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A061550 a(n) = (2*n+1)*(2*n+3)*(2*n+5).

%I A061550 #44 Dec 28 2024 00:06:38
%S A061550 15,105,315,693,1287,2145,3315,4845,6783,9177,12075,15525,19575,24273,
%T A061550 29667,35805,42735,50505,59163,68757,79335,90945,103635,117453,132447,
%U A061550 148665,166155,184965,205143,226737,249795,274365,300495,328233,357627
%N A061550 a(n) = (2*n+1)*(2*n+3)*(2*n+5).
%C A061550 sum(1/a(k), k=0..n) = 1/12 - 1/((8*n+12)*(2*n+5)). Jolley equation 209 (offset adjusted).  - _Gary Detlefs_, Sep 20 2011
%D A061550 L. B. W. Jolley, "Summation of Series", Dover Publications, 1961, p. 40
%H A061550 Harry J. Smith, <a href="/A061550/b061550.txt">Table of n, a(n) for n = 0..1000</a>
%H A061550 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).
%F A061550 a(n) = A162540(n)/3.
%F A061550 1/15 + 1/105 + 1/315...= 1/12 [Jolley, eq. 209]
%F A061550 sum_{i=0..n-1} a(i) = A196506(n), partial sums [Jolley eq (43)]. - _R. J. Mathar_, Mar 24 2011
%F A061550 sum_{i=0..infinity} (-1)^i/a(i) = Pi/8-1/3 = 0.0593657... [Jolley eq 240]
%F A061550 a(n)=(-1)^(n+1)*(4*n^2+12*n+7)/Integral_{x=0..Pi/2} (cos((2*n+3)*x))*(sin(x))^2 dx. - _Francesco Daddi_, Aug 03 2011
%F A061550 G.f. ( 15+45*x-15*x^2+3*x^3 ) / (x-1)^4. - _R. J. Mathar_, Oct 03 2011
%p A061550 For n from 0 to 100 do (2*n+1)*(2*n+3)*(2*n+5) end do;
%t A061550 f[n_] := n/GCD[n, 4]; Table[ f[n] f[n + 2] f[n + 4], {n, 1, 70, 2}] (* _Robert G. Wilson v_, Jan 14 2011 *)
%t A061550 Times@@@(#+{1,3,5}&)/@(2Range[0,35]) (* _Harvey P. Dale_, Feb 13 2011 *)
%t A061550 Table[(2*n + 1)*(2*n + 3)*(2*n + 5), {n,35}] (* _T. D. Noe_, Feb 13 2011 *)
%o A061550 (PARI) a(n) = { (2*n + 1)*(2*n + 3)*(2*n + 5) } \\ _Harry J. Smith_, Jul 24 2009
%Y A061550 Cf. A005408.
%K A061550 easy,nonn
%O A061550 0,1
%A A061550 _Jason Earls_, Jun 12 2001
%E A061550 Better description and more terms from Larry Reeves (larryr(AT)acm.org), Jun 19 2001