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A057769 a(n) = 4*n^4 + 8*n^3 - 4*n - 1 = (2*n^2 - 1)*(2*n^2 + 4*n + 1).

%I A057769 #36 Jul 02 2025 16:02:00
%S A057769 -1,7,119,527,1519,3479,6887,12319,20447,32039,47959,69167,96719,
%T A057769 131767,175559,229439,294847,373319,466487,576079,703919,851927,
%U A057769 1022119,1216607,1437599,1687399,1968407,2283119,2634127,3024119,3455879,3932287,4456319,5031047,5659639,6345359,7091567
%N A057769 a(n) = 4*n^4 + 8*n^3 - 4*n - 1 = (2*n^2 - 1)*(2*n^2 + 4*n + 1).
%C A057769 It may be seen that the terms of the (signed) sequence consist of a subset of the odd squares minus two.
%C A057769 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 A057769 Albert H. Beiler, Recreations in the theory of numbers, New York: Dover, 2nd ed., 1966, p. 106, table 53.
%H A057769 Harvey P. Dale, <a href="/A057769/b057769.txt">Table of n, a(n) for n = 0..1000</a>
%H A057769 <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1).
%F A057769 a(n) = 4*b(n)^2 - 4*b(n) - 1 where b(n) = A002378(n).
%F A057769 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 A057769 G.f.: (x*(x*((x-12)*x-74)-12)+1)/(x-1)^5. - _Harvey P. Dale_, Oct 20 2011
%F A057769 Sum_{n>=0} 1/a(n) = cot(Pi/sqrt(2))*Pi/(2*sqrt(2)). - _Amiram Eldar_, Jan 22 2024
%F A057769 E.g.f.: exp(x)*(-1 + 8*x + 52*x^2 + 32*x^3 + 4*x^4). - _Stefano Spezia_, Apr 27 2025
%t A057769 Table[4n^4+8n^3-4n-1, {n,0,40}] (* _Harvey P. Dale_, Oct 20 2011 *)
%o A057769 (PARI) a(n)=(2*n^2-1)*(2*n^2+4*n+1) \\ _Charles R Greathouse IV_, Oct 07 2015
%Y A057769 Cf. A002378, A007204.
%K A057769 easy,sign
%O A057769 0,2
%A A057769 Stuart M. Ellerstein (ellerstein(AT)aol.com), Nov 01 2000
%E A057769 More terms from _James Sellers_, Nov 02 2000