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 1-7 of 7 results.

A092440 a(n) = 2^(2n+1) - 2^(n+1) + 1.

Original entry on oeis.org

1, 5, 25, 113, 481, 1985, 8065, 32513, 130561, 523265, 2095105, 8384513, 33546241, 134201345, 536838145, 2147418113, 8589803521, 34359476225, 137438429185, 549754765313, 2199021158401, 8796088827905, 35184363700225
Offset: 0

Views

Author

Christopher Hanusa (chanusa(AT)math.washington.edu), Mar 24 2004

Keywords

Comments

Arises from enumeration of domino tilings of Aztec Pillow-like regions.
Each beginning with 1, the subsequences of A046899 are 1; 1,2; 1,3,6; 1,4,10,20 and so forth. Create triangles with the sides being equal to each of these subsequences; the interior members T(i,j)=T(i-1,j-1) + T(i-1,j). The sum of all members for each triangle will reproduce the terms of this sequence. Example using the fourth subsequence 1,4,10,20 will give row(1)=1; row(2)=4,4; row(3)=10,8,10; row(4)=20,18,18,20 giving a sum for all members of 113, the fourth term in the sequence. - J. M. Bergot, Oct 17 2012
Also, the number of active (ON, black) cells at stage 2^n-1 of the two-dimensional cellular automaton defined by "Rule 510", based on the 5-celled von Neumann neighborhood. - Robert Price, May 04 2016
Let M be some square matrix of rank 2^n, containing the positive real value X everywhere except on the diagonal; let Y be some complex value with phase 3*Pi/4 everywhere else (thus all coefficients on the diagonal). Then, for M to be a unitary matrix, X must be 1/sqrt(a(n)). - Thomas Baruchel, Aug 10 2020

References

  • J. Propp, Enumeration of matchings: problems and progress, pp. 255-291 in L. J. Billera et al., eds, New Perspectives in Algebraic Combinatorics, Cambridge, 1999 (see Problem 13).
  • S. Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 170.

Links

Crossrefs

Cf. A092437, A092438, A092439, A092441, A092442, A092443.

Programs

Formula

a(n) = 2^(2n+1) - 2^(n+1) + 1.
From Colin Barker, Nov 22 2012: (Start)
a(n) = 7*a(n-1) - 14*a(n-2) + 8*a(n-3).
G.f.: -(4*x^2-2*x+1)/((x-1)*(2*x-1)*(4*x-1)). (End)
a(n) = A000225(n)^2 + (A000225(n) + 1)^2. - César Aguilera, May 28 2023

A092437 Triangle read by rows, arising from enumeration of domino tilings of Aztec Pillow-like regions.

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 5, 6, 6, 1, 1, 5, 13, 26, 30, 20, 1, 1, 5, 13, 41, 90, 140, 140, 70, 1, 1, 5, 13, 41, 121, 302, 560, 742, 630, 252
Offset: 0

Views

Author

Christopher Hanusa (chanusa(AT)math.washington.edu), Mar 24 2004

Keywords

Comments

The rows are of lengths 1, 3, 5, 7, ...
Call the first row row 0 and entries starting from 0. Then entries i=0 through k in row k are A046717(i).
In row k, entry k+1 is sequence A092438 and entry k+2 is sequence A092439.
In row k, entry 2k-1 is A002457(k-1) and entry 2k is A000984(k).

Examples

			Triangle begins:
  1;
  1, 1, 2;
  1, 1, 5, 6, 6;
  1, 1, 5, 13, 26, 30, 20;
  ...

References

  • James Propp, Enumeration of matchings: problems and progress, pp. 255-291 in L. J. Billera et al., eds, New Perspectives in Algebraic Combinatorics, Cambridge, 1999 (see Problem 13).

Links

Crossrefs

Cf. A092438-A092443.

A092438 Sequence arising from enumeration of domino tilings of Aztec Pillow-like regions.

Original entry on oeis.org

0, 2, 6, 26, 90, 302, 966, 3026, 9330, 28502, 86526, 261626, 788970, 2375102, 7141686, 21457826, 64439010, 193448102, 580606446, 1742343626, 5228079450, 15686335502, 47063200806, 141197991026, 423610750290, 1270865805302
Offset: 0

Views

Author

Christopher Hanusa (chanusa(AT)math.washington.edu), Mar 24 2004

Keywords

Examples

			a(3) = (3^4+(-1)^4)/2-2^4+1 = 26.

References

  • J. Propp, Enumeration of matchings: problems and progress, pp. 255-291 in L. J. Billera et al., eds, New Perspectives in Algebraic Combinatorics, Cambridge, 1999 (see Problem 13).

Links

Crossrefs

Cf. A046717, A092437-A092443, A140420.

Formula

a(n) = A092437(n, n+1).
a(n) = A046717(n+1)-2^(n+1)+1.
a(n) = (3^(n+1)+(-1)^(n+1))/2-2^(n+1)+1.
From R. J. Mathar, Apr 21 2010: (Start)
a(n) = +5*a(n-1) -5*a(n-2) -5*a(n-3) +6*a(n-4) = 2*A140420(n).
G.f.: -2*x*(1-2*x+3*x^2) / ( (x-1)*(3*x-1)*(2*x-1)*(1+x) ). (End)

A092439 Sequence arising from enumeration of domino tilings of Aztec Pillow-like regions.

Original entry on oeis.org

0, 0, 6, 30, 140, 560, 2058, 7098, 23472, 75372, 237182, 735878, 2260596, 6896136, 20933778, 63325170, 191089112, 575626052, 1731858246, 5206059774, 15640198620, 46966732320, 140996664986, 423191320490, 1269993390720
Offset: 0

Views

Author

Christopher Hanusa (chanusa(AT)math.washington.edu), Mar 24 2004

Keywords

Examples

			a(3) = (3^5+(-1)^5)/2 - 2^5 - 5*(2^4-1) + 4^2 = 30.

References

  • James Propp, Enumeration of matchings: problems and progress, pp. 255-291 in L. J. Billera et al., eds, New Perspectives in Algebraic Combinatorics, Cambridge, 1999 (see Problem 13).

Links

Crossrefs

Cf. A046717, A092437-A092443.

Programs

  • Mathematica
    Table[(3^(n+2)+(-1)^(n+2))/2-2^(n+2)-(n+2)(2^(n+1)-1)+(n+1)^2,{n,0,30}] (* or *) LinearRecurrence[{9,-30,42,-9,-39,40,-12},{0,0,6,30,140,560,2058},30] (* Harvey P. Dale, Nov 27 2011 *)

Formula

a(n) = (3^(n+2)+(-1)^(n+2))/2-2^(n+2)-(n+2)*(2^(n+1)-1)+(n+1)^2.
a(n) = A092437(n, n+2), for n >= 2.
a(n) = A046717(n+2)-2^(n+2)-(n+2)*(2^(n+1)-1)+(n+1)^2.
a(n) = 9*a(n-1)-30*a(n-2)+42*a(n-3)-9*a(n-4)-39*a(n-5)+40*a(n-6)-12*a(n-7). - Harvey P. Dale, Nov 27 2011
G.f.: 2*x^2*(6*x^4-26*x^3+25*x^2-12*x+3)/((x-1)^3*(x+1)*(2*x-1)^2*(3*x-1)). - Colin Barker, Nov 22 2012
E.g.f.: exp(x)*(4*x + x^2 - 4*(2 + x)*cosh(x) - 4*(2 + x)*sinh(x) + 2*(2*cosh(x) + sinh(x))^2). - Stefano Spezia, Sep 01 2025

A092441 Sequence arising from enumeration of domino tilings of Aztec Pillow-like regions.

Original entry on oeis.org

1, 10, 65, 346, 1637, 7218, 30529, 126034, 513125, 2072698, 8335505, 33439914, 133972165, 536346850, 2146369793, 8587575586, 34354757957, 137428468074, 549733794193, 2198977118650, 8795996553701, 35184170762770
Offset: 0

Views

Author

Christopher Hanusa (chanusa(AT)math.washington.edu), Mar 24 2004

Keywords

Examples

			a(3) = 2^9-2^5-10(2^4-1)+4^2 = 346.

References

  • J. Propp, Enumeration of matchings: problems and progress, pp. 255-291 in L. J. Billera et al., eds, New Perspectives in Algebraic Combinatorics, Cambridge, 1999 (see Problem 13).

Links

Crossrefs

Cf. A092437-A092443.

Programs

  • Mathematica
    LinearRecurrence[{11,-47,101,-116,68,-16},{1,10,65,346,1637,7218},30] (* Harvey P. Dale, Nov 26 2022 *)

Formula

a(n) = 2^(2n+3)-2^(n+2)-2(n+2)(2^(n+1)-1)+(n+1)^2.
G.f.: -(8*x^4-2*x^2+x-1)/((x-1)^3*(2*x-1)^2*(4*x-1)). [Colin Barker, Nov 22 2012]

A092442 Sequence arising from enumeration of domino tilings of Aztec Pillow-like regions.

Original entry on oeis.org

0, 1, 5, 19, 59, 161, 405, 967, 2231, 5029, 11153, 24443, 53091, 114505, 245549, 524047, 1113839, 2358989, 4980393, 10485379, 22019675, 46136881, 96468485, 201326039, 419429799, 872414581, 1811938625, 3758095627, 7784627411
Offset: 0

Views

Author

Christopher Hanusa (chanusa(AT)math.washington.edu), Mar 24 2004

Keywords

Comments

Differences give A066368.

Examples

			a(3)=4(2^3-1)-3^2=19.

References

  • J. Propp, Enumeration of matchings: problems and progress, pp. 255-291 in L. J. Billera et al., eds, New Perspectives in Algebraic Combinatorics, Cambridge, 1999 (see Problem 13).

Links

Crossrefs

Cf. A092437-A092443.

Formula

a(n)=(n+1)(2^n-1)-n^2.
G.f.:(x*(4*x^3-3*x^2+2*x-1))/((2*x-1)^2*(x-1)^3) [From Maksym Voznyy (voznyy(AT)mail.ru), Jul 26 2009]

A375177 a(n) = 1/(n + 1)^2 * Sum_{k = 1..n+1} (k^4)*binomial(n+1, k)^2.

Original entry on oeis.org

1, 5, 26, 134, 670, 3262, 15540, 72732, 335478, 1528670, 6894316, 30820660, 136736236, 602610764, 2640266600, 11508115320, 49928451750, 215717144670, 928515985980, 3983029119300, 17032882625220, 72631992447300, 308911087394520, 1310670689270280, 5548646191175100
Offset: 1

Views

Author

Peter Bala, Aug 02 2024

Keywords

Comments

Compare with the identity 1/(n+1)^2 * Sum_{k = 1..n+1} (k^2)*binomial(n+1, k)^2 = binomial(2*n, n) = A000984(n).
The central binomial coefficients satisfy the supercongruence binomial(2*p, p) == 2 (mod p^3) for all primes p >= 5 (Wolstenholme's theorem).

Links

Crossrefs

Cf. A000984, A092443.

Programs

  • Maple
    seq( 1/(n+1)^2 *add( (k^4)*binomial(n+1, k)^2, k = 1..n+1), n = 0..25);
# faster program for large n
a := proc (n) option remember; if n = 0 then 1 else (4*n-6)*(n^3+4*n^2+2*n-2)*a(n-1)/(n*(n^3+n^2-3*n-1)) end if; end:
seq(a(n), n = 0..25);

Formula

a(n) = hypergeom([2, 2, -n, -n], [1, 1, 1], 1).
a(n) = (1/2) * P(n)/(2*n - 1) * binomial(2*n, n), where P(n) = n^3 + 4*n^2 + 2*n - 2.
P-recursive: n*P(n-1)*a(n) = 2*(2*n - 3)*P(n)*a(n-1) with a(1) = 1.
a(p) == 2*p + 2 (mod p^3) for all primes p >= 5.
Showing 1-7 of 7 results.

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