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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A144097 The 4-Schroeder numbers: a(n) = number of lattice paths (Schroeder paths) from (0,0) to (3n,n) with unit steps N=(0,1), E=(1,0) and D=(1,1) staying weakly above the line y = 3x.

Original entry on oeis.org

1, 2, 14, 134, 1482, 17818, 226214, 2984206, 40503890, 561957362, 7934063678, 113622696470, 1646501710362, 24098174350986, 355715715691350, 5289547733908510, 79163575684710818, 1191491384838325474
Offset: 0

Views

Author

Joachim Schroeder (schroderjd(AT)qwa.uovs.ac.za), Sep 10 2008

Keywords

Comments

a(n) is also the number of lattice path from (0,0) to (4n,0) with unit steps (1,3), (2,2) and (1,-1) staying weakly above the x-axis.
Also, the number of planar rooted trees with n non-leaf vertices such that each non-leaf vertex has either 3 or 4 children. - Cameron Marcott, Sep 18 2013
a(n) equals the number of ordered complete 4-ary trees with 3*n + 1 leaves, where the internal vertices come in two colors and such that each vertex and its rightmost child have different colors. See Drake, Example 1.6.9. - Peter Bala, Apr 30 2023

Examples

			a(2)=14, because
  01: NNNENNNE,
  02: NNDNNNE,
  03: NNNENND,
  04: NNDNND,
  05: NNNDNNE,
  06: NNNDND,
  07: NNNNENNE,
  08: NNNNEND,
  09: NNNNDNE,
  10: NNNNDD,
  11: NNNNNENE,
  12: NNNNNED,
  13: NNNNNDE,
  14: NNNNNNEE
are all the paths from (0,0) to (2,6) with steps N,E and D weakly above y=3x.
		

References

  • Sheng-Liang Yang and Mei-yang Jiang, The m-Schröder paths and m-Schröder numbers, Disc. Math. (2021) Vol. 344, Issue 2, 112209. doi:10.1016/j.disc.2020.112209. See Table 1.

Crossrefs

Cf. A027307 (the case y=2x), A008288 (Delannoy numbers), A008412 (4-dimensional coordination numbers).
This appears to equal 2*A243675. - N. J. A. Sloane, Mar 28 2021
The sequences listed in Yang-Jiang's Table 1 appear to be A006318, A001003, A027307, A034015, A144097, A243675, A260332, A243676. - N. J. A. Sloane, Mar 28 2021

Programs

  • Maple
    Schr:=proc(n,m,l)(n-l*m+1)/m*sum(2^v*binomial(m,v)*binomial(n,v-1),v=1..m) end proc; where n=3m and l=3, also
    Schr:=proc(n,m,l)(n-l*m+1)/(n+1)*sum(2^v*binomial(m-1,v-1)*binomial(n+1,v),v=0..m) end proc; where n=3m and l=3, also
    Schr:=proc(n,m,l)(n-l*m+1)/m*sum(binomial(m,v)*binomial(n+v,m-1),v=0..m) end proc; where n=3m and l=3, also
    Schr:=proc(n,m,l)(n-l*m+1)/(n+1)*sum(binomial(n+1,v)*binomial(m-1+v,n),v=0..n+1) end proc; where n=3m and l=3.
    # alternative Maple program:
    a:= proc(n) option remember; `if`(n<2, n+1,
          ((15610*n^5 -67123*n^4 +106824*n^3 -77633*n^2
           +25514*n-3000)*a(n-1) -(3*(n-2))*(3*n-4)*
           (3*n-5)*(35*n^2-28*n+5)*a(n-2)) / ((3*(3*n-1))
           *(3*n+1)*n*(35*n^2-98*n+68)))
        end:
    seq(a(n), n=0..20);  # Alois P. Heinz, May 26 2015
  • Mathematica
    d[n_, k_] := Binomial[n+k, k] Hypergeometric2F1[-k, -n, -n-k, -1]; a[0] = 1; a[n_] = d[3n, n] - 3d[3n+1, n-1] - 2d[3n, n-1]; Table[a[n], {n, 0, 20}] (* Jean-François Alcover, Feb 25 2017 *)
  • PARI
    {a(n) = sum(k=0, n, binomial(n, k) * binomial(3*n+k+1, n)/(3*n+k+1))} \\ Seiichi Manyama, Jul 25 2020
    
  • PARI
    {a(n) = if(n==0, 1, sum(k=1, n, 2^k*binomial(n, k) * binomial(3*n, k-1)/n))} \\ Seiichi Manyama, Jul 25 2020

Formula

G.f. A(z) satisfies A(z) = 1 + z(A(z)^3 + A(z)^4) a(n)= S_{3n+1}(n) - 3S_n(3n + 1), where S_a(b) are coordination numbers, i.e., the number of points in the a-dimensional cubic lattice Z^a having distance b in the L_1 norm.
Also a(n) = D(3n,n) - 3D(3n + 1,n-1) - 2D(3n,n-1), where D(a,b) are the Delannoy numbers, i.e., the number of paths with N, E and D steps from (0,0) to (a,b).
D-finite with recurrence 3*n*(3*n-1)*(3*n+1)*(35*n^2-98*n+68) *a(n) +(-15610*n^5+67123*n^4-106824*n^3+77633*n^2-25514*n+3000)*a(n-1) +3*(n-2) *(3*n-4) *(3*n-5) *(35*n^2-28*n+5) *a(n-2)=0. - R. J. Mathar, Sep 06 2016
From Seiichi Manyama, Jul 25 2020: (Start)
a(n) = Sum_{k=0..n} binomial(n,k) * binomial(3*n+k+1, n)/(3*n+k+1).
a(n) = (1/n) * Sum_{k=1..n} 2^k * binomial(n,k) * binomial(3*n,k-1) for n > 0. (End)
a(n) ~ sqrt(12160 + 3853*sqrt(10)) * 3^(3*n - 9/2) / (2*sqrt(5*Pi) * n^(3/2) * (223 - 70*sqrt(10))^(n - 1/2)). - Vaclav Kotesovec, Jul 31 2021
Series reversion of x*(1 - x^3)/(1 + x^3) = x + 2*x^4 + 14*x^7 + 134*x^10 + ... = Sum_{n >= 0} a(n)*x^(3*n+1). - Peter Bala, Apr 30 2023
From Peter Bala, Jun 16 2023: (Start)
The g.f. A(x) = 1 + 2*x + 14*x^2 + 134*x^3 + ... satisfies A(x)^3 = (1/x) * the series reversion of ((1 - x)/(1 + x))^3.
Define b(n) = [x^(3*n)] ( (1 + x)/(1 - x) )^n = (1/3) * [x^n] ((1 + x)/(1 - x))^(3*n) = A333715(n). Then A(x) = exp( Sum_{n >= 1} b(n)*x^n/n ).
a(n) = 2*hypergeom([1 - n, -3*n], [2], 2) for n >= 1. (End)
a(n) = (1/n) * Sum_{k=0..n-1} (-1)^k * 2^(n-k) * binomial(n,k) * binomial(4*n-k,n-1-k) for n > 0. - Seiichi Manyama, Aug 09 2023

A144078 a(n) = the number of digits in the binary representation of n that differ from the corresponding digit in the binary reversal of n. (I.e., a(n) = number of 1's in n XOR A030101(n).)

Original entry on oeis.org

0, 2, 0, 2, 0, 2, 0, 2, 0, 4, 2, 4, 2, 2, 0, 2, 0, 4, 2, 2, 0, 4, 2, 4, 2, 2, 0, 4, 2, 2, 0, 2, 0, 4, 2, 4, 2, 6, 4, 4, 2, 6, 4, 2, 0, 4, 2, 4, 2, 2, 0, 6, 4, 4, 2, 6, 4, 4, 2, 4, 2, 2, 0, 2, 0, 4, 2, 4, 2, 6, 4, 2, 0, 4, 2, 4, 2, 6, 4, 4, 2, 6, 4, 2, 0, 4, 2, 4, 2, 6, 4, 2, 0, 4, 2, 4, 2, 2, 0, 6, 4, 4, 2, 4, 2
Offset: 1

Views

Author

Leroy Quet, Sep 09 2008

Keywords

Comments

a(n) + A144079(n) = A070939(n), the number of binary digits in n.

Examples

			20 in binary is 10100. Compare this with its digit reversal, 00101. XOR each pair of corresponding digits: 1 XOR 0 = 1, 0 XOR 0 = 0, 1 XOR 1 = 0, 0 XOR 0 = 0, 0 XOR 1 = 1. There are two bit pairs that differ, so a(20) = 2.
		

Crossrefs

Programs

  • Maple
    A144078 := proc(n) local a,dgs,i; a := 0 ; dgs := convert(n,base,2) ; for i from 1 to nops(dgs) do if op(i,dgs)+op(-i,dgs) = 1 then a := a+1 ; fi; od; RETURN(a) ; end: for n from 1 to 240 do printf("%d,",A144078(n)) ; od: # R. J. Mathar, Sep 14 2008
  • Mathematica
    brd[n_]:=Module[{idn2=IntegerDigits[n,2]},Count[Transpose[{idn2, Reverse[ idn2]}], ?(#[[1]]!=#[[2]]&)]]; Array[brd,110] (* _Harvey P. Dale, May 09 2016 *)
  • PARI
    a(n) = hammingweight(bitxor(n, fromdigits(Vecrev(binary(n)),2))) \\ Rémy Sigrist, Oct 07 2018

Formula

From Rémy Sigrist, Oct 07 2018: (Start)
a(n) = 0 iff n is a binary palindrome (A006995).
a(A143960(n)) = 2*n (in fact A143960(n) is the least k such that a(k) = 2*n).
(End)

Extensions

More terms from R. J. Mathar, Sep 14 2008
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