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

-1, 7, 119, 527, 1519, 3479, 6887, 12319, 20447, 32039, 47959, 69167, 96719, 131767, 175559, 229439, 294847, 373319, 466487, 576079, 703919, 851927, 1022119, 1216607, 1437599, 1687399, 1968407, 2283119, 2634127, 3024119, 3455879, 3932287, 4456319, 5031047, 5659639, 6345359, 7091567

Offset

0,2

Comments

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

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

References

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

Links

Harvey P. Dale, Table of n, a(n) for n = 0..1000

Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Formula

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

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

G.f.: (x*(x*((x-12)*x-74)-12)+1)/(x-1)^5. - Harvey P. Dale, Oct 20 2011

Sum_{n>=0} 1/a(n) = cot(Pi/sqrt(2))*Pi/(2*sqrt(2)). - Amiram Eldar, Jan 22 2024

Mathematica

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

Prog

(PARI) a(n)=(2*n^2-1)*(2*n^2+4*n+1) \\ Charles R Greathouse IV, Oct 07 2015

Crossrefs

Cf. A002378, A007204.

Sequence in context: A266482 A368631 A076283 * A221031 A221323 A268300

Adjacent sequences: A057766 A057767 A057768 * A057770 A057771 A057772

Keyword

easy,sign

Author

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

Extensions

More terms from James A. Sellers, Nov 02 2000

Status

approved