A369518 Array read by downward antidiagonals: A(n,k) = Sum_{j=0..k + (k mod 3) + 1} A(n-1,j) with A(0,k) = 1, n >= 0, k >= 0.
1, 1, 2, 1, 4, 6, 1, 6, 17, 23, 1, 5, 33, 80, 103, 1, 7, 24, 184, 408, 511, 1, 9, 41, 121, 1054, 2208, 2719, 1, 8, 63, 235, 643, 6196, 12486, 15205, 1, 10, 51, 411, 1363, 3571, 37244, 72992, 88197, 1, 12, 74, 309, 2625, 8057, 20543, 228092, 437821, 526018
Offset: 0
Examples
Array begins: ================================================== n\k| 0 1 2 3 4 5 6 ... ---+---------------------------------------------- 0 | 1 1 1 1 1 1 1 ... 1 | 2 4 6 5 7 9 8 ... 2 | 6 17 33 24 41 63 51 ... 3 | 23 80 184 121 235 411 309 ... 4 | 103 408 1054 643 1363 2625 1861 ... 5 | 511 2208 6196 3571 8057 16701 11296 ... 6 | 2719 12486 37244 20543 48540 106560 69376 ... ...
Links
- Terence Tao, Elegant recursion for A301897, answer to question on MathOverflow (2023).
Crossrefs
Column k=0 is A301897 (with different offset).
Programs
-
PARI
A(m, n=m)={my(r=vectorv(m+1), v=vector(n+3*m+1, k, 1)); r[1] = v[1..n+1]; for(i=1, m, v=vector(#v-3, k, sum(j=1, k + (k-1)%3 + 1, v[j])); r[1+i] = v[1..n+1]); Mat(r)} { A(6) }
Formula
A(n,3k) = A(n,3k-1) - A(n-1,3k+2), A(n,3k+1) = A(n,3k) + A(n-1,3k+2) + A(n-1,3k+3), A(n,3k+2) = A(n, 3k+1) + A(n-1,3k+4) + A(n-1,3k+5) with A(n,0) = A(n-1,0) + A(n-1,1), A(0,k) = 1. - Mikhail Kurkov, Nov 24 2024