A100231 -id:A100231 - cp's OEIS frontend

A100232 Triangle, read by rows, of the coefficients of [x^k] in G100231(x)^n such that the row sums are 5^n-1 for n>0, where G100231(x) is the g.f. of A100231.

1, 1, 3, 1, 6, 17, 1, 9, 39, 75, 1, 12, 70, 220, 321, 1, 15, 110, 470, 1165, 1363, 1, 18, 159, 852, 2895, 5922, 5777, 1, 21, 217, 1393, 5943, 16807, 29267, 24475, 1, 24, 284, 2120, 10822, 38536, 93468, 141688, 103681, 1, 27, 360, 3060, 18126, 77274, 236748

Offset: 0

Paul D. Hanna, Nov 29 2004

The main diagonal forms A100233. Secondary diagonal is: T(n+1,n) = (n+1)*A033887(n) = (n+1)*Fibonacci(3*n+1). More generally, if g.f. F(x) satisfies: m^n-b^n = Sum_{k=0..n} [x^k]F(x)^n, then F(x) also satisfies: (m+z)^n - (b+z)^n + z^n = Sum_{k=0..n} [x^k](F(x)+z*x)^n for all z and F(x)=(1+(m-1)*x+sqrt(1+2*(m-2*b-1)*x+(m^2-2*m+4*b+1)*x^2))/2; the triangle formed from powers of F(x) will have the g.f.: G(x,y)=(1-2*x*y+m*x^2*y^2)/((1-x*y)*(1-(m-1)*x*y-x^2*y^2-x*(1-x*y))).

			Rows begin:
[1],
[1,3],
[1,6,17],
[1,9,39,75],
[1,12,70,220,321],
[1,15,110,470,1165,1363],
[1,18,159,852,2895,5922,5777],
[1,21,217,1393,5943,16807,29267,24475],
[1,24,284,2120,10822,38536,93468,141688,103681],...
where row sums form 5^n-1 for n>0:
5^1-1 = 1+3 = 4
5^2-1 = 1+6+17 = 24
5^3-1 = 1+9+39+75 = 124
5^4-1 = 1+12+70+220+321 = 624
5^5-1 = 1+15+110+470+1165+1363 = 3124.
The main diagonal forms A100233 = [1,3,17,75,321,1363,5777,...],
where Sum_{n>=1} A100233(n)/n*x^n = log((1-x)/(1-4*x-x^2)).

Cf. A100231, A100233, A033887, A100229.

PARI
```
T(n,k,m=5)=if(n
				
```

G.f.: A(x, y)=(1-2*x*y+5*x^2*y^2)/((1-x*y)*(1-4*x*y-x^2*y^2-x*(1-x*y))).

A100234 G.f. A(x) satisfies: 6^n - 1 = Sum_{k=0..n} [x^k]A(x)^n and also satisfies: (6+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.

Original entry on oeis.org

1, 4, 5, -15, 20, 90, -695, 1785, 3895, -53985, 196255, 121635, -4907130, 23332140, -13181145, -470127465, 2866898820, -4455872910, -44776087145, 356263904235, -873534120380, -3988869806010, 44179467566755, -147200296896765, -293052319462105, 5409366658571715

Offset: 0

Paul D. Hanna, Nov 29 2004

sign

More generally, if g.f. A(x) satisfies: m^n-b^n = Sum_{k=0..n} [x^k]A(x)^n, then A(x) also satisfies: (m+z)^n - (b+z)^n + z^n = Sum_{k=0..n} [x^k](A(x)+z*x)^n for all z and A(x)=(1+(m-1)*x+sqrt(1+2*(m-2*b-1)*x+(m^2-2*m+4*b+1)*x^2))/2.

			From the table of powers of A(x) (A100235), we see that
6^n-1 = Sum of coefficients [x^0] through [x^n] in A(x)^n:
A^1=[1,4],5,-15,20,90,-695,1785,...
A^2=[1,8,26],10,-55,190,-245,-1690,...
A^3=[1,12,63,139],15,-120,635,-2130,...
A^4=[1,16,116,436,726],20,-210,1480,...
A^5=[1,20,185,965,2830,3774],25,-325,...
A^6=[1,24,270,1790,7335,17634,19601],30,...

Cf. A100235, A100236, A100231.

a(n)=if(n==0,1,(6^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)

a(n)=if(n==0,1,if(n==1,4,if(n==2,5,-(3*(2*n-3)*a(n-1)+29*(n-3)*a(n-2))/n)))

a(n)=polcoeff((1+5*x+sqrt(1+6*x+29*x^2+x^2*O(x^n)))/2,n)

a(n)=-(3*(2*n-3)*a(n-1)+29*(n-3)*a(n-2))/n for n>2, with a(0)=1, a(1)=4, a(2)=5. G.f.: A(x) = (1+5*x+sqrt(1+6*x+29*x^2))/2.

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A100232 Triangle, read by rows, of the coefficients of [x^k] in G100231(x)^n such that the row sums are 5^n-1 for n>0, where G100231(x) is the g.f. of A100231.

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A100234 G.f. A(x) satisfies: 6^n - 1 = Sum_{k=0..n} [x^k]A(x)^n and also satisfies: (6+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.

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