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 1-4 of 4 results.

A099586 Constant term in (1+x)^n mod (1+x^4).

Original entry on oeis.org

1, 1, 1, 0, -4, -14, -34, -68, -116, -164, -164, 0, 560, 1912, 4616, 9232, 15760, 22288, 22288, 0, -76096, -259808, -627232, -1254464, -2141504, -3028544, -3028544, 0, 10340096, 35303296, 85229696, 170459392, 290992384, 411525376, 411525376, 0, -1405035520
Offset: 1

Views

Author

Ralf Stephan, Oct 24 2004

Keywords

Comments

Equals real part of term (1,1) in M^n, where M = a 2 X 2 matrix [1,1; i,1], where i = sqrt(-1). - Gary W. Adamson, Mar 25 2009
{A099586, A099587, A099588, A099589} is the difference analog of the trigonometric functions {k_1(x), k_2(x), k_3(x), k_4(x)} of order 4.
For the definition, see [Erdelyi] and the Shevelev link. - Vladimir Shevelev, Jul 03 2017
a(n) = 0 if and only if n == 4 (mod 8). - Robert Israel, Jul 04 2017

References

  • A. Erdelyi, Higher Transcendental Functions, McGraw-Hill, 1955, Vol. 3, Chapter XVIII.

Crossrefs

Programs

  • Maple
    seq(eval(rem((1+x)^n, 1+x^4,x),x=0),n=1..40); # Robert Israel, Jul 03 2017
  • Mathematica
    RecurrenceTable[{a[n] == 4 * a[n - 1] - 6 * a[n - 2] + 4 * a[n - 3] - 2 * a[n - 4], a[1] = 1, a[2] = 1, a[3] = 1, a[4] = 0}, a, {n, 50}] (* G. C. Greubel, Nov 10 2015 *)
    a[n_] := HypergeometricPFQ[{(1-n)/4, (2-n)/4, (3-n)/4, -n/4}, {1/4, 1/2, 3/4}, -1]; Array[a, 40] (* Jean-François Alcover, Jul 20 2017, from Vladimir Shevelev's first formula *)
  • PARI
    a(n) = polcoeff(((1+x)^n)%(x^4+1),0)
    
  • PARI
    Vec(-x*(2*x-1)*(x^2-x+1)/(2*x^4-4*x^3+6*x^2-4*x+1) + O(x^100)) \\ Colin Barker, Nov 08 2015
    
  • PARI
    a(n) = real(([1,1; I,1])^n)[1,1]; \\ Michel Marcus, Nov 08 2015

Formula

From Colin Barker, Nov 08 2015: (Start)
a(n) = 4*a(n-1)-6*a(n-2)+4*a(n-3)-2*a(n-4) for n>4.
G.f.: -x*(2*x-1)*(x^2-x+1) / (2*x^4-4*x^3+6*x^2-4*x+1).
(End)
a(n) = (1/2)*((2+sqrt(2))^(n/2)*cos(n*Pi/8) + (2-sqrt(2))^(n/2)*cos(3*n*Pi/8)). - G. C. Greubel, Nov 10 2015
From Vladimir Shevelev, Jun 29 2017: (Start)
a(n) = Sum_{k >= 0}(-1)^k*binomial(n,4*k).
a(n) = round((2+sqrt(2))^(n/2)*cos(Pi*(n)/8)/2), where round(x) is the integer nearest to x.
a(n+m) = a(n)*a(m) - K_4(n)*K_2(m) - K_3(n)*K_3(m) - K_2(n)*K_4(m), where K_2 is A099587, K_3 is A099588 and K_4 is A099589.
(End)

A099587 a(n) = coefficient of x in (1+x)^n mod (1+x^4).

Original entry on oeis.org

0, 1, 2, 3, 4, 4, 0, -14, -48, -116, -232, -396, -560, -560, 0, 1912, 6528, 15760, 31520, 53808, 76096, 76096, 0, -259808, -887040, -2141504, -4283008, -7311552, -10340096, -10340096, 0, 35303296, 120532992, 290992384, 581984768
Offset: 0

Views

Author

Ralf Stephan, Oct 24 2004

Keywords

Comments

{A099586, A099587, A099588, A099589} is the difference analog of the trigonometric functions {k_1(x), k_2(x), k_3(x), k_4(x)} of order 4.
For the definition, see [Erdelyi] and the Shevelev link. - Vladimir Shevelev, Jul 03 2017

References

  • A. Erdelyi, Higher Transcendental Functions, McGraw-Hill, 1955, Vol. 3, Chapter XVIII.

Crossrefs

Programs

  • Mathematica
    RecurrenceTable[{a[1]=1, a[2]=2, a[3]=3, a[4]=4, a[n] = 4*a[n-1] - 6*a[n-2] + 4*a[n-3] - 2*a[n-4]}, a, {n, 1, 100}] (* G. C. Greubel, Nov 09 2015 *)
    a[n_] := n*HypergeometricPFQ[{(1-n)/4, (2-n)/4, (3-n)/4, (4-n)/4}, {1/2, 3/4, 5/4}, -1]; Array[a, 40, 0] (* Jean-François Alcover, Jul 20 2017, from Vladimir Shevelev's first formula *)
    LinearRecurrence[{4,-6,4,-2},{0,1,2,3},50] (* Harvey P. Dale, Mar 27 2022 *)
  • PARI
    a(n) = polcoeff(((1+x)^n)%(x^4+1),1)

Formula

G.f.: x*(x-1)^2 / (2*x^4-4*x^3+6*x^2-4*x+1). - Colin Barker, Jul 15 2013
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - 2*a(n-4). - G. C. Greubel, Nov 09 2015
From Vladimir Shevelev, Jun 29 2017: (Start)
a(n) = Sum_{k >= 0}(-1)^k*binomial(n,4*k+1).
a(n) = round((2+sqrt(2))^(n/2)*cos(Pi*(n-2)/8)/2), where round(x) is the integer nearest to x.
a(n+m) = a(n)*K_1(m) + K_1(n)*a(m) - K_4(n)*K_3(m) - K_3(n)*K_4(m), where K_1 is A099586, K_3=A099588, and K_4=A099589.
(End)
a(n) = A099589(n+2)-2*A099589(n+1)+A099589(n). - R. J. Mathar, Jun 28 2025

Extensions

a(0)=0 added by N. J. A. Sloane, Jun 30 2017

A099589 Expansion of x^3 / (1 - 4*x + 6*x^2 - 4*x^3 + 2*x^4).

Original entry on oeis.org

0, 0, 0, 1, 4, 10, 20, 34, 48, 48, 0, -164, -560, -1352, -2704, -4616, -6528, -6528, 0, 22288, 76096, 183712, 367424, 627232, 887040, 887040, 0, -3028544, -10340096, -24963200, -49926400, -85229696, -120532992, -120532992, 0, 411525376, 1405035520, 3392055808
Offset: 0

Views

Author

Ralf Stephan, Oct 24 2004

Keywords

Comments

{A099586, A099587, A099588, A099589} is the difference analog of the trigonometric functions {k_1(x), k_2(x), k_3(x), k_4(x)} of order 4. For the definition, see [Erdelyi] and the Shevelev link. - Vladimir Shevelev, Jul 04 2017

References

  • A. Erdelyi, Higher Transcendental Functions, McGraw-Hill, 1955, Vol. 3, Chapter XVIII.

Crossrefs

Programs

  • Mathematica
    Round@Table[(1/(2*Sqrt[2]))*((2-Sqrt[2])^(n/2)*(Cos[3*Pi*n/8] + Sin[3*Pi*n/8]) + (2+Sqrt[2])^(n/2)*(Sin[Pi*n/8] - Cos[Pi*n/8])), {n, 0, 40}] (* G. C. Greubel, Nov 07 2015 *)
    RecurrenceTable[{a[n] == 4*a[n-1] - 6*a[n-2] + 4*a[n-3] - 2*a[n-4], a[0]==0, a[1]==0, a[2]==0, a[3]==1}, a, {n, 0, 40}] (* G. C. Greubel, Nov 10 2015 *)
    Table[Sum[(-1)^k*Binomial[n, 4 k + 3], {k, 0, n}], {n, 0, 37}] (* Michael De Vlieger, Jun 30 2017 *)
    a[n_] := n*(n-1)*(n-2)/6 HypergeometricPFQ[{(3-n)/4, (4-n)/4, (5-n)/4, (6-n)/4}, {5/4, 3/2, 7/4}, -1]; Array[a, 40, 0] (* Jean-François Alcover, Jul 20 2017, from Vladimir Shevelev's first formula *)
  • PARI
    a(n) = polcoeff(((1+x)^n)%(x^4+1),3)
    
  • PARI
    concat([0, 0], Vec(x^3/((1-x)^4+x^4) + O(x^50))) \\ Altug Alkan, Nov 08 2015
    
  • PARI
    a(n) = sum(t=0, (n-3)\4, (-1)^t*binomial(n,4*t+3)); \\ Michel Marcus, Jun 30 2017

Formula

G.f.: x^3/((1-x)^4 + x^4), the binomial transform of x^3/(1+x^4). - Paul Barry, Apr 01 2005
Coefficient of x^3 in (1+x)^n mod (1 + x^4).
a(n) = (1/(2*sqrt(2)))*((2-sqrt(2))^(n/2)*(cos(3*Pi*n/8) + sin(3*Pi*n/8)) + (2+sqrt(2))^(n/2)*(sin(Pi*n/8) - cos(Pi*n/8))). - Paul Barry, Apr 01 2005
From Colin Barker, Nov 08 2015: (Start)
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - 2*a(n-4) for n > 4.
G.f.: x^3 / (2*x^4 - 4*x^3 + 6*x^2 - 4*x + 1). (End)
From Vladimir Shevelev, Jul 04 2017: (Start)
a(n) = Sum_{t >= 0} (-1)^t*binomial(n,4*t+3).
a(n) = round((2+sqrt(2))^(n/2)*cos(Pi*(n-6)/8)/2), where round(x) is the integer nearest to x.
a(n+m) = a(n)*K_1(m) + K_3(n)*K_2(m) + K_2(n)*K_3(m) + K_1(n)*a(m), where
K_1 is A099586, K_2 is A099587, K_3 is A099588. (End)

Extensions

a(0)=0 added by N. J. A. Sloane, Jul 04 2017

A290286 Determinant of circulant matrix of order 4 with entries in the first row (-1)^j*Sum_{k>=0}(-1)^k*binomial(n, 4*k+j), j=0,1,2,3.

Original entry on oeis.org

1, 0, 0, 0, -1008, -37120, -473600, 0, 63996160, 702013440, 2893578240, 0, -393379835904, -12971004067840, -160377313820672, 0, 21792325059543040, 239501351489372160, 987061897553510400, 0, -134124249770961666048, -4422152303189489090560
Offset: 0

Views

Author

Keywords

Comments

In the Shevelev link the author proved that, for odd N>=3 and every n>=1, the determinant of circulant matrix of order N with entries in the first row (-1)^j*Sum{k>=0}(-1)^k*binomial(n, N*k+j), j=0..N-1, is 0.
This sequence shows what happens for the first even N>3.

Crossrefs

Cf. A099586 (prefixed by a(0)=1), A099587, A099588, A099589, A290285.

Programs

  • Maple
    seq(LinearAlgebra:-Determinant(Matrix(4,shape=Circulant[seq((-1)^j*
    add((-1)^k*binomial(n, 4*k+j),k=0..n/4),j=0..3)])),n=0..50); # Robert Israel, Jul 26 2017
  • Mathematica
    ro[n_] := Table[Sum[(-1)^(j+k) Binomial[n, 4k+j], {k, 0, n/4}], {j, 0, 3}];
    M[n_] := Table[RotateRight[ro[n], m], {m, 0, 3}];
    a[n_] := Det[M[n]];
    Table[a[n], {n, 0, 21}] (* Jean-François Alcover, Aug 09 2018 *)
  • Python
    from sympy.matrices import Matrix
    from sympy import binomial
    def mj(j, n): return (-1)**j*sum((-1)**k*binomial(n, 4*k + j) for k in range(n//4 + 1))
    def a(n):
        m=Matrix(4, 4, lambda i,j: mj((i-j)%4,n))
        return m.det()
    print([a(n) for n in range(22)]) # Indranil Ghosh, Jul 31 2017

Formula

a(n) = 0 for n == 3 (mod 4).
G.f. (empirical): (1/8)*(68*x^2+1)/(16*x^4+136*x^2+1)+(1/4)*(68*x^2-8*x+1)/(16*x^4+64*x^3+128*x^2-16*x+1)+(1/2)*(12*x^2+1)/(16*x^4+24*x^2+1)+3/(8*(4*x^2+1))-(1/4)*(12*x^2-4*x+1)/(16*x^4-32*x^3+32*x^2-8*x+1)-(1/4)*(4*x^2+1)/(16*x^4+1)+(1/4)*(12*x^2+4*x+1)/(16*x^4+32*x^3+32*x^2+8*x+1). - Robert Israel, Jul 26 2017
Showing 1-4 of 4 results.