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.

A307039 Square array A(n,k), n >= 0, k >= 1, read by antidiagonals, where column k is the expansion of g.f. ((1-x)^(k-1))/((1-x)^k+x^k).

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, -2, 0, 1, 1, 1, 0, -4, 0, 1, 1, 1, 1, -3, -4, 0, 1, 1, 1, 1, 0, -9, 0, 0, 1, 1, 1, 1, 1, -4, -18, 8, 0, 1, 1, 1, 1, 1, 0, -14, -27, 16, 0, 1, 1, 1, 1, 1, 1, -5, -34, -27, 16, 0, 1, 1, 1, 1, 1, 1, 0, -20, -68, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, -6, -55, -116, 81, -32, 0
Offset: 0

Views

Author

Seiichi Manyama, Mar 21 2019

Keywords

Examples

			Square array begins:
   1,  1,   1,    1,    1,   1,   1,  1, ...
   0,  1,   1,    1,    1,   1,   1,  1, ...
   0,  0,   1,    1,    1,   1,   1,  1, ...
   0, -2,   0,    1,    1,   1,   1,  1, ...
   0, -4,  -3,    0,    1,   1,   1,  1, ...
   0,  0, -18,  -14,   -5,   0,   1,  1, ...
   0,  8, -27,  -34,  -20,  -6,   0,  1, ...
   0, 16, -27,  -68,  -55, -27,  -7,  0, ...
   0, 16,   0, -116, -125, -83, -35, -8, ...
		

Crossrefs

Programs

  • Mathematica
    T[n_, k_] := Sum[(-1)^j * Binomial[n, k*j], {j, 0, Floor[n/k]}]; Table[T[n-k, k], {n, 0, 13}, {k, n, 1, -1}] // Flatten (* Amiram Eldar, May 20 2021 *)

Formula

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

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

A289389 a(n) = Sum_{k>=0} (-1)^k*binomial(n,5*k+4).

Original entry on oeis.org

0, 0, 0, 0, 1, 5, 15, 35, 70, 125, 200, 275, 275, 0, -1000, -3625, -9500, -21250, -42500, -76875, -124375, -171875, -171875, 0, 621875, 2250000, 5890625, 13171875, 26343750, 47656250, 77109375, 106562500, 106562500, 0, -385546875, -1394921875, -3651953125
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 + 4], {k, 0, n}], {n, 0, 36}] (* or *)
    CoefficientList[Series[(-x^4)/((-1 + x)^5 - x^5), {x, 0, 36}], x] (* Michael De Vlieger, Jul 10 2017 *)
  • PARI
    a(n) = sum(k=0, (n-4)\5, (-1)^k*binomial(n, 5*k+4)); \\ Michel Marcus, Jul 05 2017

Formula

G.f.: (-x^4)/((-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-8)/10) + (phi-1)^n*cos (3* Pi*(n-8)/10)), where phi is the golden ratio;
a(n+m) = a(n)*K_1(m) + K_4(n)*K_2(m) + K_3(n)*K_3(m) + K_2(n)*K_4(m) + K_1(n)*a(m), where K_1 is A289306, K_2 is A289321, K_3 is A289387, K_4 is A289388.
a(n) = 0 if and only if n=0,1,2 or n==3 (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

A348310 a(n) = Sum_{k=0..floor(n/10)} (-1)^k * binomial(n-5*k,5*k).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, -5, -20, -55, -125, -251, -461, -791, -1286, -2001, -3001, -4356, -6121, -8281, -10626, -12500, -12340, -6885, 10110, 49875, 131626, 286921, 565781, 1044971, 1838626, 3110751, 5087561, 8064366, 12395461, 18444251, 26451625, 36249035, 46692715, 54618710
Offset: 0

Views

Author

Seiichi Manyama, Oct 11 2021

Keywords

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{5, -10, 10, -5, 1, 0, 0, 0, 0, -1}, {1, 1, 1, 1, 1, 1, 1, 1, 1, 1}, 45] (* Amiram Eldar, Oct 11 2021 *)
  • PARI
    a(n) = sum(k=0, n\10, (-1)^k*binomial(n-5*k, 5*k));
    
  • PARI
    my(N=66, x='x+O('x^N)); Vec((1-x)^4/((1-x)^5+x^10))

Formula

G.f.: (1-x)^4/((1-x)^5 + x^10).
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) - a(n-10).
Showing 1-6 of 6 results.