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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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A291219 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - S - S^3.

Original entry on oeis.org

1, 1, 3, 5, 11, 21, 42, 83, 163, 323, 635, 1255, 2473, 4880, 9625, 18985, 37451, 73869, 145715, 287421, 566954, 1118331, 2205947, 4351307, 8583091, 16930447, 33395857, 65874464, 129939569, 256310161, 505580371, 997274197, 1967156763, 3880282533, 7653987242
Offset: 0

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Author

Clark Kimberling, Aug 24 2017

Keywords

Comments

Suppose s = (c(0), c(1), c(2), ...) is a sequence and p(S) is a polynomial. Let S(x) = c(0)*x + c(1)*x^2 + c(2)*x^3 + ... and T(x) = (-p(0) + 1/p(S(x)))/x. The p-INVERT of s is the sequence t(s) of coefficients in the Maclaurin series for T(x). Taking p(S) = 1 - S gives the "INVERT" transform of s, so that p-INVERT is a generalization of the "INVERT" transform (e.g., A033453).
In the following guide to p-INVERT sequences using s = (1,0,1,0,1,...) = A000035, in some cases t(1,0,1,0,1,...) is a shifted version of the indicated sequence.
p(S) t(1,0,1,0,1,...)
1 - S A000045 (Fibonacci numbers)
1 - S^2 A147600
1 - S^3 A291217
1 - S^5 A291218
1 - S - S^2 A289846
1 - S - S^3 A291219
1 - S - S^4 A291220
1 - S^3- S^6 A291221
1 - S^2- S^3 A291222
1 - S^3- S^4 A291223
1 - 2S A052542
1 - 3S A006190
(1 - S)^2 A239342
(1 - S)^3 A276129
(1 - S)^4 A291224
(1 - S)^5 A291225
(1 - S)^6 A291226
1 - S - 2 S^2 A291227
1 - 2 S - 2 S^2 A291228
1 - 3 S - 2 S^2 A060801
(1 - S)(1 - 2 S) A291229
(1 - S)(1 - 2 S)(1 - 3 S) A291230
(1 - S)(1 - 2 S)(1 - 3 S)( 1 - 4 S) A291231
(1 - 2 S)^2 A291264
(1 - 3 S)^2 A291232
1 - S - S^2 - S^3 A291233
1 - S - S^2 - S^3 - S^4 A291234
1 - S - S^2 - S^3 - S^4 - S^5 A291235
(1 - S)(1 - 3 S) A291236
(1 - S)(1 - 2S)( 1 - 4S) A291237
(1 - S)^2 (1 - 2S) A291238
(1 - S^2) (1 - 2S) A291239
(1 - S^3)^2 A291240
1 - S - S^2 + S^3 A291241
1 - 2 S - S^2 + S^3 A291242
1 - 3 S + S^2 A291243
1 - 4 S + S^2 A291244
1 - 5 S + S^2 A291245
1 - 6 S + S^2 A291246
1 - S - S^2 - S^3 + S^4 A291247
1 - S - S^2 - S^3 - S^4 + S^5 A291248
1 - S - S^2 - S^3 + S^4 + S^5 A291249
1 - S - 2 S^2 + 2 S^3 A291250
1 - 3 S^2 + 2 S^3 A291251 (includes negative terms)
(1 - S^3)^3 A291252
(1 - S - S^2)^2 A291253
(1 - 2 S - S^2)^2 A291254
(1 - S - 2 S^2)^2 A291255

Links

Crossrefs

Cf. A000035, A290890, A291000.

Programs

  • Magma
    I:=[1,1,3,5,11,21]; [n le 6 select I[n] else Self(n-1)+3*Self(n-2)-Self(n-3)-3*Self(n-4)+Self(n-5)+Self(n-6): n in [1..45]]; // Vincenzo Librandi, Aug 25 2017
  • Mathematica
    z = 60; s = x/(1 - x^2); p = 1 - s - s^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291219 *)
LinearRecurrence[{1, 3, -1, -3, 1, 1}, {1, 1, 3, 5, 11, 21}, 50] (* Vincenzo Librandi, Aug 25 2017 *)

Formula

G.f.: -(1 - x^2 + x^4)/(-1 + x + 3*x^2 - x^3 - 3*x^4 + x^5 + x^6).
a(n) = a(n-1) + 3*a(n-2) - a(n-3) - 3*a(n-4) + a(n-5) + a(n-6) for n >= 7.

A276129 a(n) is the number of ordered ways to tile a strip of length n+2 with white tiles of odd lengths summing to length n and two red squares.

Original entry on oeis.org

1, 3, 6, 13, 27, 54, 106, 204, 387, 725, 1344, 2469, 4500, 8145, 14652, 26213, 46665, 82704, 145982, 256722, 449937, 786109, 1369494, 2379447, 4123944, 7130895, 12303714, 21186013, 36411399, 62466906, 106987282, 182946888, 312367887, 532587461, 906840060
Offset: 0

Views

Author

Gregory L. Simay, Aug 21 2016

Keywords

Comments

a(n) is a specific case of b(r,n), the number of ordered ways to rearrange a tiling of length n + r, with odd(1,3,5...) white tiles summing to n and r red squares.
Define the following summation: b(0,r,n) = b(r,n); b(1,r,n) = b(r, n-2) + b(r, n-4) + b(r, n-6) + ..; b(s, r, n) = b(s-1, r, n-2) + b(s-1, r, n-4) + b(r-1, s, n-6) + ...
The number of compositions of n with exactly r even numbers is b(r, r, n-2r).
Except for the initial 1, this is the p-INVERT transform of (1,0,1,0,1,0,...) for p(S) = (1 - S)^3. See A291219. - Clark Kimberling, Sep 04 2017

Examples

			Let 1,3 be the lengths of the odd tiles summing to 3 and let r,r be the two odd squares. Then the resulting number of compositions is a(3) = 13. The 6 compositions are 3,r,r; r,3,r; r,r,3; 1,1,1,r,r; 1,1,r,r,1; 1,r,r,1,1; r,r,1,1,1; 1,1,r,1,r; 1,r,1,r,1; r,1,r,1,1; r,1,1,r,1; 1,r,1,1,r; r,1,1,1,r.

Crossrefs

Cf. A000045, A029907, A239242.

Programs

  • Maple
    b:= proc(n, m) option remember;
      `if`(n+m=0, 1, `if`(m>0, b(n, m-1), 0)+
      add(`if`(j::odd, b(n-j, m), 0), j=1..n))
    end:
a:= n-> b(n, 2):
seq(a(n), n=0..50);  # Alois P. Heinz, Aug 29 2016
  • Mathematica
    a[0] = 1; a[n_] := Sum[Binomial[n - 2*k + 2, 2]*Binomial[n - k - 1, k], {k, 0, (n - 1)/2}];
Table[a[n], {n, 0, 50}] (* Jean-François Alcover, Dec 27 2017, after Andrew Howroyd *)
  • PARI
    a(n)={if(n<=0, n==0, sum(k=0, (n-1)\2, binomial(n-2*k+2, 2)*binomial(n-k-1, k)))} \\ Andrew Howroyd, Dec 26 2017

Formula

a(n) = Sum_{k=0..floor((n-1)/2)} binomial(n-2*k+2, 2)*binomial(n-k-1, k) for n > 0. - Andrew Howroyd, Dec 26 2017
b(0,0)=1. For n>=1, b(0,n) = b(0,0,n) = the n-th Fibonacci number, A000045(n).
b(1,0)=1. For n>=1, b(1,n) = A239342(n+1).
b(2,n) = a(n) = a(n-1) + a(n-2) + A239342(n+1) + A239342(n-1).
G.f. for b(2,n): ((1-x^2)/(1-x-x^2))^3.
G.f. for b(r,n): ((1-x^2)/(1-x-x^2))^(r+1).
b(1,1,n) = A029907(n+1).
b(r,n) = b(r, n-1) + b(r, n-2) + b(r-1, n) - b(r-1, n-2).
b(r,r,n) = b(r-1, r-1, n) + b(r, r, n-1) + b(r, r, n-2).
G.f. for b(r,r,n): (1-x^2)/((1-x-x^2)^(r+1)).

Extensions

More terms from Alois P. Heinz, Aug 29 2016
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