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.

A320770 a(n) = (-1)^floor(n/4) * 2^floor(n/2).

Original entry on oeis.org

1, 1, 2, 2, -4, -4, -8, -8, 16, 16, 32, 32, -64, -64, -128, -128, 256, 256, 512, 512, -1024, -1024, -2048, -2048, 4096, 4096, 8192, 8192, -16384, -16384, -32768, -32768, 65536, 65536, 131072, 131072, -262144, -262144, -524288, -524288, 1048576, 1048576
Offset: 0

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Author

Michael Somos, Oct 20 2018

Keywords

Examples

			G.f. = 1 + x + 2*x^2 + 2*x^3 - 4*x^4 - 4*x^5 - 8*x^6 - 8*x^7 + ...
		

Crossrefs

Cf. A016116.
The following sequences are all essentially the same, in the sense that they are simple transformations of each other, with A029744 = {s(n), n>=1}, the numbers 2^k and 3*2^k, as the parent: A029744 (s(n)); A052955 (s(n)-1), A027383 (s(n)-2), A354788 (s(n)-3), A347789 (s(n)-4), A209721 (s(n)+1), A209722 (s(n)+2), A343177 (s(n)+3), A209723 (s(n)+4); A060482, A136252 (minor differences from A354788 at the start); A354785 (3*s(n)), A354789 (3*s(n)-7). The first differences of A029744 are 1,1,1,2,2,4,4,8,8,... which essentially matches eight sequences: A016116, A060546, A117575, A131572, A152166, A158780, A163403, A320770. The bisections of A029744 are A000079 and A007283. - N. J. A. Sloane, Jul 14 2022

Programs

  • Magma
    [(-1)^Floor(n/4)* 2^Floor(n/2): n in [0..50]]; // G. C. Greubel, Oct 27 2018
    
  • Mathematica
    a[ n_] := (-1)^Quotient[n, 4] * 2^Quotient[n, 2];
  • PARI
    {a(n) = (-1)^floor(n/4) * 2^floor(n/2)};
    
  • Python
    def A320770(n): return -(1<<(n>>1)) if n&4 else 1<<(n>>1) # Chai Wah Wu, Jan 18 2023

Formula

G.f.: (1 + x) * (1 + 2*x^2) / (1 + 4*x^4).
G.f.: A(x) = 1/(1 - x/(1 - x/(1 + 2*x/(1 - 4*x/(1 + 3*x/(1 + 5*x/(3 - 2*x))))))).
a(n) = (-1)^floor(n/2) * 2 * a(n-2) = -4 * a(n-4) for all n in Z.
a(n) = c(n) * (-2)^n * a(-n) for all n in Z where c(4*k+2) = -1 else 1.
a(n) = a(n+1) = (1+I)^n * (-I)^(n/2) * (-1)^floor(n/4) if n = 2*k.
a(n) = (-1)^floor(n/4) * A016116(n).
E.g.f.: cosh(x)*(cos(x) + sin(x)) + sin(x)*sinh(x). - Stefano Spezia, Feb 04 2023