A327916 Triangle T(k, n) read by rows: Array A(k, n) = 2^k*(k + 1 + 2*n), k >= 0, n >= 0, read by antidiagonals upwards.
1, 4, 3, 12, 8, 5, 32, 20, 12, 7, 80, 48, 28, 16, 9, 192, 112, 64, 36, 20, 11, 448, 256, 144, 80, 44, 24, 13, 1024, 576, 320, 176, 96, 52, 28, 15, 2304, 1280, 704, 384, 208, 112, 60, 32, 17, 5120, 2816, 1536, 832, 448, 240, 128, 68, 36, 19, 11264, 6144, 3328, 1792, 960, 512, 272, 144, 76, 40, 21
Offset: 0
Examples
The triangle T(k, n) begins: k\n 0 1 2 3 4 5 6 7 8 9 10 ... ----------------------------------------------------- 0: 1 1: 4 3 2: 12 8 5 3: 32 20 12 7 4: 80 48 28 16 9 5: 192 112 64 36 20 11 6: 448 256 144 80 44 24 13 7: 1024 576 320 176 96 52 28 15 8: 2304 1280 704 384 208 112 60 32 17 9: 5120 2816 1536 832 448 240 128 68 36 19 10: 11264 6144 3328 1792 960 512 272 144 76 40 21 ...
Crossrefs
Column sequences without leading zeros are for n=0..9: A001787(n+1), A001792(n+1), A045623(n+2), A045891(n+3), A034007(n+4), A111297(n+3), A159694(n+1), A159695(n+1), A159696(n+1), A159697(n+1).
The sequence of (sub)diagonal k, for k >= 0, is the row k sequence of array A: {(k + 2*n + 1)*2^k}_{k >= 0}.
Row sums: A213569(k+1), k >= 0 (see the J. M. Bergot comments there).
Programs
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Mathematica
Table[2^#*(# + 1 + 2 n) &[k - n], {k, 0, 10}, {n, 0, k}] // Flatten (* Michael De Vlieger, Oct 03 2019 *)
Formula
Array A(k, n) = Sum_{j=0..k} binomial(k, j)*(2*(n+j) + 1) = 2^k*(k + 1+ 2*n), for k >= 0 and n >= 0.
Triangle T(k, n) = A(k-n, n) = 2^(k-n)*(k + n + 1), n >= 0, k = 0..n.
Recurrence: T(k, 0) = (k+1)*2^k = A001787(k+1), for k >= 0, and T(k, n) = T(k, n-1) - T(k-1, n-1), for n >= 1, k >= 1, with T(k, n) = 0 if k < n.
O.g.f. for row polynomials: G(z,x) = Sum_{n=0..k} R(k, x)*z^n =
(1 + x*z*(1 - 4*z))/((1 - 2*z)^2*(1 - x*z)^2).
T(k, 0) = Sum_{n=0..k} binomial(k,n)*T(n, n), k >= 0 (binomial transform).
Extensions
Definition corrected by Georg Fischer, Jul 13 2023
Comments