A084174 -id:A084174 - cp's OEIS frontend

A112340 Triangle read by rows of numbers b_{n,k}, n>=1, 1<=k<=n such that Product_{n,k} 1/(1-q^n t^k)^{b_{n,k}} = 1 + Sum_{i,j>=1} S_{i,j} q^i t^j where S_{i,j} are entries in the table A008277 (the inverse Euler transformation of the table of Stirling numbers of the second kind).

1, 1, 0, 1, 2, 0, 1, 5, 3, 0, 1, 13, 16, 4, 0, 1, 28, 67, 34, 5, 0, 1, 60, 249, 229, 65, 6, 0, 1, 123, 853, 1265, 609, 107, 7, 0, 1, 251, 2787, 6325, 4696, 1376, 168, 8, 0, 1, 506, 8840, 29484, 31947, 14068, 2772, 244, 9, 0, 1, 1018, 27503, 131402, 199766, 124859, 36252

Offset: 1

Mike Zabrocki, Sep 05 2005; Aug 06 2006

Row sums equal to A085686, second column = A084174 - 1

The number of set partitions of size n length k which are 'Lyndon,' that is, since all set partitions are isomorphic to sequences of atomic set partitions (A087903), those which are smallest of all rotations of these sequences in lex order (with respect to some ordering on the atomic set partitions) are Lyndon. 1; 1, 0; 1, 2, 0; 1, 5, 3, 0; 1, 13, 16, 4, 0;

			There are 6 set partitions of size 4 and length 3, {12|3|4}, {13|2|4}, {14|2|3}, {1|23|4}, {1|24|3}, {1|2|34} and the sequences the correspond to are ({12},{1},{1}), ({13|2}, {1}), ({14|2|3}), ({1},{12},{1}), ({1},{13|2}), ({1},{1},{12}). Now there are three {({12},{1},{1}), ({1},{12},{1}), ({1},{1},{12})} that are rotations of each other and ({1}, {1}, {12}) is the smallest of these, {({13|2}, {1}), ({1},{13|2})} are rotations of each other and ({1},{13|2}) is the smallest and ({14|2|3}) is atomic and all atomic s.p. are Lyndon. Hence {1|2|34}, {1|24|3}, {14|2|3} are Lyndon and a(4,3) = 3
Triangle begins:
  1;
  1,  0;
  1,  2,  0;
  1,  5,  3,  0;
  1, 13, 16,  4, 0;
  1, 28, 67, 34, 5, 0;
  ...

N. Bergeron, M. Zabrocki, The Hopf algebras of symmetric functions and quasisymmetric functions in non-commutative variables are free and cofree, arXiv:math/0509265 [math.CO], 2005.
M. Rosas and B. Sagan, Symmetric Functions in Noncommuting Variables, Transactions of the American Mathematical Society, 358 (2006), no. 1, 215-232.
M. C. Wolf, Symmetric functions for non-commutative elements, Duke Math. J., 2 (1936), 626-637.

Cf. A008277, A085686, A112339.

Cf. A087903, A000110.

EULERitable:=proc(tbl) local ser,out,i,j,tmp; ser:=1+add(add(q^i*t^j*tbl[i][j], j=1..nops(tbl[i])), i=1..nops(tbl)); out:=[]; for i from 1 to nops(tbl) do tmp:=coeff(ser,q,i); ser:=expand(ser*mul(add((-q^i*t^j)^k*choose(abs(coeff(tmp,t,j)),k),k=0..nops(tbl)/i), j = 1..degree(tmp,t))); ser:=subs({seq(q^j=0,j=nops(tbl)+1..degree(ser,q))},ser); out:=[op(out),[seq(abs(coeff(tmp,t,j)), j=1..degree(tmp,t))]]; end do; out; end: EULERitable([seq([seq(combinat[stirling2](n,k),k=1..n)],n=1..10)]);

nmax = 11; b[n_, k_] /; k < 1 || k > n = 0;
coes[m_] := Product[1/(1 - q^n t^k)^b[n, k], {n, 1, m}, {k, 1, m}] - 1 - Sum[ StirlingS2[i, j] q^i t^j, {i, 1, m}, {j, 1, m}] + O[t]^m + O[q]^m // Normal // CoefficientList[#, {t, q}]&;
sol[1] = {b[1, 1] -> 1};
Do[sol[m] = Solve[Thread[(coes[m] /. sol[m - 1]) == 0]], {m, 2, nmax + 1}];
bb = Flatten[Table[sol[m], {m, 1, nmax + 1}]];
Table[b[n, k] /. bb, {n, 1, nmax}, {k, 1, n}] // Flatten (* Jean-François Alcover, Dec 11 2017 *)

A112339 Triangle read by rows of numbers b_{n,k}, n >= 2, 1 <= k < n such that (1/(1-qt))Product_{n,k} 1/(1 - q^nt^k)^b_{n,k} = Sum_{i,j>=1} S_{i,j} q^it^j where S_{i,j} are entries in the table A008277 (the inverse Euler transformation of the table of Stirling numbers of the second kind).

Original entry on oeis.org

1, 1, 2, 1, 5, 3, 1, 13, 16, 4, 1, 28, 67, 34, 5, 1, 60, 249, 229, 65, 6, 1, 123, 853, 1265, 609, 107, 7, 1, 251, 2787, 6325, 4696, 1376, 168, 8, 1, 506, 8840, 29484, 31947, 14068, 2772, 244, 9, 1, 1018, 27503, 131402, 199766, 124859, 36252, 5118, 345, 10

Offset: 2

Mike Zabrocki, Sep 05 2005

Row sums equal to A085686 second column = A084174 - 1.

			Triangle begins:
  1;
  1,  2;
  1,  5,  3;
  1, 13, 16,  4;
  ...

Cf. A008277, A085686, A112340.

EULERitable:=proc(tbl) local ser,out,i,j,tmp; ser:=1+add(add(q^i*t^j*tbl[i][j], j=1..nops(tbl[i])), i=1..nops(tbl)); out:=[]; for i from 1 to nops(tbl) do tmp:=coeff(ser,q,i); ser:=expand(ser*mul(add((-q^i*t^j)^k*choose(abs(coeff(tmp,t,j)),k),k=0..nops(tbl)/i), j = 1..degree(tmp,t))); ser:=subs({seq(q^j=0,j=nops(tbl)+1..degree(ser,q))},ser); out:=[op(out),[seq(abs(coeff(tmp,t,j)), j=1..degree(tmp,t))]]; end do; out; end: EULERitable([seq([seq(combinat[stirling2](n,k),k=1..n)],n=1..11)]);

A131087 Triangle read by rows: T(n,k) = 2*binomial(n,k) - (1 + (-1)^(n-k))/2 (0 <= k <= n).

Original entry on oeis.org

1, 2, 1, 1, 4, 1, 2, 5, 6, 1, 1, 8, 11, 8, 1, 2, 9, 20, 19, 10, 1, 1, 12, 29, 40, 29, 12, 1, 2, 13, 42, 69, 70, 41, 14, 1, 1, 16, 55, 112, 139, 112, 55, 16, 1, 2, 17, 72, 167, 252, 251, 168, 71, 18, 1, 1, 20, 89, 240, 419, 504, 419, 240, 89, 20, 1, 2, 21, 110, 329, 660, 923, 924

Offset: 0

Gary W. Adamson, Jun 14 2007

Row sums = A084174: (1, 3, 6, 14, 29, ...).

2*A007318 - A128174 as infinite lower triangular matrices. - Emeric Deutsch, Jun 21 2007

			First few rows of the triangle:
  1;
  2,  1;
  1,  4,  1;
  2,  5,  6,  1;
  1,  8, 11,  8,  1;
  2,  9, 20, 19, 10,  1;
  1, 12, 29, 40, 29, 12,  1;
  ...

Cf. A128174, A084174.

T := proc (n, k) options operator, arrow; 2*binomial(n, k)-1/2-(1/2)*(-1)^(n-k) end proc; for n from 0 to 11 do seq(T(n, k), k = 0 .. n) end do; # yields sequence in triangular form - Emeric Deutsch, Jun 21 2007

G.f.: G(t,z) = (1 + z - tz - 2z^2 + 2tz^3)/((1-z^2)*(1-tz)*(1-z-tz)). - Emeric Deutsch, Jun 21 2007

More terms from Emeric Deutsch, Jun 21 2007

A134520 Triangle read by rows: A007318 + A128174 - A000012 as infinite lower triangular matrices.

Original entry on oeis.org

1, 0, 1, 1, 1, 1, 0, 3, 2, 1, 1, 3, 6, 3, 1, 0, 5, 9, 10, 4, 1, 1, 5, 15, 19, 15, 5, 1, 0, 7, 20, 35, 34, 21, 6, 1, 1, 7, 28, 55, 70, 55, 28, 7, 1, 0, 9, 35, 84, 125, 126, 83, 36, 8, 1, 1, 9, 45, 119, 210, 251, 210, 119, 45, 9, 1, 0, 11, 54, 165, 329, 462, 461, 330, 164, 55, 10, 1

Offset: 0

Gary W. Adamson, Oct 29 2007

Row sums = A084174: (1, 1, 3, 6, 14, 29, 61, ...).

			First few rows of the triangle:
  1;
  0,  1;
  1,  1,  1;
  0,  3,  2,  1;
  1,  3,  6,  3,  1;
  0,  5,  9, 10,  4,  1;
  1,  5, 15, 19, 15,  5,  1;
  0,  7, 20, 35, 34, 21,  6,  1;
  1,  7, 28, 55, 70, 55, 28,  7,  1;
  ...

Cf. A084174, A128174.

a(45) = 1 corrected and more terms from Georg Fischer, Jun 07 2023

cp's OEIS Frontend

A112340 Triangle read by rows of numbers b_{n,k}, n>=1, 1<=k<=n such that Product_{n,k} 1/(1-q^n t^k)^{b_{n,k}} = 1 + Sum_{i,j>=1} S_{i,j} q^i t^j where S_{i,j} are entries in the table A008277 (the inverse Euler transformation of the table of Stirling numbers of the second kind).

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A112339 Triangle read by rows of numbers b_{n,k}, n >= 2, 1 <= k < n such that (1/(1-qt))Product_{n,k} 1/(1 - q^nt^k)^b_{n,k} = Sum_{i,j>=1} S_{i,j} q^it^j where S_{i,j} are entries in the table A008277 (the inverse Euler transformation of the table of Stirling numbers of the second kind).

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Maple

A131087 Triangle read by rows: T(n,k) = 2*binomial(n,k) - (1 + (-1)^(n-k))/2 (0 <= k <= n).

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A134520 Triangle read by rows: A007318 + A128174 - A000012 as infinite lower triangular matrices.

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cp's OEIS Frontend

A112340 Triangle read by rows of numbers b_{n,k}, n>=1, 1<=k<=n such that Product_{n,k} 1/(1-q^n t^k)^{b_{n,k}} = 1 + Sum_{i,j>=1} S_{i,j} q^i t^j where S_{i,j} are entries in the table A008277 (the inverse Euler transformation of the table of Stirling numbers of the second kind).

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Author

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Examples

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Maple

Mathematica

A112339 Triangle read by rows of numbers b_{n,k}, n >= 2, 1 <= k < n such that (1/(1-q*t))*Product_{n,k} 1/(1 - q^n*t^k)^b_{n,k} = Sum_{i,j>=1} S_{i,j} q^i*t^j where S_{i,j} are entries in the table A008277 (the inverse Euler transformation of the table of Stirling numbers of the second kind).

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Author

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Examples

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Programs

Maple

A131087 Triangle read by rows: T(n,k) = 2*binomial(n,k) - (1 + (-1)^(n-k))/2 (0 <= k <= n).

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Author

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Programs

Maple

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A134520 Triangle read by rows: A007318 + A128174 - A000012 as infinite lower triangular matrices.

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Comments

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Extensions

A112339 Triangle read by rows of numbers b_{n,k}, n >= 2, 1 <= k < n such that (1/(1-qt))Product_{n,k} 1/(1 - q^nt^k)^b_{n,k} = Sum_{i,j>=1} S_{i,j} q^it^j where S_{i,j} are entries in the table A008277 (the inverse Euler transformation of the table of Stirling numbers of the second kind).