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A090666
Number of repetitions (defined as the number of appearances minus one) of L quantum number for a given value of N=2*nb+tau=0,1,2,... principal quantum number for the 5 dimensional harmonic oscillator (connected to the solution of Bohr equation in 5 dimensional). For each tau, nu=0,1,..,[tau/3] and K=tau-2*nu. Finally L=K,K+1,K+2,...,2*K-2,2*K (or alternatively from K to 2*K with the exception of 2*K-1).
0
0, 0, 0, 0, 2, 3, 7, 11, 18, 26, 36, 48, 63, 79, 99, 121, 146, 174, 206, 240, 279, 321, 367, 417, 472, 530, 594, 662, 735, 813, 897
OFFSET
0,5
COMMENTS
Nuclear Phyics, collective model: classification of states based on quantum numbers and group theory.
REFERENCES
L. Fortunato, Compendium on the solutions of the Bohr hamiltonian, in preparation (Dec. 2003).
LINKS
L. Fortunato, Solutions of the Bohr hamiltonian, a compendium, arXiv:nucl-th/0411087, 2004; Eur. Phys. J. A26S1 (2005) 1-30. (arxiv preprint)
FORMULA
N = 2*nb+tau = 0, 1, 2, ..... nb=0, 1, 2, .... tau=0, 1, 2, .... nu=0, 1, .., [tau/3] K=tau-2*nu L=K, K+1, K+2, ..., 2*K-2, 2*K
G.f.: x^4*(x^2-x+1)*(x^5-x^4-2x^3+x+2)/((x-1)^4*(x+1)*(x^2+x+1)). - conjectured by Jean-François Alcover, Feb 18 2019
EXAMPLE
a(N=0)=0 because N=0 implies nb=0 and tau=0. Hence nu=0, K=0 and L=0. There are no repetitions.
a(N=4)=2 because N=2 implies (nb,tau)= (0,4),(1,2),(2,0). At the end L=0,2,4,2,4,4,5,6,8, so that only 2 repetitions are found.
PROG
(FORTRAN 77) implicit integer(a-z) dimension mrep(0:100) do N=0, 30 do L=0, 100 mrep(L)=0 enddo do nb=0, nint(real(N)/2.-0.01) tau=N-2*nb numax=int(tau/3) do nu=0, numax K=tau-3*nu do L=K, 2*K if(L.eq.(2*K-1)) goto 100 mrep(L)=mrep(L)+1 100 enddo enddo enddo sum=0 do L=0, 100 if(mrep(L).gt.0) then sum=sum+mrep(L)-1 endif enddo print *, N, sum enddo end
CROSSREFS
Sequence in context: A060341 A114345 A077165 * A140409 A201011 A108541
KEYWORD
nonn,more
AUTHOR
Lorenzo Fortunato (fortunat(AT)pd.infn.it), Dec 16 2003
STATUS
approved