login
A340660
A000079 is the first row. For the second row, subtract A001045. For the third row, subtract A001045 from the second one, etc. The resulting array is read by ascending antidiagonals.
1
1, 1, 2, 1, 1, 4, 1, 0, 3, 8, 1, -1, 2, 5, 16, 1, -2, 1, 2, 11, 32, 1, -3, 0, -1, 6, 21, 64, 1, -4, -1, -4, 1, 10, 43, 128, 1, -5, -2, -7, -4, -1, 22, 85, 256, 1, -6, -3, -10, -9, -12, 1, 42, 171, 512, 1, -7, -4, -13, -14, -23, -20, -1, 86, 341, 1024
OFFSET
0,3
COMMENTS
Every row has the signature (1,2).
(Among consequences: a(n) read by antidiagonals is
1,
1, 2,
1, 1, 4,
1, 0, 3, 8,
1, -1, 2, 5, 16
1, -2, 1, 2, 11, 32,
1, -3, 0, -1, 6, 21, 64,
... .
The row sums and their first two difference table terms are
1, 3, 6, 12, 23, 45, 88, ... = A086445(n+1) - 1
2, 3, 6, 11, 22, 43, 86, ... = A005578(n+2)
1, 3, 5, 11, 21, 43, 85, ... = A001045(n+2).
The antidiagonal sums are
b(n) = 1, 1, 3, 2, 5, 3, 9, 4, 15, 5, 27, 6, 49, 7, ... .)
FORMULA
A(n,k) = 2^k - n*round(2^k/3).
EXAMPLE
Square array:
1, 2, 4, 8, 16, 32, 64, 128, ... = A000079(n)
1, 1, 3, 5, 11, 21, 43, 85, ... = A001045(n+1)
1, 0, 2, 2, 6, 10, 22, 42, ... = A078008(n)
1, -1, 1, -1, 1, -1, 1, -1, ... = A033999(n)
1, -2, 0, -4, -4, -12, -20, -44, ... = -A084247(n)
1, -3, -1, -7, -9, -23, -41, -87, ... = (-1)^n*A140966(n+1)
1, -4, -2, -10, -14, -34, -62, -130, ... = -A135440(n)
1, -5, -3, -13, -19, -45, -83, -173, ... = -A155980(n+3) or -A171382(n+1)
...
MAPLE
A:= (n, k)-> (<<0|1>, <2|1>>^k. <<1, 2-n>>)[1$2]:
seq(seq(A(d-k, k), k=0..d), d=0..12); # Alois P. Heinz, Jan 21 2021
MATHEMATICA
A340660[m_, n_] := LinearRecurrence[{1, 2}, {1, m}, {n}]; Table[Reverse[Table[A340660[m, n + m - 2] // First, {m, 2, -n + 3, -1}]], {n, 1, 11}] // Flatten (* Robert P. P. McKone, Jan 28 2021 *)
PROG
(PARI) T(n, k) = 2^k - n*(2^k - (-1)^k)/3;
matrix(10, 10, n, k, T(n-1, k-1)) \\ Michel Marcus, Jan 19 2021
KEYWORD
sign,tabl,easy
AUTHOR
Paul Curtz, Jan 15 2021
STATUS
approved