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.

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A279196 Number of polynomials of the form P(x,y) = 1 + (x+y-1) * Q(x,y) such that P(1,1) = n and both polynomials P and Q have nonnegative integer coefficients.

Original entry on oeis.org

1, 1, 2, 5, 13, 36, 102, 295, 864, 2557, 7624, 22868, 68920, 208527, 632987, 1926752, 5878738, 17973523, 55050690, 168881464, 518818523, 1595878573, 4914522147, 15150038699, 46747391412, 144370209690, 446214862158, 1380161749537, 4271808447154, 13230257155092, 40999697820032
Offset: 1

Views

Author

N. J. A. Sloane, Dec 15 2016

Keywords

Comments

Original definition did not have the requirement for Q to have nonnegative coefficients. However, this results in different terms given by A363933. We have a(n) <= A363933(n), which is strict for n >= 5. - Max Alekseyev, Jun 28 2023
In colorful terms, one can view a polynomial as a configuration made of piles of tokens located at lattice points (i>=0, j>=0). One introduces the notion of "degradation of a configuration": to degrade a configuration, choose a nonempty pile of tokens in it, say that at (i,j); remove one token from that pile; then add one token at (i+1,j) and one token at (i,j+1). This is a nondeterministic process. a(n) is the number of distinct configurations one can possibly get after (n-1) degradations of the initial configuration consisting of just one token at (0,0). In this metaphor, the P's are the resulting configurations and the Q's are records of where the tokens have been taken. - Luc Rousseau, Jun 30 2023

Examples

			From _Peter Kagey_, Feb 03 2017 (Start):
For n = 1 the a(1) = 1 solution is:
  1 = 0(x + y - 1) + 1.
For n = 2 the a(2) = 1 solution is:
  x + y = (x + y - 1) + 1.
For n = 3 the a(3) = 2 solutions are:
  xy + x + y^2 = (y + 1)(x + y - 1) + 1;
  xy + y + x^2 = (x + 1)(x + y - 1) + 1.
For n = 4 the a(4) = 5 solutions are:
  x^2 + 2xy + y^2       = (x + y + 1)(x + y - 1) + 1;
  x^2y + x^2 + xy^2 + y = (xy + x + 1)(x + y - 1) + 1;
  x^2y + xy^2 + x + y^2 = (xy + y + 1)(x + y - 1) + 1;
  xy^2 + xy + x + y^3   = (y^2 + y + 1)(x + y - 1) + 1;
  x^3 + x^2y + xy + y   = (x^2 + x + 1)(x + y - 1) + 1.
(End) [Corrected by _Luc Rousseau_, Jun 30 2023]
		

Crossrefs

Cf. A363933.

Programs

  • Java
    // See Rousseau link.
    (Java + Prolog) // See Rousseau link.
    
  • Prolog
    % See Rousseau link.

Extensions

Definition corrected by Max Alekseyev, Jun 28 2023
a(10)-a(18) from Luc Rousseau, Jun 30 2023
a(19)-a(25) from Max Alekseyev, Jul 04 2023
a(26)-a(29) from Luc Rousseau, Jul 31 2023
a(30) from Luc Rousseau, Nov 10 2023
a(31) from Luc Rousseau, Dec 18 2023
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