A335903 Column 1 in the matrix of A279212 (whose indexing starts at 0).
2, 6, 15, 37, 88, 204, 464, 1040, 2304, 5056, 11008, 23808, 51200, 109568, 233472, 495616, 1048576, 2211840, 4653056, 9764864, 20447232, 42729472, 89128960, 185597952, 385875968, 801112064, 1660944384, 3439329280, 7113539584, 14696841216, 30333206528, 62545461248, 128849018880, 265214230528, 545460846592
Offset: 1
Examples
a(17) = a(A233328(2)) = 1048576 = 2^20 = T(16, 1) = T(21, 0) in terms of matrix T of A279212; 2^20 is in column 1 of the 17th antidiagonal and in column 0 of the 21st antidiagonal of the matrix of A279212. A search for duplicates in A279212 through antidiagonal 2000 produced only pairs of powers of 2 in columns 0 and 1 of the matrix of A279212. Let k_0 and k_1 be the antidiagonals in columns 0 and 1, respectively, for the pair of the n-th duplicates. Since k_0 = 2 and k_1 = 1 for the duplicates of 2, the first pair in both columns, then k_0 = k_1 + 3*n - 2 for the n-th pair, n >=1. Table of duplicates in column 1 of the matrix of A279212 (the values for k_0 are one larger than the exponents in the left column of the table below because column 0 is sequence A011782): value of number of index in number antidiagonal A279212 ------------------------------------- 2^1 1 2 2^20 17 154 2^151 145 10586 2^1178 1169 683866 2^9373 9361 43818842 2^74912 74897 2804817754 2^599203 599185 179511631706 ... ... ... The central column of the table is A233328. The values for the first 4 antidiagonals were computed using sequence A279212, the ones larger than antidiagonal 2000 were determined by computing those n for which 7*n + 9 is a power of 2. The right column is n*(n+1)/2 + 1, where n is the number in the central column.
Links
- Index entries for linear recurrences with constant coefficients, signature (4,-4).
Programs
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Mathematica
a335903[1] = 2; a335903[2] = 6; a335903[n_] := (7n+9)*2^(n-4) Map[a335903, Range[35]] (* data *)
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PARI
Vec(x*(1 - x)*(2 - x^2) / (1 - 2*x)^2 + O(x^30)) \\ Colin Barker, Jun 29 2020
Formula
a(1) = 2, a(2) = 6, a(3) = 15, a(n) = 2 * a(n-1) + 7 * 2^(n-4), for n >= 4 (recursion for column 1 in the matrix of A279212).
a(1) = 2, a(2) = 6, a(n) = (7*n + 9) * 2^(n - 4), for n >= 3.
From Colin Barker, Jun 29 2020: (Start)
G.f.: x*(1 - x)*(2 - x^2) / (1 - 2*x)^2.
a(n) = 4*a(n-1) - 4*a(n-2) for n > 4.
(End)
Comments