Fibonacci coeff. and Lucas numbers

Discussion in 'General Math' started by alexandru.lupas, Oct 30, 2005.

  1. In the following , let

    F_1=F_2=1 , F_{k+1}=F_k+F_{k-1} , k in {2,3,...},


    L_0=2,L_1=1 , L_{j+1}=L_j+L_{j-1} , j in {1,2,...} ,


    C(n,0;F)=C(n,n;F):=1 for n in {0,1,...} ,

    C(n,k;F)=(F_nF_{n-1}...F_{n-k+1})/(F_1F_2...F_k)

    when k is in {1,2,...,n-1} .


    =====================================
    It's true that for m in {1,2,...} , the identities


    L_1L_2...L_{2m} =
    (1)

    =SUM_{k=0 to k=m}(-1)^{k(k-1)/2}C(2m+1,k;F)L_{m+2m(m-k)}

    and


    L_1L_2...L_{2m-1} = (-1)^{m(m-1)/2}C(2m,m;F)+
    (2)

    +SUM_{k=0 to k=m-1}(-1)^{k(k-1)/2}C(2m,k;F)L_{(m-k)(2m-1)} ,


    are true ??
    =====================================
     
    alexandru.lupas, Oct 30, 2005
    #1
    1. Advertisements

  2. On 29 Oct 2005 22:21:17 -0700, <>
    wrote in
    I haven't looked at (1), but (2) is false as stated. Take m
    = 2. The left-hand side is 1*3*4 = 12. The right-hand side
    is

    (-1)^0 * C(4,0;F) * L_6 + (-1)^0 * C(4,1;F) * L_3 =
    L_6 + (F_4/F_1) * L_3 =
    18 + 3*4 = 30.

    Brian
     
    Brian M. Scott, Oct 31, 2005
    #2
    1. Advertisements

  3. Indeed , it's seems to be a misprint.Thanks !
    However the RHS of (2) has three terms, more precisely
    -(1)^1*C(4,2;F)+
    (-1)^0 * C(4,0;F) * L_6 + (-1)^0 * C(4,1;F) * L_3 =
    = - 6+ L_6 + (F_4/F_1) * L_3 =24
    Of course, false !
     
    alexandru.lupas, Nov 1, 2005
    #3
    1. Advertisements

Ask a Question

Want to reply to this thread or ask your own question?

You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.