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A146160
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Period 4 sequence (1, 4, 1, 16)
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3
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1, 4, 1, 16, 1, 4, 1, 16, 1, 4, 1, 16, 1, 4, 1, 16, 1, 4, 1, 16, 1, 4, 1, 16, 1, 4, 1, 16, 1, 4, 1, 16, 1, 4, 1, 16, 1, 4, 1, 16, 1, 4, 1, 16, 1, 4, 1, 16, 1, 4, 1, 16, 1, 4, 1, 16, 1, 4, 1, 16, 1, 4, 1, 16, 1, 4, 1, 16, 1, 4, 1, 16, 1, 4, 1, 16, 1, 4, 1, 16, 1, 4, 1, 16, 1, 4, 1, 16, 1, 4, 1, 16, 1
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,2
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FORMULA
| 1) Continued fraction of (8 + Sqrt[78])/14 2) GCD[4 k - k^2, 5 k^2, 20 k - 20 k^2, 16 - 32 k + 16 k^2] k=1,2,3,...
Contribution from Artur Jasinski (grafix(AT)csl.pl), Oct 29 2008: (Start)
a(n)=1 when n congruent to 1 or 3 mod 4
a(n)=4 when n congruent to 2 mod 4
a(n)=16 when n congruent to 0 mod 4
(End)
a(n+4)=a(n); a(n)=4.5*(-1)^(n+1) + 5.5 + 6*cos(Pi*(n+1)/2) ; o.g.f f(z)=a(0)+a(1)*z+...=(1+4*+z^2+16*z^3)/(1-z^4) [From Richard Choulet (richardchoulet(AT)yahoo.fr), Nov 03 2008]
a(n)=(1/6)*{28*(n mod 4)-17*[(n+1) mod 4]+10*[(n+2) mod 4]+[(n+3) mod 4]}, with n>=0 [From Paolo P. Lava (paoloplava(AT)gmail.com), Nov 06 2008]
a(n)=(11/2)+3*I^(n+1)-(9/2)*(-1)^n-3*I^(1-n), with n>=0 and I=sqrt(-1) [From Paolo P. Lava (paoloplava(AT)gmail.com), May 04 2010]
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MATHEMATICA
| Table[GCD[4 k - k^2, 5 k^2, 20 k - 20 k^2, 16 - 32 k + 16 k^2], {k, 1, 100}]
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CROSSREFS
| A010156
A145996 [From Artur Jasinski (grafix(AT)csl.pl), Oct 29 2008]
Sequence in context: A056920 A123382 A197653 * A059222 A117292 A062780
Adjacent sequences: A146157 A146158 A146159 * A146161 A146162 A146163
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KEYWORD
| nonn,mult
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AUTHOR
| Artur Jasinski (grafix(AT)csl.pl), Oct 27 2008
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