A102051 Matrix inverse of triangle A101275 (number of Schröder paths).
1, -1, 1, 3, -4, 1, -9, 15, -7, 1, 31, -58, 36, -10, 1, -113, 229, -170, 66, -13, 1, 431, -924, 775, -372, 105, -16, 1, -1697, 3795, -3481, 1939, -691, 153, -19, 1, 6847, -15822, 15542, -9674, 4072, -1154, 210, -22, 1, -28161, 66801, -69276, 47012, -22446, 7606, -1788, 276, -25, 1
Offset: 0
Examples
Rows begin: [1], [ -1,1], [3,-4,1], [ -9,15,-7,1], [31,-58,36,-10,1], [ -113,229,-170,66,-13,1], [431,-924,775,-372,105,-16,1], [ -1697,3795,-3481,1939,-691,153,-19,1], [6847,-15822,15542,-9674,4072,-1154,210,-22,1],... Matrix inverse equals triangle A101275: [1], [1,1], [1,4,1], [1,13,7,1], [1,44,34,10,1],...
Programs
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Maxima
T(n,m):=(-1)^(n-m)*(2*m+1)*(sum((binomial(k,n-k)*binomial(2*k,k-m))/(m+k+1),k,0,n)); /* Vladimir Kruchinin, Apr 18 2015 */
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PARI
{T(n,k)=polcoeff(polcoeff(2/(2*y+(1-y)*(1+sqrt(1+4*x-4*x^2+x*O(x^n)))),n)+y*O(y^k),k)}
Formula
G.f.: 2/(1+y+(1-y)*sqrt(1+4*x-4*x^2)).
T(n,m) = (-1)^(n-m)*(2*m+1)*Sum_{k=0..n} C(k,n-k)*C(2*k,k-m)/(m+k+1). - Vladimir Kruchinin, Apr 18 2015
Comments