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a(n) = 4*n^4 + 8*n^3 - 4*n - 1 = (2*n^2 - 1)*(2*n^2 + 4*n + 1).
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%I #30 Jan 22 2024 01:50:51

%S -1,7,119,527,1519,3479,6887,12319,20447,32039,47959,69167,96719,

%T 131767,175559,229439,294847,373319,466487,576079,703919,851927,

%U 1022119,1216607,1437599,1687399,1968407,2283119,2634127,3024119,3455879,3932287,4456319,5031047,5659639,6345359,7091567

%N a(n) = 4*n^4 + 8*n^3 - 4*n - 1 = (2*n^2 - 1)*(2*n^2 + 4*n + 1).

%C It may be seen that the terms of the (signed) sequence consist of a subset of the odd squares minus two.

%C One leg of Pythagorean triangles with hypotenuse a square: a(n)^2 + A069074(n-1)^2 = A007204(n)^2. - _Martin Renner_, Nov 12 2011

%D Albert H. Beiler, Recreations in the theory of numbers, New York: Dover, 2nd ed., 1966, p. 106, table 53.

%H Harvey P. Dale, <a href="/A057769/b057769.txt">Table of n, a(n) for n = 0..1000</a>

%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) = 4*b(n)^2 - 4*b(n) - 1 where b(n) = n-th pronic number A002378(n).

%F a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5); a(0)=-1, a(1)=7, a(2)=119, a(3)=527, a(4)=1519. - _Harvey P. Dale_, Oct 20 2011

%F G.f.: (x*(x*((x-12)*x-74)-12)+1)/(x-1)^5. - _Harvey P. Dale_, Oct 20 2011

%F Sum_{n>=0} 1/a(n) = cot(Pi/sqrt(2))*Pi/(2*sqrt(2)). - _Amiram Eldar_, Jan 22 2024

%t Table[4n^4+8n^3-4n-1, {n,0,40}] (* _Harvey P. Dale_, Oct 20 2011 *)

%o (PARI) a(n)=(2*n^2-1)*(2*n^2+4*n+1) \\ _Charles R Greathouse IV_, Oct 07 2015

%Y Cf. A002378, A007204.

%K easy,sign

%O 0,2

%A Stuart M. Ellerstein (ellerstein(AT)aol.com), Nov 01 2000

%E More terms from _James A. Sellers_, Nov 02 2000