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

Author

Stanislav Sykora, Jun 11 2012

Keywords

• sign
• easy

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

Crossrefs

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

Programs

Maple

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

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]));

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)