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A203648
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Period length of sum(k^2, k=1..n) mod 2n, divided by 4.
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1
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1, 2, 9, 4, 5, 18, 7, 8, 27, 10, 11, 36, 13, 14, 45, 16, 17, 54, 19, 20, 63, 22, 23, 72, 25, 26, 81, 28, 29, 90, 31, 32, 99, 34, 35, 108, 37, 38, 117, 40, 41, 126, 43, 44, 135, 46, 47, 144, 49, 50, 153, 52, 53, 162, 55, 56, 171
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OFFSET
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1,2
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COMMENTS
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This sequence lists the period lengths of the sum of the first n squares mod 2*n. In most cases, sum(k^(2*j), k=1..n) mod 2*n will produce the same sequence. The periods appear to always end in 2 zeros.
The sum(k^j, k=1..n) mod 2 has period 4 repeating [1,1,0,0] for any j.
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LINKS
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Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index to sequences with linear recurrences with constant coefficients, signature (0,0,2,0,0,-1).
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FORMULA
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a(n) = 3*n if n mod 3 = 0, else n.
a(n) = n*(1+2*floor(((n+2) mod 3)/2))).
Contribution from Bruno Berselli, Jan 04 2012: (Start)
G.f.: x*(1+2*x+9*x^2+2*x^3+x^4))/((1-x)^2*(1+x+x^2)^2).
a(n) = 2n + 2n((-1)^(-2n/3)+(-1)^(2n/3)-1/2)/3.
a(n) = -a(-n) = 2*a(n-3)-a(n-6). (End)
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EXAMPLE
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sum(k^2,k=1..n) mod 4 has period 8 repeating [1,1,2,2,3,3,0,0] so a(2)= 2
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MAPLE
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seq(n*(1+floor(((n+2) mod 3)/2))), n= 1..57);
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MATHEMATICA
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CoefficientList[Series[(1+2*x+9*x^2+2*x^3+x^4)/((1-x)^2*(1+x+x^2)^2), {x, 0, 60}], x] (* Vincenzo Librandi, Mar 19 2012 *)
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PROG
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(MAGMA) [n*(1+2*Floor(((n+2) mod 3)/2)): n in [1..60]]; // Vincenzo Librandi, Mar 19 2012
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CROSSREFS
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Sequence in context: A016642 A070700 A222239 * A202324 A199267 A115290
Adjacent sequences: A203645 A203646 A203647 * A203649 A203650 A203651
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KEYWORD
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nonn,easy
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AUTHOR
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Gary Detlefs, Jan 04 2012
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STATUS
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approved
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