2n-1)(k) - 3^k*(2n-1), as defined in Comments. 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). 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

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Last modified May 27 16:48 EDT 2022. Contains 354110 sequences. (Running on oeis4.)