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.

A100233 a(n) = Lucas(3*n) - 1.

Original entry on oeis.org

1, 3, 17, 75, 321, 1363, 5777, 24475, 103681, 439203, 1860497, 7881195, 33385281, 141422323, 599074577, 2537720635, 10749957121, 45537549123, 192900153617, 817138163595, 3461452808001, 14662949395603, 62113250390417, 263115950957275, 1114577054219521
Offset: 0

Views

Author

Paul D. Hanna, Nov 29 2004

Keywords

Comments

Main diagonal of triangle A100232.

Links

Crossrefs

Cf. A000032, A100230, A100232, A016064.

Programs

  • Magma
    I:=[1, 3, 17]; [n le 3 select I[n] else 5*Self(n-1) -3*Self(n-2) -Self(n-3): n in [1..30]]; // G. C. Greubel, Dec 21 2017
  • Mathematica
    Table[LucasL[3*n] - 1, {n,0,50}] (* or *) LinearRecurrence[{5,-3,-1}, {1,3,17}, 30] (* G. C. Greubel, Dec 21 2017 *)
  • PARI
    a(n)=if(n==0,1,n*polcoeff(log((1-x)/(1-4*x-x^2)+x*O(x^n)),n))
  • PARI
    Vec((1-2*x+5*x^2)/((1-x)*(1-4*x-x^2)) + O(x^40)) \\ Colin Barker, Jun 02 2016

Formula

a(n) = A014448(n) - 1.
a(n) = 4*a(n-1) + a(n-2) + 4 for n>1, with a(0)=1, a(1)=3.
G.f.: Sum_{n>=1} a(n)*x^n/n = log((1-x)/(1-4*x-x^2)).
a(n) = [x^n] ( 1 + 2*x + sqrt(1 + 2*x + 5*x^2) )^n. Cf. A016064. - Peter Bala, Jun 23 2015
From Colin Barker, Jun 02 2016: (Start)
a(n) = -1+(2-sqrt(5))^n+(2+sqrt(5))^n.
a(n) = 5*a(n-1)-3*a(n-2)-a(n-3) for n>2.
G.f.: (1-2*x+5*x^2) / ((1-x)*(1-4*x-x^2)).
(End)

Extensions

New definition from Ralf Stephan, Dec 01 2004

A100235 Triangle, read by rows, of the coefficients of [x^k] in G100234(x)^n such that the row sums are 6^n-1 for n>0, where G100234(x) is the g.f. of A100234.

Original entry on oeis.org

1, 1, 4, 1, 8, 26, 1, 12, 63, 139, 1, 16, 116, 436, 726, 1, 20, 185, 965, 2830, 3774, 1, 24, 270, 1790, 7335, 17634, 19601, 1, 28, 371, 2975, 15505, 52444, 106827, 101784, 1, 32, 488, 4584, 28860, 124424, 358748, 633952, 528526, 1, 36, 621, 6681, 49176, 256194
Offset: 0

Views

Author

Paul D. Hanna, Nov 29 2004

Keywords

Comments

The main diagonal forms A100236. Secondary diagonal is: T(n+1,n) = (n+1)*A100237(n). 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],
[1,4],
[1,8,26],
[1,12,63,139],
[1,16,116,436,726],
[1,20,185,965,2830,3774],
[1,24,270,1790,7335,17634,19601],
[1,28,371,2975,15505,52444,106827,101784],
[1,32,488,4584,28860,124424,358748,633952,528526],...
where row sums form 6^n-1 for n>0:
6^1-1 = 1+4 = 5
6^2-1 = 1+8+26 = 35
6^3-1 = 1+12+63+139 = 215
6^4-1 = 1+16+116+436+726 = 1295
6^5-1 = 1+20+185+965+2830+3774 = 7775.
The main diagonal forms A100236 = [1,4,26,139,726,3774,...],
where Sum_{n>=1} A100236(n)/n*x^n = log((1-x)/(1-5*x-x^2)).

Crossrefs

Cf. A100234, A100236, A100237, A100232.

Programs

  • Mathematica
    row[n_] := CoefficientList[ Series[ (1 + 5*x + Sqrt[1 + 6*x + 29*x^2])^n/2^n, {x, 0, n}], x]; Flatten[ Table[ row[n], {n, 0, 9}]](* Jean-François Alcover, May 11 2012, after PARI *)
  • PARI
    T(n,k,m=6)=if(n

Formula

G.f.: A(x, y)=(1-2*x*y+6*x^2*y^2)/((1-x*y)*(1-5*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

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) (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,...

Crossrefs

Cf. A100232, A100233, A100228.

Programs

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

Formula

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.
Showing 1-3 of 3 results.

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