A213421 Real part of Q^n, Q being the quaternion 2+i+j+k.

1, 2, 1, -10, -47, -118, -143, 254, 2017, 6290, 11041, 134, -76751, -307942, -694511, -622450, 2371777, 13844258, 38774593, 58188566, -38667887, -561991510, -1977290831, -3975222754, -2059855199, 19587138482

Offset

0,2

Links

Stanislav Sykora, Table of n, a(n) for n = 0..1000

Beata Bajorska-Harapińska, Barbara Smoleń, Roman Wituła, On Quaternion Equivalents for Quasi-Fibonacci Numbers, Shortly Quaternaccis, Advances in Applied Clifford Algebras Vol. 29, No. 3 (2019), Article 54.

Wikipedia, Lucas sequence

Formula

Conjecture: G.f. (1-2x)/(1-4x+7x^2). a(n) = A168175(n)-2*A168175(n-1). - R. J. Mathar, Jun 25 2012

From Peter Bala, Mar 29 2015: (Start)

The above o.g.f. is correct; this is the Lucas sequence V_n(4,7).

a(n) = Re( (2 + sqrt(3)*i)^n )= 1/2*( (2 + sqrt(3)*i)^n + (2 - sqrt(3)*i)^n ).

a(n) = 1/2 * trace( [ 2 + i, 1 + i; -1 + i, 2 - i ]^n ) = 1/2 * trace( [ 2 , sqrt(3)*i ; sqrt(3)*i, 2 ]^n ).

a(n) = 4*a(n-1) - 7*a(n-2) with a(0) = 1, a(1) = 2. (End)

Maple

#A213421
seq(simplify(1/2*((2+I*sqrt(3))^n+(2-I*sqrt(3))^n)), n = 0 .. 25); # Peter Bala, Mar 29 2015

Prog

(PARI)
QuaternionToN(a, b, c, d, nmax) = {local (C); C = matrix(nmax+1, 4); C[1, 1]=1; for(n=2, nmax+1, C[n, 1]=a*C[n-1, 1]-b*C[n-1, 2]-c*C[n-1, 3]-d*C[n-1, 4]; C[n, 2]=b*C[n-1, 1]+a*C[n-1, 2]+d*C[n-1, 3]-c*C[n-1, 4]; C[n, 3]=c*C[n-1, 1]-d*C[n-1, 2]+a*C[n-1, 3]+b*C[n-1, 4]; C[n, 4]=d*C[n-1, 1]+c*C[n-1, 2]-b*C[n-1, 3]+a*C[n-1, 4]; ); return (C); }
Q=QuaternionToN(2, 1, 1, 1, 1000);
for(n=1, #Q[, 1], write("A213421.txt", n-1, " ", Q[n, 1]));

Crossrefs

Cf. A168175, A087455, A088138, A128018, A146559.

Sequence in context: A336729 A304359 A054781 * A370068 A239070 A271042

Adjacent sequences: A213418 A213419 A213420 * A213422 A213423 A213424

Keyword

sign,easy

Author

Stanislav Sykora, Jun 11 2012

Status

approved