A190577 - OEIS

%I #45 Nov 28 2024 13:34:10

%S 105,384,945,1920,3465,5760,9009,13440,19305,26880,36465,48384,62985,

%T 80640,101745,126720,156009,190080,229425,274560,326025,384384,450225,

%U 524160,606825,698880,801009,913920,1038345,1175040,1324785

%N a(n) = n*(n+2)*(n+4)*(n+6).

%C Each term is the difference between a square and a fourth power:

%C n*(n+2)*(n+4)*(n+6) = ((n+2)*(n+4)-2^2)^2-2^4. More generally,

%C 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.

%D Miguel de Guzmán Ozámiz, Para Pensar Mejor, Editions Pyramid, 2001, p. 294-295.

%H Vincenzo Librandi, <a href="/A190577/b190577.txt">Table of n, a(n) for n = 1..1000</a>

%H Rafael Parra Machío, <a href="http://hojamat.es/parra/prop2011.pdf">Propiedades del 2011: Un paseo a través de los números primos</a>, section 7.3, p. 22.

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1).

%F a(n) = ((n+2)*(n+4)-2^2)^2-2^4.

%F G.f.: 3*x*(5*x^3-25*x^2+47*x-35)/(x-1)^5.

%F 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

%F From _Amiram Eldar_, Oct 03 2024: (Start)

%F Sum_{n>=1} 1/a(n) = 7/480.

%F Sum_{n>=1} (-1)^(n+1)/a(n) = 11/1440. (End)

%e a(3) = 945 = 3*(3+2)*(3+4)*(3+6) = ((3+2)*(3+2*2)-2^2)^2-2^4 = 31^2-2^4.

%e a(13) = 62985 = 13*(13+2)*(13+4)*(13+6) = ((13+2)*(13+2*2)+2^2)^2-2^4 = 251^2-2^4.

%t Table[n (n + 2) (n + 4) (n + 6), {n, 1, 15}]

%t Table[((n + 2) (n + 4) - 2^2)^2 - 2^4, {n, 1, 15}]

%t LinearRecurrence[{5,-10,10,-5,1},{105,384,945,1920,3465},40] (* _Harvey P. Dale_, Nov 28 2024 *)

%o (Magma) [((n+2)*(n+4)-2^2)^2-2^4: n in [1..40]]; // _Vincenzo Librandi_, May 23 2011

%o (Python)

%o def A190577(n): return n*(n*(n*(n + 12) + 44) + 48) # _Chai Wah Wu_, Mar 06 2024

%K nonn,easy

%O 1,1

%A _Rafael Parra Machio_, May 18 2011