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URL: https://oeis.org/A190577

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A190577
a(n) = n*(n+2)*(n+4)*(n+6).
6
105, 384, 945, 1920, 3465, 5760, 9009, 13440, 19305, 26880, 36465, 48384, 62985, 80640, 101745, 126720, 156009, 190080, 229425, 274560, 326025, 384384, 450225, 524160, 606825, 698880, 801009, 913920, 1038345, 1175040, 1324785
OFFSET
1,1
COMMENTS
Each term is the difference between a square and a fourth power:
n*(n+2)*(n+4)*(n+6) = ((n+2)*(n+4)-2^2)^2-2^4. More generally,
n*(n+k)*(n+2*k)*(n+3*k) = ((n+k)*(n+2*k)-k^2)^2-k^4 for any k; in this case, k=2.
REFERENCES
Miguel de Guzmán Ozámiz, Para Pensar Mejor, Editions Pyramid, 2001, p. 294-295.
FORMULA
a(n) = ((n+2)*(n+4)-2^2)^2-2^4.
G.f.: 3*x*(5*x^3-25*x^2+47*x-35)/(x-1)^5.
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5). - Wesley Ivan Hurt, May 15 2023
From Amiram Eldar, Oct 03 2024: (Start)
Sum_{n>=1} 1/a(n) = 7/480.
Sum_{n>=1} (-1)^(n+1)/a(n) = 11/1440. (End)
EXAMPLE
a(3) = 945 = 3*(3+2)*(3+4)*(3+6) = ((3+2)*(3+2*2)-2^2)^2-2^4 = 31^2-2^4.
a(13) = 62985 = 13*(13+2)*(13+4)*(13+6) = ((13+2)*(13+2*2)+2^2)^2-2^4 = 251^2-2^4.
MATHEMATICA
Table[n (n + 2) (n + 4) (n + 6), {n, 1, 15}]
Table[((n + 2) (n + 4) - 2^2)^2 - 2^4, {n, 1, 15}]
LinearRecurrence[{5, -10, 10, -5, 1}, {105, 384, 945, 1920, 3465}, 40] (* Harvey P. Dale, Nov 28 2024 *)
PROG
(Magma) [((n+2)*(n+4)-2^2)^2-2^4: n in [1..40]]; // Vincenzo Librandi, May 23 2011
(Python)
def A190577(n): return n*(n*(n*(n + 12) + 44) + 48) # Chai Wah Wu, Mar 06 2024
CROSSREFS
Sequence in context: A145752 A195266 A113480 * A102792 A013594 A160340
KEYWORD
nonn,easy
AUTHOR
Rafael Parra Machio, May 18 2011
STATUS
approved