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

15, 105, 315, 693, 1287, 2145, 3315, 4845, 6783, 9177, 12075, 15525, 19575, 24273, 29667, 35805, 42735, 50505, 59163, 68757, 79335, 90945, 103635, 117453, 132447, 148665, 166155, 184965, 205143, 226737, 249795, 274365, 300495, 328233, 357627

Offset: 0

Author

Jason Earls, Jun 12 2001

Keywords

• easy
• nonn

Comments

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

References

• L. B. W. Jolley, "Summation of Series", Dover Publications, 1961, p. 40

Links

• Harry J. Smith, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Crossrefs

Cf. A005408.

Programs

Maple

For n from 0 to 100 do (2*n+1)*(2*n+3)*(2*n+5) end do;

Mathematica

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 *)
Times@@@(#+{1,3,5}&)/@(2Range[0,35]) (* Harvey P. Dale, Feb 13 2011 *)
Table[(2*n + 1)*(2*n + 3)*(2*n + 5), {n,35}] (* T. D. Noe, Feb 13 2011 *)

PARI

a(n) = { (2*n + 1)*(2*n + 3)*(2*n + 5) } \\ Harry J. Smith, Jul 24 2009

Formula

a(n) = A162540(n)/3.

1/15 + 1/105 + 1/315...= 1/12 [Jolley, eq. 209]

sum_{i=0..n-1} a(i) = A196506(n), partial sums [Jolley eq (43)]. - R. J. Mathar, Mar 24 2011

sum_{i=0..infinity} (-1)^i/a(i) = Pi/8-1/3 = 0.0593657... [Jolley eq 240]

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

G.f. ( 15+45*x-15*x^2+3*x^3 ) / (x-1)^4. - R. J. Mathar, Oct 03 2011

Extensions

Better description and more terms from Larry Reeves (larryr(AT)acm.org), Jun 19 2001