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%I #25 Feb 03 2023 11:32:40
%S 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 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 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 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 The empty bit string is used as binary expansion of 0, so A(0,k) = 0.
%H Alois P. Heinz, <a href="/A360099/b360099.txt">Antidiagonals n = 0..200, flattened</a>
%F G.f. for column k satisfies g_k(x) = k*(x+1)*g_k(x^2) + x/(1+x).
%F A(n,k) = k*A(floor(n/2),k)+2*(n mod 2)-1 for n>0, A(0,k)=0.
%F A(n,k) mod 2 = A057427(n) if k is even.
%F A(n,k) mod 2 = A030300(n) if k is odd and n>=1.
%F A(2^(n+1),1) + n = 0.
%e Square array A(n,k) begins:
%e 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...
%e 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...
%e -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, ...
%e 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, ...
%e -1, -1, 1, 5, 11, 19, 29, 41, 55, 71, 89, ...
%e 1, 1, 3, 7, 13, 21, 31, 43, 57, 73, 91, ...
%e -1, 1, 5, 11, 19, 29, 41, 55, 71, 89, 109, ...
%e 1, 3, 7, 13, 21, 31, 43, 57, 73, 91, 111, ...
%e -1, -2, 1, 14, 43, 94, 173, 286, 439, 638, 889, ...
%e 1, 0, 3, 16, 45, 96, 175, 288, 441, 640, 891, ...
%e -1, 0, 5, 20, 51, 104, 185, 300, 455, 656, 909, ...
%p A:= proc(n, k) option remember; local m;
%p `if`(n=0, 0, k*A(iquo(n, 2, 'm'), k)+2*m-1)
%p end:
%p seq(seq(A(n, d-n), n=0..d), d=0..12);
%p # second Maple program:
%p A:= (n, k)-> (l-> add((2*l[i]-1)*k^(i-1), i=1..nops(l)))(Bits[Split](n)):
%p seq(seq(A(n, d-n), n=0..d), d=0..12);
%Y Columns k=0-6, 10 give: A062157, A145037, A006257, A147991, A147992, A153777, A147993, A359925.
%Y Rows n=0-10 give: A000004, A000012, A023443, A000027(k+1), A165900, A002061, A165900(k+1), A002061(k+1), A083074, A152618, A062158.
%Y Main diagonal gives A360096.
%Y Cf. A030300, A057427.
%K sign,tabl,look,base
%O 0,14
%A _Alois P. Heinz_, Jan 25 2023
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