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A143622
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a(n) = (-1)^binomial(n,8): Periodic sequence 1,1,1,1,1,1,1,1,-1,-1,-1,-1,-1,-1,-1,-1,... .
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3
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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, 1
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,1
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COMMENTS
| Periodic sequence with period 16. 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 A1436221 (r = 2).
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FORMULA
| a(n) = (-1)^binomial(n,8) = (-1)^floor(n/8), since sum {k = 1..n-7} k(k+1)...(k+6)/7! = binomial(n,8) == floor(n/8) (mod 2) for n = 0,1,...,15 by calculation and both sides increase by an even number if we substitute n+16 for n. a(n) = 1/8*((n+8) mod 16 - n mod 16). O.g.f.: (1+x+x^2+x^3+x^4+x^5+x^6+x^7)/(1+x^8 ) = (1+x)*(1+x^2)*(1+x^4)/(1+x^8) = (1-x^8)/((1-x)*(1+x^8)).
Define E(k) = sum {n = 0..inf} a(n)*n^k/n! for k = 0,1,2,... . Then E(k) is a an integral linear combination of E(0),E(1),...,E(7) (a Dobinski-type relation).
a(n)=(-1)^A180969(3,n)
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MAPLE
| with(combinat):
a := n -> (-1)^binomial(n, 8):
seq(a(n), n = 0..95);
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CROSSREFS
| A033999, A057077, A130151, A143621.
Sequence in context: A165326 A143621 A098417 * A076479 A155040 A033999
Adjacent sequences: A143619 A143620 A143621 * A143623 A143624 A143625
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KEYWORD
| easy,sign
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AUTHOR
| Peter Bala (pbala(AT)toucansurf.com), Aug 30 2008
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