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.

A104722 Self-convolution of repeated Catalan numbers.

Original entry on oeis.org

1, 2, 3, 4, 7, 10, 19, 28, 56, 84, 174, 264, 561, 858, 1859, 2860, 6292, 9724, 21658, 33592, 75582, 117572, 266798, 416024, 950912, 1485800, 3417340, 5348880, 12369285, 19389690, 45052515, 70715340, 165002460, 259289580, 607283490
Offset: 0

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Author

Paul Barry, Mar 20 2005

Keywords

Comments

This is the same as A059348 after the first term. [Proof by James Sellers, seqfan 19 May 2008: The generating functions are the same, ignoring the constant terms which cause the difference between the two sequences. If the g.f. in the formula here is expanded, the constant term ignored, we obtain ( 1 + 2x - x^2 - 4x^3 - (x+1)^2*sqrt(1-4*x^2) )/(2*x^4) ].
From the Bernhart Reference in A059348 we see that A059348 originates from A000108 padded with zeros, 1 0 1 0 2 0 5 0 14 0 42 0 132 ... with g.f. C(x^2). Taking the sum of each pair of consecutive values we get the auxiliary sequence 1 1 1 2 2 5 5 14 14 42 42 132 132 ... with g.f. ((1+x)C(x^2) - 1)/x . Sum pairs of consecutive values once more to obtain 2 2 3 4 7 10 19 28 56 ... which is A059348.
So this generating function is (1+x)*(((1+x)*C(x^2) - 1)/x -1) / x again ignoring the constant term. Straightforward algebraic manipulations show that this quantity equals (1+2x-...)/(2x^4) above, again ignoring the constant term.

Links

Crossrefs

Cf. A000108, A059348, A104721.

Programs

  • Mathematica
    CoefficientList[Series[( (1 + x)*(1 - Sqrt[1 - 4*x^2])/(2*x^2))^2, {x, 0, 100}], x] (* G. C. Greubel, Jan 07 2017 *)
  • PARI
    Vec( ((1 + x)*(1 - sqrt(1 - 4*x^2))/(2*x^2))^2 + O(x^20)) \\ G. C. Greubel, Jan 07 2017

Formula

G.f.: (1+x)^2*c(x^2)^2, c(x) the g.f. of the Catalan numbers A000108;
Let b(n) = (binomial(n-1, (n-1)/2)/((n-1)/2+1))*(1-(-1)^n)/2 + (binomial(n, n/2)/(n/2+1))*(1+(-1)^n)/2, then a(n) = Sum_{k=0..n} b(k)*b(n-k).
Conjecture: (n+4)*a(n) + (n+1)*a(n-1) - 4*(n+1)*a(n-2) + 4*(2-n)*a(n-3) = 0. - R. J. Mathar, Nov 09 2012
a(n) ~ 2^(n + 1/2) * (9 + (-1)^n) / (sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, Mar 10 2018

A084568 a(0)=1, a(1)=5, a(n+2)=4a(n), n>0.

Original entry on oeis.org

1, 5, 8, 20, 32, 80, 128, 320, 512, 1280, 2048, 5120, 8192, 20480, 32768, 81920, 131072, 327680, 524288, 1310720, 2097152, 5242880, 8388608, 20971520, 33554432, 83886080, 134217728, 335544320, 536870912, 1342177280, 2147483648
Offset: 0

Views

Author

Paul Barry, May 30 2003

Keywords

Comments

Binomial transform is A080926 (without leading zero).

Links

Crossrefs

Cf. A080926, A104721.

Programs

  • Mathematica
    Join[{1},LinearRecurrence[{0,4},{5,8},30]] (* Harvey P. Dale, Sep 12 2013 *)

Formula

G.f.: (1+5*x+4*x^2)/((1+2*x)*(1-2*x)).
E.g.f.: (9*exp(2*x)-exp(-2*x))/4-exp(0).
a(n) = (9*2^n-(-2)^n)/4-0^n.

A209518 Triangle by rows, reversal of A104712.

Original entry on oeis.org

1, 1, 3, 1, 4, 6, 1, 5, 10, 10, 1, 6, 15, 20, 15, 1, 7, 21, 35, 35, 21, 1, 8, 28, 56, 70, 56, 28, 1, 9, 36, 84, 126, 126, 84, 36, 1, 10, 45, 120, 210, 252, 210, 120, 45, 1, 11, 55, 165, 330, 462, 462, 330, 165, 55
Offset: 0

Views

Author

Gary W. Adamson, Mar 09 2012

Keywords

Comments

The offset is chosen as "0" to match the generalized or compositional Bernoulli numbers.
Following [Blandin and Diaz], we can generalize a subset of Bernoulli numbers to comply with the origin of the triangle (the Pascal matrix A007318 beheaded once: (A074909), twice: (this triangle), and so on...); and a corresponding Bernoulli sequence that equals the inverse of the triangle, extracting the left border. This procedure done with A074909 results in The Bernoulli numbers (A027641/A026642) starting (1, -1/2, 1/6,...). Done with this triangle we obtain A006568/A006569: (1, -1/3, 1/18, 1/90,...).
A generalized algebraic property of the subset of such triangles and compositional Bernoulli numbers is that the triangle M * [corresponding Bernoulli sequence considered as a vector, V] = [1, 0, 0, 0,...].
The infinite set of generalized Bernoulli number sequences thus generated from variants of Pascal's triangle begins: [(1, -1/2, 1/6,...); (1, -1/3, 1/18,...); (1, -1/4, 1/40,...); (1, -1/5, 1/75,...); where the third term denominators = A002411 (1, 6, 18, 40, 75,...) after the "1".
Row sums of the triangle = A000295 starting (1, 4, 11, 26, 57,...).

Examples

			First few rows of the triangle =
1;
1, 3;
1, 4, 6;
1, 5, 10, 10;
1, 6, 15, 20, 15;
1, 7, 21, 35, 35, 21;
1, 8, 28, 56, 70, 56, 28;
1, 9, 36, 84, 126, 126, 84, 36;
1, 10, 45, 120, 210, 252, 210, 120, 45;
1, 11, 55, 165, 330, 462, 462, 30, 165, 55;
...

Links

  • Hector Blandin and Rafael Diaz, Compositional Bernoulli numbers, arXiv:0708.0809 [math.CO], 2007-2008, Page 7, 2nd table is identical to A006569/A006568.

Crossrefs

Cf. A000295, A104721, A074909, A027641, A027642, A006568, A006569, A007318, A002411.

Programs

  • Mathematica
    Table[Binomial[n+2, k+2], {n, 0, 9}, {k, n , 0, -1}] // Flatten (* Jean-François Alcover, Aug 08 2018 *)

Formula

Doubly beheaded variant of Pascal's triangle in which two rightmost diagonals are deleted.
T(n,k)=T(n-1,k)+3*T(n-1,k-1)-2*T(n-2,k-1)-3*T(n-2,k-2)+T(n-3,k-2)+T(n-3,k-3), T(0,0)=1, T(n,k)=0 if k<0 or if k>n. - Philippe Deléham, Jan 11 2014
Showing 1-3 of 3 results.

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