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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A084137 Binomial transform of A084136.

Original entry on oeis.org

1, 2, 8, 32, 144, 672, 3200, 15360, 73984, 356864, 1722368, 8314880, 40144896, 193830912, 935886848, 4518838272, 21818834944, 105350561792, 508677324800, 2456111022080, 11859152338944, 57261051346944, 276480810549248
Offset: 0

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Author

Paul Barry, May 16 2003

Keywords

Comments

Exponential self-convolution of companion Pell numbers (A002203), divided by 4. - Vladimir Reshetnikov, Oct 07 2016

Crossrefs

Programs

  • Magma
    A002203:= func< n | Round((1+Sqrt(2))^n + (1-Sqrt(2))^n) >;
    [2^(n-2)*(2+A002203(n)): n in [0..40]]; // G. C. Greubel, Oct 13 2022
    
  • Mathematica
    Table[2^(n-2)*(2+LucasL[n,2]), {n,0,20}] (* Vladimir Reshetnikov, Oct 07 2016 *)
  • PARI
    a(n)=if(n<0,0,polsym(4+4*x-x^2,n)[n+1]/4+2^(n-1))
    
  • SageMath
    [2^(n-2)*(2+lucas_number2(n, 2, -1)) for n in range(41)] # G. C. Greubel, Oct 13 2022

Formula

G.f.: (1-4*x)/((1-2*x)*(1-4*x-4*x^2)).
E.g.f.: exp(2*x)*cosh(sqrt(2)*x)^2 = (exp(x)*cosh(sqrt(2)*x))^2.
a(n) = ((2+sqrt(8))^n + (2-sqrt(8))^n + 2^(n+1))/4.
a(n) = (A084128(n) + 2^n)/2.
a(n) = 2^(n-2)*(2 + A002203(n)). - Vladimir Reshetnikov, Oct 07 2016
a(n) = 6*a(n-1) - 4*a(n-2) - 8*a(n-3). - G. C. Greubel, Oct 13 2022

A267652 a(n) = 4*a(n - 1) + 4*a(n - 2) for n>1, a(0)=2, a(1)=3.

Original entry on oeis.org

2, 3, 20, 92, 448, 2160, 10432, 50368, 243200, 1174272, 5669888, 27376640, 132186112, 638251008, 3081748480, 14879997952, 71846985728, 346907934720, 1675019681792, 8087710466048, 39050920591360, 188554524229632, 910421779283968, 4395905214054400, 21225307973353472, 102484852749631488
Offset: 0

Views

Author

Ilya Gutkovskiy, Jan 19 2016

Keywords

Comments

Generalized Fibonacci sequence.

Crossrefs

Programs

  • Mathematica
    Table[2^(n - 5/2) ((1 + 4 Sqrt[2]) (1 - Sqrt[2])^n - (1 - 4 Sqrt[2]) (1 + Sqrt[2])^n), {n, 0, 25}]
    RecurrenceTable[{a[0] == 2, a[1] == 3, a[n] == 4 a[n - 1] + 4 a[n - 2]}, a, {n, 25}] (* Bruno Berselli, Jan 19 2016 *)
    LinearRecurrence[{4, 4}, {2, 3}, 20] (* Vincenzo Librandi, Jan 19 2016 *)
  • PARI
    Vec((2-5*x)/(1-4*x-4*x^2) + O(x^100)) \\ Altug Alkan, Jan 19 2016

Formula

G.f.: (2 - 5*x)/(1 - 4*x - 4*x^2).
a(n) = 2^(n-5/2)*((1+4*sqrt(2))*(1-sqrt(2))^n - (1-4*sqrt(2))*(1+sqrt(2))^n).
Lim_{n -> infinity} a(n)/a(n - 1) = 2 + 2*sqrt(2) = 2*A014176 = 4.82842712...
a(n) = 2*A057087(n)-5*A057087(n-1). - R. J. Mathar, Jun 07 2016

A320660 Number of business cards required to build an origami level n Jerusalem cube.

Original entry on oeis.org

12, 72, 672, 6048, 55488, 511872, 4738560, 43943424, 407890944, 3787941888, 35186122752, 326885842944, 3037038034944, 28217571901440, 262178452930560, 2436006721486848, 22634041833160704, 210303674768424960, 1954034324430913536, 18155901427591938048
Offset: 0

Views

Author

Keywords

Comments

The actual Jerusalem cube fractal cannot be built using a simple integer grid. However, one can create an approximate one by choosing the cube side length to be a Pell number (see link).
In practice, the first two terms represent the level 0 because they both consist of cubes (1 X 1 X 1 and 2 X 2 X 2, respectively). The "cross" shape appears at index 2, which is usually considered as the first iteration (for example, the "hole" shape in the Menger Sponge is visible at level 1).
The limit of a(n+1)/a(n) is equal to 2*(2+sqrt(7)) as n approaches infinity.

Examples

			a(2) = 672 because 456 business cards are needed for the squeleton and 216 more for the panels.
		

References

  • Eric Baird, L'art fractal, Tangente 150 (2013), 45.
  • Thomas Hull, Project Origami: Activities for Exploring Mathematics, A K Peters/CRC Press, 2006.

Crossrefs

At the n-th level, the cube side length is A000129(n+1), the squeleton requires 6*A239549(n+1) business cards, and each face requires A057087(n) units for the panels.
Cf. A212596 (Origami Menger sponge), A304960 (Origami Mosely snowflake sponge).

Programs

  • Mathematica
    LinearRecurrence[{12, -16, -80, -48}, {12, 72, 672, 6048}, 20]
  • Maxima
    makelist((3/14)*(7*(2 - 2*sqrt(2))^n + 7*(2 + 2*sqrt(2))^n + (21 - 5*sqrt(7))*(4 - 2*sqrt(7))^n + (21 + 5*sqrt(7))*(4 + 2*sqrt(7))^n), n, 0, 20), ratsimp;

Formula

a(n) = (3/14)*(7*(2 - 2*sqrt(2))^n + 7*(2 + 2*sqrt(2))^n + (21 - 5*sqrt(7))*(4 - 2*sqrt(7))^n + (21 + 5*sqrt(7))*(4 + 2*sqrt(7))^n).
a(n) = 12*a(n-1) - 16*a(n-2) - 80*a(n-3) - 48*a(n-4), n > 4.
G.f.: 12*(1 - 6*x + 8*x^3)/((1-4*x-4*x^2)*(1-8*x-12*x^2)) .
E.g.f.: (3/14)*(7*exp((2 - 2*sqrt(2))*x) + 7*exp((2 + 2*sqrt(2))*x) + (21 - 5*sqrt(7))*exp((4 - 2*sqrt(7))*x) + (21 + 5*sqrt(7))*exp((4 + 2*sqrt(7))*x)).
a(n) = 3*( A084128(n) -2*A239549(n) +3*A239549(n+1) ). - R. J. Mathar, Mar 06 2022
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