A100228 G.f. A(x) satisfies: 4^n - 1 = Sum_{k=0..n} [x^k] A(x)^n and also satisfies: (4+z)^n - (1+z)^n + z^n = Sum_{k=0..n} [x^k](A(x)+z*x)^n for all z, where [x^k] A(x)^n denotes the coefficient of x^k in A(x)^n.
1, 2, 3, -3, -6, 24, 3, -183, 273, 1131, -4407, -3081, 48360, -54750, -396195, 1282551, 1860186, -17122944, 11240049, 166745823, -432682314, -1054472016, 6822994737, -835915197, -76044224139, 152526011235, 587055710271, -2871405804783, -1378878506592, 36081844133766
Offset: 0
Keywords
Examples
From the table of powers of A(x) (A100229), we see that 4^n-1 = Sum of coefficients [x^0] through [x^n] in A(x)^n: A^1=[1,2],3,-3,-6,24,3,-183,273,... A^2=[1,4,10],6,-15,6,75,-174,-276,... A^3=[1,6,21,35],9,-36,63,72,-612,... A^4=[1,8,36,92,118],12,-66,192,-147,... A^5=[1,10,55,185,380,392],15,-105,420,... A^6=[1,12,78,322,879,1506,1297],18,-153,...
Programs
-
Mathematica
CoefficientList[Series[(1+3x+Sqrt[1+2x+13x^2])/2,{x,0,30}],x] (* or *) Join[{1},RecurrenceTable[{a[1]==2,a[2]==3,a[n]==-((2n-3)a[n-1]+ 13(n-3)a[n-2])/n},a,{n,30}]] (* Harvey P. Dale, Feb 29 2012 *)
-
PARI
{a(n) = if(n==0,1,(4^n-1-sum(k=0,n,polcoeff(sum(j=0,min(k,n-1),a(j)*x^j)^n+x*O(x^k),k)))/n)} for(n=0,20,print1(a(n),", "))
-
PARI
{a(n) = if(n==0,1,if(n==1,2,if(n==2,3,-((2*n-3)*a(n-1)+13*(n-3)*a(n-2))/n)))} for(n=0,30,print1(a(n),", "))
-
PARI
{a(n) = polcoeff((1+3*x+sqrt(1+2*x+13*x^2+x^2*O(x^n)))/2,n)} for(n=0,30,print1(a(n),", "))
Formula
a(n) = -((2*n-3)*a(n-1) + 13*(n-3)*a(n-2))/n for n>2, with a(0)=1, a(1)=2, a(3)=3.
G.f.: (1+3*x + sqrt(1 + 2*x + 13*x^2))/2.
Comments