OFFSET
1,1
COMMENTS
LINKS
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)
EXAMPLE
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.
MATHEMATICA
a335903[1] = 2; a335903[2] = 6; a335903[n_] := (7n+9)*2^(n-4)
Map[a335903, Range[35]] (* data *)
PROG
(PARI) Vec(x*(1 - x)*(2 - x^2) / (1 - 2*x)^2 + O(x^30)) \\ Colin Barker, Jun 29 2020
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Hartmut F. W. Hoft, Jun 29 2020
STATUS
approved