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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A073370 Convolution triangle of A001045(n+1) (generalized (1,2)-Fibonacci), n>=0.

Original entry on oeis.org

1, 1, 1, 3, 2, 1, 5, 7, 3, 1, 11, 16, 12, 4, 1, 21, 41, 34, 18, 5, 1, 43, 94, 99, 60, 25, 6, 1, 85, 219, 261, 195, 95, 33, 7, 1, 171, 492, 678, 576, 340, 140, 42, 8, 1, 341, 1101, 1692, 1644, 1106, 546, 196, 52, 9, 1
Offset: 0

Views

Author

Wolfdieter Lang, Aug 02 2002

Keywords

Comments

The g.f. for the row polynomials P(n,x) = Sum_{m=0..n} T(n,m)*x^m is 1/(1-(1+x+2*z)*z). See Shapiro et al. reference and comment under A053121 for such convolution triangles.
Riordan array (1/(1-x-2*x^2), x/(1-x-2*x^2)). - Paul Barry, Mar 15 2005
Subtriangle (obtained by dropping the first column) of the triangle given by (0, 1, 2, -2, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (1, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Feb 19 2013
The number of ternary words of length n having k letters equal 2 and 0,1 avoid runs of odd lengths. - Milan Janjic, Jan 14 2017

Examples

			Triangle begins as:
    1;
    1,   1;
    3,   2,   1;
    5,   7,   3,   1;
   11,  16,  12,   4,   1;
   21,  41,  34,  18,   5,   1;
   43,  94,  99,  60,  25,   6,   1;
   85, 219, 261, 195,  95,  33,   7,   1;
  171, 492, 678, 576, 340, 140,  42,   8,   1;
The triangle (0, 1, 2, -2, 0, 0, ...) DELTA (1, 0, 0, 0, 0, ...) begins:
  1;
  0,  1;
  0,  1,  1;
  0,  3,  2,  1;
  0,  5,  7,  3,  1;
  0, 11, 16, 12,  4,  1;
  0, 21, 41, 34, 18,  5,  1; - _Philippe Deléham_, Feb 19 2013
		

Crossrefs

Columns: A001045 (k=0), A073371 (k=1), A073372 (k=2), A073373 (k=3), A073374 (k=4), A073375 (k=5), A073376 (k=6), A073377 (k=7), A073378 (k=8), A073379 (k=9).
Cf. A002605 (row sums), A006130 (diagonal sums), A073399, A073400.

Programs

  • Magma
    A073370:= func< n,k | (&+[Binomial(n-j,k)*Binomial(n-k-j,j)*2^j: j in [0..Floor((n-k)/2)]]) >;
    [A073370(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Oct 01 2022
    
  • Maple
    # Uses function PMatrix from A357368. Adds a row above and a column to the left.
    PMatrix(10, n -> (2^n - (-1)^n) / 3); # Peter Luschny, Oct 07 2022
  • Mathematica
    T[n_, k_]:= T[n, k]= Sum[Binomial[n-j,k]*Binomial[n-k-j,j]*2^j, {j,0,Floor[(n- k)/2]}];
    Table[T[n, k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Oct 01 2022 *)
  • SageMath
    def A073370(n,k): return binomial(n,k)*sum( 2^j * binomial(2*j,j) * binomial(n-k,2*j)/binomial(n,j) for j in range(1+(n-k)//2))
    flatten([[A073370(n,k) for k in range(n+1)] for n in range(12)]) # G. C. Greubel, Oct 01 2022

Formula

T(n, m) = Sum_{k=0..floor((n-m)/2)} binomial(n-k, m)*binomial(n-m-k, k)*2^k, if n > m, else 0.
Sum_{k=0..n} T(n, k) = A002605(n+1).
T(n, m) = (1*(n-m+1)*T(n, m-1) + 2*2*(n+m)*T(n-1, m-1))/((1^2 + 4*2)*m), n >= m >= 1, T(n, 0) = A001045(n+1), n >= 0, else 0.
T(n, m) = (p(m-1, n-m)*1*(n-m+1)*T(n-m+1) + q(m-1, n-m)*2*(n-m+2)*T(n-m))/(m!*9^m), n >= m >= 1, with T(n) = T(n, m=0) = A001045(n+1), else 0; p(k, n) = Sum_{j=0..k} (A(k, j)*n^(k-j) and q(k, n) = Sum_{j=0..k} B(k, j)*n^(k-j), with the number triangles A(k, m) = A073399(k, m) and B(k, m) = A073400(k, m).
G.f.: 1/(1-(1+2*x)*x)^(m+1) = 1/((1+x)*(1-2*x))^(m+1), m >= 0, for column m (without leading zeros).
T(n, 0) = A001045(n), T(1, 1) = 1, T(n, k) = 0 if k>n, T(n, k) = T(n-1, k-1) + 2*T(n-2, k) + T(n-1, k) otherwise. - Paul Barry, Mar 15 2005
G.f.: (1+x)*(1-2*x)/(1-x-2*x^2-x*y) for the triangle including the 1, 0, 0, 0, 0, ... column. - R. J. Mathar, Aug 11 2015
From Peter Bala, Oct 07 2019: (Start)
Recurrence for row polynomials: R(n,x) = (1 + x)*R(n-1,x) + 2*R(n-2,x) with R(0,x) = 1 and R(1,x) = 1 + x.
The row reverse polynomial x^n*R(n,1/x) is equal to the numerator polynomial of the finite continued fraction 1 + x/(1 - 2*x/(1 + ... + x/(1 - 2*x/(1)))) (with 2*n partial numerators). Cf. A110441. (End)
From G. C. Greubel, Oct 01 2022: (Start)
T(n, k) = binomial(n,k)*Sum_{j=0..floor((n-k)/2)} 2^j*binomial(2*j, j)*binomial(n-k, 2*j)/binomial(n, j).
T(n, k) = binomial(n, k)*Hypergeometric2F1([(k-n)/2, (k-n+1)/2], [-2*n], -8).
Sum_{k=0..n} (-1)^k * T(n, k) = A077957(n).
Sum_{k=0..floor(n/2)} T(n-k, k) = A006130(n).
Sum_{k=0..floor(n/2)} (-1)^k * T(n-k, k) = A000045(n+1). (End)

A062111 Upper-right triangle resulting from binomial transform calculation for nonnegative integers.

Original entry on oeis.org

0, 1, 1, 4, 3, 2, 12, 8, 5, 3, 32, 20, 12, 7, 4, 80, 48, 28, 16, 9, 5, 192, 112, 64, 36, 20, 11, 6, 448, 256, 144, 80, 44, 24, 13, 7, 1024, 576, 320, 176, 96, 52, 28, 15, 8, 2304, 1280, 704, 384, 208, 112, 60, 32, 17, 9, 5120, 2816, 1536, 832, 448, 240, 128, 68, 36, 19, 10
Offset: 0

Views

Author

Henry Bottomley, May 30 2001

Keywords

Comments

From Philippe Deléham, Apr 15 2007: (Start)
This triangle can be found in the Laisant reference in the following form:
.......................5...11..
...................4...9...20..
...............3...7..16...36..
...........2...5..12..28.......
.......1...3...8..20..48.......
...0...1...4..12..32..80....... (End)
Triangle A152920 reversed. - Philippe Deléham, Apr 21 2009

Examples

			As a lower triangle (T(n, k)):
    0;
    1,   1;
    4,   3,   2;
   12,   8,   5,  3;
   32,  20,  12,  7,  4;
   80,  48,  28, 16,  9,  5;
  192, 112,  64, 36, 20, 11,  6;
  448, 256, 144, 80, 44, 24, 13, 7;
		

Crossrefs

Rows include (essentially) A001787, A001792, A034007, A045623, A045891.
Diagonals include (essentially) A001477, A005408, A008586, A008598, A017113.
Column sums are A058877.

Programs

  • Magma
    [2^(n-k-1)*(n+k): k in [0..n], n in [0..12]]; // G. C. Greubel, Sep 28 2022
    
  • Mathematica
    Table[2^(n-k-1)*(n+k), {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Sep 28 2022 *)
  • SageMath
    def A062111(n,k): return 2^(n-k-1)*(n+k)
    flatten([[A062111(n,k) for k in range(n+1)] for n in range(12)]) # G. C. Greubel, Sep 28 2022

Formula

A(n, k) = A(n, k-1) + A(n+1, k) if k > n with A(n, n) = n.
A(n, k) = (k+n)*2^(k-n-1) if k >= n.
T(2*n, n) = 3*n*2^(n-1) = 3*A001787(n). - Philippe Deléham, Apr 21 2009
From G. C. Greubel, Sep 28 2022: (Start)
T(n, k) = 2^(n-k-1)*(n+k) for 0 <= k <= n, n >= 0.
T(m*n, n) = 2^((m-1)*n-1)*(m+1)*A001477(n), m >= 1.
T(2*n-1, n-1) = A130129(n-1).
T(2*n+1, n-1) = 12*A001787(n).
Sum_{k=0..n} T(n, k) = A058877(n+1).
Sum_{k=0..n} (-1)^k*T(n, k) = 3*A073371(n-2), n >= 2.
T(n, k) = A152920(n, n-k). (End)

A127978 a(n) = ((15*n + 34)/54)*2^(n-1) - (-1)^(n-1)*(6*n + 5)/27.

Original entry on oeis.org

3, 5, 15, 31, 75, 163, 367, 799, 1747, 3771, 8119, 17367, 37019, 78579, 166271, 350735, 737891, 1548587, 3242823, 6776903, 14136363, 29437795, 61205775, 127071871, 263464435, 545570203, 1128423127, 2331411639, 4811954107
Offset: 2

Views

Author

Artur Jasinski, Feb 09 2007

Keywords

Comments

In the Bosma's paper there is an error (see table of the first few values at p. 37): for n=1 ((15*n+34)/54)*2^(n-1)-(-1)^(n-1)*(6*n+5)/27 is 1/2 and not 1.

Crossrefs

Programs

  • Magma
    I:=[3, 5, 15, 31]; [n le 4 select I[n] else 2*Self(n-1) + 3*Self(n-2) -4*Self(n-3) -4*Self(n-4): n in [1..30]]; // G. C. Greubel, May 07 2018
  • Mathematica
    Table[((15n+34)/54)2^(n-1) -((-1)^(n-1))(6n+5)/27, {n, 2, 50}]
    LinearRecurrence[{2,3,-4,-4}, {3, 5, 15, 31}, 50] (* G. C. Greubel, May 07 2018 *)
  • PARI
    x='x+O('x^30); Vec(x^2*(3-x-4*x^2-2*x^3)/((1+x)^2*(1-2*x)^2)) \\ G. C. Greubel, May 07 2018
    

Formula

G.f.: x^2*(3-x-4*x^2-2*x^3)/((1+x)^2*(1-2*x)^2). - Colin Barker, Apr 02 2012

Extensions

Offset changed from 1 to 2 (according to Bosma's Proposition 5) from Bruno Berselli, Apr 02 2012

A127979 a(n) = (5*n/18 + 19/54)*2^n - (-1)^(n-1)*(3*n + 4)/27.

Original entry on oeis.org

1, 4, 9, 24, 55, 130, 293, 660, 1459, 3206, 6977, 15096, 32463, 69482, 148061, 314332, 665067, 1402958, 2951545, 6194368, 12971271, 27107634, 56545429, 117751204, 244823075, 508287510, 1053857713, 2182280840, 4513692479, 9325646586
Offset: 1

Views

Author

Artur Jasinski, Feb 09 2007

Keywords

Crossrefs

Programs

  • Magma
    [(5*n/18 +19/54)*2^n -(-1)^(n-1)*(3*n+4)/27: n in [1..50]]; // G. C. Greubel, May 08 2018
  • Mathematica
    Table[(5n/18 + 19/54)2^(n) - ((-1)^(n - 1) )(3n + 4)/27, {n, 1, 100}]
    LinearRecurrence[{2,3,-4,-4}, {1,4,9,24}, 50] (* G. C. Greubel, May 08 2018 *)
  • PARI
    for(n=1, 50, print1((5*n/18 +19/54)*2^n -(-1)^(n-1)*(3*n+4)/27, ", ")) \\ G. C. Greubel, May 08 2018
    

Formula

G.f.: x*(1+2*x-2*x^2-2*x^3)/((1+x)^2*(1-2x)^2). - Bruno Berselli, Apr 02 2012
a(n) = 2*a(n-1)+3*a(n-2)-4*a(n-3)-4*a(n-4). - Wesley Ivan Hurt, May 07 2021

A073372 Second convolution of A001045(n+1) (generalized (1,2)-Fibonacci), n>=0, with itself.

Original entry on oeis.org

1, 3, 12, 34, 99, 261, 678, 1692, 4149, 9959, 23568, 55014, 127031, 290457, 658602, 1482240, 3314025, 7365915, 16285300, 35832810, 78500811, 171293293, 372412782, 806963364, 1743173469, 3754782351, 8066319768, 17285917742, 36957928479, 78847115649
Offset: 0

Views

Author

Wolfdieter Lang, Aug 02 2002

Keywords

Crossrefs

Third (m=2) column of triangle A073370.

Programs

  • Magma
    [(2^(n+3)*(16+15*n+3*n^2) +(-1)^n*(34+21*n+3*n^2))/162: n in [0..40]]; // G. C. Greubel, Sep 28 2022
    
  • Mathematica
    CoefficientList[Series[-(-1+x+2x^2)^(-3),{x,0,78}],x] (* or *) Table[(-3*(-1)^n*n^2+3*2^(n+2)*n^2-15*(-1)^n*n+9*2^(n+2)*n-16*(-1)^n+2^(n+4))/162,{n,42}] (* Vladimir Joseph Stephan Orlovsky, Feb 01 2012 *)
  • SageMath
    def A073372(n): return (2^(n+3)*(16+15*n+3*n^2) +(-1)^n*(34+21*n+3*n^2))/162
    [A073372(n) for n in range(40)] # G. C. Greubel, Sep 28 2022

Formula

a(n) = Sum_{k=0..n} b(k)*c(n-k), with b(k) = A001045(k+1) and c(k) = A073371(k).
a(n) = Sum_{k=0..floor(n/2)} binomial(n-k+2, 2) * binomial(n-k, k) * 2^k.
a(n) = ((30+9*n)*(n+1)*U(n+1) + 2*(33+9*n)*(n+2)*U(n))/162 with U(n) = A001045(n+1), n>=0.
G.f.: 1/(1 - (1+2*x)*x)^3.
E.g.f.: (1/162)*(32*(4 + 9*x + 3*x^2)*exp(2*x) + (34 - 24*x + 3*x^2)*exp(-x)). - G. C. Greubel, Sep 28 2022

A127980 a(n) = (n + 2/3)*2^(n-1) - 1/2 - (-1)^(n-1)*(1/6).

Original entry on oeis.org

1, 5, 14, 37, 90, 213, 490, 1109, 2474, 5461, 11946, 25941, 55978, 120149, 256682, 546133, 1157802, 2446677, 5155498, 10835285, 22719146, 47535445, 99265194, 206918997, 430615210, 894784853, 1856678570, 3847574869, 7963585194
Offset: 1

Views

Author

Artur Jasinski, Feb 09 2007

Keywords

Crossrefs

Programs

  • Magma
    I:=[1,5,14,37]; [n le 4 select I[n] else 4*Self(n-1)-3*Self(n-2)-4*Self(n-3)+4*Self(n-4): n in [1..30]]; // G. C. Greubel, May 08 2018
  • Mathematica
    Table[(n+2/3)2^(n-1) - 1/2 -(-1)^(n-1)*(1/6), {n, 1, 50}]
    LinearRecurrence[{4,-3,-4,4}, {1,5,14,37}, 50] (* G. C. Greubel, May 08 2018 *)
  • PARI
    x='x+O('x^30); Vec(x*(1+x-3*x^2)/((1-x)*(1+x)*(1-2*x)^2)) \\ G. C. Greubel, May 08 2018
    

Formula

G.f.: x*(1+x-3*x^2)/((1-x)*(1+x)*(1-2*x)^2). - Colin Barker, Apr 02 2012
E.g.f.: ((1 + 3*x)*cosh(2*x) - 2*sinh(x) + cosh(x)*((2 + 6*x)*sinh(x) - 1))/3. - Stefano Spezia, May 25 2023

A127984 a(n) = (n/3 + 7/9)*2^(n - 1) + (-1)^n/9.

Original entry on oeis.org

1, 3, 7, 17, 39, 89, 199, 441, 967, 2105, 4551, 9785, 20935, 44601, 94663, 200249, 422343, 888377, 1864135, 3903033, 8155591, 17010233, 35418567, 73633337, 152859079, 316902969, 656175559, 1357090361, 2803659207, 5786275385, 11930464711, 24576757305
Offset: 1

Views

Author

Artur Jasinski, Feb 09 2007

Keywords

Comments

a(n) is the number of runs of strictly increasing parts in all compositions of n. a(3) = 7: (1)(1)(1), (12), (2)(1), (3). - Alois P. Heinz, Apr 30 2017
From Hugo Pfoertner, Feb 19 2020: (Start)
a(n)/2^(n-2) apparently is the expected number of flips of a fair coin to completion of a game where the player advances by 1 for heads and by 2 for tails, starting at position 0 and repeating to flip until the target n+1 is exactly reached. If the position n (1 below the target) is reached, the player stays at this position and continues to flip the coin and count the flips until he can advance by 1.
The expected number of flips for targets 1, 2, 3,... , found by inversion of the corresponding Markov matrices, is 2, 2, 3, 7/2, 17/4, 39/8, 89/16, 199/32, 441/64, ...
Target 1 needs an expected number of 2 flips and would require a(0) = 1/2.
n=1, target n+1 = 2: 1 / 2^(1-2) = 2;
n=2, target n+1 = 3: 3 / 2^(2-2) = 3;
n=3, target n+1 = 4: 7 / 2^(3-2) = 7/2.
(End)

Crossrefs

Programs

  • Magma
    [(n/3+7/9)*2^(n-1)+(-1)^n/9: n in [1..35]]; // Vincenzo Librandi, Jun 15 2017
  • Maple
    A127984:=n->(n/3 + 7/9)*2^(n - 1) + (-1)^n/9; seq(A127984(n), n=1..50); # Wesley Ivan Hurt, Mar 14 2014
  • Mathematica
    Table[(n/3 + 7/9)2^(n - 1) + (-1)^n/9, {n, 50}] (* Artur Jasinski *)
    CoefficientList[Series[(1 - 2 x^2) / ((-1 + 2 x)^2 (1 + x)), {x, 0, 40}], x] (* Vincenzo Librandi, Jun 15 2017 *)

Formula

a(n) = (n/3 + 7/9)*2^(n - 1) + (-1)^n/9.
From R. J. Mathar, Apr 04 2008: (Start)
O.g.f.: -x*(-1+2x^2)/((-1+2x)^2*(1+x)).
a(n) = 3*a(n-1) - 4*a(n-3). (End)
a(n) + a(n+1) = A087447(n+1). - R. J. Mathar, Feb 21 2009
A172481(n) = a(n) + 2^(n-1). Application: Problem 11623, AMM 119 (2012) 161. - Stephen J. Herschkorn, Feb 11 2012
From Wolfdieter Lang, Jun 14 2017: (Start)
a(n) = f(n+1)*2^(n-1), where f(n) is a rational Fibonacci type sequence based on fuse(a,b) = (a+b+1)/2 with f(0) = 0, f(1) = 1 and f(n) = fuse(f(n-1),f(n-2)), for n >= 2. For fuse(a,b) see the Jeff Erickson link under A188545. Proof: f(n) = (3*n+4 - (-1)^n/2^(n-2))/9, n >= 0, by induction.
a(n) = a(n-1) + a(n-2) + 2^(n-2), n >= 1, with inputs a(-1) = 0, a(0) = 1/2.
(End)
E.g.f.: (2*exp(-x) + exp(2*x)*(7 + 6*x) - 9)/18. - Stefano Spezia, Feb 19 2020

A127981 a(n) = (n + 1/3)*2^(n-1) - 1/2 + (-1)^(n-1)*(1/6).

Original entry on oeis.org

1, 4, 13, 34, 85, 202, 469, 1066, 2389, 5290, 11605, 25258, 54613, 117418, 251221, 535210, 1135957, 2402986, 5068117, 10660522, 22369621, 46836394, 97867093, 204122794, 425022805, 883600042, 1834308949, 3802835626, 7874106709
Offset: 1

Views

Author

Artur Jasinski, Feb 09 2007

Keywords

Crossrefs

Programs

  • Magma
    [(n+1/3)*2^(n-1) -1/2 +(-1)^(n-1)*(1/6): n in [1..50]]; // G. C. Greubel, May 08 2018
  • Mathematica
    Table[(n+1/3)*2^(n-1) -1/2 +(-1)^(n-1)*(1/6), {n, 1, 50}]
    LinearRecurrence[{4,-3,-4,4}, {1,4,13,34}, 50] (* G. C. Greubel, May 08 2018 *)
  • PARI
    for(n=1,50, print1((n+1/3)*2^(n-1) -1/2 +(-1)^(n-1)*(1/6), ", ")) \\ G. C. Greubel, May 08 2018
    

Formula

G.f.: x*(1-2*x^3)/((1-x)*(1+x)*(1-2*x)^2). - Colin Barker, Apr 02 2012

A127982 a(n) = (n - 1/3)*2^n - n/2 + 1/4 + (-1)^n/12.

Original entry on oeis.org

1, 6, 20, 57, 147, 360, 850, 1959, 4433, 9894, 21840, 47781, 103759, 223908, 480590, 1026723, 2184525, 4631202, 9786700, 20621985, 43341131, 90876576, 190141770, 397060767, 827675977, 1722460830, 3579139400, 7426714269, 15390299463
Offset: 1

Views

Author

Artur Jasinski, Feb 09 2007

Keywords

Crossrefs

Programs

  • Magma
    [(n-1/3)*2^n -n/2 +1/4 +(-1)^n/12: n in [1..50]]; // G. C. Greubel, May 08 2018
  • Mathematica
    Table[(n-1/3)*2^n -n/2 +1/4 +(-1)^n/12, {n, 1, 50}]
    LinearRecurrence[{5,-7,-1,8,-4}, {1,6,20,57,147}, 50] (* G. C. Greubel, May 08 2018 *)
  • PARI
    a(n) = (n-1/3)*2^n -n/2 +1/4 +(-1)^n/12 \\ G. C. Greubel, May 08 2018
    

Formula

a(n) = (n - 1/3)*2^n - n/2 + 1/4 + (-1)^n/12.
G.f.: -x*(3*x^2-x-1)/((1+x)*(2*x-1)^2*(x-1)^2). - Maksym Voznyy (voznyy(AT)mail.ru), Aug 14 2009 [checked and corrected by R. J. Mathar, Sep 16 2009]

A127983 a(n) = (n - 2/3)*2^n - n/2 + 3/4 - (-1)^n/12.

Original entry on oeis.org

1, 5, 18, 52, 137, 339, 808, 1874, 4263, 9553, 21158, 46416, 101029, 218447, 469668, 1004878, 2140835, 4543821, 9611938, 20272460, 42642081, 89478475, 187345568, 391468362, 816491167, 1700091209, 3534400158, 7337235784, 15211342493
Offset: 1

Views

Author

Artur Jasinski, Feb 09 2007

Keywords

Crossrefs

Programs

  • Magma
    [(n-2/3)*2^n -n/2 +3/4 -(-1)^n/12: n in [1..50]]; // G. C. Greubel, May 08 2018
  • Mathematica
    Table[(n-2/3)*2^n -n/2 +3/4 -(-1)^n/12, {n, 1, 50}]
    LinearRecurrence[{5,-7,-1,8,-4}, {1,5,18,52,137}, 50] (* G. C. Greubel, May 08 2018 *)
  • PARI
    a(n) = (n-2/3)*2^n -n/2 +3/4 -(-1)^n/12 \\ G. C. Greubel, May 08 2018
    

Formula

a(n) = (n - 2/3)*2^n - n/2 + 3/4 - (-1)^n/12.
G.f.: x*(1-2*x^3)/(1+x)/((2*x-1)^2*(x-1)^2). - Maksym Voznyy (voznyy(AT)mail.ru), Aug 14 2009 [checked and corrected by R. J. Mathar, Sep 16 2009]
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