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Search: id:A143621
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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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+0
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
(list; graph; listen)
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OFFSET
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0,1
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COMMENT
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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).
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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..inf} 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).
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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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CROSSREFS
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A033999, A057077, A130151, A143622.
Sequence in context: A097807 A014077 A165326 this_sequence A098417 A143622 A076479
Adjacent sequences: A143618 A143619 A143620 this_sequence A143622 A143623 A143624
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KEYWORD
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easy,sign
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AUTHOR
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Peter Bala (pbala(AT)toucansurf.com), Aug 30 2008
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