A100233 -id:A100233 - cp's OEIS frontend

A100236 Main diagonal of triangle A100235.

1, 4, 26, 139, 726, 3774, 19601, 101784, 528526, 2744419, 14250626, 73997554, 384238401, 1995189564, 10360186226, 53796120699, 279340789726, 1450500069334, 7531841136401, 39109705751344, 203080369893126

Offset: 0

Paul D. Hanna, Nov 30 2004

nonn

Cf. A100233, A100235.

Equals A087130(n) - 1.

a(n)=if(n==0,1,n*polcoeff(log((1-x)/(1-5*x-x^2)+x*O(x^n)),n))

a(n) = 5*a(n-1) + a(n-2) + 5 for n>1, with a(0)=1, a(1)=4. G.f.: Sum_{n>=1} a(n)*x^n/n = log((1-x)/(1-5*x-x^2)).

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.

Original entry on oeis.org

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))).

A100231 G.f. A(x) satisfies: 5^n - 1 = Sum_{k=0..n} [x^k]A(x)^n and also satisfies: (5+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, 3, 4, -8, 0, 64, -192, -128, 2816, -7680, -13312, 157696, -352256, -1179648, 9748480, -16220160, -99614720, 630456320, -651427840, -8218214400, 41481666560, -13191086080, -667334737920, 2724661821440, 1460876083200, -53446942130176, 175607589634048, 286761410363392

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) (A100232), we see that
5^n-1 = Sum of coefficients [x^0] through [x^n] in A(x)^n:
A^1=[1,3],4,-8,0,64,-192,-128,...
A^2=[1,6,17],8,-32,64,64,-896,...
A^3=[1,9,39,75],12,-72,256,-384,...
A^4=[1,12,70,220,321],16,-128,640,...
A^5=[1,15,110,470,1165,1363],20,-200,...
A^6=[1,18,159,852,2895,5922,5777],24,...

Cf. A100232, A100233, A100228.

a(n)=if(n==0,1,(5^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==1,3,if(n==2,4,-((4*n-6)*a(n-1)+20*(n-3)*a(n-2))/n)))

a(n)=polcoeff((1+4*x+sqrt(1+4*x+20*x^2+x^2*O(x^n)))/2,n)

a(n)=-((4*n-6)*a(n-1)+20*(n-3)*a(n-2))/n for n>2, with a(0)=1, a(1)=3, a(3)=4. G.f.: A(x) = (1+4*x+sqrt(1+4*x+20*x^2))/2.

cp's OEIS Frontend

A100236 Main diagonal of triangle A100235.

Views

Author

Keywords

Crossrefs

Programs

PARI

Formula

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.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

PARI

Formula

A100231 G.f. A(x) satisfies: 5^n - 1 = Sum_{k=0..n} [x^k]A(x)^n and also satisfies: (5+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.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

PARI

PARI

PARI

Formula