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A203648
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a(n) = (1/4) * period of repeating sequence {S(j) mod 2n}, where S(j) is the sum of the first j squares.
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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, 58, 59, 180, 61, 62, 189, 64, 65
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OFFSET
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1,2
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COMMENTS
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This sequence lists the periods of the sum of the first n squares mod 2*n. In most cases, (Sum_{k=1..n} k^(2*j)) mod 2*n will produce the same sequence. The repeating sequences appear to always end in 2 zeros.
(Sum_{k=1..n} k^j) mod 2 has period 4 repeating [1,1,0,0] for any j.
It appears that a(n) is the number of n-colorings of the trefoil knot. - Tsuyoshi Miezaki, May 01 2022
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LINKS
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FORMULA
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a(n) = 3*n if n mod 3 = 0, otherwise n.
a(n) = n*(1 + 2*floor(((n+2) mod 3)/2)).
G.f.: x*(1 + 2*x + 9*x^2 + 2*x^3 + x^4)/((1-x)^2*(1 + x + x^2)^2).
a(n) = 2*n + 2*n*((-1)^(-2*n/3) + (-1)^(2*n/3)-1/2)/3.
a(n) = -a(-n) = 2*a(n-3) - a(n-6). (End)
a(n) = numerator(3n/((3 + 2*(-1 + n))*(1 + n))). - Andres Cicuttin, Jan 12 2017
a(n) is multiplicative with a(p^e) = p^(e+1) if p = 3, a(p^e) = p^e otherwise. - Michael Somos, Jan 18 2017
a(n) = n*(5 + 4*cos((2*Pi*n)/3)) / 3. - Colin Barker, Mar 06 2017
Dirichlet g.f.: zeta(s-1)*(3^(-s)*(6 + 3^s)).
Sum_{k=1..n} a(k) ~ (5/6) * n^2. (End)
Sum_{n>=1} (-1)^(n+1)/a(n) = 7*log(2)/9. - Amiram Eldar, Aug 21 2023
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EXAMPLE
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G.f. = x + 2*x^2 + 9*x^3 + 4*x^4 + 5*x^5 + 18*x^6 + 7*x^7 + 8*x^8 + 27*x^9 + ...
(Sum_{k=1..n} k^2) 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
(PARI) {a(n) = if( n%3, n, 3*n)}; /* Michael Somos, Jan 18 2017 */
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CROSSREFS
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KEYWORD
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nonn,easy,mult
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AUTHOR
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STATUS
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approved
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