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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A256311 Number T(n,k) of length 3n words such that all letters of the k-ary alphabet occur at least once and are introduced in ascending order and which can be built by repeatedly inserting triples of identical letters into the initially empty word; triangle T(n,k), n>=0, 0<=k<=n, read by rows.

Original entry on oeis.org

1, 0, 1, 0, 1, 3, 0, 1, 18, 12, 0, 1, 97, 198, 55, 0, 1, 530, 2520, 1820, 273, 0, 1, 2973, 29886, 42228, 15300, 1428, 0, 1, 17059, 347907, 859180, 564585, 122094, 7752, 0, 1, 99657, 4048966, 16482191, 17493938, 6577494, 942172, 43263
Offset: 0

Views

Author

Alois P. Heinz, Mar 25 2015

Keywords

Examples

			T(0,0) = 1: (the empty word).
T(1,1) = 1: aaa.
T(2,1) = 1: aaaaaa.
T(2,2) = 3: aaabbb, aabbba, abbbaa.
T(3,1) = 1: aaaaaaaaa.
T(3,2) = 18: aaaaaabbb, aaaaabbba, aaaabbbaa, aaabaaabb, aaabbaaab, aaabbbaaa, aaabbbbbb, aabaaabba, aabbaaaba, aabbbaaaa, aabbbabbb, aabbbbbba, abaaabbaa, abbaaabaa, abbbaaaaa, abbbaabbb, abbbabbba, abbbbbbaa.
T(3,3) = 12: aaabbbccc, aaabbcccb, aaabcccbb, aabbbaccc, aabbbccca, aabbcccba, aabcccbba, abbbaaccc, abbbaccca, abbbcccaa, abbcccbaa, abcccbbaa.
Triangle T(n,k) begins:
  1;
  0, 1;
  0, 1,     3;
  0, 1,    18,     12;
  0, 1,    97,    198,     55;
  0, 1,   530,   2520,   1820,    273;
  0, 1,  2973,  29886,  42228,  15300,   1428;
  0, 1, 17059, 347907, 859180, 564585, 122094, 7752;
		

Crossrefs

Row sums give A321031.
Main diagonal gives A001764.
T(2n,n) gives A321041.

Programs

  • Maple
    A:= (n, k)-> `if`(n=0, 1,
        k/n *add(binomial(3*n, j) *(n-j) *(k-1)^j, j=0..n-1)):
    T:= (n, k)-> add((-1)^i*A(n, k-i)/(i!*(k-i)!), i=0..k):
    seq(seq(T(n, k), k=0..n), n=0..10);

Formula

T(n,k) = Sum_{i=0..k} (-1)^i * A213028(n,k-i) / (i!*(k-i)!).

A213027 Number A(n,k) of 3n-length k-ary words, either empty or beginning with the first letter of the alphabet, that can be built by repeatedly inserting triples of identical letters into the initially empty word; square array A(n,k), n>=0, k>=0, by antidiagonals.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 4, 1, 0, 1, 1, 7, 19, 1, 0, 1, 1, 10, 61, 98, 1, 0, 1, 1, 13, 127, 591, 531, 1, 0, 1, 1, 16, 217, 1810, 6101, 2974, 1, 0, 1, 1, 19, 331, 4085, 27631, 65719, 17060, 1, 0, 1, 1, 22, 469, 7746, 82593, 441604, 729933, 99658, 1, 0
Offset: 0

Views

Author

Alois P. Heinz, Jun 03 2012

Keywords

Comments

In general, column k > 1 is asymptotic to a(n) ~ 3^(3*n+1/2) * (k-1)^(n+1) / (sqrt(Pi) * (2*k-3)^2 * 4^n * n^(3/2)). - Vaclav Kotesovec, Aug 31 2014

Examples

			A(0,k) = 1: the empty word.
A(n,1) = 1: (aaa)^n.
A(2,2) = 4: there are 4 words of length 6 over alphabet {a,b}, either empty or beginning with the first letter of the alphabet, that can be built by repeatedly inserting triples of identical letters into the initially empty word: aaaaaa, aaabbb, aabbba, abbbaa.
A(2,3) = 7: aaaaaa, aaabbb, aaaccc, aabbba, aaccca, abbbaa, acccaa.
A(3,2) = 19: aaaaaaaaa, aaaaaabbb, aaaaabbba, aaaabbbaa, aaabaaabb, aaabbaaab, aaabbbaaa, aaabbbbbb, aabaaabba, aabbaaaba, aabbbaaaa, aabbbabbb, aabbbbbba, abaaabbaa, abbaaabaa, abbbaaaaa, abbbaabbb, abbbabbba, abbbbbbaa.
Square array A(n,k) begins:
  1, 1,    1,     1,      1,       1,       1, ...
  0, 1,    1,     1,      1,       1,       1, ...
  0, 1,    4,     7,     10,      13,      16, ...
  0, 1,   19,    61,    127,     217,     331, ...
  0, 1,   98,   591,   1810,    4085,    7746, ...
  0, 1,  531,  6101,  27631,   82593,  195011, ...
  0, 1, 2974, 65719, 441604, 1751197, 5153626, ...
		

Crossrefs

Rows n=0-3 give: A000012, A057427, A016777(k-1), A127854(k-1).
Main diagonal gives: A218472.

Programs

  • Maple
    A:= (n, k)-> `if`(n=0, 1, `if`(k<2, k,
        1/n *add(binomial(3*n, j) *(n-j) *(k-1)^j, j=0..n-1))):
    seq(seq(A(n, d-n), n=0..d), d=0..12);
  • Mathematica
    a[0, ] = 1; a[, k_ /; k < 2] := k; a[n_, k_] := 1/n*Sum[Binomial[3*n, j]*(n-j)*(k-1)^j, {j, 0, n-1}]; Table[a[n-k, k], {n, 0, 12}, {k, n, 0, -1}] // Flatten (* Jean-François Alcover, Dec 11 2013 *)

Formula

A(n,k) = 1/n * Sum_{j=0..n-1} C(3*n,j) * (n-j) * (k-1)^j if n>0, k>1; A(0,k) = 1; A(n,k) = k if n>0, k<2.
A(n,k) = 1/k * A213028(n,k) if n>0, k>1; else A(n,k) = A213028(n,k).

A047098 a(n) = 2*binomial(3*n, n) - Sum_{k=0..n} binomial(3*n, k).

Original entry on oeis.org

1, 2, 8, 38, 196, 1062, 5948, 34120, 199316, 1181126, 7080928, 42860534, 261542752, 1607076200, 9934255472, 61732449648, 385393229460, 2415935640198, 15200964233864, 95962904716402, 607640599286276, 3858198001960438, 24559243585545644, 156692889782067712
Offset: 0

Views

Author

Clark Kimberling, Aug 15 1998

Keywords

Comments

T(2n,n), array T as in A047089. [Corrected Dec 08 2006]
Let B_3^+ denote the semigroup with presentation . Let D=aba be the 'fundamental word'. Then this sequence is also equal to the number of words in B_3^+ equal in B_3^+ to D^n, n >= 0. - Stephen P. Humphries, Jan 20 2004
In the language of Riordan arrays, row sums of (1/(1+x), x/(1+x)^3)^-1, where (1/(1+x), x/(1+x)^3) has general term (-1)^(n-k)*binomial(n+2k, 3k). - Paul Barry, May 09 2005
Hankel transform is 2^n*A051255(n) where A051255 is the Hankel transform of C(3n,n)/(2n+1). - Paul Barry, Jan 21 2007

Crossrefs

Column k=2 of A213028.

Programs

  • Maple
    A047098 := n -> 2*binomial(3*n, n)-add(binomial(3*n, k), k=0..n);
  • Mathematica
    Table[2Binomial[3n,n]-Sum[Binomial[3n,k],{k,0,n}],{n,0,35}] (* Harvey P. Dale, Jul 27 2011 *)
  • PARI
    a(n)=if(n<0,0,polcoeff((((1+10*x-2*x^2)+(1-4*x)*sqrt(1-4*x+x*O(x^n)))/2)^n,n))
    
  • PARI
    a(n)=if(n<0,0, 2*binomial(3*n,n)-sum(k=0,n,binomial(3*n,k)))

Formula

G.f. A(x)=y satisfies (8x-1)y^3-y^2+y+1=0. - Michael Somos, Jan 28 2004
Coefficient of x^n in ((1+10x-2x^2+(1-4x)^(3/2))/2)^n. - Michael Somos, Sep 25 2003
a(n) = Sum_{k = 0..n} A109971(k)*2^k; a(0) = 1, a(n) = Sum_{k = 0..n} 2^k*C(3n-k,n-k)*2*k/(3*n-k), n > 0. - Paul Barry, Jan 21 2007
Conjecture: 2*n*(2*n-1)*a(n) +(-71*n^2+112*n-48)*a(n-1) +3*(131*n^2-391*n+296)*a(n-2) -72*(3*n-7)*(3*n-8)*a(n-3)=0. - R. J. Mathar, Nov 30 2012
a(n) = A321957(n) + 2*binomial(3*n, n) - 8^n. - Peter Luschny, Nov 22 2018
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for all primes p and positive integers n and k. - Peter Bala, Mar 05 2022

Extensions

Clark Kimberling, Dec 08 2006, changed "T(3n,2n)" to "T(2n,n)" in the comment line, but observes that some of the other comments seem to apply to the sequence T(3n,2n) rather than to the sequence T(2n,n).
Edited by N. J. A. Sloane, Dec 21 2006, replacing the old definition in terms of A047089 by an explicit formula supplied by Benoit Cloitre, Oct 25 2003.
Showing 1-3 of 3 results.