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A144454
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First trisection of A061039.
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8
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0, 1, 8, 5, 8, 35, 16, 7, 80, 11, 40, 143, 56, 65, 224, 85, 32, 323, 40, 133, 440, 161, 176, 575, 208, 75, 728, 87, 280, 899, 320, 341, 1088, 385, 136, 1295, 152, 481, 1520, 533, 560, 1763, 616, 215, 2024, 235, 736, 2303, 800, 833, 2600, 901, 312, 2915, 336, 1045
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OFFSET
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1,3
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COMMENTS
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Numerator of (n^2-1)/(9n^2). Denominator is A147650(n).
Terms alternate between even and odd. The sequence modulo 9 reads (0, 1, 8, 5, 8, 8, 7, 7, 8, 2, 4, 8, 2, 2, 8, 4, 5, ...) (Is there a meaning to the interpretation as the constant 0.1858877824822845...?) The first appearance of 3 (mod 9) is at a(26)=75, the second at a(55)=336. The first appearance of 6 (mod 9) is at a(28)=87, the second at a(53)=312.
a(n) also gives the numerator of (n^2 - 1)/(3*((2*n)^2 - 1)) =: r(n-1), with denominators A300295(n-1), for n >= 1. For the proof see a comment in A300295; also for details on r(n) with the Jolley reference. - Wolfdieter Lang, Mar 15 2018
a(n) is also the numerator of Sum_{k=0..n} (1/((2*k-3)(2*k-1)*(2*k+1))). This summation is an offset adjusted form of formula 209 in Jolley's "Summation of Series". The closed form is given in the Lang comment above. - Gary Detlefs, Mar 15 2018
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REFERENCES
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L. B. W. Jolley, "Summation of Series", Dover Publications, 1961, pp. 40, 41.
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LINKS
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Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,0,0,3,0,0,0,0,0,0,0,0,-3,0,0,0,0,0,0,0,0,1).
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FORMULA
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For n > 27, a(n) = 3*a(n-9) - 3*a(n-18) + a(n-27). - Harvey P. Dale, Jan 16 2013
a(n) = (n^2 - 1)/9 if n == 1 (mod 9) or == 8 (mod 9). For other n: a(n) = (n^2 - 1)/3 if n == 1 (mod 3) or == 2(mod 3), and a(n) = n^2 - 1 if n == 0 (mod 3). The proof uses the first comment and gcd(n^2-1, n^2) = 1. - Wolfdieter Lang, Mar 15 2018
G.f.: x^2*(1 + 8*x + 5*x^2 + 8*x^3 + 35*x^4 + 16*x^5 + 7*x^6 + 80*x^7 + 11*x^8 + 37*x^9 + 119*x^10 + 41*x^11 + 41*x^12 + 119*x^13 + 37*x^14 + 11*x^15 + 83*x^16 + 7*x^17 + 16*x^18 + 35*x^19 + 8*x^20 + 5*x^21 + 8*x^22 + x^23 - x^25) / ((1 - x)^3*(1 + x + x^2)^3*(1 + x^3 + x^6)^3). - Colin Barker, Mar 15 2018
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EXAMPLE
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The rationals (n^2 - 1)/(9*n^2) begin: 0/1, 1/12, 8/81, 5/48, 8/75, 35/324, 16/147, 7/64, 80/729, 11/100, 40/363, 143/1296, 56/507, 65/588, ... - Wolfdieter Lang, Mar 15 2018
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MAPLE
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P:=n-> sum(1/((2*k-3)*(2*k-1)*(2*k+1)), k = 0 n); seq(numer(P(i))i, =1..50) # Gary Detlefs, Mar 15 2018
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MATHEMATICA
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Numerator[Table[((n-1)(n+1))/(9n^2), {n, 60}]] (* or *) LinearRecurrence[ {0, 0, 0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0, -3, 0, 0, 0, 0, 0, 0, 0, 0, 1}, {0, 1, 8, 5, 8, 35, 16, 7, 80, 11, 40, 143, 56, 65, 224, 85, 32, 323, 40, 133, 440, 161, 176, 575, 208, 75, 728}, 60] (* Harvey P. Dale, Jan 16 2013 *)
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PROG
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(PARI) concat(0, Vec(x^2*(1 + 8*x + 5*x^2 + 8*x^3 + 35*x^4 + 16*x^5 + 7*x^6 + 80*x^7 + 11*x^8 + 37*x^9 + 119*x^10 + 41*x^11 + 41*x^12 + 119*x^13 + 37*x^14 + 11*x^15 + 83*x^16 + 7*x^17 + 16*x^18 + 35*x^19 + 8*x^20 + 5*x^21 + 8*x^22 + x^23 - x^25) / ((1 - x)^3*(1 + x + x^2)^3*(1 + x^3 + x^6)^3) + O(x^60))) \\ Colin Barker, Mar 15 2018
(PARI) a(n) = numerator(1/9-1/(3*n)^2); \\ Altug Alkan, Mar 15 2018
(Sage) [numerator((1 - 1/n^2)/9) for n in (1..100)] # G. C. Greubel, Mar 07 2022
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CROSSREFS
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KEYWORD
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nonn,frac,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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