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.

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A100226 Triangle, read by rows, of the coefficients of [x^k] in G100225(x)^n such that the row sums are 3^n-1 for n>0, where G100225(x) is the g.f. of A100225.

Original entry on oeis.org

1, 1, 1, 1, 2, 5, 1, 3, 9, 13, 1, 4, 14, 28, 33, 1, 5, 20, 50, 85, 81, 1, 6, 27, 80, 171, 246, 197, 1, 7, 35, 119, 301, 553, 693, 477, 1, 8, 44, 168, 486, 1064, 1724, 1912, 1153, 1, 9, 54, 228, 738, 1854, 3600, 5220, 5193, 2785, 1, 10, 65, 300, 1070, 3012, 6730, 11760
Offset: 0

Views

Author

Paul D. Hanna, Nov 28 2004

Keywords

Comments

Main diagonal forms A100227. Secondary diagonal is: T(n+1,n) = (n+1)*A001333(n), where A001333 is the numerators of continued fraction convergents to sqrt(2). 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],
  [1,1],
  [1,2,5],
  [1,3,9,13],
  [1,4,14,28,33],
  [1,5,20,50,85,81],
  [1,6,27,80,171,246,197],
  [1,7,35,119,301,553,693,477],
  [1,8,44,168,486,1064,1724,1912,1153],...
where row sums form 3^n-1 for n>0:
3^1-1 = 1+1
3^2-1 = 1+2+5
3^3-1 = 1+3+9+13
3^4-1 = 1+4+14+28+33
3^5-1 = 1+5+20+50+85+81.
The main diagonal forms A100227 = [1,1,5,13,33,81,197,477,...], where Sum_{n>=1} (A100227(n)/n)*x^n = log((1-x)/(1-2*x-x^2)).

Crossrefs

Cf. A100225, A100227, A001333.

Programs

  • PARI
    T(n,k,m=3)=if(n

Formula

G.f.: A(x, y)=(1-2*x*y+3*x^2*y^2)/((1-x*y)*(1-2*x*y-x^2*y^2-x*(1-x*y))).

A100223 G.f. A(x) satisfies: 2^n - 1 = Sum_{k=0..n} [x^k]A(x)^n and also satisfies: (2+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, 0, 1, 1, 0, -2, -3, 1, 11, 15, -13, -77, -86, 144, 595, 495, -1520, -4810, -2485, 15675, 39560, 6290, -159105, -324805, 87075, 1592843, 2616757, -2136539, -15726114, -20247800, 32296693, 152909577, 145139491, -417959049, -1460704685, -885536173, 4997618808, 13658704994, 3223741399
Offset: 0

Views

Author

Paul D. Hanna, Nov 28 2004

Keywords

Comments

The current sequence and A007440, A214649, A343773 form a cluster. The base of the cluster is A343773. We get the current sequence if we shift all the terms in A343773 to the right by 2 and take a(0) = 1. - Gennady Eremin, May 15 2021

Examples

			G.f. = 1 + x^2 + x^3 - 2*x^5 - 3*x^6 + x^7 + 11*x^8 + 15*x^9 - 13*x^10 - 77*x^11 - 86*x^12 + 144*x^13 + ...
From the table of powers of A(x) (A100224), we see that
2^n-1 = Sum of coefficients [x^0] through [x^n] in A(x)^n:
A^1=[1,0],1,1,0,-2,-3,1,11,...
A^2=[1,0,2],2,1,-2,-5,-2,12,...
A^3=[1,0,3,3],3,0,-5,-6,6,...
A^4=[1,0,4,4,6],4,-2,-8,-3,...
A^5=[1,0,5,5,10,10],5,-5,-10,...
A^6=[1,0,6,6,15,18,17],6,-9,...
A^7=[1,0,7,7,21,28,35,28],7,...
A^8=[1,0,8,8,28,40,60,64,46],...
the main diagonal of which is A001610 = [0,2,3,6,10,17,...],
where Sum_{n>=1} A001610(n-1)/n*x^n = log((1-x)/(1-x-x^2)).

Links

Crossrefs

Cf. A001006, A001610, A007440, A100224, A100225, A214649, A343773.

Programs

  • Mathematica
    CoefficientList[Series[(1+x+Sqrt[1-2*x+5*x^2])/2, {x, 0, 30}], x] (* Vaclav Kotesovec, Feb 11 2015 *)
  • PARI
    {a(n) = if( n<=0, n==0, (2^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, n==0, if( n==1, 0, if( n==2, 1, ((2*n-3) * a(n-1) - 5 * (n-3) * a(n-2)) / n)))}
  • PARI
    {a(n) = if( n<0, 0, polcoeff((1 + x + sqrt(1 - 2*x + 5*x^2 + x * O(x^n))) / 2, n))}
  • Python
    A100223 = [1, 0, 1]
for n in range(3, 801):
    A100223.append( ((2*n-3)*A100223[-1]
      - 5*(n-3)*A100223[-2])//n )  # Gennady Eremin, Apr 24 2021

Formula

G.f.: A(x) = (1+x+sqrt(1-2*x+5*x^2))/2.
G.f.: A(x) = x/(series_reversion[x*(1-x)/(1-x-x^2)]).
D-finite with recurrence a(n) = ((2*n-3)*a(n-1) - 5*(n-3)*a(n-2))/n (for n>2), with a(0)=1, a(1)=0, a(2)=1.
Given g.f. A(x), then B(x) = A(x) - 1 + x series reversion is -B(-x). - Michael Somos, Sep 07 2005
Given g.f. A(x) and C(x) = reversion of x + x^2, then B(x) = A(x) - 1 + x satisfies B(x) = x + C(x * B(x)). - Michael Somos, Sep 07 2005
From Paul Barry, Nov 07 2010: (Start)
a(n+1) has Hankel transform (-1)^n*F(n).
a(n+2) has Hankel transform (-1)^comb(n+1,2).
a(n+3) has Hankel transform (-1)^comb(n+1,2)*F(n+2).
Hankel transform of a(n+4) is sum{k=0..n, (-1)^(n-k+1)*(n-k+1)(F(2k+2)-C(1,k)-C(0,k))}. (End)
G.f.: A(x) = x + G(0) = 1 + x^2/G(0) where G(k) = 1 - x + x^2/G(k+1); (continued fraction). - Sergei N. Gladkovskii, Dec 25 2011
Lim sup n->infinity |a(n)|^(1/n) = sqrt(5).
From Gennady Eremin, May 15 2021: (Start)
a(n+2) = A343773(n), n >= 0.
G.f.: A(x) = 1 + x^2*B(x), where B(x) is the g.f. of A343773.
Lim_{n->infinity} a(n)/A001006(n) = 0. (End)

Extensions

Entry revised by Paul D. Hanna, Mar 19 2013

A100227 Main diagonal of triangle A100226.

Original entry on oeis.org

1, 1, 5, 13, 33, 81, 197, 477, 1153, 2785, 6725, 16237, 39201, 94641, 228485, 551613, 1331713, 3215041, 7761797, 18738637, 45239073, 109216785, 263672645, 636562077, 1536796801, 3710155681, 8957108165, 21624372013, 52205852193, 126036076401, 304278004997
Offset: 0

Views

Author

Paul D. Hanna, Nov 29 2004

Keywords

Comments

Specify that a triangle has T(n,0) = T(n,n) = (n+1)*(n+2)/2. The interior terms T(r,c) = T(r-1,c) + T(r-1,c-1) + T(r-2,c-1). The difference between the sum of the terms in row(n+1) and those in row(n) is a(n+2). - J. M. Bergot, Mar 15 2013
Starting with offset 1 the sequence is A001333: (1, 3, 7, 17, 41, ...), convolved with (1, 2, 0, 2, 0, 2, ...). - Gary W. Adamson, Aug 10 2016
Number of ways to tile a bracelet of length n with single-color squares, and two colors of k-ominoes for k > 1. Compare to A001333 as mentioned in the previous comment: A001333 can be thought of as the number of ways to tile a strip of length n with single-color squares and two-color k-ominoes for k > 1. - Greg Dresden, Feb 26 2020

Links

Crossrefs

Cf. A000129, A001333, A002203, A100225, A100226.

Programs

  • Mathematica
    LucasL[Range[0,35], 2] - 1 (* G. C. Greubel, Feb 26 2020 *)
  • Maxima
    a(n):=if n=0 then 1 else n*sum(sum(binomial(k, i)*binomial(n-i-1, k-1),i,0,n-k)/k,k,1,n); /* Vladimir Kruchinin, May 13 2011 */
  • PARI
    a(n)=if(n==0,1,n*polcoeff(log((1-x)/(1-2*x-x^2)+x*O(x^n)),n))
  • PARI
    a(n)=polcoeff((1-2*x+3*x^2)/(1-3*x+x^2+x^3)+x*O(x^n),n)

Formula

a(n) = A002203(n) - 1.
a(n) = 2*a(n-1) + a(n-2) + 2 for n > 1, with a(0)=1, a(1)=1.
G.f.: Sum_{n>=1} a(n)*x^n/n = log((1-x)/(1-2*x-x^2)).
G.f.: (1-2*x+3*x^2)/((1-x)*(1-2*x-x^2)). - Paul D. Hanna, Feb 22 2005
a(n) = n*Sum_{k=1..n} (1/k)*Sum_{i=0..n-k} binomial(k, i)*binomial(n-i-1, k-1), n > 0, a(0)=1. - Vladimir Kruchinin, May 13 2011
a(n) = -1 + (1-sqrt(2))^n + (1+sqrt(2))^n. - Colin Barker, Mar 16 2016
E.g.f.: (2*cosh(sqrt(2)*x) - 1)*exp(x). - Ilya Gutkovskiy, Aug 22 2016
a(n) = A000129(n+1) + A000129(n-1)-1 for n > 0. - Rigoberto Florez, Jul 12 2020
a(n) = 3*a(n-1) - a(n-2) - a(n-3). - Wesley Ivan Hurt, Jul 13 2020
Showing 1-3 of 3 results.

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