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 1-3 of 3 results.

A215007 a(n) = 7*a(n-1) - 14*a(n-2) + 7*a(n-3), a(0)=1, a(1)=3, a(2)=9.

Original entry on oeis.org

1, 3, 9, 28, 91, 308, 1078, 3871, 14161, 52479, 196196, 737793, 2785160, 10540390, 39955041, 151615947, 575723785, 2187128524, 8311078307, 31587815308, 120069510526, 456434707519, 1735184512425, 6596692255391, 25079305566420
Offset: 0

Views

Author

Roman Witula, Jul 31 2012

Keywords

Comments

The sequence {a(n)} we shall call the Berndt-type sequence of type 1 for the argument 2*Pi/7; our motivation comes from Berndt's et al. and my papers (see the first formula below, which is in agreement with the respective identities discussed in these papers).
We note that a(n) = A105849(n) for n=0,1,...,5, and A105849(6) - a(6) = 1. Moreover we have a(n) = 2*A215008(n) - A215008(n+1).

References

  • R. Witula, E. Hetmaniok, and D. Slota, Sums of the powers of any order roots taken from the roots of a given polynomial, Proceedings of the Fifteenth International Conference on Fibonacci Numbers and Their Applications, Eger, Hungary, 2012.

Crossrefs

Programs

  • GAP
    a:=[1,3,9];; for n in [4..30] do a[n]:=7*(a[n-1]-2*a[n-2]+a[n-3]); od; a; # G. C. Greubel, Oct 03 2019
  • Magma
    R:=PowerSeriesRing(Rationals(), 30); Coefficients(R!((1-4*x+2*x^2)/(1-7*x+14*x^2-7*x^3))); // G. C. Greubel, Feb 01 2018
    
  • Maple
    seq(coeff(series((1-4*x+2*x^2)/(1-7*x+14*x^2-7*x^3), x, n+1), x, n), n = 0..30); # G. C. Greubel, Oct 03 2019
  • Mathematica
    LinearRecurrence[{7,-14,7}, {1,3,9}, 30] (* G. C. Greubel, Feb 01 2018 *)
  • PARI
    Vec((1-4*x+2*x^2)/(1-7*x+14*x^2-7*x^3)+O(x^30)) \\ Charles R Greathouse IV, Sep 27 2012
    
  • Sage
    def A215007_list(prec):
        P. = PowerSeriesRing(ZZ, prec)
        return P((1-4*x+2*x^2)/(1-7*x+14*x^2-7*x^3)).list()
    A215007_list(30) # G. C. Greubel, Oct 03 2019
    

Formula

a(n) = (1/sqrt(7))*(cot(8*Pi/7)*(s(1))^2n + cot(4*Pi/7)*(s(4))^2n + cot(2*Pi/7)*(s(2))^2n), where s(j) := 2*sin(2Pi*j/7).
G.f.: (1-4*x+2*x^2)/(1-7*x+14*x^2-7*x^3).

A215494 a(n) = 7*a(n-1) - 14*a(n-2) + 7*a(n-3) with a(1)=7, a(2)=21, a(3)=70.

Original entry on oeis.org

7, 21, 70, 245, 882, 3234, 12005, 44933, 169099, 638666, 2417807, 9167018, 34790490, 132119827, 501941055, 1907443237, 7249766678, 27557748813, 104759610858, 398257159370, 1514069805269, 5756205681709, 21884262613787, 83201447389466, 316323894905207
Offset: 1

Views

Author

Roman Witula, Aug 13 2012

Keywords

Comments

The Berndt-type sequence number 5 for the argument 2*Pi/7; see also A215007, A215008, A215143, A215493 and A215510.
We note that if we set:
x(n) := s(2)*s(1)^n + s(4)*s(2)^n + s(1)*s(4)^n,
y(n) := s(4)*s(1)^n + s(1)*s(2)^n + s(2)*s(4)^n,
z(n) := s(1)^(n+1) + s(2)^(n+1) + s(4)^(n+1),
for every n=0,1,..., where s(j) := 2*sin(2*Pi*j/7), then the following system of recurrence equations holds true:
x(n+2)=2*x(n)-y(n), y(n+2)=2*y(n)-x(n)+z(n), z(n+2)=y(n)+3*z(n).
Moreover we have a(n)=z(2*n-1), A215493(n)=z(2*n-2), A094429(n)=y(2n-1)-x(2n-1)=-x(2*n+2)/sqrt(7), A094430(n)=-x(2*n+3), y(2*n-2)=sqrt(7)*A215143(n), y(2*n-1)=A215510(n) and x(11)=-(y(10)+z(10))/sqrt(7)=-1078.
We can also deduce the following relations:
x(n-1) = c(1)*s(1)^n + c(2)*s(2)^n + c(4)*s(4)^n,
-y(n-1)-z(n-1) = c(2)*s(1)^n + c(4)*s(2)^n + c(1)*s(4)^n,
y(n-1)-x(n-1) = c(4)*s(1)^n + c(1)*s(2)^n + c(2)*s(4)^n,
for every n=1,2,..., where x(0)=y(0)=z(0)=sqrt(7), and c(j) := 2*cos(2*Pi*j/7).
All these sequences satisfy the following recurrence equation: Z(n+6)-7*Z(n+4)+14*Z(n+2)-7*Z(n)=0. The characteristic polynomial of this equation (after rescaling) has the form (X-s(1)^2)*(X-s(2)^2)*(X-s(3)^2)=X^3-7*X^2+14*X-7 and was recognized by Johannes Kepler (1571-1630); see the Savio-Suryanarayan paper.
We also have the following decomposition: (X-s(1)^(n+1))*(X-s(2)^(n+1))*(X-s(4)^(n+1)) = X^3 - z(n)*X^2 + (1/2)*(z(n)^2-z(2n+1))*X - (-sqrt(7))^(n+1).
Further we have a(n)=A146533(n) for n=1,...,6, and A146533(7)-a(7)=7. We note that all numbers 7^(-1-floor(n/3))*a(n) are integers.

Examples

			We have a(3)=5*7^2 and a(6)=5*7^4, which implies that s(1)^12 + s(2)^12 + s(4)^12 = 49*(s(1)^6 + s(2)^6 + s(4)^6). We also have a(9) = (a(1) + a(3))*7^49.
		

Crossrefs

See A122068.

Programs

  • Magma
    I:=[7,21,70]; [n le 3 select I[n] else 7*Self(n-1)-14*Self(n-2)+7*Self(n-3): n in [1..30]]; // Vincenzo Librandi, Dec 01 2016
    
  • Mathematica
    LinearRecurrence[{7,-14,7}, {7,21,70}, 50]
  • PARI
    polsym(x^3 - 7*x^2 + 14*x - 7, 30) \\ (includes a(0)=3) Joerg Arndt, May 31 2017
    
  • PARI
    x='x+O('x^30); Vec((7-28*x+21*x^2)/(1-7*x+14*x^2-7*x^3)) \\ G. C. Greubel, Apr 23 2018

Formula

Equals 7*A122068. - M. F. Hasler, Aug 25 2012
a(n) = s(1)^(2n) + s(2)^(2n) + s(4)^(2n), where s(j) := 2*Sin(2*Pi*j/7) (for the sums of the respective odd powers see A215493, see also A094429, A115146). For the proof of these formula see Witula-Slota's paper.
G.f.: (7 - 28*x + 21*x^2)/(1 - 7*x + 14*x^2 - 7*x^3) = -d(log(1 - 7*x + 14*x^2 - 7*x^3))/dx.

A376499 Array read by ascending antidiagonals: A(n,k) = A376484/(2*n+1).

Original entry on oeis.org

1, 1, 3, 1, 3, 9, 1, 3, 10, 27, 1, 3, 10, 35, 81, 1, 3, 10, 35, 125, 243, 1, 3, 10, 35, 126, 450, 729, 1, 3, 10, 35, 126, 462, 1625, 2187, 1, 3, 10, 35, 126, 462, 1715, 5875, 6561, 1, 3, 10, 35, 126, 462, 1716, 6419, 21250, 19683, 1, 3, 10, 35, 126, 462, 1716, 6435, 24157, 76875, 59049
Offset: 1

Views

Author

Cheng-Jun Li, Sep 25 2024

Keywords

Comments

It is only a conjecture that the A(n,k) are always integers.
Values repeated as a staircase for all A(n+x,2*n) (x > 0 and are equal to A(n,2*n)).

Examples

			First ten rows start as follows:
  1 3  9 27  81 243  729 2187  6561 19683  59049  177147  531441  1594323  4782969
  1 3 10 35 125 450 1625 5875 21250 76875 278125 1006250 3640625 13171875 47656250
  1 3 10 35 126 462 1715 6419 24157 91238 345401 1309574 4970070 18874261 71705865
  1 3 10 35 126 462 1716 6435 24309 92358 352485 1350054 5185350 19960020 76964985
  1 3 10 35 126 462 1716 6435 24310 92378 352715 1352054 5199975 20055024 77531355
  1 3 10 35 126 462 1716 6435 24310 92378 352716 1352078 5200299 20058272 77558325
  1 3 10 35 126 462 1716 6435 24310 92378 352716 1352078 5200300 20058300 77558759
  1 3 10 35 126 462 1716 6435 24310 92378 352716 1352078 5200300 20058300 77558760
  1 3 10 35 126 462 1716 6435 24310 92378 352716 1352078 5200300 20058300 77558760
  1 3 10 35 126 462 1716 6435 24310 92378 352716 1352078 5200300 20058300 77558760
		

Crossrefs

All of these are conjectures. Rows: A000244, A081567, A122068. Columns: A000012, A000012 * 3, A095049 for n >= 20. A(1,k) = A000244, A(2,k) = A081567, A(3,k) = A122068 (First 3 rows of the array).A(n,1) = A(n,2) / 3 = A000012, A(n,3) = A095049 for n >= 20 (First 3 columns of the array). When k increases, the row of A(n,k) gets closer to A001700.

Formula

A(n,k) = A376484(n,k)/(2*n+1)
Showing 1-3 of 3 results.