A215636 a(n) = - 12*a(n-1) - 54*a(n-2) - 112*a(n-3) - 105*a(n-4) - 36*a(n-5) - 2*a(n-6) with a(0)=a(1)=a(2)=0, a(3)=-3, a(4)=24, a(5)=-135.

0, 0, 0, -3, 24, -135, 660, -3003, 13104, -55689, 232500, -958617, 3916440, -15890355, 64127700, -257698347, 1032023136, -4121456625, 16421256420, -65301500577, 259259758056, -1027901275131, 4070632899300, -16104283594083, 63657906293520, -251447560563465, 992593021410900

Offset

0,4

Comments

The Berndt-type sequence number 4 for the argument 2Pi/9 defined by the relation: X(n) = b(n) + a(n)*sqrt(2), where X(n) := ((cos(Pi/24))^(2*n) + (cos(7*Pi/24))^(2*n) + ((cos(3*Pi/8))^(2*n))*(-4)^n. We have b(n)=A215635(n) (see also section "Example" below). For more details - see comments to A215635, A215634 and Witula-Slota's reference.

Links

Table of n, a(n) for n=0..26.

R. Witula and D. Slota, On modified Chebyshev polynomials, J. Math. Anal. Appl., 324 (2006), 321-343.

Index entries for linear recurrences with constant coefficients, signature (-12,-54,-112,-105,-36,-2).

Formula

G.f.: (-3*x^3-12*x^4-9*x^5)/(1+12*x+54*x^2+112*x^3+105*x^4+36*x^5+2*x^6).

Example

We have X(1)=-6, X(2)=18 and X(3)=-60-3*sqrt(2), which implies the equality: (cos(Pi/24))^6 + (cos(7*Pi/24))^6 + ((cos(3*Pi/8))^6 = (60+3*sqrt(2))/64.

Mathematica

LinearRecurrence[{-12, -54, -112, -105, -36, -2}, {0, 0, 0, -3, 24, -135}, 50]

Crossrefs

Cf. A215455, A215634, A215635, A215664, A215885, A215665, A215666, A215829, A215831, A215917, A215919, A215945, A216034, A215948, A216757.

Sequence in context: A359884 A069515 A206949 * A056350 A056344 A201231

Adjacent sequences: A215633 A215634 A215635 * A215637 A215638 A215639

Keyword

sign,easy

Author

Roman Witula, Aug 18 2012

Status

approved