|
Search: id:A108299
|
|
|
| A108299 |
|
Triangle read by rows, 0 <= k <= n: T(n,k)=binomial(n-[(k+1)/2],[k/2])*(-1)^[(k+1)/2]. |
|
+0 39
|
|
| 1, 1, -1, 1, -1, -1, 1, -1, -2, 1, 1, -1, -3, 2, 1, 1, -1, -4, 3, 3, -1, 1, -1, -5, 4, 6, -3, -1, 1, -1, -6, 5, 10, -6, -4, 1, 1, -1, -7, 6, 15, -10, -10, 4, 1, 1, -1, -8, 7, 21, -15, -20, 10, 5, -1, 1, -1, -9, 8, 28, -21, -35, 20, 15, -5, -1, 1, -1, -10, 9, 36, -28, -56, 35, 35, -15, -6, 1, 1, -1, -11, 10, 45, -36, -84, 56, 70
(list; table; graph; listen)
|
|
|
OFFSET
|
0,9
|
|
|
COMMENT
|
Matrix inverse of A124645 .
Let L(n,x) = Sum(T(n,k)*x^(n-k): 0<=k<=n) and Pi=3.14...:
L(n,x) = Prod(x - 2*cos((2*k-1)*Pi/(2*n+1)): 1<=k<=n);
Sum(T(n,k): 0<=k<=n) = L(n,1) = A010892(n+1);
Sum(abs(T(n,k)): 0<=k<=n) = A000045(n+2);
abs(T(n,k))=A065941(n,k), T(n,k)=A065941(n,k)*A087960(k);
T(2*n,k) + T(2*n+1,k+1) = 0 for 0<=k<=2*n;
T(n,0)=A000012(n)=1; T(n,1)=-1 for n>0;
T(n,2)=-(n-1) for n>1; T(n,3)=A000027(n)=n for n>2;
T(n,4)=A000217(n-3) for n>3; T(n,5)=-A000217(n-4) for n>4;
T(n,6)=-A000292(n-5) for n>5; T(n,7)=A000292(n-6) for n>6;
T(n,n-3)=A058187(n-3)*(-1)^[n/2] for n>2;
T(n,n-2)=A008805(n-2)*(-1)^[(n+1)/2] for n>1;
T(n,n-1)=A008619(n-1)*(-1)^[n/2] for n>0;
T(n,n) = L(n,0) = (-1)^[(n+1)/2];
L(n,1) = A010892(n+1); L(n,-1) = A061347(n+2);
L(n,2) = 1; L(n,-2) = A005408(n)*(-1)^n;
L(n,3) = A001519(n); L(n,-3) = A002878(n)*(-1)^n;
L(n,4) = A001835(n+1); L(n,-4) = A001834(n)*(-1)^n;
L(n,5) = A004253(n); L(n,-5) = A030221(n)*(-1)^n;
L(n,6) = A001653(n); L(n,-6) = A002315(n)*(-1)^n;
L(n,7) = A049685(n); L(n,-7) = A033890(n)*(-1)^n;
L(n,8) = A070997(n); L(n,-8) = A057080(n)*(-1)^n;
L(n,9) = A070998(n); L(n,-9) = A057081(n)*(-1)^n;
L(n,10) = A072256(n+1); L(n,-10) = A054320(n)*(-1)^n;
L(n,11) = A078922(n+1); L(n,-11) = A097783(n)*(-1)^n;
L(n,12) = A077417(n); L(n,-12) = A077416(n)*(-1)^n;
L(n,13) = A085260(n);
L(n,14) = A001570(n); L(n,-14) = A028230(n)*(-1)^n;
L(n,n) = A108366(n); L(n,-n) = A108367(n).
Row n of the matrix inverse (A124645) has g.f.: x^[n/2]*(1-x)^(n-[n/2]). - Paul D. Hanna (pauldhanna(AT)juno.com), Jun 12 2005
|
|
REFERENCES
|
Friedrich L. Bauer, 'De Moivre und Lagrange: Cosinus eines rationalen Vielfachen von Pi', Informatik Spektrum 28 (Springer, 2005).
|
|
FORMULA
|
T(n+1, k) = if sign(T(n, k-1))=sign(T(n, k)) then T(n, k-1)+T(n, k) else -T(n, k-1) for 0<k<n, T(n, 0) = 1, T(n, n) = (-1)^[(n+1)/2].
G.f.: A(x, y) = (1 - x*y)/(1 - x + x^2*y^2). - Paul D. Hanna (pauldhanna(AT)juno.com), Jun 12 2005
|
|
EXAMPLE
|
Triangle begins :
1 ;
1, -1 ;
1, -1, -1 ;
1, -1, -2, 1 ;
1, -1, -3, 2, 1 ;
1, -1, -4, 3, 3, -1 ;
1, -1, -5, 4, 6, -3, -1 ;
1, -1, -6, 5, 10, -6, -4, 1 ;
1, -1, -7, 6, 15, -10, -10, 4, 1 ;
1, -1, -8, 7, 21, -15, -20, 10, 5, -1 ;
1, -1, -9, 8, 28, -21, -35, 20, 15, -5, -1 ;
1, -1, -10, 9, 36, -28, -56, 35, 35, -15, -6, 1 ;...
Matrix inverse (A124645) begins:
1;
1,-1;
0,1,-1;
0,1,-2,1;
0,0,1,-2,1;
0,0,1,-3,3,-1;
0,0,0,1,-3,3,-1; ...
|
|
PROGRAM
|
(PARI) {T(n, k)=polcoeff(polcoeff((1-x*y)/(1-x+x^2*y^2+x^2*O(x^n)), n, x)+y*O(y^k), k, y)} (Hanna)
|
|
CROSSREFS
|
Cf. A049310, A039961, A124645.
Sequence in context: A136568 A152157 A039961 this_sequence A065941 A123320 A054123
Adjacent sequences: A108296 A108297 A108298 this_sequence A108300 A108301 A108302
|
|
KEYWORD
|
sign,tabl
|
|
AUTHOR
|
Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Jun 01 2005
|
|
EXTENSIONS
|
Corrected and edited. - Philippe DELEHAM (kolotoko(AT)wanadoo.fr), Oct 20 2008
|
|
|
Search completed in 0.003 seconds
|