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 1-6 of 6 results.

A267633 Expansion of (1 - 4t)/(1 - x + t x^2): a Fibonacci-type sequence of polynomials.

Original entry on oeis.org

1, -4, 1, -4, 1, -5, 4, 1, -6, 8, 1, -7, 13, -4, 1, -8, 19, -12, 1, -9, 26, -25, 4, 1, -10, 34, -44, 16, 1, -11, 43, -70, 41, -4, 1, -12, 53, -104, 85, -20, 1, -13, 64, -147, 155, -61, 4, 1, -14, 76, -200, 259, -146, 24
Offset: 0

Views

Author

Tom Copeland, Jan 18 2016

Keywords

Examples

			Row polynomials:
P(0,t) = 1 - 4t
P(1,t) = 1 - 4t = [-t(0) + (1-4t)] = -t(0) + P(0,t)
P(2,t) = 1 - 5t + 4t^2 = [-t(1-4t) + (1-4t)] = -t P(0,t) + P(1,t)
P(3,t) = 1 - 6t + 8t^2 = [-t(1-4t) + (1-5t+4t^2)] = -t P(1,t) + P(2,t)
P(4,t) = 1 - 7t + 13t^2 - 4t^3 = [-t(1-5t+4t^2) + (1-6t+8t^2)]
P(5,t) = 1 - 8t + 19t^2 - 12t^3 = [-t(1-6t+8t^2) + (1-7t+13t^2)]
P(6,t) = 1 - 9t + 26t^2 - 25t^3 + 4t^4
P(7,t) = 1 - 10t + 34t^2 - 44t^3 + 16t^4
P(8,t) = 1 - 11t + 43t^2 - 70t^3 + 41t^4 - 4t^5
P(9,t) = 1 - 12t + 53t^2 - 104t^3 + 85t^4 - 20t^5
P(10,t) = 1 - 13t + 64t^2 - 147t^3 + 155t^4 - 61t^5 + 4t^6
P(11,t) = 1 - 14t + 76t^2 - 200t^3 + 259t^4 - 146t^5 + 24t^6
...
Apparently: The odd rows for n>1 are reversed rows of A140882 (mod signs), and the even rows for n>0, the 9th, 15th, 21st, 27th, etc. rows of A228785 (mod signs). The diagonals are reverse rows of A202241.
		

Crossrefs

Programs

  • Mathematica
    p = (1 - 4 t) / (1 - x + t x^2) + O[x]^12 // CoefficientList[#, x] &;
    CoefficientList[#, t] & /@ p // Flatten (* Andrey Zabolotskiy, Mar 07 2024 *)

Formula

O.g.f. G(x,t) = (1 - 4t)/(1 - x + t x^2) = a / [t (x - (1+sqrt(a))/(2t))(x - (1-sqrt(a))/(2t))] with a = 1-4t.
Recursion P(n,t) = -t P(n-2,t) + P(n-1,t) with P(-1,t)=0 and P(0,t) = 1-4t.
Convolution of the Fibonacci polynomials of signed A011973 Fb(n,-t) with coefficients of (1-4t), e.g., (1-4t)Fb(5,-t) = (1-4t)(1-3t+t^2) = 1-7t+13t^2-4t^3, so, for n>=1 and k<=(n-1), T(n,k) = (-1)^k [-4*binomial(n-(k-1),k-1) - binomial(n-k,k)] with the convention that 1/(-m)! = 0 for m>=1, i.e., let binomial(n,k) = nint[n!/((k+c)!(n-k+c)!)] for c sufficiently small in magnitude.
Third column is A034856, and the fourth, A000297. Embedded in the coefficients of the highest order term of the polynomials is A008586 (cf. also A008574).
With P(0,t)=0, the o.g.f. is H(x,t) = (1-4t) x(1-tx)/[1-x(1-tx)] = (1-4t) Linv(Cinv(tx)/t), where Linv(x) = x/(1-x) with inverse L(x) = x/(1+x) and Cinv(x) = x (1-x) is the inverse of the o.g.f. of the shifted Catalan numbers A000108, C(x) = [1-sqrt(1-4x)]/2. Then Hinv(x,t) = C[t L(x/(1-4t))]/t = {1 - sqrt[1-4t(x/(1-4t))/[1+x/(1-4t)]]}/2t = {1-sqrt[1-4tx/(1-4t+x)]}/2t = 1/(1-4t) + (-1+t)/(1-4t)^2 x + (1-2t+2t^2)/(1-4t)^3 x^ + (-1+3t-6t^2+5t^3)/(1-4t)^4 + ..., where the numerators are the signed polynomials of A098474, reverse of A124644.
Row sums (t=1) are periodic -3,-3,0,3,3,0, repeat the six terms ... with o.g.f. -3 - 3x (1-x) / [1-x(1-x)]. Cf. A084103.
Unsigned row sums (t=-1) are shifted A022088 with o.g.f. 5 + 5x(1+x) / [x(1+x)].

Extensions

Data corrected by Andrey Zabolotskiy, Mar 07 2024

A055035 Degree of minimal polynomial of sin(Pi/n) over the rationals.

Original entry on oeis.org

1, 1, 2, 2, 4, 1, 6, 4, 6, 2, 10, 4, 12, 3, 8, 8, 16, 3, 18, 8, 12, 5, 22, 8, 20, 6, 18, 12, 28, 4, 30, 16, 20, 8, 24, 12, 36, 9, 24, 16, 40, 6, 42, 20, 24, 11, 46, 16, 42, 10, 32, 24, 52, 9, 40, 24, 36, 14, 58, 16, 60, 15, 36, 32, 48, 10, 66, 32, 44, 12, 70, 24, 72
Offset: 1

Views

Author

Shawn Cokus (Cokus(AT)math.washington.edu)

Keywords

Comments

Also degree of minimal polynomial of function F(n)=(gamma(1/n)*gamma((n-1)/n))/Pi over the rationals. Roots of minimal polynomials of F(n) belonging to algebraic extension of sin(n/Pi) and vice versa (e.g. gamma(1/11)*gamma(10/11)/Pi = 20*sin(Pi/11) - 112*sin(Pi/11)^3 + 256*sin(Pi/11)^5 - 256*sin(Pi/11)^7 + (1024*sin(Pi/11)^9)/11). - Artur Jasinski, Oct 17 2011
The algebraic numbers sin(Pi/(2*l)) are given in A228783 in the power basis of the number field Q(2*cos(Pi/(2*l))) if n is even and of Q(2*cos(Pi/l)) if l is odd. In A228785, sin(Pi/(2*l+1)) is given in the power basis of Q(2*cos(Pi/(2*(2*l+1)))) (only odd powers appear). The minimal polynomials for 2*sin(Pi/n), n>=1, are given in A228786. - Wolfdieter Lang, Oct 10 2013

Crossrefs

Cf. A000010, A228786 (row length), A093819.

Programs

  • Mathematica
    a[n_] := If[n==2, 1, EulerPhi[n]/{1, 1, 2, 1}[[Mod[n, 4]+1]]]; Table[a[n], {n, 80}]
    a[n_] := Exponent[ MinimalPolynomial[Sin[Pi/n]][x], x]; Array[a, 75] (* Robert G. Wilson v, Jul 28 2014 *)

Formula

a(1)=1, a(2)=1, a(n)=phi(n)/(1, 1, 2, 1 for n=0, 1, 2, 3 mod 4) for n>2, where phi is Euler's totient, A000010
a(n) = A093819(2*n), n >= 1.- Wolfdieter Lang, Oct 29 2019

A228787 Decimal expansion of 2*sin(Pi/17), the ratio side/R in the regular 17-gon inscribed in a circle of radius R.

Original entry on oeis.org

3, 6, 7, 4, 9, 9, 0, 3, 5, 6, 3, 3, 1, 4, 0, 6, 6, 3, 1, 4, 8, 8, 1, 7, 6, 7, 9, 2, 4, 1, 4, 5, 5, 1, 6, 4, 9, 7, 8, 2, 7, 7, 0, 4, 7, 6, 8, 8, 9, 9, 8, 8, 1, 1, 7, 0, 1, 3, 0, 1, 7, 1, 5, 4, 9, 7, 8, 2, 9, 8, 5, 6, 5, 0, 6, 1, 0, 0, 3, 4, 6, 0, 6, 1, 2, 0, 2, 3, 9, 0, 2, 4, 2, 1, 4, 6, 0, 9, 7, 1, 8, 5, 9, 3, 5, 9, 5
Offset: 0

Views

Author

Wolfdieter Lang, Oct 07 2013

Keywords

Comments

s(17) := 2*sin(Pi/17) is an algebraic integer of degree 16 (over the rationals). Its minimal polynomial is 17 - 204*x^2 + 714*x^4 - 1122*x^6 + 935*x^8 - 442*x^10 + 119*x^12 - 17*x^14 + x^16. Its coefficients in the power basis of the algebraic number field Q(2*cos(Pi/34)) are [0, -15, 0, 140, 0, -378, 0, 450, 0, -275, 0, 90, 0, -15, 0, 1] (see row l = 8 of A228785). The decimal expansion of 2*cos(Pi/34) is given in A228788.
The continued fraction expansion starts with 0; 2, 1, 2, 1, 1, 2, 2, 2, 1, 5, 1, 3, 2, 2, 2, 1, 1, 43, 3, 1, 5, 2, 17, 2, ...
Gauss' formula for cos(2*Pi/17), given in A210644, can be inserted into s(17) = sqrt(2*(1 - cos(2*Pi/17))).
Since 17 is a Fermat prime, this number is constructible and can be written as an expression containing just integers, the basic four arithmetic operations, and square roots. See A003401 for more details. - Stanislav Sykora, May 02 2016

Examples

			0.367499035633140663148817679...
		

Crossrefs

Programs

Formula

s(17) = 2*sin(Pi/17) = 2*A241243.
Equals sqrt(34-2*sqrt(17)-2*sqrt(34-2*sqrt(17))-4*sqrt(17+3*sqrt(17)-sqrt(34-2*sqrt(17))-2*sqrt(34+2*sqrt(17))))/4. - Stanislav Sykora, May 02 2016

Extensions

Offset corrected by Rick L. Shepherd, Jan 01 2014

A228786 Table of coefficients of the minimal polynomials of 2*sin(Pi/n), n >= 1.

Original entry on oeis.org

0, 1, -2, 1, -3, 0, 1, -2, 0, 1, 5, 0, -5, 0, 1, -1, 1, -7, 0, 14, 0, -7, 0, 1, 2, 0, -4, 0, 1, -3, 0, 9, 0, -6, 0, 1, -1, 1, 1, -11, 0, 55, 0, -77, 0, 44, 0, -11, 0, 1, 1, 0, -4, 0, 1, 13, 0, -91, 0, 182, 0, -156, 0, 65, 0, -13, 0, 1, 1, -2, -1, 1, 1, 0, -8, 0, 14, 0, -7, 0, 1, 2, 0, -16, 0, 20, 0, -8, 0, 1, 17, 0, -204, 0, 714, 0, -1122, 0, 935, 0, -442, 0, 119, 0, -17, 0, 1, 1, -3, 0, 1
Offset: 1

Views

Author

Wolfdieter Lang, Oct 07 2013

Keywords

Comments

s(n) := 2*sin(Pi/n) is, for n >= 2, the length ratio side/R of the regular n-gon inscribed in a circle of radius R. This algebraic number s(n), n >= 1, has the degree gamma(n) := A055035(n), and the row length of this table is gamma(n) + 1.
s(n) has been given in the power basis of the relevant algebraic number field in A228783 for even n (bisected into n == 0 (mod 4) and n == 2 (mod 4)), and in A228785 for odd n.
For the computation of the minimal polynomials ps(n,x), using the coefficients of s(n) in the relevant number field, and the conjugates of the corresponding algebraic numbers rho (giving the length ratios (smallest diagonal)/side in the relevant regular polygons see a comment on A228781. Note that the product of the gamma(n) linear factors (x - conjugates) has to be computed modulo the minimal polynomial of the relevant rho(k) = 2*cos(Pi/k) (called C(k,x=rho(k)) in A187360).
Thanks go to Seppo Mustonen, who asked a question about the square of the sum of all lengths in the regular n-gon, which led to this computation of s(n) and its minimal polynomial.
It would be interesting to find out which length ratios in the regular n-gon give the other positive zeros of the minimal polynomial ps(n,x). See some examples below.
The zeros of the row polynomials ps(n,x) are 2*cos(2*Pi*k/c(2*n))) for gcd(k, c(2*n)) = 1, where c(n) = A178182(n), and k from {0, ..., floor(c(2*n)/2)}, for n >= 1. The number of these solutions is gamma(n) = A055035(n). See the formula section. This results from the zeros of the minimal polynomials of sin(2*Pi/n), with coefficients given in A181872/A181873. - Wolfdieter Lang, Oct 30 2019

Examples

			The table a(n, m) starts:
n\m   0  1    2 3   4 5     6 7   8 9   10 12  13 14  15 16 17
1:    0  1
2:   -2  1
3:   -3  0    1
4:   -2  0    1
5:    5  0   -5 0   1
6:   -1  1
7:   -7  0   14 0  -7 0     1
8:    2  0   -4 0   1
9:   -3  0    9 0  -6 0     1
10:  -1  1    1
11: -11  0   55 0 -77 0    44 0 -11 0    1
12:   1  0   -4 0   1
13:  13  0  -91 0 182 0  -156 0  65 0  -13 0   1
14:   1 -2   -1 1
15:   1  0   -8 0  14 0    -7 0   1
16:   2  0  -16 0  20 0    -8 0   1
17:  17  0 -204 0 714 0 -1122 0 935 0 -442  0 119  0 -17  0  1
18:   1 -3    0 1
...
n = 19: [-19, 0, 285, 0, -1254, 0, 2508, 0, -2717, 0, 1729, 0, -665, 0, 152, 0, -19, 0, 1],
n = 20: [1, 0, -12, 0, 19, 0, -8, 0, 1]
n = 5: ps(5,x) = 5 -5*x^2 +1*x^4, with the zeros s(5) = sqrt(3 - tau), sqrt(2 + tau) = tau*s(5) and their negative values, where tau =rho(5) is the golden section. tau*s(5) is the length ratio diagonal/radius in the pentagon.
n = 7: ps(7,x) = -7 + 14*x^2 -7*x^4 + 1*x^6, with the positive zeros s(7) (side/R) about 0.868, s(7)*rho(7) (smallest diagonal/R) about 1.564, and s(7)*(rho(7)^2-1) (longer diagonal/R) about 1.950 in the heptagon inscribed in a circle with radius R.
n = 8: ps(8,x) = 2 -4*x^2 + x^4, with the positive zeros s(8) = sqrt(2-sqrt(2)) and rho(8) = sqrt(2+sqrt(2)) (smallest diagonal/side).
n = 10: ps(10,x) = -1 + x + x^2 with the positive zero s(10) = tau - 1 (the negative solution is -tau).
		

Crossrefs

Formula

a(n, m) = [x^m](minimal polynomial ps(n, x) of 2*sin(Pi/n) over the rationals), n >= 1, m = 0, ..., gamma(n), with gamma(n) = A055035(n).
ps(n,x) = Product_{k=0..floor(c(2*n)/n) and gcd(k, c(2*n)) = 1} (x - 2*cos(2*Pi*k/c(2*n)), with c(2*n) = A178182(2*n), for n >= 1. There are gamma(n) = A055035(n) zeros. - Wolfdieter Lang, Oct 30 2019

A228783 Table of coefficients of the algebraic number s(2*l) = 2*sin(Pi/2*l) as a polynomial in powers of rho(2*l) = 2*cos(Pi/(2*l)) if l is even and of rho(l) = 2*cos(Pi/l) if l is odd (reduced version).

Original entry on oeis.org

2, 0, 1, 1, 0, -3, 0, 1, -1, 1, 0, 4, 0, -1, -1, -1, 1, 0, -7, 0, 14, 0, -7, 0, 1, 2, 1, -1, 0, 8, 0, -18, 0, 8, 0, -1, 1, 2, -3, -1, 1, 0, -8, 0, 6, 0, -1, 0, 0, -1, 3, 3, -4, -1, 1, 0, 12, 0, -67, 0, 96, 0, -52, 0, 12, 0, -1, -2, 3, 1, -1, 0, -15, 0, 140, 0, -378, 0, 450, 0, -275, 0, 90, 0, -15, 0, 1
Offset: 1

Views

Author

Wolfdieter Lang, Oct 06 2013

Keywords

Comments

In the regular (2*l)-gon inscribed in a circle of radius R the length ratio side/R is s(2*l) = 2*sin(Pi/(2*l)). This can be written as a polynomial in the length ratio (smallest diagonal)/side given by rho(2*l) = 2*cos(Pi/(2*l)). (For the 2-gon there is no such diagonal and rho(2) = 0). This leads, in a first step, to the triangle A127672 (see the Oct 05 2013 comment there referring also to the bisections signed A111125 and A127677). Because the minimal polynomial of the algebraic number rho(2*l) of degree delta(2*l) = A055034(2*l), called C(2*l,x) (with coefficients given in A187360) one can eliminate all powers rho(2*l)^k with k >= delta(2*l) by using C(2*l,rho(2*l)) = 0. Furthermore, because for odd l only even powers of rho(2*l) appear, but rho(2*l)^2 = 2 + rho(l), one will obtain a reduced table for s(2*l) with powers rho(2*l)^(2*k+1), k= 0, ..., (delta(2*l)-2)/2 if l is even, and with powers rho(l)^m, m=0, ... , delta(l)-1 if l is odd.
This leads to a reduction of the triangle A127672, when applied for the s(2*l) computation, giving the present table with row length delta(4*L) = A055034(4*L) = phi(8*L)/2 if l =2*L, if L >= 1, and phi(2*L+1)/2 = A055035(4*L+2), if l = 2*L + 1, L >= 1, where phi(n) = A000010(n) (Euler totient).
This table gives the coefficients of s(2*l) in the power basis of the algebraic number field Q(rho(2*l)) of degree delta(2*l) = A055034(2*l) if l is even, and in Q(rho(l)) of degree delta(2*l)/2 if l is odd. s(2) and s(6) are rational integers of degree 1.
Thanks go to Seppo Mustonen whose question about the square of the sum of all length in a regular n-gon, led me to this computation.
If l = 2*L+1, L >= 0, that is n == 2 (mod 4), one can obtain s(2*l) more directly in powers of rho(l) by s(2*l) = R(l-1, rho(l)) (mod C(l,rho(l))), with the monic (except for l=1) Chebyshev T-polynomials, called R, in A127672, and the C polynomials from A187360. - Wolfdieter Lang, Oct 10 2013

Examples

			The table a(l,m), with n = 2*l, begins:
n,  l \m  0   1   2    3   4   5   6    7   8   9  10  11 ...
2   1:    2
4   2:    0   1
6   3:    1
8   4:    0  -3   0    1
10  5:   -1   1
12  6:    0   4   0   -1
14  7:   -1  -1   1
16  8:    0  -7   0   14   0  -7   0    1
18  9:    2   1  -1
20 10:    0   8   0  -18   0   8   0   -1
22 11:    1   2  -3   -1   1
24 12:    0  -8   0    6   0  -1   0    0
26 13:   -1   3   3   -4  -1   1
28 14:    0  12   0  -67   0  96   0  -52  0  12  0  -1
30 15:   -2   3   1   -1
...
n = 8, l = 4:  s(8)  = -3*rho(8) + rho(8)^3 = -3*sqrt(2 + sqrt(2)) + (sqrt(2 + sqrt(2)))^3 = (sqrt(2) - 1)*sqrt(2 + sqrt(2)).
n = 10, l = 5:  s(10) =  -1 + rho(5), where rho(5) = tau = (1 + sqrt(5))/2, the golden section.
		

Crossrefs

Cf. A127672, A111125, A127677, A055034, A187360, A228785 (odd n case), A228786 (minimal polynomials).

Formula

a(2*L,m) = [x^m](s(4*L,x)(mod C(4*L,x))), with s(4*L,x) = sum((-1)^(L-1-s)*A111125(L-1,s)*x^(2*s+1),s=0..L-1), L >= 1, m =0, ..., delta(4*L)-1, and
a(2*L+1,m) = [x^m](s(4*L+2,x)(mod C(2*L+1,x))), with s(4*L+2,x) = sum(A127677(L,s)*(2+x)^(L-s)),s=0..L) (with s(2,x) = 2 for L = 0), L >= 0, m = 0, ..., delta(4*L+2)/2, with delta(n) = A055034(2*l).

A140882 Triangle by rows with row n formed by coefficients of the characteristic polynomial of the n X n tridiagonal matrix with m_{i,i} = 2 for i=1..n, m_{i,i-1} = m_{i,i+1} = -1 for i=2..n-1, and m_{1,2} = m_{n,n-1} = -2.

Original entry on oeis.org

1, 2, -1, 0, -4, 1, 0, -8, 6, -1, 0, -12, 19, -8, 1, 0, -16, 44, -34, 10, -1, 0, -20, 85, -104, 53, -12, 1, 0, -24, 146, -259, 200, -76, 14, -1, 0, -28, 231, -560, 606, -340, 103, -16, 1, 0, -32, 344, -1092, 1572, -1210, 532, -134, 18, -1, 0, -36, 489, -1968, 3630, -3652, 2171, -784, 169, -20, 1
Offset: 0

Views

Author

Roger L. Bagula and Gary W. Adamson, Jul 22 2008

Keywords

Examples

			  1;
  2,  -1;
  0,  -4,   1;
  0,  -8,   6,    -1;
  0, -12,  19,    -8,    1;
  0, -16,  44,   -34,   10,    -1;
  0, -20,  85,  -104,   53,   -12,    1;
  0, -24, 146,  -259,  200,   -76,   14,   -1;
  0, -28, 231,  -560,  606,  -340,  103,  -16,   1;
  0, -32, 344, -1092, 1572, -1210,  532, -134,  18,  -1;
  0, -36, 489, -1968, 3630, -3652, 2171, -784, 169, -20, 1;
  ...
		

References

  • Kemeny, Snell and Thompson, Introduction to Finite Mathematics, 1966, Prentice-Hall, New Jersey, Section 3, Chapter VII, page 407.

Crossrefs

Programs

  • Maple
    # Assume T(1, 0) = 0 instead of 2.
    # Then a slightly modified form of Copeland's second comment gives
    # T(n, k) = [t^k] [x^n] gf where
    gf := ((t - 4)*t*x^2) / ((t - 2)*x + x^2 + 1) - t*x + 1:
    ser := series(gf, x, 12): cx := n -> coeff(ser, x, n):
    for n from 0 to 10 do lprint(seq(coeff(cx(n), t, k), k = 0..n)) od;
    # Peter Luschny, Apr 27 2024
  • Mathematica
    T[n_, m_, d_] := If[ n == m, 2, If[(n == d && m == d - 1) || ( n == 1 && m == 2), -2, If[(n == m - 1 || n == m + 1), -1, 0]]];
    M[d_] := Table[T[n, m, d], {n, 1, d}, {m, 1, d}];
    a = Join[{{1}}, Table[CoefficientList[Det[M[ d] - x*IdentityMatrix[d]], x], {d, 1, 10}]];
    Flatten[a]

Formula

Starting with the third row (0, -4, 1), these coefficients appear to occur in the odd row polynomials in inverse powers of u for the o.g.f. (u - 4)/(u - u*x + x^2) of A267633, which generates a Fibonacci-type running average of adjacent polynomials of the o.g.f. involving these polynomials and others embedded in A228785. - Tom Copeland, Jan 16 2016
A generating function for the third and later shifted, unsigned rows is (4 + t) / (1 - (2 + t)*x + x^2) = Sum_{n >= 0} (4+t)*U_n((2 + t)/2)*x^n = 4 + t + (8 + 6*t + t^2)*x + ..., where U_n(t) are the Chebyshev polynomials of the second kind of A133156. U_n((2 + t)/2) = V_n((2 + t)), where V_n(t) are the Chebyshev polynomials of A049310. A formula for the coefficients is given by Eqn. 8 in Haukkanen et al. - Tom Copeland, Apr 26 2024

Extensions

Edited by the editors of the OEIS, Apr 27 2024
Showing 1-6 of 6 results.