A133476 -id:A133476 - cp's OEIS frontend

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

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

{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.

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

A138003 Binomial transform of 1, 1, 0, -1, -1 (periodically continued).

Original entry on oeis.org

1, 2, 3, 3, 0, -8, -21, -34, -34, 0, 89, 233, 377, 377, 0, -987, -2584, -4181, -4181, 0, 10946, 28657, 46368, 46368, 0, -121393, -317811, -514229, -514229, 0, 1346269, 3524578, 5702887, 5702887, 0, -14930352, -39088169, -63245986, -63245986

Offset: 0

Paul Curtz, May 01 2008

Shares many elements with A103311, as already indicated by the similarity of the two generating functions. First differences are essentially in A105371. - R. J. Mathar, May 02 2008

The longer of the two recurrences ensures that the sequence (like A133476) equals its 5th differences. - R. J. Mathar, May 02 2008

Paolo Xausa, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-4,2,-1).

Cf. A129929.

LinearRecurrence[{3,-4,2,-1},{1,2,3,3},50] (* Paolo Xausa, Dec 05 2023 *)

a=[1,2,3,3];for(i=1,99,a=concat(a,3*a[#a]-4*a[#a-1]+2*a[#a-2]-a[#a-3]));a \\ Charles R Greathouse IV, Jun 02 2011

From R. J. Mathar, May 02 2008: (Start)
O.g.f.: (x^2-x+1)/(x^4-2*x^3+4*x^2-3*x+1).
a(n) = 5a(n-1)-10a(n-2)+10a(n-3)-5a(n-4)+2a(n-5).
a(n) = 3a(n-1)-4a(n-2)+2a(n-3)-a(n-4). (End)

Edited by R. J. Mathar, May 02 2008

A307078 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, 3, 1, 2, 7, 1, 2, 4, 15, 1, 2, 3, 8, 31, 1, 2, 3, 5, 16, 63, 1, 2, 3, 4, 10, 32, 127, 1, 2, 3, 4, 6, 21, 64, 255, 1, 2, 3, 4, 5, 12, 43, 128, 511, 1, 2, 3, 4, 5, 7, 28, 86, 256, 1023, 1, 2, 3, 4, 5, 6, 14, 64, 171, 512, 2047, 1, 2, 3, 4, 5, 6, 8, 36, 136, 341, 1024, 4095

Offset: 0

Seiichi Manyama, Mar 22 2019

			Square array begins:
     1,   1,   1,   1,  1,  1,  1,  1, 1, ...
     3,   2,   2,   2,  2,  2,  2,  2, 2, ...
     7,   4,   3,   3,  3,  3,  3,  3, 3, ...
    15,   8,   5,   4,  4,  4,  4,  4, 4, ...
    31,  16,  10,   6,  5,  5,  5,  5, 5, ...
    63,  32,  21,  12,  7,  6,  6,  6, 6, ...
   127,  64,  43,  28, 14,  8,  7,  7, 7, ...
   255, 128,  86,  64, 36, 16,  9,  8, 8, ...
   511, 256, 171, 136, 93, 45, 18, 10, 9, ...

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

Columns 1-6 give A126646, A000079, A024494(n+1), A038504(n+1), A133476(n+1), A119336.

Cf. A306846, A306915, A307079.

T[n_, k_] := Sum[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)} binomial(n+1,k*j+1).

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

cp's OEIS Frontend

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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A138003 Binomial transform of 1, 1, 0, -1, -1 (periodically continued).

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Mathematica

PARI

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Extensions

A307078 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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A289389 a(n) = Sum_{k>=0} (-1)^kbinomial(n,5k+4).

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