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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A008795 Molien series for 3-dimensional representation of dihedral group D_6 of order 6.

Original entry on oeis.org

1, 0, 3, 1, 6, 3, 10, 6, 15, 10, 21, 15, 28, 21, 36, 28, 45, 36, 55, 45, 66, 55, 78, 66, 91, 78, 105, 91, 120, 105, 136, 120, 153, 136, 171, 153, 190, 171, 210, 190, 231, 210, 253, 231, 276, 253, 300, 276, 325, 300, 351, 325, 378, 351, 406, 378, 435, 406, 465, 435, 496, 465, 528, 496
Offset: 0

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Author

Keywords

Comments

a(n-3) is the number of ordered triples of positive integers which are the side lengths of a nondegenerate triangle of perimeter n. - Rob Pratt, Jul 12 2004
a(n) is the number of ways to distribute n identical objects into 3 distinguishable bins so that no bin contains an absolute majority of objects. - Geoffrey Critzer, Mar 17 2010
From Omar E. Pol, Feb 05 2012: (Start)
Also terms of A000217 and A000217-shifted interleaved.
Also 0 together with this sequence give the first row of the square array A194801. (End)
a(n) is the number of coins left after packing 3-curves coins patterns into fountain of coins base n. Refer to A005169: "A fountain is formed by starting with a row of coins, then stacking additional coins on top so that each new coin touches two in the previous row". See illustration in links. - Kival Ngaokrajang, Oct 12 2013

Crossrefs

Cf. A005044.
First differences of A053307.

Programs

  • GAP
    a := [1,0,3,1,6];; for n in [6..70] do a[n] := a[n-1] + 2*a[n-2] -2*a[n-3] -a[n-4] +a[n-5]; od; a; # Muniru A Asiru, Feb 01 2018
    
  • Magma
    [(2*n^2+6*n+7)/16+3*(2*n+3)*(-1)^n/16: n in [0..70] ]; // Vincenzo Librandi, Aug 21 2011
    
  • Maple
    a:= n-> binomial(n/2+2-3*irem(n, 2)/2, 2):
    seq(a(n), n=0..70); # Muniru A Asiru, Feb 01 2018
  • Mathematica
    Table[If[EvenQ[n], Binomial[n/2+2, 2], Binomial[(n+1)/2, 2]], {n, 0, 70}]
    CoefficientList[Series[(1+x^3)/(1-x^2)^3, {x, 0, 70}], x] (* Robert G. Wilson v, Feb 05 2012 *)
    a[ n_]:= Binomial[ Quotient[n, 2] + 2 - Mod[n, 2], 2]; (* Michael Somos, Feb 01 2018 *)
    a[ n_]:= With[ {m = If[ n < 0, -3 - n, n]}, SeriesCoefficient[ (1 - x + x^2) / ((1 - x)^3 (1 + x)^2), {x, 0, m}]]; (* Michael Somos, Feb 01 2018 *)
    LinearRecurrence[{1,2,-2,-1,1}, {1,0,3,1,6}, 70] (* Robert G. Wilson v, Feb 01 2018 *)
  • PARI
    a(n)=(2*n^2+6*n+7)/16+3*(2*n+3)*(-1)^n/16 \\ Charles R Greathouse IV, Oct 22 2015
    
  • PARI
    {a(n) = binomial(n\2 + 2 - n%2, 2)}; /* Michael Somos, Feb 01 2018 */
    
  • Sage
    [(2*n^2 +6*n +7 +3*(2*n+3)*(-1)^n)/16 for n in (0..70)] # G. C. Greubel, Sep 11 2019

Formula

The signed version with g.f. (1-x^3)/(1-x^2)^3 is the inverse binomial transform of A084861. - Paul Barry, Jun 12 2003
a(n) = binomial(n/2+2, 2) for n even, binomial((n+1)/2, 2) for n odd. - Rob Pratt, Jul 12 2004
From Paul Barry, Jul 29 2004: (Start)
a(n-2) interleaves n(n+1)/2 and n(n-1)/2.
G.f.: (1-x+x^2)/((1+x)^2*(1-x)^3).
a(n) = (2*n^2 + 6*n + 7 + 3*(2*n+3)*(-1)^n)/16. (End)
a(n) = n*(n+1)/2, n = +- 1, +- 2... - Omar E. Pol, Feb 05 2012
From Michael Somos, Feb 01 2018: (Start)
Euler transform of length 6 sequence [0, 3, 1, 0, 0, -1].
G.f.: (1 + x^3) / (1 - x^2)^3.
a(n) = a(-3-n) for all in Z. (End)

Extensions

Definition clarified by N. J. A. Sloane, Feb 02 2018

A070084 Greatest common divisor of sides of integer triangles [A070080(n), A070081(n), A070082(n)], sorted by perimeter, sides lexicographically ordered.

Original entry on oeis.org

1, 1, 2, 1, 1, 1, 1, 1, 3, 2, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 3, 1, 1, 5, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 6, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 2, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 3, 1, 7, 2, 1, 2, 1, 1, 1
Offset: 1

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Author

Reinhard Zumkeller, May 05 2002

Keywords

Comments

a(n)>1 iff there exists a smaller similar triangle [A070080(k), A070081(k), A070082(k)] with kA070080(n)=A070080(k)*a(n), A070081(n)=A070081(k)*a(n) and A070082(n)=A070082(k)*a(n).

Crossrefs

Programs

  • Mathematica
    maxPer = 22; maxSide = Floor[(maxPer - 1)/2]; order[{a_, b_, c_}] := (a + b + c)*maxPer^3 + a*maxPer^2 + b*maxPer + c; triangles = Reap[Do[If[a + b + c <= maxPer && c - b < a < c + b && b - a < c < b + a && c - a < b < c + a, Sow[{a, b, c}]], {a, 1, maxSide}, {b, a, maxSide}, {c, b, maxSide}]][[2, 1]]; GCD @@@ Sort[triangles, order[#1] < order[#2] &] (* Jean-François Alcover, May 27 2013 *)

Formula

a(n) = GCD(A070080(n), A070081(n), A070082(n)).

A070093 Number of acute integer triangles with perimeter n.

Original entry on oeis.org

0, 0, 1, 0, 1, 1, 1, 1, 2, 2, 2, 2, 3, 2, 4, 3, 5, 4, 5, 5, 5, 6, 6, 6, 7, 7, 9, 8, 10, 9, 10, 10, 11, 12, 12, 12, 14, 13, 16, 14, 17, 16, 17, 18, 18, 20, 20, 20, 22, 22, 24, 23, 25, 26, 26, 27, 28, 30, 30, 29, 32, 31, 35, 33, 36, 36, 38, 39, 40, 40
Offset: 1

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Author

Reinhard Zumkeller, May 05 2002

Keywords

Comments

An integer triangle [A070080(k) <= A070081(k) <= A070082(k)] is acute iff A070085(k) > 0.

Examples

			For n=9 there are A005044(9)=3 integer triangles: [1,4,4], [2,3,4] and [3,3,3]; two of them are acute, as 2^2+3^2<16=4^2, therefore a(9)=2.
		

Crossrefs

Programs

  • Mathematica
    Table[Sum[Sum[(1 - Sign[Floor[(n - i - k)^2/(i^2 + k^2)]]) Sign[Floor[(i + k)/(n - i - k + 1)]], {i, k, Floor[(n - k)/2]}], {k, Floor[n/3]}], {n, 100}] (* Wesley Ivan Hurt, May 12 2019 *)

Formula

a(n) = A005044(n) - A070101(n) - A024155(n);
a(n) = A042154(n) + A070098(n).
a(n) = Sum_{k=1..floor(n/3)} Sum_{i=k..floor((n-k)/2)} (1-sign(floor((n-i-k)^2/(i^2+k^2)))) * sign(floor((i+k)/(n-i-k+1))). - Wesley Ivan Hurt, May 12 2019

A070083 Perimeters of integer triangles, sorted by perimeter, sides lexicographically ordered.

Original entry on oeis.org

3, 5, 6, 7, 7, 8, 9, 9, 9, 10, 10, 11, 11, 11, 11, 12, 12, 12, 13, 13, 13, 13, 13, 14, 14, 14, 14, 15, 15, 15, 15, 15, 15, 15, 16, 16, 16, 16, 16, 17, 17, 17, 17, 17, 17, 17, 17, 18, 18, 18, 18, 18, 18, 18, 19, 19, 19, 19, 19, 19, 19, 19, 19
Offset: 1

Views

Author

Reinhard Zumkeller, May 05 2002

Keywords

Comments

A005044(p) is the number of all integer triangles having perimeter p.

Programs

  • Mathematica
    maxPer = 19; maxSide = Floor[(maxPer-1)/2]; order[{a_, b_, c_}] := (a+b+c)*maxPer^3 + a*maxPer^2 + b*maxPer + c; triangles = Reap[Do[If[ a+b+c <= maxPer && c-b < a < c+b && b-a < c < b+a && c-a < b < c+a, Sow[{a, b, c}]], {a, 1, maxSide}, {b, a, maxSide}, {c, b, maxSide}]][[2, 1]]; Total /@ Sort[triangles, order[#1] < order[#2] &] (* Jean-François Alcover, Jun 12 2012 *)
    maxPer = m = 22; sides[per_] := Select[Reverse /@ IntegerPartitions[per, {3}, Range[Ceiling[per/2]]], #[[1]] < per/2 && #[[2]] < per/2 && #[[3]] < per/2&]; triangles = DeleteCases[Table[ sides[per], {per, 3, m}], {}] // Flatten[#, 1]& // SortBy[Total[#] m^3 + #[[1]] m^2 + #[[2]] m + #[[1]]&]; Total /@ triangles (* Jean-François Alcover, Jul 09 2017 *)

Formula

a(n) = A070080(n) + A070081(n) + A070082(n).

A070088 Number of integer-sided triangles with perimeter n and prime sides.

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 2, 1, 2, 0, 1, 0, 1, 0, 1, 1, 2, 0, 3, 1, 3, 0, 2, 0, 2, 0, 3, 1, 3, 0, 5, 1, 5, 0, 4, 0, 3, 0, 5, 1, 5, 0, 4, 0, 4, 0, 2, 0, 3, 0, 5, 1, 3, 0, 6, 1, 8, 0, 5, 0, 5, 0, 4, 0, 3, 0, 5, 1, 6, 0, 6, 0, 4, 0, 7, 1, 7, 0, 9, 1, 10, 0
Offset: 1

Views

Author

Reinhard Zumkeller, May 05 2002

Keywords

Examples

			For n=15 there are A005044(15)=7 integer triangles: [1,7,7], [2,6,7], [3,5,7], [3,6,6], [4,4,7], [4,5,6] and [5,5,5]: two of them consist of primes, therefore a(15)=2.
		

Crossrefs

Programs

  • Mathematica
    triangleQ[sides_] := With[{s = Total[sides]/2}, AllTrue[sides, # < s&]];
    a[n_] := Select[IntegerPartitions[n, {3}, Select[Range[Ceiling[n/2]], PrimeQ]], triangleQ] // Length; Array[a, 90] (* Jean-François Alcover, Jul 09 2017 *)
    Table[Sum[Sum[(PrimePi[i] - PrimePi[i - 1]) (PrimePi[k] - PrimePi[k - 1]) (PrimePi[n - i - k] - PrimePi[n - i - k - 1]) Sign[Floor[(i + k)/(n - i - k + 1)]], {i, k, Floor[(n - k)/2]}], {k, Floor[n/3]}], {n, 100}] (* Wesley Ivan Hurt, May 13 2019 *)

Formula

a(n) = A070090(n) + A070092(n) = A070095(n) + A070103(n).
a(n) = Sum_{k=1..floor(n/3)} Sum_{i=k..floor((n-k)/2)} sign(floor((i+k)/(n-i-k+1))) * c(i) * c(k) * c(n-i-k), where c = A010051. - Wesley Ivan Hurt, May 13 2019

A024155 Number of integer-sided triangles with sides a,b,c, a

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 2, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0
Offset: 1

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Author

Keywords

Comments

Also number of right integer triangles with perimeter n having integral inradius. - Reinhard Zumkeller, May 05 2002
Every integer-sided right triangle has integer inradius. If the triple is [p^2-q^2,2pq,p^2+q^2] then inradius = pq-q^2. - Michael Somos, Sep 13 2005

Crossrefs

Formula

a(n) = A070201(n) - A070205(n) - A070206(n).

A051493 Triangles with perimeter n and relatively prime integer side lengths.

Original entry on oeis.org

0, 0, 1, 0, 1, 0, 2, 1, 2, 1, 4, 2, 5, 2, 5, 4, 8, 4, 10, 6, 9, 6, 14, 8, 15, 9, 16, 12, 21, 11, 24, 16, 22, 16, 27, 18, 33, 20, 31, 24, 40, 23, 44, 30, 39, 30, 52, 32, 54, 35, 52, 42, 65, 38, 65, 48, 64, 49, 80, 48, 85, 56, 77, 64, 90, 58, 102, 72, 93, 69, 114, 72, 120, 81
Offset: 1

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Author

Keywords

Comments

From Peter Munn, Jul 26 2017: (Start)
The triangles that meet the conditions are listed by nondecreasing n in A070110.
Without the requirement for relatively prime side lengths, this sequence becomes A005044.
Counting the triangles by longest side instead of perimeter, this sequence becomes A123323.
a(n) = A070094(n) + A070102(n) + A070109(n).
(End)

Examples

			There are 3 triangles with integer-length sides and perimeter 9: 1-4-4, 2-3-4, 3-3-3. 3-3-3 is omitted because isomorphic to 1-1-1, so a(9)=2.
		

Crossrefs

Equivalent sequences, restricted to subsets: A070091 (isosceles), A070094 (acute), A070102 (obtuse), A070109 (right-angled), A070138 (with integer area), A070202 (with integer inradius).

Programs

  • Mathematica
    nmax = 100;
    A005044[n_] := Quotient[n^2 + 6n Mod[n, 2] + 24, 48];
    A = Array[A005044, nmax];
    mob[m_, n_] := If[ Mod[m, n] == 0, MoebiusMu[m/n], 0];
    Reap[Do[Sow[Sum[mob[n, d] A[[d]], {d, 1, n}]], {n, 1, nmax}]][[2, 1]] (* Jean-François Alcover, Oct 05 2021 *)

Formula

Moebius transform of A005044.

Extensions

Corrected and extended with formula by Christian G. Bower, Nov 15 1999
Formula updated due to change to referenced sequence, and definition clarified by Peter Munn, Jul 26 2017

A070101 Number of obtuse integer triangles with perimeter n.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 2, 0, 2, 2, 3, 2, 3, 3, 5, 3, 7, 4, 8, 5, 9, 7, 10, 8, 11, 9, 14, 11, 16, 12, 18, 14, 19, 17, 21, 18, 23, 21, 27, 22, 30, 24, 32, 27, 34, 30, 37, 33, 40, 35, 44, 37, 47, 40, 50, 44, 53, 49, 56, 52, 60, 55, 64, 57, 68
Offset: 1

Views

Author

Reinhard Zumkeller, May 05 2002

Keywords

Comments

An integer triangle [A070080(k) <= A070081(k) <= A070082(k)] is obtuse iff A070085(k) < 0.

Examples

			For n=14 there are A005044(14)=4 integer triangles: [2,6,6], [3,5,6], [4,4,6] and [4,5,5]; two of them are obtuse, as 3^2+5^2<36=6^2 and 4^2+4^2<36=6^2, therefore a(14)=2.
		

Crossrefs

Programs

  • Mathematica
    Table[Sum[Sum[(1 - Sign[Floor[(i^2 + k^2)/(n - i - k)^2]]) Sign[Floor[(i + k)/(n - i - k + 1)]], {i, k, Floor[(n - k)/2]}], {k, Floor[n/3]}], {n, 100}] (* Wesley Ivan Hurt, May 12 2019 *)

Formula

a(n) = A005044(n) - A070093(n) - A024155(n).
a(n) = A024156(n) + A070106(n).
a(n) = Sum_{k=1..floor(n/3)} Sum_{i=k..floor((n-k)/2)}
(1-sign(floor((i^2 + k^2)/(n-i-k)^2))) * sign(floor((i+k)/(n-i-k+1))). - Wesley Ivan Hurt, May 12 2019

A062890 Number of quadrilaterals that can be formed with perimeter n. In other words, number of partitions of n into four parts such that the sum of any three is more than the fourth.

Original entry on oeis.org

0, 0, 0, 0, 1, 1, 1, 2, 3, 4, 5, 7, 8, 11, 12, 16, 18, 23, 24, 31, 33, 41, 43, 53, 55, 67, 69, 83, 86, 102, 104, 123, 126, 147, 150, 174, 177, 204, 207, 237, 241, 274, 277, 314, 318, 358, 362, 406, 410, 458, 462, 514, 519, 575, 579, 640, 645, 710
Offset: 0

Views

Author

Amarnath Murthy, Jun 29 2001

Keywords

Comments

Partition sets of n into four parts (sides) such that the sum of any three is more than the fourth do not uniquely define a quadrilateral, even if it is further constrained to be cyclic. This is because the order of adjacent sides is important. E.g. the partition set [1,1,2,2] for a perimeter n=6 can be reordered to generate two non-congruent cyclic quadrilaterals, [1,2,1,2] and [1,1,2,2], where the first is a rectangle and the second a kite. - Frank M Jackson, Jun 29 2012

Examples

			a(7) = 2 as the two partitions are (1,2,2,2), (1,1,2,3) and in each sum of any three is more than the fourth.
		

Crossrefs

Number of k-gons that can be formed with perimeter n: A005044 (k=3), this sequence (k=4), A069906 (k=5), A069907 (k=6), A288253 (k=7), A288254 (k=8), A288255 (k=9), A288256 (k=10).

Programs

  • Mathematica
    CoefficientList[Series[x^4*(1+x+x^5)/((1-x^2)*(1-x^3)*(1-x^4)*(1-x^6)), {x, 0, 60}], x] (* Frank M Jackson, Jun 09 2017 *)

Formula

G.f.: x^4*(1+x+x^5)/((1-x^2)*(1-x^3)*(1-x^4)*(1-x^6)).
a(2*n+6) = A026810(2*n+6) - A000601(n), a(2*n+7) = A026810(2*n+7) - A000601(n) for n >= 0. - Seiichi Manyama, Jun 08 2017
From Wesley Ivan Hurt, Jan 01 2021: (Start)
a(n) = a(n-1) + a(n-2) - a(n-3) + a(n-4) - a(n-5) - a(n-8) + a(n-9) - a(n-10) + a(n-11) + a(n-12) - a(n-13).
a(n) = Sum_{k=1..floor(n/4)} Sum_{j=k..floor((n-k)/3)} Sum_{i=j..floor((n-j-k)/2)} sign(floor((i+j+k)/(n-i-j-k+1))). (End)

Extensions

More terms from Vladeta Jovovic and Dean Hickerson, Jul 01 2001

A008667 Expansion of g.f.: 1/((1-x^2)*(1-x^3)*(1-x^4)*(1-x^5)).

Original entry on oeis.org

1, 0, 1, 1, 2, 2, 3, 3, 5, 5, 7, 7, 10, 10, 13, 14, 17, 18, 22, 23, 28, 29, 34, 36, 42, 44, 50, 53, 60, 63, 71, 74, 83, 87, 96, 101, 111, 116, 127, 133, 145, 151, 164, 171, 185, 193, 207, 216, 232, 241, 258, 268, 286, 297, 316, 328, 348, 361, 382, 396, 419, 433, 457
Offset: 0

Views

Author

Keywords

Comments

Also, Molien series for invariants of finite Coxeter group A_4. The Molien series for the finite Coxeter group of type A_k (k >= 1) has g.f. = 1/Product_{i=2..k+1} (1-x^i). Note that this is the root system A_k not the alternating group Alt_k. - N. J. A. Sloane, Jan 11 2016
Number of partitions into parts 2, 3, 4, and 5. - Joerg Arndt, Apr 29 2014

Examples

			a(4)=2 because f''''(x)/4!=2 at x=0 for f=1/((1-x^2)(1-x^3)(1-x^4)(1-x^5)).
G.f. = 1 + x^2 + x^3 + 2*x^4 + 2*x^5 + 3*x^6 + 3*x^7 + 5*x^8 + 5*x^9 + 7*x^10 + 7*x^11 + ... .
		

References

  • J. E. Humphreys, Reflection Groups and Coxeter Groups, Cambridge, 1990. See Table 3.1, page 59.
  • L. Smith, Polynomial Invariants of Finite Groups, Peters, 1995, p. 199 (No. 32).

Crossrefs

Molien series for finite Coxeter groups A_1 through A_12 are A059841, A103221, A266755, A008667, A037145, A001996, and A266776-A266781.
Cf. A005044, A001401 (partial sums).

Programs

  • Magma
    R:=PowerSeriesRing(Integers(), 65); Coefficients(R!( 1/((1-x^2)*(1-x^3)*(1-x^4)*(1-x^5)) )); // G. C. Greubel, Sep 08 2019
    
  • Maple
    seq(coeff(series(1/((1-x^2)*(1-x^3)*(1-x^4)*(1-x^5)), x, n+1), x, n), n = 0..65); # G. C. Greubel, Sep 08 2019
  • Mathematica
    SeriesCoefficient[1/((1-x^2)(1-x^3)(1-x^4)(1-x^5)),{x,0,#}]&/@Range[0,100] (* or *) a[k_]=SeriesCoefficient[1/((1-x^2)(1-x^3)(1-x^4) (1-x^5)),{x,0,k}] (* Peter Pein (petsie(AT)dordos.net), Sep 09 2006 *)
    CoefficientList[Series[1/Times@@Table[(1-x^n),{n,2,5}],{x,0,70}],x] (* Harvey P. Dale, Feb 22 2018 *)
  • PARI
    {a(n) = if( n<-13, -a(-14 - n), polcoeff( prod( k=2, 5, 1 / (1 - x^k), 1 + x * O(x^n)), n))} /* Michael Somos, Oct 14 2006 */
    
  • Sage
    def A008667_list(prec):
        P. = PowerSeriesRing(ZZ, prec)
        return P(1/((1-x^2)*(1-x^3)*(1-x^4)*(1-x^5))).list()
    A008667_list(65) # G. C. Greubel, Sep 08 2019

Formula

Euler transform of length 5 sequence [ 0, 1, 1, 1, 1]. - Michael Somos, Sep 23 2006
a(-14 - n) = -a(n). - Michael Somos, Sep 23 2006
a(n) ~ 1/720*n^3. - Ralf Stephan, Apr 29 2014
a(n) = a(n-2) + a(n-3) + a(n-4) - a(n-6) - 2*a(n-7) - a(n-8) + a(n-10) + a(n-11) + a(n-12) - a(n-14). - David Neil McGrath, Sep 13 2014
From R. J. Mathar, Jun 23 2021: (Start)
a(n)-a(n-2) = A008680(n).
a(n)-a(n-3) = A025802(n).
a(n)-a(n-4) = A025795(n).
a(n)-a(n-5) = A005044(n+3). (End)
a(n)= floor((n^3 + 21*n^2 + 156*n - 45*n*(n mod 2) + 720)/720 - [(n mod 10)=1]/5). - Hoang Xuan Thanh, Aug 20 2025
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