A153777 -id:A153777 - cp's OEIS frontend

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.

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, 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, 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

Offset: 0

Alois P. Heinz, Jan 25 2023

The empty bit string is used as binary expansion of 0, so A(0,k) = 0.

			Square array A(n,k) begins:
   0,  0, 0,  0,  0,   0,   0,   0,   0,   0,   0, ...
   1,  1, 1,  1,  1,   1,   1,   1,   1,   1,   1, ...
  -1,  0, 1,  2,  3,   4,   5,   6,   7,   8,   9, ...
   1,  2, 3,  4,  5,   6,   7,   8,   9,  10,  11, ...
  -1, -1, 1,  5, 11,  19,  29,  41,  55,  71,  89, ...
   1,  1, 3,  7, 13,  21,  31,  43,  57,  73,  91, ...
  -1,  1, 5, 11, 19,  29,  41,  55,  71,  89, 109, ...
   1,  3, 7, 13, 21,  31,  43,  57,  73,  91, 111, ...
  -1, -2, 1, 14, 43,  94, 173, 286, 439, 638, 889, ...
   1,  0, 3, 16, 45,  96, 175, 288, 441, 640, 891, ...
  -1,  0, 5, 20, 51, 104, 185, 300, 455, 656, 909, ...

Alois P. Heinz, Antidiagonals n = 0..200, flattened

Columns k=0-6, 10 give: A062157, A145037, A006257, A147991, A147992, A153777, A147993, A359925.
Rows n=0-10 give: A000004, A000012, A023443, A000027(k+1), A165900, A002061, A165900(k+1), A002061(k+1), A083074, A152618, A062158.
Main diagonal gives A360096.
Cf. A030300, A057427.

A:= proc(n, k) option remember; local m;
     `if`(n=0, 0, k*A(iquo(n, 2, 'm'), k)+2*m-1)
    end:
seq(seq(A(n, d-n), n=0..d), d=0..12);
# second Maple program:
A:= (n, k)-> (l-> add((2*l[i]-1)*k^(i-1), i=1..nops(l)))(Bits[Split](n)):
seq(seq(A(n, d-n), n=0..d), d=0..12);

G.f. for column k satisfies g_k(x) = k*(x+1)*g_k(x^2) + x/(1+x).
A(n,k) = k*A(floor(n/2),k)+2*(n mod 2)-1 for n>0, A(0,k)=0.
A(n,k) mod 2 = A057427(n) if k is even.
A(n,k) mod 2 = A030300(n) if k is odd and n>=1.
A(2^(n+1),1) + n = 0.

A359925 Numbers with easy multiplication table - the first 9 multiples of these numbers can be derived by either incrementing or decrementing the corresponding digits from the previous multiple.

Original entry on oeis.org

1, 9, 11, 89, 91, 109, 111, 889, 891, 909, 911, 1089, 1091, 1109, 1111, 8889, 8891, 8909, 8911, 9089, 9091, 9109, 9111, 10889, 10891, 10909, 10911, 11089, 11091, 11109, 11111, 88889, 88891, 88909, 88911, 89089, 89091, 89109, 89111, 90889, 90891, 90909, 90911

Offset: 1

Kiran Ananthpur Bacche, Jan 25 2023

This is also the list of numbers having exactly one dot or one antidot in each box in the Decimal Exploding Dots notation.

			a(4) = 89. The first nine multiples of 89 are {089, 178, 267, 356, 445, 534, 623, 712, 801}. The digits in the hundreds place increment by 1, while the digits in the tens and units place decrement by 1. In the Decimal Exploding Dots notation, 89 is represented as DOT-ANTIDOT-ANTIDOT = 100 - 10 - 1 = 89

Global Math Week, Exploding Dots.
Gevorg Hmayakyan, Generalizing a Trig Identity [mentions this sequence]

Cf. A006257, A147991, A147992, A153777, A147993, A256290.

Column k=10 of A360099.

a:= proc(n) option remember;
      `if`(n=0, 0, 10*a(iquo(n, 2, 'm'))+2*m-1)
    end:
seq(a(n), n=1..44);   # Alois P. Heinz, Jan 25 2023

a(n) = 10*a(floor(n/2))+2*(n mod 2)-1 for n>0, a(0)=0. - Alois P. Heinz, Jan 25 2023

a(n) = 2*A256290(n-1) + 1 for n>1. - Hugo Pfoertner, Jan 28 2023

cp's OEIS Frontend

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.

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