A057681 a(n) = Sum_{j=0..floor(n/3)} (-1)^j*binomial(n,3*j).

1, 1, 1, 0, -3, -9, -18, -27, -27, 0, 81, 243, 486, 729, 729, 0, -2187, -6561, -13122, -19683, -19683, 0, 59049, 177147, 354294, 531441, 531441, 0, -1594323, -4782969, -9565938, -14348907, -14348907, 0, 43046721, 129140163, 258280326, 387420489, 387420489

Offset

0,5

Comments

Let M be any endomorphism on any vector space, such that M^3 = 1 (identity). Then (1-M)^n = a(n)-A057682(n)*M+z(n)*M^2, where z(0)=z(1)=0 and, apparently, z(n+2)=A057083(n). - Stanislav Sykora, Jun 10 2012

Pisano period lengths: 1, 3, 1, 6, 24, 3, 6, 12, 1, 24, 60, 6, 12, 6, 24, 24, 96, 3, 18, 24, ... . - R. J. Mathar, Aug 10 2012

{A057681, A057682, A*}, where A* is A057083 prefixed by two 0's, is the difference analog of the trigonometric functions of order 3, {k_1(x), k_2(x), k_3(x)}. For a definition see [Erdelyi] and the Shevelev link. - Vladimir Shevelev, Jun 25 2017

References

A. Erdelyi, Higher Transcendental Functions, McGraw-Hill, 1955, Vol. 3, Chapter XVIII.

Links

G. C. Greubel, Table of n, a(n) for n = 0..1000

T. Alden Gassert, Discriminants of simplest 3^n-tic extensions, arXiv:1409.7829 [math.NT], 2014.

Mark W. Coffey, Reductions of particular hypergeometric functions 3F2 (a, a+1/3, a+2/3; p/3, q/3; +-1), arXiv:1506.09160 [math.CA], 2015.

John B. Dobson, A matrix variation on Ramus's identity for lacunary sums of binomial coefficients, arXiv:1610.09361 [math.NT], 2016.

Ira Gessel, The Smith College diploma problem.

Vladimir Shevelev, Combinatorial identities generated by difference analogs of hyperbolic and trigonometric functions of order n, arXiv:1706.01454 [math.CO], 2017.

Index entries for linear recurrences with constant coefficients, signature (3,-3).

Formula

From Paul Barry, Feb 26 2004: (Start)

G.f.: (1-x)^2/((1-x)^3+x^3).

a(n) = 0^n/3 + 2*3^((n-2)/2)*cos(Pi*n/6). (End)

From Paul Barry, Feb 27 2004: (Start)

Binomial transform of (1, 0, 0, -1, 0, 0, 1, 0, 0, -1, 0, ...).

E.g.f.: 2*exp(3x/2)*cos(sqrt(3)*x/2)/3+1/3.

a(n) = (((3+sqrt(-3))/2)^n+((3-sqrt(-3))/2)^n)/3+0^n/3. (End)

a(n) = 6*a(n-1)-15*a(n-2)+20*a(n-3)-15*a(n-4)+6*a(n-5). - Paul Curtz, Jan 02 2008

Start with x(0)=1,y(0)=0,z(0)=0 and set x(n+1)=x(n)-z(n), y(n+1)=y(n)-x(n),z(n+1)=z(n)-y(n). Then a(n)=x(n). But this recurrence falls into a repetitive cycle of length 6 and multiplicative factor -27, so that a(n) = -27*a(n-6) for any n>6. - Stanislav Sykora, Jun 10 2012

E.g.f.: (1+2*exp(3*z/2)*cos(z*sqrt(3/4)))/3. - Peter Luschny, Jul 10 2012

a(0)=a(1)=a(2)=1, a(n)=3*a(n-1)-3*a(n-2), n>=3. - Wesley Ivan Hurt, Nov 11 2014

For n>=1, a(n) = 2*3^((n-2)/2)*cos(Pi*n/6). - Vladimir Shevelev, Jun 25 2017

a(n+m) = a(n)*a(m)-A057682(n)*A*057083(m)-A*057083(n)*A057682(m), where A*057083 is A057083 prefixed by two 0's. - Vladimir Shevelev, Jun 25 2017

Example

If M^3=1 then (1-M)^6 = a(6)-A057682(6)*M+A057083(4)*M^2 = -18+9*M+9*M^2.

Maple

A057681 := n->add((-1)^j*binomial(n, 3*j), j=0..floor(n/3)); seq(A057681(n), n=0..50);
A057681_list := proc(n) local i; series((1+2*exp(3*z/2)*cos(z*sqrt(3/4)))/3, z, n+2): seq(i!*coeff(%, z, i), i=0..n) end: A057681_list(38); # Peter Luschny, Jul 10 2012

Mathematica

Join[{1}, LinearRecurrence[{3, -3}, {1, 1}, 40]] (* Harvey P. Dale, Aug 19 2014 *)

Prog

(PARI) x='x+O('x^40); Vec((1-x)^2/((1-x)^3+x^3)) \\ G. C. Greubel, Oct 23 2018
(Magma) I:=[1, 1]; [1] cat [n le 2 select I[n] else 3*Self(n-1) - 3*Self(n-2): n in [1..40]]; // G. C. Greubel, Oct 23 2018
(GAP) a:=[1, 1];; for n in [3..40] do a[n]:=3*a[n-1]-3*a[n-2]; od; Concatenation([1], a); # Muniru A Asiru, Oct 24 2018

Crossrefs

Cf. A009116, A009545, A057682, A057083, A103312.

Sequence in context: A203594 A325751 A123877 * A103312 A325729 A159794

Adjacent sequences: A057678 A057679 A057680 * A057682 A057683 A057684

Keyword

sign,easy

Author

N. J. A. Sloane, Oct 20 2000

Status

approved