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%I #39 Mar 06 2024 11:18:37
%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
%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}]
%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
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