A069074 a(n) = (2*n+2)*(2*n+3)*(2*n+4) = 24*A000330(n+1).

24, 120, 336, 720, 1320, 2184, 3360, 4896, 6840, 9240, 12144, 15600, 19656, 24360, 29760, 35904, 42840, 50616, 59280, 68880, 79464, 91080, 103776, 117600, 132600, 148824, 166320, 185136, 205320, 226920, 249984, 274560, 300696, 328440, 357840

Offset: 0

Author

Benoit Cloitre, Apr 05 2002

Keywords

• easy
• nonn

Comments

sqrt((Sum_{k=0..n} 2*a(k)) + 1) = A056220(n+2). - Doug Bell, Mar 09 2009

Second leg of Pythagorean triangles with hypotenuse a square: A057769(n)^2 + a(n-1)^2 = A007204(n)^2. - Martin Renner, Nov 12 2011

Numbers which are both the sum of 2*n + 4 consecutive odd integers and the sum of the 2*n + 2 immediately higher consecutive odd integers. In general, let f(k,n) = 3*k^3*A000330(n). Then f(k,n) is both the sum of k*n + k consecutive terms from the arithmetic progression with first term A000217(k) and constant difference k and the immediately higher k*n terms from the same progression. When k = 1, f(k,n) = A059270(n). - Charlie Marion, Aug 23 2021

References

• Albert H. Beiler, Recreations in the theory of numbers, New York: Dover, (2nd ed.) 1966, p. 106, table 53.
• T. J. I'a. Bromwich, Introduction to the Theory of Infinite Series, Macmillan, 2nd. ed. 1949, p. 190.
• Jolley, Summation of Series, Dover (1961).
• Konrad Knopp, Theory and application of infinite series, Dover, p. 269

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..10000
• Konrad Knopp, Theorie und Anwendung der unendlichen Reihen, Berlin, J. Springer, 1922. (Original german edition of "Theory and Application of Infinite Series")
• Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Crossrefs

Cf. A001844. A001844(n+1)^2 - a(n) and A001844(n+1)^2 + a(n) are both square numbers. - Doug Bell, Mar 08 2009

Cf. A000466. a(n) = Sum_{k=0..2n+3} (A000466(n+1) + 2k) which is the sum of 2n+4 consecutive odd integers starting at A000466(n+1). - Doug Bell, Mar 08 2009

Programs

Magma

[(2*n+2)*(2*n+3)*(2*n+4): n in [0..40]]; // Vincenzo Librandi, Oct 04 2011

Mathematica

LinearRecurrence[{4,-6,4,-1},{24,120,336,720},40] (* Harvey P. Dale, Apr 10 2017 *)

PARI

a(n)=6*binomial(2*n+4,3) \\ Charles R Greathouse IV, Mar 21 2015

Formula

Sum_{n>=0} (-1)^n/a(n) = (Pi-3)/4 = 0.03539816339... [Jolley, eq. 244]

Sum_{n>=0} 1/a(n) = 3/4 - log(2) = 0.05685281... [Jolley, eq. 249]

G.f.: ( 24+24*x ) / (x-1)^4. - R. J. Mathar, Oct 03 2011