Search: id:A101032 Results 1-1 of 1 results found. %I A101032 %S A101032 1,1,1,1,1,2,1,6,17,6,1,14,83,142,24,1,25,265,1235,2314,120,1,39,655,5565, %T A101032 24184,41556,720,1,56,1372,18200,137599,556304,944628,5040,1,76,2562,48664, %U A101032 560049,3884524,15021068,24875376,40320,1,99,4398,113022,1829793,19043451 %V A101032 1,1,1,1,-1,2,1,-6,17,6,1,-14,83,-142,24,1,-25,265,-1235,2314,120,1,-39, 655,-5565, %W A101032 24184,-41556,720,1,-56,1372,-18200,137599,-556304,944628,5040,1,-76,2562, -48664, %X A101032 560049,-3884524,15021068,-24875376,40320,1,-99,4398,-113022,1829793,-19043451 %N A101032 Table (read by rows) giving the coefficients of sum formulae 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 A101032 R. D. Knott, The Lucas Numbers in Pascal's Triangle. %H A101032 A. F. Labossiere, Sobalian Coefficients. %H A101032 A. F. Labossiere, Miscellaneous. %e A101032 L(13)=521; substituting n=13 in the formula of the k-th row we obtain k=7 and the coefficients %e A101032 T(i,7) will be the following: 1,-39,655,-5565,24184,-41556,720, %e A101032 => L(13) = [13^6-39*13^5+655*13^4-5565*13^3+24184*13^2-41556*13+720]/ 6! = 521. %Y A101032 Cf. A101033, A000204, A100492, A099731, A000045, A094216, A094638, A000108. %Y A101032 Sequence in context: A049951 A025263 A097947 this_sequence A025271 A153804 A100404 %Y A101032 Adjacent sequences: A101029 A101030 A101031 this_sequence A101033 A101034 A101035 %K A101032 sign,tabl %O A101032 1,6 %A A101032 Andre F. Labossiere (boronali(AT)laposte.net), Nov 30 2004 Search completed in 0.001 seconds