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Triangle read by rows giving the coefficients of general sum formulas of n-th Lucas numbers (A000204). The k-th row (k>=1) contains T(i,k) for i=1 to 2*k-1, where T(i,k) satisfies L(n) = Sum_{k=1..n} Sum_{i=1..2*k-1} T(i,k) * C(n-k,i-1) * n^(n-k) / (n-1)!.
5

%I #11 Jul 07 2016 23:54:48

%S 1,1,-2,-3,2,15,51,65,27,6,-148,-945,-2292,-2776,-1680,-405,24,2290,

%T 19580,71965,145525,175244,125950,50085,8505,120,-41676,-473072,

%U -2340400,-6676835,-12132890,-14587261,-11619692,-5290005,-1752030,-229635,720,943908,13132532,81977672,303352938,740797855

%N Triangle read by rows giving the coefficients of general sum formulas of n-th Lucas numbers (A000204). The k-th row (k>=1) contains T(i,k) for i=1 to 2*k-1, where T(i,k) satisfies L(n) = Sum_{k=1..n} Sum_{i=1..2*k-1} T(i,k) * C(n-k,i-1) * n^(n-k) / (n-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(7)= (1/(7-1)!) * [ 7^(7-1) -{-1+2*(7-2)+3*C(7-2,2)}*7^(7-2) +{2+15*(7-3)+51*C(7-3,2)+65*C(7-3,3) +27*C(7-3,4)}*7^(7-3) -{-6+148*(7-4)+945*C(7-4,2)+2292*C(7-4,3)}*7^(7-4) +... ]

%e = (1/6!) * [ 7^6 -{-1+10+30}*7^5 +{2+60+306+260+27}*7^4 -{-6+444+2835+2292}*7^3 +{24+4580+19580}*7^2 -{-120+41676}*7 +{720} ] = (1/6!) * [ 7^6 -39*7^5 +655*7^4 -5565*7^3 +24184*7^2 -41556*7 +720 ]

%e = (1/720) * [ 117649 -655473 +1572655 -1908795 +1185016 -290892 +720 ] = 20880/720 = 29.

%Y Cf. A101032, A000204, A100492, A099731, A000045, A094216, A094638, A000108.

%K easy,sign,tabl

%O 1,3

%A _André F. Labossière_, Nov 30 2004