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 1-2 of 2 results.

A052886 Expansion of e.g.f.: (1/2) - (1/2)*(5 - 4*exp(x))^(1/2).

Original entry on oeis.org

0, 1, 3, 19, 207, 3211, 64383, 1581259, 45948927, 1541641771, 58645296063, 2494091717899, 117258952478847, 6038838138717931, 338082244882740543, 20443414320405268939, 1327850160592214291967, 92200405122521276427691, 6815359767190023358085823, 534337135055010788022858379
Offset: 0

Views

Author

encyclopedia(AT)pommard.inria.fr, Jan 25 2000

Keywords

Comments

Previous name was: A simple grammar.
From the symmetry present in Vladeta Jovovic's Feb 02 2005 formula, it is easy to see that there are no positive even numbers in this sequence. Question: are there any multiples of 5 after the initial zero? Compare also to the comments in A366884. - Antti Karttunen, Jan 02 2024

Links

Crossrefs

Cf. A000108, A054867, A277120, A293379, A366377, A366884.
Cf. A371316, A371317.
Cf. A371329, A376036, A376037.

Programs

  • Maple
    spec := [S,{C=Set(Z,1 <= card),S=Prod(B,C),B=Sequence(S)},labeled]: seq(combstruct[count](spec,size=n), n=0..20);
  • Mathematica
    CoefficientList[Series[1/2-1/2*(5-4*E^x)^(1/2), {x, 0, 20}], x]* Range[0, 20]! (* Vaclav Kotesovec, Sep 30 2013 *)
a[n_] := Sum[k! StirlingS2[n, k] CatalanNumber[k - 1], {k, 1, n}];
Array[a, 20, 0] (* Peter Luschny, Apr 30 2020 *)
  • PARI
    N=66; x='x+O('x^N);
concat([0],Vec(serlaplace(serreverse(log(1+x-x^2)))))
\\ Joerg Arndt, May 25 2011
  • PARI
    lista(nn) = {my(va = vector(nn)); va[1] = 1; for (n=2, nn, va[n] = 1+ sum(k=1, n-1, binomial(n,k)*va[k]*va[n-k]);); concat(0, va);} \\ Michel Marcus, Apr 30 2020
  • PARI
    A000108(n) = binomial(2*n, n)/(n+1);
A052886(n) = sum(k=1,n,k!*stirling(n,k,2)*A000108(k-1)); \\ Antti Karttunen, Jan 02 2024

Formula

E.g.f.: (1/2) - (1/2)*(5 - 4*exp(x))^(1/2).
a(n) = 1 + Sum_{k=1..n-1} binomial(n,k)*a(k)*a(n-k). - Vladeta Jovovic, Feb 02 2005
a(n) = Sum_{k=1..n} k!*Stirling2(n,k)*C(k-1), where C(k) = Catalan numbers (A000108). - Vladimir Kruchinin, Sep 15 2010
a(n) ~ sqrt(5/2)/2 * n^(n-1) / (exp(n)*(log(5/4))^(n-1/2)). - Vaclav Kotesovec, Sep 30 2013
Equals the logarithmic derivative of A293379. - Paul D. Hanna, Oct 22 2017
O.g.f.: Sum_{k>=1} C(k-1)*Product_{r=1..k} r*x/(1-r*x), where C = A000108. - Petros Hadjicostas, Jun 12 2020
a(n) = A366377(A000040(n)) = A366884(A098719(n)). - Antti Karttunen, Jan 02 2024
From Seiichi Manyama, Sep 09 2024: (Start)
E.g.f. satisfies A(x) = (exp(x) - 1) / (1 - A(x)).
E.g.f.: Series_Reversion( log(1 + x * (1 - x)) ). (End)

Extensions

New name using e.g.f. by Vaclav Kotesovec, Sep 30 2013

A331406 Array read by antidiagonals: A(n,m) is the number of ways to place non-adjacent counters on the black squares of a 2n-1 X 2m-1 checker board.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 4, 4, 1, 1, 8, 17, 8, 1, 1, 16, 73, 73, 16, 1, 1, 32, 314, 689, 314, 32, 1, 1, 64, 1351, 6556, 6556, 1351, 64, 1, 1, 128, 5813, 62501, 139344, 62501, 5813, 128, 1, 1, 256, 25012, 596113, 2976416, 2976416, 596113, 25012, 256, 1
Offset: 0

Views

Author

Andrew Howroyd, Jan 16 2020

Keywords

Comments

The array has been extended with A(n,0) = A(0,m) = 1 for consistency with recurrences and existing sequences.
The checker board is such that the black squares are in the corners and adjacent means diagonally adjacent, since the white squares are not included.
Equivalently, A(n,m) is the number of independent sets in the generalized Aztec diamond graph E(L_{2n-1}, L_{2m-1}). The E(L_{2n-1},L_{2m-1}) Aztec diamond is the graph with vertices {(a,b) : 1<=a<=2n-1, 1<=b<=2m-1, a+b even} and edges between (a,b) and (c,d) if and only if |a-b|=|c-d|=1.
All rows (or columns) are linear recurrences with constant coefficients. For n > 0 an upper bound on the order of the recurrence is A005418(n-1), which is the number of binary words of length n up to reflection.
A stronger upper bound on the recurrence order is A005683(n+2). This upper bound is exact for at least 1 <= n <= 10. This bound follows from considerations about which patterns of counters in a row are redundant because they attack the same points in adjacent rows. For example, the pattern of counters 1101101 is equivalent to 1111111 because they each attack all points in the neighboring rows.
It appears that the denominators for the recurrences are the same as those for the rows and columns of A254414. This suggests there is a connection.

Examples

			Array begins:
===========================================================
n\m | 0  1    2      3        4          5            6
----+------------------------------------------------------
  0 | 1  1    1      1        1          1            1 ...
  1 | 1  2    4      8       16         32           64 ...
  2 | 1  4   17     73      314       1351         5813 ...
  3 | 1  8   73    689     6556      62501       596113 ...
  4 | 1 16  314   6556   139344    2976416     63663808 ...
  5 | 1 32 1351  62501  2976416  142999897   6888568813 ...
  6 | 1 64 5813 596113 63663808 6888568813 748437606081 ...
  ...
Case A(2,2): the checker board has 5 black squares as shown below.
      __    __
     |__|__|__|
      __|__|__
     |__|  |__|
If a counter is placed on the central square then a counter cannot be placed on the other 4 squares, otherwise counters can be placed in any combination. The total number of arrangements is then 1 + 2^4 = 17, so A(2, 2) = 17.

Links

Crossrefs

Rows n=0..5 are A000012, A000079, A018902, A254150, A254151, A254152.
Main diagonal is A054867.
Cf. A005418, A005683, A089934, A254414.

Programs

  • PARI
    step1(v)={vector(#v/2, t, my(i=t-1); sum(j=0, #v-1, if(!bitand(i, bitor(j, j>>1)), v[1+j])))}
step2(v)={vector(#v*2, t, my(i=t-1); sum(j=0, #v-1, if(!bitand(i, bitor(j, j<<1)), v[1+j])))}
T(n,k)={if(n==0||k==0, 1, my(v=vector(2^k, i, 1)); for(i=2, n, v=step2(step1(v))); vecsum(v))}
{ for(n=0, 7, for(k=0, 7, print1(T(n,k), ", ")); print) }

Formula

A(n,m) = A(m,n).
Showing 1-2 of 2 results.

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