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A360099 To get A(n,k), replace 0's in the binary expansion of n with (-1) and interpret the result in base k; square array A(n,k), n>=0, k>=0, read by antidiagonals.

%I A360099 #25 Feb 03 2023 11:32:40
%S A360099 0,0,1,0,1,-1,0,1,0,1,0,1,1,2,-1,0,1,2,3,-1,1,0,1,3,4,1,1,-1,0,1,4,5,
%T A360099 5,3,1,1,0,1,5,6,11,7,5,3,-1,0,1,6,7,19,13,11,7,-2,1,0,1,7,8,29,21,19,
%U A360099 13,1,0,-1,0,1,8,9,41,31,29,21,14,3,0,1,0,1,9,10,55,43,41,31,43,16,5,2,-1
%N A360099 To get A(n,k), replace 0's in the binary expansion of n with (-1) and interpret the result in base k; square array A(n,k), n>=0, k>=0, read by antidiagonals.
%C A360099 The empty bit string is used as binary expansion of 0, so A(0,k) = 0.
%H A360099 Alois P. Heinz, <a href="/A360099/b360099.txt">Antidiagonals n = 0..200, flattened</a>
%F A360099 G.f. for column k satisfies g_k(x) = k*(x+1)*g_k(x^2) + x/(1+x).
%F A360099 A(n,k) = k*A(floor(n/2),k)+2*(n mod 2)-1 for n>0, A(0,k)=0.
%F A360099 A(n,k) mod 2 = A057427(n) if k is even.
%F A360099 A(n,k) mod 2 = A030300(n) if k is odd and n>=1.
%F A360099 A(2^(n+1),1) + n = 0.
%e A360099 Square array A(n,k) begins:
%e A360099    0,  0, 0,  0,  0,   0,   0,   0,   0,   0,   0, ...
%e A360099    1,  1, 1,  1,  1,   1,   1,   1,   1,   1,   1, ...
%e A360099   -1,  0, 1,  2,  3,   4,   5,   6,   7,   8,   9, ...
%e A360099    1,  2, 3,  4,  5,   6,   7,   8,   9,  10,  11, ...
%e A360099   -1, -1, 1,  5, 11,  19,  29,  41,  55,  71,  89, ...
%e A360099    1,  1, 3,  7, 13,  21,  31,  43,  57,  73,  91, ...
%e A360099   -1,  1, 5, 11, 19,  29,  41,  55,  71,  89, 109, ...
%e A360099    1,  3, 7, 13, 21,  31,  43,  57,  73,  91, 111, ...
%e A360099   -1, -2, 1, 14, 43,  94, 173, 286, 439, 638, 889, ...
%e A360099    1,  0, 3, 16, 45,  96, 175, 288, 441, 640, 891, ...
%e A360099   -1,  0, 5, 20, 51, 104, 185, 300, 455, 656, 909, ...
%p A360099 A:= proc(n, k) option remember; local m;
%p A360099      `if`(n=0, 0, k*A(iquo(n, 2, 'm'), k)+2*m-1)
%p A360099     end:
%p A360099 seq(seq(A(n, d-n), n=0..d), d=0..12);
%p A360099 # second Maple program:
%p A360099 A:= (n, k)-> (l-> add((2*l[i]-1)*k^(i-1), i=1..nops(l)))(Bits[Split](n)):
%p A360099 seq(seq(A(n, d-n), n=0..d), d=0..12);
%Y A360099 Columns k=0-6, 10 give: A062157, A145037, A006257, A147991, A147992, A153777, A147993, A359925.
%Y A360099 Rows n=0-10 give: A000004, A000012, A023443, A000027(k+1), A165900, A002061, A165900(k+1), A002061(k+1), A083074, A152618, A062158.
%Y A360099 Main diagonal gives A360096.
%Y A360099 Cf. A030300, A057427.
%K A360099 sign,tabl,look,base
%O A360099 0,14
%A A360099 _Alois P. Heinz_, Jan 25 2023