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Search: id:A090666
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| A090666 |
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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-2nu. Finally L=K,K+1,K+2,...,2K-2,2K (or alternatively from K to 2K with the exception of 2K-1). |
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+0
1
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| 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
(list; graph; listen)
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OFFSET
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0,5
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COMMENT
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Nuclear Phyics, collective model: classification of states based on quantum numbers and group theory.
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REFERENCES
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L. Fortunato, Compendium on the solutions of the Bohr hamiltonian, in preparation (Dec. 2003).
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FORMULA
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N = 2*nb+tau = 0, 1, 2, ..... nb=0, 1, 2, .... tau=0, 1, 2, .... nu=0, 1, .., [tau/3] K=tau-2nu L=K, K+1, K+2, ..., 2K-2, 2K
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EXAMPLE
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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.
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PROGRAM
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(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
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CROSSREFS
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Sequence in context: A060341 A114345 A077165 this_sequence A140409 A108541 A038937
Adjacent sequences: A090663 A090664 A090665 this_sequence A090667 A090668 A090669
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KEYWORD
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nonn,uned
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AUTHOR
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Lorenzo Fortunato (fortunat(AT)pd.infn.it), Dec 16 2003
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