cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

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

Original entry on oeis.org

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

Views

Author

Peter Bala, Sep 10 2022

Keywords

Comments

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.

Examples

			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;
...
		

Crossrefs

Programs

  • Maple
    # 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);

Formula

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

A356988 a(n) = n - a^[2](n - a^[3](n-1)) with a(1) = 1, where a^[2](n) = a(a(n)) and a^[3](n) = a(a(a(n))).

Original entry on oeis.org

1, 1, 2, 3, 3, 4, 5, 5, 6, 7, 8, 8, 8, 9, 10, 11, 12, 13, 13, 13, 13, 14, 15, 16, 17, 18, 19, 20, 21, 21, 21, 21, 21, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 34, 34, 34, 34, 34, 34, 34, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 55, 55, 55, 55, 55
Offset: 1

Views

Author

Peter Bala, Sep 08 2022

Keywords

Comments

This is the second sequence in a family of nested-recurrent sequences with apparently similar structure defined as follows. Given a sequence s = {s(n) : n >= 1} we define the k-th iterated sequence s^[k] by putting s^[1](n) = s(n) and setting s^[k](n) = s^[k-1](s(n)) for k >= 2. For k >= 1, we define a nested-recurrent sequence, dependent on k, by putting u(1) = 1 and setting u(n) = n - u^[k](n - u^[k+1](n-1)) for n >= 2. This is the case k = 2. For other cases see A006165 (k = 1), A356989 (k = 3) and A356990 (k = 4).
The sequence is slow, that is, for n >= 1, a(n+1) - a(n) is either 0 or 1. The sequence is unbounded.
The line graph of the sequence {a(n)} thus consists of a series of plateaus (where the value of the ordinate a(n) remains constant as n increases) joined by lines of slope 1.
The sequence of plateau heights begins 3, 5, 8, 13, 21, 34, 55, ..., the Fibonacci numbers A000045.
The plateaus start at abscissa values n = 4, 7, 11, 18, 29, 47, 76, ..., the Lucas numbers A000032, and finish at abscissa values n = 5, 8, 13, 21, 34, 55, 89, ..., the Fibonacci numbers. The sequence of plateau lengths 1, 1, 2, 3, 5, 8, 13, ... is thus the Fibonacci sequence.
The iterated sequences{a^[k](n) : n >= 1}, k = 2, 3,..., share similar properties to the present sequence. See the Example section below.

Examples

			Related sequences:
1) The square of the sequence: {a^[2](n) : n >= 1} = {a(a(n)) : n >= 1}. The first few terms are
  1, 1, 1, 2, 2, 3, 3, 3, 4, 5, 5, 5, 5, 6, 7, 8, 8, 8, 8, 8, 8, 9, 10, 11, 12, 13, 13, 13, 13, 13, 13, 13, 13, 13, 14, 15, 16, 17, 18, 19, 20, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 35, 36, 37, 38, 39, ...
The sequence is slow. The line graph of the sequence has plateaus of height Fibonacci(k), k >= 2, starting at abscissa value 2*Fibonacci(k) and ending at abscissa Fibonacci(k+2).
2) The cube of the sequence: {a^[3](n) : n >= 1} = {a(a(a(n))) : n >= 1}. The first few terms are
  1, 1, 1, 1, 1, 2, 2, 2, 3, 3, 3, 3, 3, 4, 5, 5, 5, 5, 5, 5, 5, 6, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 9, 10, 11, 12, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 14, 15, 16, 17, 18, 19, 20, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 22, 23, 24, 25, 26, 27, ...
The line graph of the sequence has plateaus of height Fibonacci(k), k >= 2, starting at abscissa value 3*Fibonacci(k) and ending at abscissa Fibonacci(k+3).
		

Crossrefs

Programs

  • Maple
    a := proc(n) option remember; if n = 1 then 1 else n - a(a(n - a(a(a(n-1))))) end if; end proc:
    seq(a(n), n = 1..100);

Formula

a(n+1) - a(n) = 0 or 1.
The terms of the sequence are completely determined by the following two results:
a) for n >= 2, a(L(n-1) + j) = F(n) for 0 <= j <= F(n-3), where F(n) = A000045(n), the n-th Fibonacci number with F(-1) = 1 and L(n) = A000032(n), the n-th Lucas number;
b) for n >= 2, a(F(n+1) + j) = F(n) + j for 0 <= j <= F(n-1).
Hence a(F(n+2)) = a(F(n+1)) + a(F(n)) for n >= 2 and a(L(n+2)) = a(L(n+1)) + a(L(n)) for n >= 0.
a(2*F(n)) = Lucas(n-1) for n >= 2;
a(3*F(n)) = 2*F(n) for n >= 1;
a(4*F(n)) = F(n+2) for n >= 2;
a(5*F(n)) = 4*F(n) - F(n-1) = A022120(n-2) for n >= 2.
a(2*L(n)) = F(n) + 3*F(n-1) = A104449(n) for n >= 0;
a(3*L(n)) = F(n+3) for n >= 3;
a(4*L(n)) = F(n+4) - L(n-3) = A022114(n-1) for n >= 3;
a(5*L(n)) = 11*F(n-1) + F(n-4) = A022367(n-1) for n >= 4.
For n >= 1, m >= 2, a(F(m*n)) = F(m*n-1) and a(L(m*n)) = F(m*n+1). Hence
a(L(m*n)) + a(F(m*n)) = L(m*n) and a(L(m*n)) - a(F(m*n)) = F(m*n).
Conjectures:
1) a(n) + a^[2](n - a^[2](n - a^[2](n))) = n for n >= 2.
2) If k >= 2 and m = 2*k - 1 then a(m*n - a(k*n)) = a(m*n - a(m*n - a(m*n - a(k*n)))).

A356990 a(n) = n - a^[4](n - a^[5](n-1)) with a(1) = 1, where a^[4](n) = a(a(a(a(n)))) and a^[5](n) = a(a(a(a(a(n))))).

Original entry on oeis.org

1, 1, 2, 3, 4, 5, 5, 6, 7, 7, 8, 9, 10, 10, 11, 12, 13, 14, 14, 15, 16, 17, 18, 19, 19, 19, 20, 21, 22, 23, 24, 25, 26, 26, 26, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 36, 36, 36, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 50, 50, 50, 50, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 69, 69, 69, 69, 69, 69, 69, 70, 71
Offset: 1

Views

Author

Peter Bala, Sep 08 2022

Keywords

Comments

This is the fourth sequence in a family of nested-recurrent sequences with apparently similar structure defined as follows. Given a sequence s = {s(n); n >= 1} we define the k-th iterated sequence s^[k] by putting s^[1](n) = s(n) and setting s^[k](n) = s^[k-1](u(n)) for k >= 2. For k >= 1, we define a nested-recurrent sequence {u(n): n >= 1}, dependent on k, by putting u(1) = 1 and setting u(n) = n - u^[k](n - u^[k+1](n-1)) for n >= 2. This is the case k = 4. For other cases see A006165 (k = 1), A356988 (k = 2) and A356989 (k = 3).
The sequence is slow, that is, for n >= 1, a(n+1) - a(n) is either 0 or 1.
The line graph of the sequence {a(n)} thus consists of a series of plateaus (where the value of the ordinate a(n) remains constant as n increases) joined by lines of slope 1.
The sequence of plateau heights begins 5, 7, 10, 14, 19, 26, 36, 50, ..., which appears to be A003269.
The plateaus start at absiccsa values n = 6, 9, 13, 18, 24, 33, 46, 64, ..., which appears to be A014101, and terminate at abscissa values 7, 10, 14, 19, 26, 36, 50, ..., conjecturally A003269.

Crossrefs

Programs

Showing 1-3 of 3 results.