A164316 -id:A164316 - cp's OEIS frontend

A356996 a(n) = b(n) - b(b(n)) - b(n - b(n)) for n >= 3, where b(n) = A356989(n).

0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 2, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 3, 2, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 3, 4, 3, 2, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0

Offset: 3

Peter Bala, Sep 10 2022

The sequence appears to consist of blocks of terms of the form 1, 2, 3, ..., A(k) - 1, A(k), A(k) - 1, ..., 3, 2, 1, where A(k) = A000930(k), separated by blocks of consecutive zeros.

The sequence of local peak values of the line graph of the sequence {a(n)} begins 1, 1, 1, 2, 3, 4, 6, 9, 13, 19, ..., conjecturally A000930; the local peaks occur at abscissa values n = 8, 12, 17, 25, 37, 54, 79, 116, 170, 249, ..., conjecturally {A179070(k): k >= 7}. Cf. A356995 and A356997.

			Sequence arranged as an irregular triangle; after the first row of zeros the row lengths are conjecturally equal to A164316(k) for k >= 2.
0, 0, 0, 0, 0;
1, 0, 0, 0;
1, 0, 0, 0, 0;
1, 0, 0, 0, 0, 0, 0;
1, 2, 1, 0, 0, 0, 0, 0, 0, 0, 0;
1, 2, 3, 2, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0;
1, 2, 3, 4, 3, 2, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0;
1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0;
...

Cf. A000930, A164316, A179070, A356989, A356995, A356997.

# b(n) = A356989
b := proc(n) option remember; if n = 1 then 1 else n - b(b(b(n - b(b(b(b(n-1))))))) end if; end proc:
seq(b(n) - b(b(n)) - b(n - b(n)), n = 3..300);

a(n+1) - a(n) belongs to {1, 0, -1}.

A365746 Table read by antidiagonals upward: T(n,k) is the number of binary strings of length k with the property that every substring of length A070939(n) is lexicographically earlier than the binary expansion of n; n, k >= 0.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 2, 2, 1, 0, 1, 2, 3, 2, 1, 0, 1, 2, 4, 5, 2, 1, 0, 1, 2, 4, 4, 8, 2, 1, 0, 1, 2, 4, 5, 4, 13, 2, 1, 0, 1, 2, 4, 6, 7, 4, 21, 2, 1, 0, 1, 2, 4, 7, 10, 11, 4, 34, 2, 1, 0, 1, 2, 4, 8, 13, 16, 16, 4, 55, 2, 1, 0, 1, 2, 4, 8, 8, 24

Offset: 0

Peter Kagey, Sep 17 2023

			Table begins:
 n\k | 0  1  2  3   4   5   6   7    8    9   10   11
-----+----------------------------------------------------
   0 | 1, 0, 0, 0,  0,  0,  0,  0,   0,   0,   0,   0, ...
   1 | 1, 1, 1, 1,  1,  1,  1,  1,   1,   1,   1,   1, ...
   2 | 1, 2, 2, 2,  2,  2,  2,  2,   2,   2,   2,   2, ...
   3 | 1, 2, 3, 5,  8, 13, 21, 34,  55,  89, 144, 233, ...
   4 | 1, 2, 4, 4,  4,  4,  4,  4,   4,   4,   4,   4, ...
   5 | 1, 2, 4, 5,  7, 11, 16, 23,  34,  50,  73, 107, ...
   6 | 1, 2, 4, 6, 10, 16, 26, 42,  68, 110, 178, 288, ...
   7 | 1, 2, 4, 7, 13, 24, 44, 81, 149, 274, 504, 927, ...
   8 | 1, 2, 4, 8,  8,  8,  8,  8,   8,   8,   8,   8, ...
   9 | 1, 2, 4, 8,  9, 11, 15, 23,  32,  43,  58,  81, ...
For (n,k) = (3,4), we see that T(3,4) = 8 because there are 8 binary strings of length k = 4 where all length A070939(3) = 2 substrings are lexicographically earlier than "11" (the binary expansion of n = 3): 0000, 0001, 0010, 0100, 0101, 1000, 1001, and 1010.

Cf. A030190, A070939, A209972.

Cf. A000045 (row 3), A164316 (row 5), A128588 (row 6), A000073 (row 7).

A365746Row[s_,
  numberOfTerms_] := (digits = If[s == 0, 1, Ceiling[Log[2, s + 1]]];
  m = 2^(digits - 1);
  transferMatrix =
   If[s == 0, {{0}},
    Table[If[(Ceiling[i/2] ==
         j) || ((i <= s - m) && (Ceiling[i/2] == j - m/2)), 1, 0], {i,
       1, m}, {j, 1, m}]];
  sequence =
   Table[2^k, {k, 0, digits - 1}] ~Join~
    Table[MatrixPower[transferMatrix, k] // Total // Total, {k, 1,
      numberOfTerms - digits}];
  Take[sequence, numberOfTerms])

G.f. for row n = 0: 1;
G.f. for row n = 1: 1/(1 - x);
G.f. for row n = 2: (1 + x)/(1 - x);
G.f. for row n = 3: (1 + x)/(1 - x - x^2);
G.f. for row n = 4: (1 + x + 2x^2)/(1 - x);
G.f. for row n = 5: (1 + x + 2x^2)/(1 - x - x^3);
G.f. for row n = 6: (1 + x + x^2)/(1 - x - x^2);
G.f. for row n = 7: (1 + x + x^2)/(1 - x - x^2 - x^3);
G.f. for row n = 8: (1 + x + 2 x^2 + 4 x^3)/(1 - x);
G.f. for row n = 9: (1 + x + 2x^2 + 4x^3)/(1 - x - x^4).

cp's OEIS Frontend

A356996 a(n) = b(n) - b(b(n)) - b(n - b(n)) for n >= 3, where b(n) = A356989(n).

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