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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.

%I A369518 #15 Nov 24 2024 10:08:47
%S A369518 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,
%T A369518 1054,2208,2719,1,8,63,235,643,6196,12486,15205,1,10,51,411,1363,3571,
%U A369518 37244,72992,88197,1,12,74,309,2625,8057,20543,228092,437821,526018
%N 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.
%H A369518 Terence Tao, <a href="https://mathoverflow.net/a/449471/231922">Elegant recursion for A301897</a>, answer to question on MathOverflow (2023).
%F A369518 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
%e A369518 Array begins:
%e A369518 ==================================================
%e A369518 n\k|    0     1     2     3     4      5     6 ...
%e A369518 ---+----------------------------------------------
%e A369518 0  |    1     1     1     1     1      1     1 ...
%e A369518 1  |    2     4     6     5     7      9     8 ...
%e A369518 2  |    6    17    33    24    41     63    51 ...
%e A369518 3  |   23    80   184   121   235    411   309 ...
%e A369518 4  |  103   408  1054   643  1363   2625  1861 ...
%e A369518 5  |  511  2208  6196  3571  8057  16701 11296 ...
%e A369518 6  | 2719 12486 37244 20543 48540 106560 69376 ...
%e A369518   ...
%o A369518 (PARI)
%o A369518 A(m, n=m)={my(r=vectorv(m+1), v=vector(n+3*m+1, k, 1)); r[1] = v[1..n+1];
%o A369518 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)}
%o A369518 { A(6) }
%Y A369518 Column k=0 is A301897 (with different offset).
%K A369518 nonn,tabl
%O A369518 0,3
%A A369518 _Mikhail Kurkov_, Jan 25 2024