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.

A306914 Square array A(n,k), n >= 0, k >= 1, read by antidiagonals, where column k is the expansion of g.f. 1/((1-x)^k+x^k).

Original entry on oeis.org

1, 1, 0, 1, 2, 0, 1, 3, 2, 0, 1, 4, 6, 0, 0, 1, 5, 10, 9, -4, 0, 1, 6, 15, 20, 9, -8, 0, 1, 7, 21, 35, 34, 0, -8, 0, 1, 8, 28, 56, 70, 48, -27, 0, 0, 1, 9, 36, 84, 126, 125, 48, -81, 16, 0, 1, 10, 45, 120, 210, 252, 200, 0, -162, 32, 0
Offset: 0

Views

Author

Seiichi Manyama, Mar 16 2019

Keywords

Examples

			Square array begins:
   1,  1,    1,    1,   1,    1,    1,    1, ...
   0,  2,    3,    4,   5,    6,    7,    8, ...
   0,  2,    6,   10,  15,   21,   28,   36, ...
   0,  0,    9,   20,  35,   56,   84,  120, ...
   0, -4,    9,   34,  70,  126,  210,  330, ...
   0, -8,    0,   48, 125,  252,  462,  792, ...
   0, -8,  -27,   48, 200,  461,  924, 1716, ...
   0,  0,  -81,    0, 275,  780, 1715, 3432, ...
   0, 16, -162, -164, 275, 1209, 2989, 6434, ...
		

Crossrefs

Columns 1-9 give A000007, A099087, A057083, A099589(n+3), A289389(n+4), A306940, (-1)^n * A049018(n), A306941, A306942.

Programs

  • Mathematica
    A[n_, k_] := SeriesCoefficient[1/((1-x)^k + x^k), {x, 0, n}];
    Table[A[n-k+1, k], {n, 0, 11}, {k, n+1, 1, -1}] // Flatten (* Jean-François Alcover, Mar 20 2019 *)

Formula

A(n,k) = Sum_{j=0..floor(n/k)} (-1)^j * binomial(n+k-1,k*j+k-1).
A(n,2*k) = Sum_{i=0..n} Sum_{j=0..n-i} (-1)^j * binomial(i+k-1,k*j+k-1) * binomial(n-i+k-1,k*j+k-1). - Seiichi Manyama, Apr 07 2019

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

A099588 Coefficient of x^2 in (1+x)^n mod 1+x^4.

Original entry on oeis.org

0, 0, 1, 3, 6, 10, 14, 14, 0, -48, -164, -396, -792, -1352, -1912, -1912, 0, 6528, 22288, 53808, 107616, 183712, 259808, 259808, 0, -887040, -3028544, -7311552, -14623104, -24963200, -35303296, -35303296, 0, 120532992, 411525376, 993510144, 1987020288, 3392055808, 4797091328, 4797091328
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

  • Maple
    f:= rectoproc({rec, a(0)=0,a(1)=0,a(2)=1,a(3)=3},a(n),remember):
    map(f, [$0..100]); # Robert Israel, Jun 30 2017
  • Mathematica
    RecurrenceTable[{a[n]==4*a[n-1] - 6*a[n-2] + 4*a[n-3] - 2*a[n-4], a[1]=0, a[2]=1, a[3]=3, a[4]=6}, a, {n, 1, 200}] (* G. C. Greubel, Nov 10 2015 *)
    Table[Sum[(-1)^k*Binomial[n, 4 k + 2], {k, 0, n}], {n, 0, 36}] (* Michael De Vlieger, Jun 30 2017 *)
    a[n_] := n*(n-1)/2 HypergeometricPFQ[{(2-n)/4, (3-n)/4, (4-n)/4, (5-n)/4}, {3/4, 5/4, 3/2}, -1]; Array[a, 40, 0] (* Jean-François Alcover, Jul 20 2017, from Vladimir Shevelev's first formula *)
  • PARI
    x='x+O('x^55); concat([0, 0], Vec(-x^2*(x-1)/(2*x^4-4*x^3+6*x^2-4*x+1))) \\ Altug Alkan, Nov 11 2015
    
  • PARI
    a(n) = sum(t=0, (n-2)\4, (-1)^t*binomial(n,4*t+2)); \\ Michel Marcus, Jun 30 2017
    
  • PARI
    a(n)=polcoeff(lift(Mod(1+x,1+x^4)^n),2); \\ Joerg Arndt, Feb 22 2018

Formula

G.f.: -x^2*(x-1) / (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 10 2015
a(n) = (1/2)*((2+sqrt(2))^(n/2) * sin(n*Pi/8) - (2-sqrt(2))^(n/2)*sin(3*n*Pi/8)). - G. C. Greubel, Nov 10 2015
From Vladimir Shevelev, Jul 04 2017 (Start)
a(n) = Sum_{k>=0}(-1)^k*binomial(n,4*k+2);
a(n) = round((2+sqrt(2))^(n/2)*cos(Pi*(n-4)/8)/2), where round(x) is the integer nearest to x;
a(n+m) = a(n)*K_1(m) + K_2(n)*K_2(m) + K_1(n)*a(m) - K_4(n)*K_4(m), where K_1 is A099586, K_2 is A099587, K_4 is A099589. (End)

Extensions

a(0)=0 added by N. J. A. Sloane, Jul 04 2017
a(673) in b-file corrected by Andrew Howroyd, Feb 21 2018

A307394 Square array A(n,k), n >= 0, k >= 1, read by antidiagonals, where column k is the expansion of g.f. ((1-x)^(k-4))/((1-x)^k+x^k).

Original entry on oeis.org

1, 1, 3, 1, 4, 6, 1, 4, 9, 10, 1, 4, 10, 14, 15, 1, 4, 10, 19, 15, 21, 1, 4, 10, 20, 28, 8, 28, 1, 4, 10, 20, 34, 28, -7, 36, 1, 4, 10, 20, 35, 48, 1, -22, 45, 1, 4, 10, 20, 35, 55, 48, -80, -21, 55, 1, 4, 10, 20, 35, 56, 75, 0, -242, 12, 66, 1, 4, 10, 20, 35, 56, 83, 75, -164, -485, 77, 78
Offset: 0

Views

Author

Seiichi Manyama, Apr 07 2019

Keywords

Examples

			Square array begins:
    1,   1,    1,    1,  1,   1,   1,   1,   1, ...
    3,   4,    4,    4,  4,   4,   4,   4,   4, ...
    6,   9,   10,   10, 10,  10,  10,  10,  10, ...
   10,  14,   19,   20, 20,  20,  20,  20,  20, ...
   15,  15,   28,   34, 35,  35,  35,  35,  35, ...
   21,   8,   28,   48, 55,  56,  56,  56,  56, ...
   28,  -7,    1,   48, 75,  83,  84,  84,  84, ...
   36, -22,  -80,    0, 75, 110, 119, 120, 120, ...
   45, -21, -242, -164,  0, 110, 154, 164, 165, ...
		

Crossrefs

Columns 1-5 give A000217(n+1), A279230, A307395, A099589(n+3), A289388(n+3).

Programs

  • Mathematica
    T[n_, k_] := Sum[(-1)^j * Binomial[n+3, k*j + 3], {j, 0, Floor[n/k]}]; Table[T[n - k, k], {n, 0, 12}, {k, n, 1, -1}] // Flatten (* Amiram Eldar, May 20 2021 *)

Formula

A(n,k) = Sum_{j=0..floor(n/k)} (-1)^j * binomial(n+3,k*j+3).
A(n,2*k) = Sum_{i=0..n} Sum_{j=0..n-i} (-1)^j * binomial(i+1,k*j+1) * binomial(n-i+1,k*j+1).

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 1-6 of 6 results.