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 1-5 of 5 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

A289306 a(n) = Sum_{k >= 0}(-1)^k*binomial(n,5*k).

Original entry on oeis.org

1, 1, 1, 1, 1, 0, -5, -20, -55, -125, -250, -450, -725, -1000, -1000, 0, 3625, 13125, 34375, 76875, 153750, 278125, 450000, 621875, 621875, 0, -2250000, -8140625, -21312500, -47656250, -95312500, -172421875, -278984375, -385546875, -385546875, 0, 1394921875
Offset: 0

Views

Author

Vladimir Shevelev, Jul 02 2017

Keywords

Comments

{A289306, A289321, A289387, A289388, A289389} is the difference analog of the trigonometric functions {k_1(x), k_2(x), k_3(x), k_4(x), k_5(x)} of order 5. For the definitions of {k_i(x)} and the difference analog {K_i (n)} see [Erdelyi] and the Shevelev link respectively. - Vladimir Shevelev, Jul 24 2017

References

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

Crossrefs

Programs

  • Mathematica
    Table[Sum[(-1)^k*Binomial[n, 5 k], {k, 0, n}], {n, 0, 36}] (* or *)
    CoefficientList[Series[-((-1 + x)^4/((-1 + x)^5 - x^5)), {x, 0, 36}], x] (* Michael De Vlieger, Jul 04 2017 *)
    LinearRecurrence[{5,-10,10,-5},{1,1,1,1,1},40] (* Harvey P. Dale, Dec 23 2018 *)
  • PARI
    a(n) = sum(k=0, n\5, (-1)^k*binomial(n,5*k)); \\ Michel Marcus, Jul 02 2017

Formula

G.f.: -((-1+x)^4/((-1+x)^5-x^5)). - Peter J. C. Moses, Jul 02 2017
For n>=1, a(n) = (2/5)*(phi+2)^(n/2)*(cos(Pi*n/10) + (phi-1)^n*cos(3 * Pi* n/10)), where phi is the golden ratio. In particular, a(n) = 0 if and only if n==5 (mod 10).
a(n+m) = a(n)*a(m) - K_5(n)*K_2(m) - K_4(n)*K_3(m) - K_3(n)*K_4(m) - K_2(n)*K_5(m), where K_2 is A289321, K_3 is A289387, K_4 is A289388, K_5 is A289389. - Vladimir Shevelev, Jul 24 2017

A289321 a(n) = Sum_{k >= 0}(-1)^k*binomial(n,5*k+1).

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 5, 0, -20, -75, -200, -450, -900, -1625, -2625, -3625, -3625, 0, 13125, 47500, 124375, 278125, 556250, 1006250, 1628125, 2250000, 2250000, 0, -8140625, -29453125, -77109375, -172421875, -344843750, -623828125, -1009375000, -1394921875
Offset: 0

Views

Author

Vladimir Shevelev, Jul 02 2017

Keywords

Comments

a(n) = 0 for n == 7 (mod 10). - Robert Israel, Jul 12 2017
{A289306, A289321, A289387, A289388, A289389} is the difference analog of the trigonometric functions {k_1(x), k_2(x), k_3(x), k_4(x), k_5(x)} of order 5. For the definitions of {k_i(x)} and the difference analog {K_i (n)} see [Erdelyi] and the Shevelev link respectively. - Vladimir Shevelev, Jul 24 2017

Crossrefs

Programs

  • Maple
    f:= gfun:-rectoproc({5*a(n)-10*a(n+1)+10*a(n+2)-5*a(n+3)+a(n+4), a(0)=0,
    a(1)=1, a(2)=2, a(3) = 3, a(4)=4}, a(n), remember):
    map(f, [$0..40]); # Robert Israel, Jul 11 2017
  • Mathematica
    Table[Sum[(-1)^k*Binomial[n, 5 k + 1], {k, 0, n}], {n, 0, 35}] (* or *)
    CoefficientList[Series[((-1 + x)^3 x)/((-1 + x)^5 - x^5), {x, 0, 35}], x] (* Michael De Vlieger, Jul 04 2017 *)
    LinearRecurrence[{5,-10,10,-5},{0,1,2,3,4},40] (* Harvey P. Dale, Dec 25 2022 *)
  • PARI
    a(n) = sum(k=0, (n-1)\5, (-1)^k*binomial(n, 5*k+1)); \\ Michel Marcus, Jul 03 2017

Formula

G.f.: ((-1+x)^3 x)/((-1+x)^5-x^5). - Peter J. C. Moses, Jul 02 2017
For n>=1, a(n) = (2/5)*(phi+2)^(n/2)*(cos(Pi*(n-2)/10) + (phi-1)^n* cos (3*Pi*(n-2)/10)), where phi is the golden ratio.
a(n+m) = a(n)*K_1(m) + K_1(n)*a(m) - K_5(n)*K_3(m) - K_4(n)*K_4(m) - K_3(n)*K_5(m), where K_1 is A289306, K_3 is A289387, K_4 is A289388, K_5 is A289389. - Vladimir Shevelev, Jul 24 2017

A289388 a(n) = Sum_{k>=0} (-1)^k*binomial(n,5*k+3).

Original entry on oeis.org

0, 0, 0, 1, 4, 10, 20, 35, 55, 75, 75, 0, -275, -1000, -2625, -5875, -11750, -21250, -34375, -47500, -47500, 0, 171875, 621875, 1628125, 3640625, 7281250, 13171875, 21312500, 29453125, 29453125, 0, -106562500, -385546875, -1009375000, -2257031250, -4514062500
Offset: 0

Views

Author

Vladimir Shevelev, Jul 05 2017

Keywords

Comments

{A289306, A289321, A289387, A289388, A289389} is the difference analog of the trigonometric functions {k_1(x), k_2(x), k_3(x), k_4(x), k_5(x)} of order 5. For the definitions of {k_i(x)} and the difference analog {K_i (n)} see [Erdelyi] and the Shevelev link respectively.

References

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

Crossrefs

Programs

  • Mathematica
    Table[Sum[(-1)^k*Binomial[n, 5 k + 3], {k, 0, n}], {n, 0, 36}] (* or *)
    CoefficientList[Series[((-1 + x) x^3)/((-1 + x)^5 - x^5), {x, 0, 36}], x] (* Michael De Vlieger, Jul 10 2017 *)
  • PARI
    a(n) = sum(k=0, (n-3)\5, (-1)^k*binomial(n, 5*k+3)); \\ Michel Marcus, Jul 05 2017

Formula

G.f.: ((-1+x)*x^3)/((-1+x)^5 - x^5). - Peter J. C. Moses, Jul 05 2017
For n>=1, a(n) = (2/5)*(phi+2)^(n/2)*(cos(Pi*(n-6)/10) + (phi-1)^n*cos (3* Pi*(n-6)/10)), where phi is the golden ratio.
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) - K_5(n)*K_5(m), where K_1 is A289306, K_2 is A289321, K_3 is A289387, K_5 is A289389.
a(n) = 0 if and only if n=0, n=2 or n==1 (mod 10). - Vladimir Shevelev, Jul 15 2017

Extensions

More terms from Peter J. C. Moses, Jul 05 2017

A289387 a(n) = Sum_{k>=0} (-1)^k*binomial(n, 5*k+2).

Original entry on oeis.org

0, 0, 1, 3, 6, 10, 15, 20, 20, 0, -75, -275, -725, -1625, -3250, -5875, -9500, -13125, -13125, 0, 47500, 171875, 450000, 1006250, 2012500, 3640625, 5890625, 8140625, 8140625, 0, -29453125, -106562500, -278984375, -623828125, -1247656250, -2257031250
Offset: 0

Views

Author

Vladimir Shevelev, Jul 05 2017

Keywords

Comments

{A289306, A289321, A289387, A289388, A289389} is the difference analog of the trigonometric functions {k_1(x), k_2(x), k_3(x), k_4(x), k_5(x)} of order 5. For the definitions of {k_i(x)} and the difference analog {K_i (n)} see [Erdelyi] and the first Shevelev link respectively.
From Robert Israel, Jul 11 2017: (Start)
a(n)=0 for n == 9 (mod 10).
A112765(a(10*k)) = (5/2)*k - 3/4 - (-1)^k/4.
A112765(a(10*k+2)) = (5/2)*k - 1/4 + (-1)^k/4.
A112765(a(10*k+3)) = A112765(a(10*k+4)) = (5/2)*k + 1/4 - (-1)^k/4.
A112765(a(10*k+5)) = A112765(a(10*k+6)) = (5/2)*k + 3/4 + (-1)^k/4.
A112765(a(10*k+7)) = A112765(a(10*k+8)) = (5/2)*k + 5/4 - (-1)^k/4. (End)
Note that from author's formula (see below) we have that, except for zeros in the sequence mentioned by Robert Israel, there are only a(0) = a(1) = 0. Indeed, otherwise for some value of n we should have the equality (phi-1)^n = -cos(Pi*(n-4)/10)/cos(3*Pi*(n-4)/10). However, the absolute value of the right hand side takes the six distinct values only: 1, phi, phi^2, phi^(-1), phi^(-2), 1/3 (the last value we have when n == 9 (mod 10), since lim_{x->Pi/2}cos(x)/cos(3*x)= -1/3). Thus for n>=3, we have (phi-1)^n = phi^(-n) < |cos(Pi*(n-4)/10)/cos(3*Pi*(n-4)/10)|. - Vladimir Shevelev, Jul 15 2017

References

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

Crossrefs

Programs

  • Maple
    f:= gfun:-rectoproc({5*a(n)-10*a(n+1)+10*a(n+2)-5*a(n+3)+a(n+4), a(0)=0,
    a(1)=0, a(2)=1, a(3) = 3,a(4)=6},a(n),remember):
    map(f, [$0..40]); # Robert Israel, Jul 11 2017
  • Mathematica
    Table[Sum[(-1)^k*Binomial[n, 5 k + 2], {k, 0, n}], {n, 0, 35}] (* or *)
    CoefficientList[Series[-((-1 + x)^2 x^2)/((-1 + x)^5 - x^5), {x, 0, 35}], x] (* Michael De Vlieger, Jul 10 2017 *)
  • PARI
    a(n) = sum(k=0, (n-2)\5, (-1)^k*binomial(n, 5*k+2)); \\ Michel Marcus, Jul 05 2017

Formula

G.f.: -((-1+x)^2*x^2)/((-1+x)^5 - x^5). - Peter J. C. Moses, Jul 05 2017
For n>=1, a(n) = (2/5)*(phi+2)^(n/2)*(cos(Pi*(n-4)/10) + (phi-1)^n*cos(3* Pi*(n-4)/10)), where phi is the golden ratio.
a(n+m) = a(n)*K_1(m) + K_2(n)*K_2(m) + K_1(n)*a(m) - K_5(n)*K_4(m) - K_4(n)*K_5(m), where K_1 is A289306, K_2 is A289321, K_4 is A289388, K_5 is A289389.
For every n>=1, the determinant of circulant matrix of order 5 (see [Wikipedia]) with the entries (-1)^(i-1)* K_i(n), i=1..5, is 0. Here K_1, K_2, K_4 and K_5 are the same as in the previous formula, while K_3(n) = a(n). For a proof and a generalization see the second Shevelev link that also contains two unsolved problems. - Vladimir Shevelev, Jul 26 2017

Extensions

More terms from Peter J. C. Moses, Jul 05 2017
Showing 1-5 of 5 results.