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%I #8 Jul 07 2016 23:54:48
%S 1,1,1,1,-1,2,1,-6,17,6,1,-14,83,-142,24,1,-25,265,-1235,2314,120,1,
%T -39,655,-5565,24184,-41556,720,1,-56,1372,-18200,137599,-556304,
%U 944628,5040,1,-76,2562,-48664,560049,-3884524,15021068,-24875376,40320,1,-99,4398,-113022,1829793,-19043451
%N Table (read by rows) giving the coefficients of sum formulas of n-th Lucas numbers (A000204). The k-th row (k>=1) contains T(i,k) for i=1 to k, where k=[2*n+1+(-1)^(n-1)]/4 and T(i,k) satisfies L(n) = Sum_{i=1..k} T(i,k) * n^(k-i) / (k-1)!.
%H Ron Knott, <a href="http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/lucasNbs.html">The Lucas Numbers in Pascal's Triangle</a>.
%e L(13)=521; substituting n=13 in the formula of the k-th row we obtain k=7 and the coefficients
%e T(i,7) will be the following: 1,-39,655,-5565,24184,-41556,720,
%e => L(13) = [13^6-39*13^5+655*13^4-5565*13^3+24184*13^2-41556*13+720]/6! = 521.
%Y Cf. A101033, A000204, A100492, A099731, A000045, A094216, A094638, A000108.
%K sign,tabl
%O 1,6
%A _André F. Labossière_, Nov 30 2004
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