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 1-7 of 7 results.

A060187 Triangle read by rows: Eulerian numbers of type B, T(n,k) (1 <= k <= n) given by T(n, 1) = T(n,n) = 1, otherwise T(n, k) = (2*n - 2*k + 1)*T(n-1, k-1) + (2*k - 1)*T(n-1, k).

Original entry on oeis.org

1, 1, 1, 1, 6, 1, 1, 23, 23, 1, 1, 76, 230, 76, 1, 1, 237, 1682, 1682, 237, 1, 1, 722, 10543, 23548, 10543, 722, 1, 1, 2179, 60657, 259723, 259723, 60657, 2179, 1, 1, 6552, 331612, 2485288, 4675014, 2485288, 331612, 6552, 1, 1, 19673, 1756340, 21707972, 69413294, 69413294, 21707972, 1756340, 19673, 1
Offset: 1

Views

Author

N. J. A. Sloane, Mar 20 2001

Keywords

Comments

Rows are expansions of p(x,n) = 2^n*(1 - x)^(1 + n)*LerchPhi(x, -n, 1/2). Row sums are A000165. - Roger L. Bagula, Sep 16 2008
Eulerian numbers of type B. The n-th row of this triangle is the h-vector of the simplicial complex dual to a permutohedron of type B_(n-1). For example, the permutohedron of type B_2 is an octagon whose dual, also an octagon, has f-polynomial f(x) = 1 + 8*x + 8*x^2 and h-polynomial given by (x-1)^2 + 8*(x-1) + 8 = 1 + 6*x + x^2, giving [1,6,1] as row 3 of this table (see Fomin and Reading, p. 21). The corresponding triangle of f-vectors for the type B permutohedra is A145901. The Hilbert transform of the current array is A145905. - Peter Bala, Oct 26 2008
From Peter Bala, Oct 13 2011: (Start)
The row polynomials count the elements of the hyperoctahedral group B_n (the group of signed permutations on n letters) according to the number of type B descents (see Chow and Gessel).
Let P denote Pascal's triangle. Then the first column of the array P*(I-t*P^2)^(-1) (I the identity array) begins [1/(1-t),(1+t)/(1-t)^2,(1+6*t+t^2)/(1-t)^3,...]. The numerator polynomials are the row polynomials of this table. Similarly, in the array (I-t*A062715)^-1, the numerator polynomials in the first column produce the row polynomials of this table (but with an extra factor of t). Cf. A145901. (End)
The Dasse-Hartaut and Hitczenko paper (section 6.1.4) shows this triangle of numbers, when suitably normalized, satisfies the central limit theorem. - Peter Bala, Mar 05 2012
These are the coefficients of the midpoint Eulerian polynomials (see Quade/Collatz and Schoenberg). In terms of the cardinal B-splines b_n(t) these polynomials can be defined as M_n(x) = 2^n*n!*Sum_{k=0..n} b_{n+1}(k+1/2)*x^k. - Peter Luschny, Apr 26 2013
The o.g.f. Godd(n, x) = Sum_{m>=0} Sodd(n, m)*x^m, with Sodd(n, m) = Sum_{j=0..m} (1+2*j)^n is Podd(n, x)/(1 - x)^(n+2) with Podd(n, x) = Sum_{k=0..n} T(n+1, k+1)*x^k. E.g., Godd(2, x) = (1 + 6*x + x^2)/(1 - x)^4; see A000447(n+1) for n >= 0. For the e.g.f.s see A282628. - Wolfdieter Lang, Mar 17 2017
Let h_0(x,y) = x*y/(x+y), and D = x*D_x - y*D_y where D_x is the partial derivative w.r.t. x, etc. Put h_{n+1}(x,y) = D(h_n)(x,y). Then h_n(x,y) = x*y/(x+y)^(n+1)*f_{n}(x,y) where f_n(x,y) = Sum_{k=0..n} (-1)^k*T(n+1,k+1)*y^(n-k)*x^k. If instead of h_0, one similarly uses g_0(x,y) = x*y/(y-x), etc., then one obtains g_n(x,y) = x*y/(y-x)^(n+1)*Sum_{k=0..n} T(n+1,k+1)*y^(n-k)*x^k. (If instead of D one considers D' = x*D_x + y*D_y, then h_0 and g_0 are fixed points of D'.) - Gregory Gerard Wojnar, Oct 28 2018
Counts coloop-free Schubert delta-matroids by cornered rank, see Remark 4.6 of the paper by Eur, Fink, Larson, Spink. - Matt Larson, May 20 2024

Examples

			The triangle T(n, k) begins:
n\k 1    2     3      4      5     6    7 8 ...
1:  1
2:  1    1
3:  1    6     1
4:  1   23    23      1
5:  1   76   230     76      1
6:  1  237  1682   1682    237     1
7:  1  722 10543  23548  10543   722    1
8:  1 2179 60657 259723 259723 60657 2179 1
...
row n = 9: 1 6552 331612 2485288 4675014 2485288 331612 6552 1,
row n = 10: 1 19673 1756340 21707972 69413294 69413294 21707972 1756340 19673 1,
row n = 11: 1 59038 9116141 178300904 906923282 1527092468 906923282 178300904 9116141 59038 1, ... reformatted. - _Wolfdieter Lang_, Mar 17 2017
		

References

  • G. Boros and V. H. Moll, Irresistible Integrals: Symbolics, Analysis and Experiments in the Evaluation of Integrals, Cambridge University Press, 2004.
  • T. K. Petersen, Eulerian Numbers, Birkhauser, 2015, Chapter 11.
  • W. Quade and L. Collatz, Zur Interpolationstheorie der reellen periodischen Funktionen. Sitzungsbericht der Preuss. Akad. der Wiss., Phys.-Math. Kl, (1938), 383-429.

Crossrefs

Diagonals give A060188, A060189, A060190. Cf. A008292.
Cf. also A000165 (row sums), A002436 (alt. row sums), A008292, A145901, A145905 (Hilbert transform). A062715.

Programs

  • GAP
    a:=Flat(List([1..11],n->List([1..n],k->Sum([1..k],i->(-1)^(k-i)*Binomial(n,k-i)*(2*i-1)^(n-1))))); # Muniru A Asiru, Feb 09 2018
    
  • Magma
    [[(&+[(-1)^(k-j)*Binomial(n,k-j)*(2*j-1)^(n-1): j in [1..k]]): k in [1..n]]: n in [1..10]]; // G. C. Greubel, Nov 08 2018
    
  • Maple
    A060187:= (n,k) -> add((-1)^(k-i)*binomial(n,k-i)*(2*i-1)^(n-1), i = 1..k):
    for n from 1 to 10 do seq(A060187(n,k),k = 1..n); end do; # Peter Bala, Oct 26 2008
    T:=proc(n,k,l) option remember; if (n=1 or k=1 or k=n) then 1 else
    (l*n-l*k+1)*T(n-1,k-1,l)+(l*k-l+1)*T(n-1,k,l); fi; end;
    for n from 1 to 10 do lprint([seq(T(n,k,2),k=1..n)]); od; # N. J. A. Sloane, May 08 2013
    P := proc(n,x) option remember; if n = 0 then 1 else
      (n*x+(1/2)*(1-x))*P(n-1,x)+x*(1-x)*diff(P(n-1,x),x);
      expand(%) fi end:
    A060187 := (n,k) -> 2^n*coeff(P(n,x),x,k):
    seq(print(seq(A060187(n,k), k=0..n)), n=0..10);  # Peter Luschny, Mar 08 2014
  • Mathematica
    p[x_, n_] = 2^n (1 - x)^(1 + n) LerchPhi[x, -n, 1/2]; Table[CoefficientList[p[x, n], x], {n, 0, 10}] // Flatten (* Roger L. Bagula, Sep 16 2008 *)
    T[n_, k_] := Sum[(-1)^(k-i)*Binomial[n, k-i]*(2*i-1)^(n-1), {i, 1, k}]; Table[T[n, k], {n, 1, 10}, {k, 1, n}] // Flatten (* Jean-François Alcover, Nov 23 2015, after Peter Bala *)
  • PARI
    {T(n, k) = if( nMichael Somos, Jan 07 2011 */
    
  • Python
    from math import isqrt, comb
    def A060187(n):
        a = (m:=isqrt(k:=n<<1))+(k>m*(m+1))
        b = n-comb(a,2)
        return sum(-comb(a,b-i)*((i<<1)-1)**(a-1) if b-i&1 else comb(a,b-i)*((i<<1)-1)**(a-1) for i in range(1,b+1)) # Chai Wah Wu, Nov 13 2024
  • Sage
    @CachedFunction
    def A060187(n, k) :
        if n == 0: return 1 if k == 0 else 0
        return (2*(n-k)+1)*A060187(n-1, k-1) + (2*k+1)*A060187(n-1, k)
    for n in (0..8): [A060187(n,k) for k in (0..n)] # Peter Luschny, Apr 26 2013
    

Formula

T(s, 2) = 3^(s-1) - s. Sum_{t=1..s} T(s, t) = 2^(s-1)*(s-1)!.
From Peter Bala, Oct 26 2008: (Start)
T(n,k) = Sum_{i = 1..k} (-1)^(k-i)*binomial(n,k-i)*(2*i-1)^(n-1).
E.g.f.: (1 - x)*exp((1 - x)*t)/(1 - x*exp(2*(1 - x)*t)) = 1 + (1 + x)*t + (1 + 6*x + x^2)*t^2/2! + ... .
The row polynomials R(n,x) satisfy R(n,x)/(1 - x)^n = Sum_{i >= 1} (2*i - 1)^(n-1)*x^i. For example, row 3 gives (x + 6*x^2 + x^3)/ (1 - x)^3 = x + 3^2*x^2 + 5 ^2*x^3 + 7^2*x^4 + ... .
The recurrence relation R(n+1,x) = [(2*n+1)*x - 1]*R(n,x) + 2*x*(1 - x)*R'(n,x) shows that the row polynomials R(n,x) have only real zeros (apply Corollary 1.2 of [Liu and Wang]).
Worpitzky-type identity: Sum_{k = 1..n} T(n,k)*binomial(x+k-1,n-1) = (2*x+1)^(n-1).
The nonzero alternating row sums are (-1)^(n-1)*A002436(n). (End)
exp(x)*(d/dx)^n [exp(x)/(1 - exp(2*x))] = R(n+1,exp(2*x))/ (1 - exp(2*x))^(n+1).
Compare with Example 12.3.1. in [Boros and Moll]. - Peter Bala, Nov 07 2008
The n-th row polynomial R(n,x) = Sum_{k = 0..n} A145901(n,k)*x^k*(1 - x)^(n-k) = Sum_{k = 0..n} A145901(n,k)*(x - 1)^(n-k). - Peter Bala, Jul 22 2014
Assuming an offset 0, the n-th row polynomial = (x - 1)^n * log(x) * Integral_{u = 0..inf} (2*floor(u) + 1)^n * x^(-u) du, provided x > 1. - Peter Bala, Feb 06 2015
The finite sums of consecutive odd integer powers is derived from this number triangle: Sum_{k=1..n}(2k-1)^m = Sum_{j=1..m+1}binomial(n+m+1-j,m+1)*T(m+1,j). - Tony Foster III, Feb 09 2018

A061980 Square array A(n,k) = A(n-1,k) + A(n-1, floor(k/2)) + A(n-1, floor(k/3)), with A(0,0) = 1, read by antidiagonals.

Original entry on oeis.org

1, 0, 3, 0, 2, 9, 0, 1, 8, 27, 0, 0, 6, 26, 81, 0, 0, 4, 23, 80, 243, 0, 0, 3, 20, 76, 242, 729, 0, 0, 3, 17, 72, 237, 728, 2187, 0, 0, 1, 17, 66, 232, 722, 2186, 6561, 0, 0, 1, 11, 66, 222, 716, 2179, 6560, 19683, 0, 0, 1, 11, 54, 222, 701, 2172, 6552, 19682, 59049
Offset: 0

Views

Author

Henry Bottomley, May 24 2001

Keywords

Examples

			Array begins as:
    1,   0,   0,   0,   0,   0,   0, ...;
    3,   2,   1,   0,   0,   0,   0, ...;
    9,   8,   6,   4,   3,   3,   1, ...;
   27,  26,  23,  20,  17,  17,  11, ...;
   81,  80,  76,  72,  66,  66,  54, ...;
  243, 242, 237, 232, 222, 222, 202, ...;
  729, 728, 722, 716, 701, 701, 671, ...;
Antidiagonal rows begin as:
  1;
  0, 3;
  0, 2, 9;
  0, 1, 8, 27;
  0, 0, 6, 26, 81;
  0, 0, 4, 23, 80, 243;
  0, 0, 3, 20, 76, 242, 729;
  0, 0, 3, 17, 72, 237, 728, 2187;
  0, 0, 1, 17, 66, 232, 722, 2186, 6561;
		

Crossrefs

Row sums are 6^n: A000400.
Columns are A000244, A024023, A060188, A061981, A061982 twice, A061983 twice, etc.

Programs

  • Mathematica
    A[n_, k_]:= A[n, k]= If[n==0, Boole[k==0], A[n-1,k] +A[n-1,Floor[k/2]] +A[n-1, Floor[k/3]]];
    T[n_, k_]:= A[k, n-k];
    Table[A[n, k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Jun 18 2022 *)
  • SageMath
    @CachedFunction
    def A(n,k):
        if (n==0): return 0^k
        else: return A(n-1, k) + A(n-1, (k//2)) + A(n-1, (k//3))
    def T(n, k): return A(k, n-k)
    flatten([[T(n, k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Jun 18 2022

Formula

A(n,k) = A(n-1,k) + A(n-1, floor(k/2)) + A(n-1, floor(k/3)), with A(0,0) = 1.
T(n, k) = A(k, n-k).
Sum_{k=0..n} A(n, k) = A000400(n).
T(n, n) = A(n, 0) = A000244(n). - G. C. Greubel, Jun 18 2022

A156918 Triangle formed by coefficients of the expansion of p(x,n) = (1+x-x^2)^(n+1)*Sum_{j >= 0} (2*j+1)^n*(-x + x^2)^j.

Original entry on oeis.org

1, 1, -1, 1, 1, -6, 7, -2, 1, 1, -23, 46, -47, 26, -3, 1, 1, -76, 306, -536, 459, -232, 82, -4, 1, 1, -237, 1919, -5046, 6965, -5995, 3109, -958, 247, -5, 1, 1, -722, 11265, -44634, 91730, -113538, 90417, -49398, 17778, -3630, 737, -6, 1, 1, -2179, 62836, -381037, 1099549, -1878718, 2123525, -1658537, 898985, -346886, 93377, -13109, 2200, -7, 1
Offset: 0

Views

Author

Roger L. Bagula, Feb 18 2009

Keywords

Comments

Row sums are one.

Examples

			Irregular triangle begins as:
  1;
  1,   -1,     1;
  1,   -6,     7,     -2,     1;
  1,  -23,    46,    -47,    26,      -3,     1;
  1,  -76,   306,   -536,   459,    -232,    82,     -4,     1;
  1, -237,  1919,  -5046,  6965,   -5995,  3109,   -958,   247,    -5,   1;
  1, -722, 11265, -44634, 91730, -113538, 90417, -49398, 17778, -3630, 737, -6, 1;
		

Crossrefs

Programs

  • Mathematica
    p[x_, n_] = (1+x-x^2)^(n+1)*Sum[(2*k+1)^n*(-x+x^2)^k, {k, 0, Infinity}];
    Table[CoefficientList[p[x, n], x], {n,0,10}]//Flatten
  • Sage
    def T(n, k): return ( (1+x-x^2)^(n+1)*sum((2*j+1)^n*(x^2-x)^j for j in (0..2*n+1)) ).series(x, 2*n+2).list()[k]
    flatten([1]+[[T(n, k) for k in (0..2*n)] for n in (1..12)]) # G. C. Greubel, Jan 07 2022

Formula

T(n, k) = coefficients of the expansion of p(x, n), where p(x,n) = (1+x-x^2)^(n + 1)*Sum_{j >= 0} (2*j+1)^n*(-x + x^2)^j.
T(n, 1) = (-1)*A060188(n), for n >= 2. - G. C. Greubel, Jan 07 2022

Extensions

Edited by G. C. Greubel, Jan 07 2022

A152975 Numerators of the redundant Stern-Brocot structure; denominators=A152976.

Original entry on oeis.org

1, 1, 3, 2, 1, 3, 2, 5, 3, 6, 3, 1, 3, 2, 5, 3, 6, 3, 7, 4, 9, 5, 10, 5, 9, 4, 1, 3, 2, 5, 3, 6, 3, 7, 4, 9, 5, 10, 5, 9, 4, 9, 5, 12, 7, 15, 8, 15, 7, 14, 7, 15, 8, 15, 7, 12, 5, 1, 3, 2, 5, 3, 6, 3, 7, 4, 9, 5, 10, 5, 9, 4, 9, 5, 12, 7, 15, 8, 15, 7, 14, 7, 15, 8, 15, 7, 12, 5, 11, 6, 15, 9, 20, 11, 21
Offset: 1

Views

Author

Reinhard Zumkeller, Dec 22 2008

Keywords

Comments

The redundant Stern-Brocot structure is constructed row by row: insert between consecutive terms of the full Stern-Brocot tree their mediant (non-reduced), where the mediant of s/t and u/v = (s+u)/(t+v);
a(2^n-n+2*k) = A007305(2^(n-1)+k+2) for 0<=k<2^(n-1);
a(2^n-n+2*k-1) = A007305(2^(n-1)+k-1+2) + A007305(2^(n-1)+k+2) for 0
the graph of this structure describes an interesting ternary representation of the positive rational numbers;
A060188(k+2) = Sum(a(i): 2^k <= i < 2^(k+1)).

Examples

			[0/1] . . . . . . . . . . . . . . . . . . . . . . . . . . . [1/0]
.............................. 1/1
............. 1/2 ............ 3/3 ............ 2/1
..... 1/3 ... 3/6 .... 2/3 ... 5/5 ... 3/2 .... 6/3 ... 3/1
. 1/4 3/9 2/5 5/10 3/5 6/9 3/4 7/7 4/3 9/6 5/3 10/5 5/2 9/3 4/1.
		

References

  • Milad Niqui, Formalising Exact Arithmetic, Ph.D. thesis, Radboud Universiteit Nijmegen, IPA Dissertation Series 2004-10, 2.6, p.65f .

A152976 Denominators of the redundant Stern-Brocot structure; numerators=A152975.

Original entry on oeis.org

1, 2, 3, 1, 3, 6, 3, 5, 2, 3, 1, 4, 9, 5, 10, 5, 9, 4, 7, 3, 6, 3, 5, 2, 3, 1, 5, 12, 7, 15, 8, 15, 7, 14, 7, 15, 8, 15, 7, 12, 5, 9, 4, 9, 5, 10, 5, 9, 4, 7, 3, 6, 3, 5, 2, 3, 1, 6, 15, 9, 20, 11, 21, 10, 21, 11, 24, 13, 25, 12, 21, 9, 18, 9, 21, 12, 25, 13, 24, 11, 21, 10, 21, 11, 20, 9, 15, 6
Offset: 1

Author

Reinhard Zumkeller, Dec 22 2008

Keywords

Comments

a(2^n-n+2*k) = A047679(2^(n-1)+k) for 0<=k<2^(n-1);
a(2^n-n+2*k-1) = A047679(2^(n-1)+k-1) + A047679(2^(n-1)+k) for 0
A060188(k+2) = Sum(a(i): 2^k <= i < 2^(k+1)).

References

  • Milad Niqui, Formalising Exact Arithmetic, Ph.D. thesis, Radboud Universiteit Nijmegen, IPA Dissertation Series 2004-10, 2.6, p.65f .

A085354 a(n) = 3*4^n - (n+4)*2^(n-1).

Original entry on oeis.org

1, 7, 36, 164, 704, 2928, 11968, 48448, 195072, 783104, 3138560, 12567552, 50298880, 201256960, 805158912, 3220914176, 12884246528, 51538231296, 206155546624, 824627691520, 3298522300416, 13194113318912, 52776503607296
Offset: 0

Author

Paul Barry, Jun 24 2003

Keywords

Comments

Binomial transform of A060188.
The depth i nodes of a perfect binary tree are numbered 2^i through 2^(i+1) - 1, so that the root has number 1, depth 1 nodes have numbers 2 and 3, depth 2 nodes have numbers 4, 5, 6 and 7 and so on. We sum all the numbers in the path connecting a leaf node to the root. For a height n tree, a(n) is the sum of these sums for all leaves nodes. So for instance a height 1 tree has paths 1, 2 and 1, 3 connecting the root to the leaves, and (1+2) + (1+3) = a(1) = 7. This interpretation suggests a recursive formula for computing a(n) by completing the paths covered in a(n-1) and adding the leaves. - Jean M. Morales, Oct 24 2013

Crossrefs

Programs

  • Magma
    [3*4^n-(n+4)*2^(n-1): n in [0..30]]; // Vincenzo Librandi, Sep 05 2011
  • Mathematica
    Table[3 * 4^n - (n + 4) * 2^(n - 1), {n, 0, 19}] (* Alonso del Arte, Oct 23 2013 *)
    LinearRecurrence[{8,-20,16},{1,7,36},30] (* Harvey P. Dale, Apr 08 2019 *)

Formula

a(n) = Sum_{m = 2^n..2^(n+1)} A005187(m). a(n) = 2^n*(2^(n+1)-1) + Sum_{k = 0..(n-1)} a(k). - Philippe Deléham, Feb 19 2004
G.f.: (1-x)/((1-4*x)*(1-2*x)^2). - Bruno Berselli, Sep 05 2011
a(n) = 2*a(n-1) + 3*2^(2n-1) - 2^(n-1), a(0) = 1. - Jean M. Morales, Oct 24 2013

A233295 Riordan array ((1+x)/(1-x)^3, 2*x/(1-x)).

Original entry on oeis.org

1, 4, 2, 9, 10, 4, 16, 28, 24, 8, 25, 60, 80, 56, 16, 36, 110, 200, 216, 128, 32, 49, 182, 420, 616, 560, 288, 64, 64, 280, 784, 1456, 1792, 1408, 640, 128, 81, 408, 1344, 3024, 4704, 4992, 3456, 1408, 256, 100, 570, 2160, 5712, 10752, 14400, 13440, 8320, 3072, 512
Offset: 0

Author

Philippe Deléham, Dec 07 2013

Keywords

Comments

Subtriangle of the triangle in A208532.
Row sums are A060188(n+2).
Diagonal sums are A000295(n+2)=A125128(n+1)=A130103(n+2).

Examples

			Triangle begins :
1
4, 2
9, 10, 4
16, 28, 24, 8
25, 60, 80, 56, 16
36, 110, 200, 216, 128, 32
49, 182, 420, 616, 560, 288, 64
64, 280, 784, 1456, 1792, 1408, 640, 128
81, 408, 1344, 3024, 4704, 4992, 3456, 1408, 256
100, 570, 2160, 5712, 10752, 14400, 13440, 8320, 3072, 512
		

Crossrefs

Cf. Columns: A000290, A006331, A112742.
Cf. Diagonal: A000079.

Formula

G.f. for the column k: 2^k*(1+x)/(1-x)^(k+3).
T(n,k) = 2^k*(binomial(n,k)+3*binomial(n,k+1)+2*binomial(n,k+2)), 0<=k<=n.
T(n,0) = 2*T(n-1,0)-T(n-2,0)+2, T(n,k)=2*T(n-1,k)+2*T(n-1,k-1)-2*T(n-2,k-1)-T(n-2,k) for k>=1, T(0,0)=1, T(1,0)=4, T(1,1)=2, T(n,k)=0 if k<0 or if k>n.
Sum_{k=0..n} T(n,k) = A060188(n+2).
Sum_{k=0..n} T(n,k)*(-1)^k = n+1.
T(n,k) = 2*sum_{j=1..n-k+1} T(n-j,k-1).
T(n,k) = 2^k*A125165(n,k).
T(n,n) = 2^n=A000079(n).
T(n,0) = (n+1)^2=A000290(n+1).
exp(2*x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(2*x)*(16 + 28*x + 24*x^2/2! + 8*x^3/3!) = 16 + 60*x + 200*x^2/2! + 616*x^3/3! + 1792*x^4/4! + .... The same property holds more generally for Riordan arrays of the form ( f(x), 2*x/(1 - x) ). Cf. A125165. - Peter Bala, Dec 21 2014
Showing 1-7 of 7 results.