cp's OEIS Frontend

A335903 Column 1 in the matrix of A279212 (whose indexing starts at 0).

%I A335903 #13 Sep 01 2023 04:07:20
%S A335903 2,6,15,37,88,204,464,1040,2304,5056,11008,23808,51200,109568,233472,
%T A335903 495616,1048576,2211840,4653056,9764864,20447232,42729472,89128960,
%U A335903 185597952,385875968,801112064,1660944384,3439329280,7113539584,14696841216,30333206528,62545461248,128849018880,265214230528,545460846592
%N A335903 Column 1 in the matrix of A279212 (whose indexing starts at 0).
%C A335903 Indexing for this sequence starts at 1 since then the index is the same as the number of the antidiagonal in the matrix for A279212 in which a number in column 1 of A279212 occurs.
%H A335903 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (4,-4).
%F A335903 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).
%F A335903 a(1) = 2, a(2) = 6, a(n) = (7*n + 9) * 2^(n - 4), for n >= 3.
%F A335903 From _Colin Barker_, Jun 29 2020: (Start)
%F A335903 G.f.: x*(1 - x)*(2 - x^2) / (1 - 2*x)^2.
%F A335903 a(n) = 4*a(n-1) - 4*a(n-2) for n > 4.
%F A335903 (End)
%e A335903 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.
%e A335903 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.
%e A335903 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):
%e A335903 value of    number of        index in
%e A335903 number      antidiagonal     A279212
%e A335903 -------------------------------------
%e A335903 2^1               1               2
%e A335903 2^20             17             154
%e A335903 2^151           145           10586
%e A335903 2^1178         1169          683866
%e A335903 2^9373         9361        43818842
%e A335903 2^74912       74897      2804817754
%e A335903 2^599203     599185    179511631706
%e A335903 ... ... ...
%e A335903 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.
%e A335903 The right column is n*(n+1)/2 + 1, where n is the number in the central column.
%t A335903 a335903[1] = 2; a335903[2] = 6; a335903[n_] := (7n+9)*2^(n-4)
%t A335903 Map[a335903, Range[35]]  (* data  *)
%o A335903 (PARI) Vec(x*(1 - x)*(2 - x^2) / (1 - 2*x)^2 + O(x^30)) \\ _Colin Barker_, Jun 29 2020
%Y A335903 Cf. A011782, A233328, A279212.
%K A335903 nonn,easy
%O A335903 1,1
%A A335903 _Hartmut F. W. Hoft_, Jun 29 2020