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%I #28 Oct 08 2023 04:45:02
%S 5,3,21,2,45,15,77,6,117,35,165,12,221,63,285,20,357,99,437,30,525,
%T 143,621,42,725,195,837,56,957,255,1085,72,1221,323,1365,90,1517,399,
%U 1677,110,1845,483,2021,132,2205,575,2397,156,2597,675,2805,182,3021,783
%N 5th diagonal of triangle defined in A051537.
%H Colin Barker, <a href="/A070262/b070262.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="/index/Rec#order_12">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,0,3,0,0,0,-3,0,0,0,1).
%F a(n) = lcm(n + 4, n) / gcd(n + 4, n).
%F From _Colin Barker_, Mar 27 2017: (Start)
%F G.f.: x*(5 + 3*x + 21*x^2 + 2*x^3 + 30*x^4 + 6*x^5 + 14*x^6 - 3*x^8 - x^9 - 3*x^10) / ((1 - x)^3*(1 + x)^3*(1 + x^2)^3).
%F a(n) = 3*a(n-4) - 3*a(n-8) + a(n-12) for n>12. (End)
%F From _Luce ETIENNE_, May 10 2018: (Start)
%F a(n) = n*(n+4)*4^((5*(n mod 4)^3 - 24*(n mod 4)^2 + 31*(n mod 4)-12)/6).
%F a(n) = n*(n+4)*(37-27*cos(n*Pi)-6*cos(n*Pi/2))/64. (End)
%F From _Amiram Eldar_, Oct 08 2023: (Start)
%F Sum_{n>=1} 1/a(n) = 11/6.
%F Sum_{n>=1} (-1)^n/a(n) = 7/6.
%F Sum_{k=1..n} a(k) ~ (37/192) * n^3. (End)
%t Table[ LCM[i + 4, i] / GCD[i + 4, i], {i, 1, 60}]
%t LinearRecurrence[{0,0,0,3,0,0,0,-3,0,0,0,1},{5,3,21,2,45,15,77,6,117,35,165,12},90] (* _Harvey P. Dale_, Jul 13 2019 *)
%o (PARI) Vec(x*(5 + 3*x + 21*x^2 + 2*x^3 + 30*x^4 + 6*x^5 + 14*x^6 - 3*x^8 - x^9 - 3*x^10) / ((1 - x)^3*(1 + x)^3*(1 + x^2)^3) + O(x^60)) \\ _Colin Barker_, Mar 27 2017
%o (PARI) a(n) = lcm(n+4,n)/gcd(n+4,n); \\ _Altug Alkan_, Sep 20 2018
%o (Magma) [LCM(n + 4, n)/GCD(n + 4, n): n in [1..50]]; // _G. C. Greubel_, Sep 20 2018
%Y Cf. A061037. [From _R. J. Mathar_, Sep 29 2008]
%Y Cf. A010873, A051537.
%Y Cf. A000466, A002378, A003185, A085027.
%K nonn,easy
%O 1,1
%A _Amarnath Murthy_, May 09 2002
%E Edited by _Robert G. Wilson v_, May 10 2002
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