A289321 -id:A289321 - cp's OEIS frontend

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

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

Vladimir Shevelev, Jul 02 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

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

Seiichi Manyama, Table of n, a(n) for n = 0..3000
John B. Dobson, A matrix variation on Ramus's identity for lacunary sums of binomial coefficients, arXiv preprint arXiv:1610.09361 [math.NT], 2016.
Vladimir Shevelev, Combinatorial identities generated by difference analogs of hyperbolic and trigonometric functions of order n, arXiv:1706.01454 [math.CO], 2017.
Index entries for linear recurrences with constant coefficients, signature (5, -10, 10, -5).

Column 5 of A307039.

Cf. A139398, A133476, A139714, A139748, A139761.

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 *)

a(n) = sum(k=0, n\5, (-1)^k*binomial(n,5*k)); \\ Michel Marcus, Jul 02 2017

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

A289388 a(n) = Sum_{k>=0} (-1)^kbinomial(n,5k+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

Vladimir Shevelev, Jul 05 2017

sign

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

Vladimir Shevelev, Combinatorial identities generated by difference analogs of hyperbolic and trigonometric functions of order n, arXiv:1706.01454 [math.CO], 2017.
Index entries for linear recurrences with constant coefficients, signature (5, -10, 10, -5).

Cf. A139398, A133476, A139714, A139748, A139761.

Cf. A289306, A289321, A289387, A289389.

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 *)

a(n) = sum(k=0, (n-3)\5, (-1)^k*binomial(n, 5*k+3)); \\ Michel Marcus, Jul 05 2017

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

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

A289389 a(n) = Sum_{k>=0} (-1)^kbinomial(n,5k+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

Vladimir Shevelev, Jul 05 2017

sign

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

Vladimir Shevelev, Combinatorial identities generated by difference analogs of hyperbolic and trigonometric functions of order n, arXiv:1706.01454 [math.CO], 2017.
Index entries for linear recurrences with constant coefficients, signature (5, -10, 10, -5).

Cf. A139398, A133476, A139714, A139748, A139761.

Cf. A289306, A289321, A289387, A289388.

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 *)

a(n) = sum(k=0, (n-4)\5, (-1)^k*binomial(n, 5*k+4)); \\ Michel Marcus, Jul 05 2017

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

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

A289387 a(n) = Sum_{k>=0} (-1)^kbinomial(n, 5k+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

Vladimir Shevelev, Jul 05 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 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

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

Vladimir Shevelev, Combinatorial identities generated by difference analogs of hyperbolic and trigonometric functions of order n, arXiv:1706.01454 [math.CO], 2017.
Vladimir Shevelev, A matrix property of sums of binomial coefficients, Seqfan, Fri Jul 21 2017.
Wikipedia, Circulant matrix

Cf. A139398, A133476, A139714, A139748, A139761, A289306, A289321, A289388, A289389.

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

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 *)

a(n) = sum(k=0, (n-2)\5, (-1)^k*binomial(n, 5*k+2)); \\ Michel Marcus, Jul 05 2017

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

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

A307079 Square array A(n,k), n >= 0, k >= 1, read by antidiagonals, where column k is the expansion of g.f. ((1-x)^(k-2))/((1-x)^k+x^k).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 2, 3, 0, 1, 1, 2, 3, 3, -4, 1, 1, 2, 3, 4, 0, -8, 1, 1, 2, 3, 4, 4, -9, -8, 1, 1, 2, 3, 4, 5, 0, -27, 0, 1, 1, 2, 3, 4, 5, 5, -14, -54, 16, 1, 1, 2, 3, 4, 5, 6, 0, -48, -81, 32, 1, 1, 2, 3, 4, 5, 6, 6, -20, -116, -81, 32, 1

Offset: 0

Seiichi Manyama, Mar 22 2019

			Square array begins:
   1,  1,   1,    1,   1,   1, 1, 1, 1, ...
   1,  2,   2,    2,   2,   2, 2, 2, 2, ...
   1,  2,   3,    3,   3,   3, 3, 3, 3, ...
   1,  0,   3,    4,   4,   4, 4, 4, 4, ...
   1, -4,   0,    4,   5,   5, 5, 5, 5, ...
   1, -8,  -9,    0,   5,   6, 6, 6, 6, ...
   1, -8, -27,  -14,   0,   6, 7, 7, 7, ...
   1,  0, -54,  -48, -20,   0, 7, 8, 8, ...
   1, 16, -81, -116, -75, -27, 0, 8, 9, ...

Seiichi Manyama, Antidiagonals n = 0..139, flattened

Columns 1-6 give A000012, A099087, A057682(n+1), A099587(n+1), A289321(n+1), A307089.

Cf. A306914, A307039, A307078.

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

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

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

cp's OEIS Frontend

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

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

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

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

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A307079 Square array A(n,k), n >= 0, k >= 1, read by antidiagonals, where column k is the expansion of g.f. ((1-x)^(k-2))/((1-x)^k+x^k).

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

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

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

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