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A069074
a(n) = (2*n+2)*(2*n+3)*(2*n+4) = 24*A000330(n+1).
6
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
0,1
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
Konrad Knopp, Theorie und Anwendung der unendlichen Reihen, Berlin, J. Springer, 1922. (Original german edition of "Theory and Application of Infinite Series")
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
MATHEMATICA
LinearRecurrence[{4, -6, 4, -1}, {24, 120, 336, 720}, 40] (* Harvey P. Dale, Apr 10 2017 *)
PROG
(Magma) [(2*n+2)*(2*n+3)*(2*n+4): n in [0..40]]; // Vincenzo Librandi, Oct 04 2011
(PARI) a(n)=6*binomial(2*n+4, 3) \\ Charles R Greathouse IV, Mar 21 2015
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
Sequence in context: A256629 A114200 A229567 * A360389 A059775 A372497
KEYWORD
easy,nonn
AUTHOR
Benoit Cloitre, Apr 05 2002
STATUS
approved