|
%I #58 Mar 07 2022 02:06:12
%S 0,1,8,5,8,35,16,7,80,11,40,143,56,65,224,85,32,323,40,133,440,161,
%T 176,575,208,75,728,87,280,899,320,341,1088,385,136,1295,152,481,1520,
%U 533,560,1763,616,215,2024,235,736,2303,800,833,2600,901,312,2915,336,1045
%N First trisection of A061039.
%C Numerator of (n^2-1)/(9n^2). Denominator is A147650(n).
%C 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.
%C 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
%C 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
%D L. B. W. Jolley, "Summation of Series", Dover Publications, 1961, pp. 40, 41.
%H Harvey P. Dale, <a href="/A144454/b144454.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="/index/Rec#order_27">Index entries for linear recurrences with constant coefficients</a>, 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).
%F a(n) = A061039(3*n).
%F For n > 27, a(n) = 3*a(n-9) - 3*a(n-18) + a(n-27). - _Harvey P. Dale_, Jan 16 2013
%F 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
%F 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
%e 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
%p 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
%t 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 *)
%o (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
%o (PARI) a(n) = numerator(1/9-1/(3*n)^2); \\ _Altug Alkan_, Mar 15 2018
%o (Sage) [numerator((1 - 1/n^2)/9) for n in (1..100)] # _G. C. Greubel_, Mar 07 2022
%Y Cf. A061039, A147650, A300295.
%K nonn,frac,easy
%O 1,3
%A _Paul Curtz_, Oct 07 2008
%E Edited and extended by _R. J. Mathar_, Oct 24 2008
|
