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A143621
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a(n) = (-1)^binomial(n,4): Periodic sequence 1,1,1,1,-1,-1,-1,-1,... .
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4
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1, 1, 1, 1, -1, -1, -1, -1, 1, 1, 1, 1, -1, -1, -1, -1, 1, 1, 1, 1, -1, -1, -1, -1, 1, 1, 1, 1, -1, -1, -1, -1, 1, 1, 1, 1, -1, -1, -1, -1, 1, 1, 1, 1, -1, -1, -1, -1, 1, 1, 1, 1, -1, -1, -1, -1, 1, 1, 1, 1, -1, -1, -1, -1, 1, 1, 1, 1, -1, -1, -1, -1, 1, 1, 1, 1, -1, -1, -1, -1, 1, 1, 1, 1, -1
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OFFSET
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0,1
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COMMENTS
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Periodic sequence with period 8. More generally, it appears that (-1)^binomial(n,2^r) gives a periodic sequence of period 2^(r+1), the period consisting of a block of 2^r plus ones followed by a block of 2^r minus ones. See A033999 (r = 0), A057077 (r = 1) and A143622 (r = 3).
Nonsimple continued fraction expansion of A188943 = 1.767591879243... - R. J. Mathar, Mar 08 2012
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LINKS
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FORMULA
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a(n) = (-1)^binomial(n,4) = (-1)^floor(n/4), since Sum_{k = 1..n-3} k*(k+1)(k+2)/3! = binomial(n,4) == floor(n/4) (mod 2) for n = 0,1,...,7 by calculation and both sides increase by an even number if we substitute n+8 for n.
a(n) = (1/4)*((n+4) mod 8 - n mod 8).
O.g.f.: (1+x+x^2+x^3)/(1+x^4) = (1+x)*(1+x^2)/(1+x^4) = (1-x^4)/((1-x)*(1+x^4)).
Define E(k) = Sum_{n>=0} a(n)*n^k/n! for k = 0,1,2,... . Then E(k) is an integral linear combination of E(0), E(1), E(2) and E(3) (a Dobinski-type relation).
Euler transform of length 8 sequence [ 1, 0, 0, -2, 0, 0, 0, 1]. - Michael Somos, Sep 30 2011
G.f.: (1 - x^4)^2 / ((1 - x) * (1 - x^8)). a(n) = -a(-1 - n) for all n in Z. - Michael Somos, Sep 30 2011
E.g.f.: sin(x/sqrt(2))*sinh(x/sqrt(2)) + (sqrt(2)*sin(x/sqrt(2)) + cos(x/sqrt(2)))*cosh(x/sqrt(2)). - Ilya Gutkovskiy, Apr 15 2016
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EXAMPLE
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G.f. = 1 + x + x^2 + x^3 - x^4 - x^5 - x^6 - x^7 + x^8 + x^9 + x^10 + ...
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MAPLE
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with(combinat):
a := n -> (-1)^binomial(n, 4):
seq(a(n), n = 0..103);
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MATHEMATICA
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PROG
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(PARI) x='x+O('x^99); Vec((1-x^4)^2/((1-x)*(1-x^8))) \\ Altug Alkan, Apr 15 2016
(Python)
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CROSSREFS
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KEYWORD
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easy,sign
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AUTHOR
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STATUS
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approved
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