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A289387
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a(n) = Sum_{k>=0} (-1)^k*binomial(n, 5*k+2).
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5
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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
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OFFSET
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0,4
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COMMENTS
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{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.
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+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
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REFERENCES
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A. Erdelyi, Higher Transcendental Functions, McGraw-Hill, 1955, Vol. 3, Chapter XVIII.
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LINKS
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FORMULA
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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
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MAPLE
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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):
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MATHEMATICA
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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 *)
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PROG
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(PARI) a(n) = sum(k=0, (n-2)\5, (-1)^k*binomial(n, 5*k+2)); \\ Michel Marcus, Jul 05 2017
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CROSSREFS
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KEYWORD
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sign,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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