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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A175886 Numbers that are congruent to {1, 12} mod 13.

Original entry on oeis.org

1, 12, 14, 25, 27, 38, 40, 51, 53, 64, 66, 77, 79, 90, 92, 103, 105, 116, 118, 129, 131, 142, 144, 155, 157, 168, 170, 181, 183, 194, 196, 207, 209, 220, 222, 233, 235, 246, 248, 259, 261, 272, 274, 285, 287, 298, 300, 311, 313, 324, 326, 337, 339, 350
Offset: 1

Views

Author

Bruno Berselli, Oct 08 2010 - Nov 17 2010

Keywords

Comments

Cf. property described by Gary Detlefs in A113801: more generally, these numbers are of the form (2*h*n+(h-4)*(-1)^n-h)/4 (h, n natural numbers), therefore ((2*h*n+(h-4)*(-1)^n-h)/4)^2-1 == 0 (mod h); in this case, a(n)^2-1 == 0 (mod 13).

Crossrefs

Programs

  • Haskell
    a175886 n = a175886_list !! (n-1)
    a175886_list = 1 : 12 : map (+ 13) a175886_list
    -- Reinhard Zumkeller, Jan 07 2012
    
  • Magma
    [n: n in [1..350] | n mod 13 in [1, 12]]; // Bruno Berselli, Feb 29 2012
    
  • Magma
    [(26*n+9*(-1)^n-13)/4: n in [1..55]]; // Vincenzo Librandi, Aug 19 2013
    
  • Mathematica
    Select[Range[1, 350], MemberQ[{1, 12}, Mod[#, 13]]&] (* Bruno Berselli, Feb 29 2012 *)
    CoefficientList[Series[(1 + 11 x + x^2) / ((1 + x) (1 - x)^2), {x, 0, 55}], x] (* Vincenzo Librandi, Aug 19 2013 *)
    LinearRecurrence[{1,1,-1},{1,12,14},60] (* Harvey P. Dale, Oct 23 2015 *)
  • PARI
    a(n)=(26*n+9*(-1)^n-13)/4 \\ Charles R Greathouse IV, Sep 24 2015

Formula

G.f.: x*(1+11*x+x^2)/((1+x)*(1-x)^2).
a(n) = (26*n+9*(-1)^n-13)/4.
a(n) = -a(-n+1) = a(n-1)+a(n-2)-a(n-3).
a(n) = a(n-2)+13.
a(n) = 13*A000217(n-1)+1 - 2*Sum_{i=1..n-1} a(i) for n>1.
Sum_{n>=1} (-1)^(n+1)/a(n) = (Pi/13)*cot(Pi/13). - Amiram Eldar, Dec 04 2021
E.g.f.: 1 + ((26*x - 13)*exp(x) + 9*exp(-x))/4. - David Lovler, Sep 04 2022
From Amiram Eldar, Nov 25 2024: (Start)
Product_{n>=1} (1 - (-1)^n/a(n)) = 2*cos(Pi/13).
Product_{n>=2} (1 + (-1)^n/a(n)) = (Pi/13)*cosec(Pi/13). (End)

A175887 Numbers that are congruent to {1, 14} mod 15.

Original entry on oeis.org

1, 14, 16, 29, 31, 44, 46, 59, 61, 74, 76, 89, 91, 104, 106, 119, 121, 134, 136, 149, 151, 164, 166, 179, 181, 194, 196, 209, 211, 224, 226, 239, 241, 254, 256, 269, 271, 284, 286, 299, 301, 314, 316, 329, 331, 344, 346, 359, 361, 374, 376, 389, 391, 404
Offset: 1

Views

Author

Bruno Berselli, Oct 08 2010 - Nov 17 2010

Keywords

Comments

Cf. property described by Gary Detlefs in A113801: more generally, these numbers are of the form (2*h*n+(h-4)*(-1)^n-h)/4 (h, n natural numbers), therefore ((2*h*n+(h-4)*(-1)^n-h)/4)^2-1==0 (mod h); in this case, a(n)^2-1==0 (mod 15).

Crossrefs

Programs

  • Haskell
    a175887 n = a175887_list !! (n-1)
    a175887_list = 1 : 14 : map (+ 15) a175887_list
    -- Reinhard Zumkeller, Jan 07 2012
    
  • Magma
    [n: n in [1..450] | n mod 15 in [1,14]];
    
  • Magma
    [(30*n+11*(-1)^n-15)/4: n in [1..55]]; // Vincenzo Librandi, Aug 19 2013
    
  • Mathematica
    Select[Range[1, 450], MemberQ[{1,14}, Mod[#, 15]]&]
    CoefficientList[Series[(1 + 13 x + x^2) / ((1 + x) (1 - x)^2), {x, 0, 55}], x] (* Vincenzo Librandi, Aug 19 2013 *)
  • PARI
    a(n)=(30*n+11*(-1)^n-15)/4 \\ Charles R Greathouse IV, Sep 28 2015

Formula

G.f.: x*(1+13*x+x^2)/((1+x)*(1-x)^2).
a(n) = (30*n+11*(-1)^n-15)/4.
a(n) = -a(-n+1) = a(n-1)+a(n-2)-a(n-3).
a(n) = 15*A000217(n-1) -2*sum(a(i), i=1..n-1) +1 for n>1.
a(n) = A047209(A047225(n+1)).
Sum_{n>=1} (-1)^(n+1)/a(n) = (Pi/15)*cot(Pi/15) = A019693 * A019976 / 10. - Amiram Eldar, Dec 04 2021
E.g.f.: 1 + ((30*x - 15)*exp(x) + 11*exp(-x))/4. - David Lovler, Sep 05 2022
From Amiram Eldar, Nov 25 2024: (Start)
Product_{n>=1} (1 - (-1)^n/a(n)) = (Pi/15)*cosec(Pi/15).
Product_{n>=2} (1 + (-1)^n/a(n)) = 2*cos(Pi/15). (End)

A094550 Numbers n such that there are integers a < b with a+(a+1)+...+(n-1) = (n+1)+(n+2)+...+b.

Original entry on oeis.org

4, 6, 9, 11, 14, 15, 16, 17, 19, 21, 22, 23, 24, 25, 26, 29, 30, 31, 32, 34, 35, 36, 38, 39, 40, 41, 43, 44, 46, 48, 49, 50, 51, 52, 53, 54, 56, 57, 59, 61, 64, 66, 68, 69, 70, 71, 72, 74, 76, 77, 79, 81, 82, 83, 84, 86, 87, 89, 91, 93, 94, 95, 96, 97, 98, 99, 100, 101, 104
Offset: 1

Views

Author

T. D. Noe, May 10 2004

Keywords

Comments

Liljestrom shows that n is in this sequence if and only if 4n^2+1 is composite.
Complement of A001912.
From Hermann Stamm-Wilbrandt, Sep 16 2014: (Start)
For n > 1, A047209 is a subset of this sequence [ 4*n^2+1 is divisible by 5 if n is (1 or 4) mod 5].
A092464 is a subset of this sequence [4*n^2+1 is divisible by 13 if n is (4 or 9) mod 13].
The above are for divisibility by 5, 13; notation (1,4,5), (4,9,13). Divisibility by p for a and p-a; notation (a,p-a,p). These are the next tuples: (2,15,17), (6,23,29), (3,34,37), (16,25,41), ... . The corresponding sequences are a subset of this sequence [ 2,15,19,32,36,49,... for (2,15,17) ]. These sequences have no entries in the OEIS yet. For any prime of the form 4*k+1 there is exactly one of these tuples/sequences [solution to 4*a^2+1=0 (mod p)].
For n>1, A000290 (squares) is a subset of this sequence (4,9,16,25,...) [ 4*(m^2)^2+1 is divisible by m^2+(m+1)^2, tuple (m^2, (m+1)^2, m^2+(m+1)^2) ].
(End)

Examples

			6 is in this sequence because 1+2+3+4+5 = 7+8.
		

Crossrefs

Programs

  • Magma
    [n: n in [1..100] |not IsPrime(4*n^2 + 1)]; // Vincenzo Librandi, Sep 27 2012
  • Mathematica
    lst={}; Do[i1=n-1; i2=n+1; s1=i1; s2=i2; While[i1>1 && s1!=s2, If[s1T. D. Noe, Nov 12 2010 *)

A352324 Decimal expansion of 4*Pi / (5*sqrt(10-2*sqrt(5))).

Original entry on oeis.org

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

Views

Author

Bernard Schott, Mar 12 2022

Keywords

Comments

Cauchy's residue theorem implies that Integral_{x=0..oo} 1/(1 + x^m) dx = (Pi/m) * csc(Pi/m); this is the case m = 5.
The area of a circle circumscribing a unit-area regular decagon.

Examples

			1.0689593321155951134251843725068826399014509252665...
		

References

  • Jean-François Pabion, Éléments d'Analyse Complexe, licence de Mathématiques, page 111, Ellipses, 1995.

Crossrefs

Integral_{x=0..oo} 1/(1+x^m) dx: A019669 (m=2), A248897 (m=3), A093954 (m=4), this sequence (m=5), A019670 (m=6), A352125 (m=8), A094888 (m=10).

Programs

  • Maple
    evalf(4*Pi / (5*(sqrt(10-2sqrt(5)))), 100);
  • Mathematica
    First[RealDigits[N[4Pi/(5Sqrt[10-2Sqrt[5]]), 93]]] (* Stefano Spezia, Mar 12 2022 *)

Formula

Equals Integral_{x=0..oo} 1/(1 + x^5) dx.
Equals (Pi/5) *csc(Pi/5).
Equals (1/2) * A019694 * A121570.
Equals 1/Product_{k>=1} (1 - 1/(5*k)^2). - Amiram Eldar, Mar 12 2022
Equals Product_{k>=2} (1 + (-1)^k/A047209(k)). - Amiram Eldar, Nov 22 2024
Equals 1/A371604 = A377405/5. - Hugo Pfoertner, Nov 22 2024

A358244 Number of n-regular, N_0-weighted pseudographs on 2 vertices with total edge weight 4, up to isomorphism.

Original entry on oeis.org

1, 6, 13, 27, 38, 55, 67, 85, 97, 115, 127, 145, 157, 175, 187, 205, 217, 235, 247, 265, 277, 295, 307, 325, 337, 355, 367, 385, 397, 415, 427, 445, 457, 475, 487, 505, 517, 535, 547, 565, 577, 595, 607, 625, 637, 655, 667, 685, 697, 715, 727, 745, 757, 775
Offset: 1

Views

Author

Lars Göttgens, Nov 04 2022

Keywords

Comments

Pseudographs are finite graphs with undirected edges without identity, where parallel edges between the same vertices and loops are allowed.

Examples

			For n = 2 the a(2) = 6 such pseudographs are:
  1. two vertices connected by a 4-edge and a 0-edge,
  2. two vertices connected by a 3-edge and a 1-edge,
  3. two vertices connected by two 2-edges,
  4. two vertices where one has a 4-loop and the other one has a 0-loop,
  5. two vertices where one has a 3-loop and the other one has a 1-loop,
  6. two vertices with a 2-loop each.
		

Crossrefs

Other total edge weights: A358243 (3), A358245 (5), A358246 (6), A358247 (7), A358248 (8), A358249 (9).
Cf. A047209.

Programs

  • Julia
    using Combinatorics
    function A(n::Int)
        sum_total = 4
        result = 0
        for num_loops in 0:div(n, 2)
            num_cross = n - 2 * num_loops
            for sum_cross in 0:sum_total
                for sum_loop1 in 0:sum_total-sum_cross
                    sum_loop2 = sum_total - sum_cross - sum_loop1
                    if sum_loop2 == sum_loop1
                        result +=
                            div(
                                npartitions_with_zero(sum_loop2, num_loops) *
                                (npartitions_with_zero(sum_loop2, num_loops) + 1),
                                2,
                            ) * npartitions_with_zero(sum_cross, num_cross)
                    elseif sum_loop2 > sum_loop1
                        result +=
                            npartitions_with_zero(sum_loop2, num_loops) *
                            npartitions_with_zero(sum_loop1, num_loops) *
                            npartitions_with_zero(sum_cross, num_cross)
                    end
                end
            end
        end
        return result
    end
    function npartitions_with_zero(n::Int, m::Int)
        if m == 0
            if n == 0
                return 1
            else
                return 0
            end
        else
            return Combinatorics.npartitions(n + m, m)
        end
    end
    print([A(n) for n in 1:54])

Formula

Apparently a(n) = 6*A047209(n-2) + 1 for n >= 6, i.e., terms satisfy the linear recurrence a(n) = a(n-1) + a(n-2) - a(n-3) for n >= 9. - Hugo Pfoertner, Dec 02 2022

A090772 Numbers that are congruent to {2, 8} mod 10.

Original entry on oeis.org

2, 8, 12, 18, 22, 28, 32, 38, 42, 48, 52, 58, 62, 68, 72, 78, 82, 88, 92, 98, 102, 108, 112, 118, 122, 128, 132, 138, 142, 148, 152, 158, 162, 168, 172, 178, 182, 188, 192, 198, 202, 208, 212, 218, 222, 228, 232, 238, 242, 248, 252, 258, 262, 268, 272, 278, 282
Offset: 1

Views

Author

Giovanni Teofilatto, Feb 07 2004

Keywords

Comments

Their square ends in the digit 4. - Kausthub Gudipati, Sep 08 2011
10*a(n) = 20, 80, 120, 180, 220, ... are the only numbers written in French ending in "vingt(s)". - Paul Curtz, Aug 02 2018

Crossrefs

Programs

  • Magma
    m:=50; R:=PowerSeriesRing(Integers(), m); Coefficients(R!(2*x*(1+3*x+x^2)/((1+x)*(1-x)^2))); // G. C. Greubel, Aug 08 2018
  • Mathematica
    Union@ Flatten@ Outer[Plus, {2, 8}, 10 Range[0, 28]] (* or *)
    CoefficientList[Series[2 (1 + 3x + x^2)/((1 + x) (1 - x)^2), {x, 0, 57}], x] (* Michael De Vlieger, Aug 02 2018 *)
    LinearRecurrence[{1, 1, -1}, {2, 8, 12}, 61] (* Robert G. Wilson v, Aug 08 2018 *)
  • PARI
    is(n) = #setintersect([2, 8], [n%10]) > 0 \\ Felix Fröhlich, Aug 02 2018
    
  • PARI
    Vec(2*x*(1+3*x+x^2)/((1+x)*(1-x)^2) + O(x^60)) \\ Felix Fröhlich, Aug 02 2018
    

Formula

a(n) = 2 * A047209(n).
a(n) = 10*n - a(n-1) - 10 (with a(1)=2). - Vincenzo Librandi, Nov 16 2010
G.f.: 2*x*(1+3*x+x^2)/((1+x)*(1-x)^2). - Bruno Berselli, Sep 08 2011
a(1) = 2. For n > 1, a(n) = a(n-1) + A226294(n). - Felix Fröhlich, Aug 02 2018
Sum_{n>=1} (-1)^(n+1)/a(n) = sqrt(1+2/sqrt(5))*Pi/10. - Amiram Eldar, Dec 28 2021
E.g.f.: 2 + ((10*x - 5)*exp(x) + exp(-x))/2. - David Lovler, Sep 03 2022
From Amiram Eldar, Nov 23 2024: (Start)
Product_{n>=1} (1 - (-1)^n/a(n)) = tan(3*Pi/10) (A019952).
Product_{n>=1} (1 + (-1)^n/a(n)) = cosec(2*Pi/5)/2 (= A179290 / 2). (End)

Extensions

Edited and extended by Ray Chandler, Feb 10 2004

A273375 Squares ending in digit 4.

Original entry on oeis.org

4, 64, 144, 324, 484, 784, 1024, 1444, 1764, 2304, 2704, 3364, 3844, 4624, 5184, 6084, 6724, 7744, 8464, 9604, 10404, 11664, 12544, 13924, 14884, 16384, 17424, 19044, 20164, 21904, 23104, 24964, 26244, 28224, 29584, 31684, 33124, 35344, 36864, 39204, 40804, 43264
Offset: 1

Views

Author

Vincenzo Librandi, May 24 2016

Keywords

Crossrefs

Cf. A017317 (numbers ending in 4), A017319 (cubes ending in 4).
Cf. similar sequences listed in A273373.

Programs

  • Magma
    /* By definition: */ [n^2: n in [0..200] | Modexp(n, 2, 10) eq 4];
    
  • Magma
    [(10*n+(-1)^n-5)^2/4: n in [1..50]];
  • Mathematica
    Table[(10 n + (-1)^n - 5)^2/4, {n, 1, 50}] (* or *) LinearRecurrence[{1, 2, -2, -1, 1}, {4, 64, 144, 324, 484}, 50]
    Select[Range[200]^2,Mod[#,10]==4&] (* or *) LinearRecurrence[{1,1,-1},{2,8,12},40]^2(* Harvey P. Dale, Aug 06 2017 *)

Formula

G.f.: 4*x*(1 + 15*x + 18*x^2 + 15*x^3 + x^4) /((1+x)^2*(1-x)^3).
a(n) = 4*A047209(n)^2 = (10*n + (-1)^n - 5)^2/4.
Sum_{n>=1} 1/a(n) = 2*Pi^2/(25*(5-sqrt(5))). - Amiram Eldar, Feb 16 2023
E.g.f.: (4 - 5*x + 25*x^2)*cosh(x) + (9 + 5*x + 25*x^2)*sinh(x) - 4. - Stefano Spezia, Feb 21 2025

Extensions

Edited by Bruno Berselli, May 24 2016

A008253 Coordination sequence for diamond.

Original entry on oeis.org

1, 4, 12, 24, 42, 64, 92, 124, 162, 204, 252, 304, 362, 424, 492, 564, 642, 724, 812, 904, 1002, 1104, 1212, 1324, 1442, 1564, 1692, 1824, 1962, 2104, 2252, 2404, 2562, 2724, 2892, 3064, 3242, 3424, 3612, 3804, 4002, 4204, 4412, 4624, 4842, 5064, 5292, 5524
Offset: 0

Views

Author

Keywords

References

  • Inorganic Crystal Structure Database: Collection Code 9327.

Crossrefs

Programs

  • Mathematica
    {1}~Join~Table[2 (2 + Sum[Floor[(5 k + 3)/2], {k, n - 1}]), {n, 50}] (* Alexander Adamchuk, May 23 2006, edited by Michael De Vlieger, May 31 2022 *)
  • PARI
    Vec((1 + 2*x + 4*x^2 + 2*x^3 + x^4) / ((1 - x)^3*(1 + x)) + O(x^60)) \\ Colin Barker, Mar 21 2017

Formula

G.f.: (1 + 2*x + 4*x^2 + 2*x^3 + x^4) / ((1 - x)^3*(1 + x)).
a(2*m) = 10*m^2+2, a(2*m+1) = 10*m^2+10*m+4 (N. J. A. Sloane).
Apart from first term, first differences of A007904(n). - Alexander Adamchuk, May 23 2006
a(n) = 2* ( 2 + Sum_{k=1..n-1} floor((5*k+3)/2) ). - Alexander Adamchuk, May 23 2006
From Colin Barker, Mar 21 2017: (Start)
a(n) = (5*n^2 + 4)/2 for n>0 and even.
a(n) = (5*n^2 + 3)/2 for n odd.
a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4) for n>4.
(End)

A032769 Numbers that are congruent to {0, 1, 2, 4} mod 5.

Original entry on oeis.org

0, 1, 2, 4, 5, 6, 7, 9, 10, 11, 12, 14, 15, 16, 17, 19, 20, 21, 22, 24, 25, 26, 27, 29, 30, 31, 32, 34, 35, 36, 37, 39, 40, 41, 42, 44, 45, 46, 47, 49, 50, 51, 52, 54, 55, 56, 57, 59, 60, 61, 62, 64, 65, 66, 67, 69, 70, 71, 72, 74, 75, 76, 77, 79, 80, 81, 82, 84, 85
Offset: 1

Views

Author

Patrick De Geest, May 15 1998

Keywords

Comments

Also, numbers m such that m*(m+1)*(m+2)*(m+3)*(m+4)/(m+(m+1)+(m+2)+(m+3)+(m+4)) is an integer.

Crossrefs

Programs

Formula

a(n) = (1/8)*(10*n-11+(-1)^n+2*(-1)^floor(n/2)). - Ralf Stephan, Jun 09 2005
a(n) = floor((5*n-4)/4). - Gary Detlefs, Mar 06 2010
G.f.: x^2*(1+x+2*x^2+x^3) / ( (1+x)*(1+x^2)*(x-1)^2 ). - R. J. Mathar, Oct 08 2011
From Wesley Ivan Hurt, May 30 2016: (Start)
a(n) = a(n-1) + a(n-4) - a(n-5) for n>5.
a(n) = (10*n-11+i^(2*n)+(1+i)*I^(-n)+(1-i)*i^n)/8 where i=sqrt(-1).
a(2k) = A047209(k), a(2k-1) = A047215(k). (End)
E.g.f.: (4 + sin(x) + cos(x) + (5*x - 6)*sinh(x) + 5*(x - 1)*cosh(x))/4. - Ilya Gutkovskiy, May 31 2016
Sum_{n>=2} (-1)^n/a(n) = log(5)/4 + 3*sqrt(5)*log(phi)/10 - sqrt(1-2/sqrt(5))*Pi/10, where phi is the golden ratio (A001622). - Amiram Eldar, Dec 10 2021

Extensions

Better description from Michael Somos, Jun 08 2000

A219259 Numbers k such that 25*k+1 is a square.

Original entry on oeis.org

0, 23, 27, 96, 104, 219, 231, 392, 408, 615, 635, 888, 912, 1211, 1239, 1584, 1616, 2007, 2043, 2480, 2520, 3003, 3047, 3576, 3624, 4199, 4251, 4872, 4928, 5595, 5655, 6368, 6432, 7191, 7259, 8064, 8136, 8987, 9063, 9960, 10040, 10983, 11067, 12056, 12144
Offset: 1

Views

Author

Bruno Berselli, Nov 19 2012

Keywords

Comments

Equivalently, numbers of the form m*(25*m+2), where m = 0,-1,1,-2,2,-3,3,...
Also, integer values of h*(h+2)/25.
Exponents in the expansion of Product_{n >= 1} (1 - q^(50*n))*(1 - q^(50*n-23))*(1 - q^(50*n-27)) = 1 - q^23 - q^27 + q^96 + q^104 - q^219 - q^231 + + - - .... - Peter Bala, Dec 18 2024

Crossrefs

Cf. similar sequences listed in A219257.

Programs

  • Magma
    [n: n in [0..13000] | IsSquare(25*n+1)];
    
  • Magma
    I:=[0,23,27,96,104]; [n le 5 select I[n] else Self(n-1)+2*Self(n-2)-2*Self(n-3)-Self(n-4)+Self(n-5): n in [1..50]]; // Vincenzo Librandi, Aug 18 2013
  • Maple
    A219259:=proc(q)
    local n;
    for n from 1 to q do if type(sqrt(25*n+1), integer) then print(n);
    fi; od; end:
    A219259(1000); # Paolo P. Lava, Feb 19 2013
  • Mathematica
    Select[Range[0, 13000], IntegerQ[Sqrt[25 # + 1]] &]
    CoefficientList[Series[x (23 + 4 x + 23 x^2)/((1 + x)^2 (1 - x)^3), {x, 0, 50}], x] (* Vincenzo Librandi, Aug 18 2013 *)

Formula

G.f.: x^2*(23 + 4*x + 23*x^2)/((1 + x)^2*(1 - x)^3).
a(n) = a(-n+1) = (50*n*(n-1) + 21*(-1)^n*(2*n - 1) + 5)/8 + 2.
25*a(n)+1 = A047209(A197652(n+1))^2.
Sum_{n>=2} 1/a(n) = 25/4 - cot(2*Pi/25)*Pi/2. - Amiram Eldar, Mar 17 2022
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