cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

A100230 Main diagonal of triangle A100229.

Original entry on oeis.org

1, 2, 10, 35, 118, 392, 1297, 4286, 14158, 46763, 154450, 510116, 1684801, 5564522, 18378370, 60699635, 200477278, 662131472, 2186871697, 7222746566, 23855111398, 78788080763, 260219353690, 859446141836, 2838557779201
Offset: 0

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Author

Paul D. Hanna, Nov 29 2004

Keywords

Comments

Let F(x) = Product_{n >= 1} (1 + x^(4*n + 1))/(1 - x^(4*n + 3)). Let alpha = (1/2)*(3 - sqrt(13)). This sequence occurs as partial numerators in the simple continued fraction expansion of the real number F(alpha) = 1.34372 29374 22358 27049 ... = 1 + 1/(2 + 1/(1 + 1/(10 + 1/(35 + 1/(1 + 1/(118 + 1/(392 + 1/(1 + ...)))))))). - Peter Bala, Oct 17 2019

Crossrefs

Equals A006497(n) - 1.

Programs

  • Mathematica
    LinearRecurrence[{4,-2,-1},{1,2,10},30] (* Harvey P. Dale, May 06 2012 *)
  • PARI
    a(n)=if(n==0,1,n*polcoeff(log((1-x)/(1-3*x-x^2)+x*O(x^n)),n))

Formula

a(n) = 3*a(n-1) + a(n-2) + 3 for n>1, with a(0)=1, a(1)=2.
G.f.: Sum_{n>=1} a(n)*x^n/n = log((1-x)/(1-3*x-x^2)).
a(0)=1, a(1)=2, a(2)=10, a(n)=4*a(n-1)-2*a(n-2)-a(n-3). [Harvey P. Dale, May 06 2012]

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.

Original entry on oeis.org

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

Views

Author

Paul D. Hanna, Nov 29 2004

Keywords

Comments

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.

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,...
		

Crossrefs

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.

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

Views

Author

Paul D. Hanna, Nov 29 2004

Keywords

Comments

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))).

Examples

			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)).
		

Crossrefs

Programs

  • PARI
    T(n,k,m=5)=if(n
    				

Formula

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))).
Showing 1-3 of 3 results.