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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A004126 a(n) = n*(7*n^2 - 1)/6.

Original entry on oeis.org

0, 1, 9, 31, 74, 145, 251, 399, 596, 849, 1165, 1551, 2014, 2561, 3199, 3935, 4776, 5729, 6801, 7999, 9330, 10801, 12419, 14191, 16124, 18225, 20501, 22959, 25606, 28449, 31495, 34751, 38224, 41921, 45849, 50015, 54426, 59089, 64011
Offset: 0

Views

Author

Albert D. Rich (Albert_Rich(AT)msn.com)

Keywords

Comments

3-dimensional analog of centered polygonal numbers.
Sum of n triangular numbers starting from T(n), where T = A000217. E.g., a(4) = T(4) + T(5) + T(6) + T(7) = 10 + 15 + 21 + 28 = 74. - Amarnath Murthy, Jul 16 2004
Also as a(n) = (1/6)*(7*n^3-n), n>0: structured heptagonal diamond numbers (vertex structure 8). Cf. A100179 = alternate vertex; A000447 = structured diamonds; A100145 for more on structured numbers. - James A. Record (james.record(AT)gmail.com), Nov 07 2004
Partial sums of A069099, centered heptagonal numbers (A000566). - Jonathan Vos Post, Mar 16 2006
Binomial transform of (0, 1, 7, 7, 0, 0, 0, ...) and third partial sum of (0, 1, 6, 7, 7, 7, ...). - Gary W. Adamson, Oct 05 2015

References

  • E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 140.

Crossrefs

Programs

Formula

a(n) = C(2*n+1,3)-C(n+1,3), n>=0. - Zerinvary Lajos, Jan 21 2007
a(n) = A000447(n) - A000292(n). - Zerinvary Lajos, Jan 21 2007
G.f.: x*(1+5*x+x^2)/(1-x)^4. - Colin Barker, Mar 02 2012
E.g.f.: (x/6)*(7*x^2 + 21*x + 6)*exp(x). - G. C. Greubel, Oct 05 2015
a(n) = Sum_{i = n..2*n-1} A000217(i). - Bruno Berselli, Sep 06 2017
a(n) = n^3 + Sum_{k=0..n-1} k*(k+1)/2. Alternately, a(n) = A000578(n) + A000292(n-1) for n>0. - Bruno Berselli, May 23 2018

A006484 a(n) = n*(n + 1)*(n^2 - 3*n + 5)/6.

Original entry on oeis.org

0, 1, 3, 10, 30, 75, 161, 308, 540, 885, 1375, 2046, 2938, 4095, 5565, 7400, 9656, 12393, 15675, 19570, 24150, 29491, 35673, 42780, 50900, 60125, 70551, 82278, 95410, 110055, 126325, 144336, 164208, 186065, 210035, 236250, 264846, 295963, 329745, 366340
Offset: 0

Views

Author

Dennis S. Kluk (mathemagician(AT)ameritech.net)

Keywords

Comments

Structured meta-pyramidal numbers, the n-th number from an n-gonal pyramidal number sequence. - James A. Record (james.record(AT)gmail.com), Nov 07 2004
The Gi4 triangle sums of A139600 are given by the terms of this sequence. For the definitions of the Gi4 and other triangle sums see A180662. - Johannes W. Meijer, Apr 29 2011

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. other meta sequences: A100177: prism; A000447: "polar" diamond; A059722: "equatorial diamond"; A100185: anti-prism; A100188: "polar" anti-diamond; and A100189: "equatorial" anti-diamond. Cf. A100145 for more on structured numbers.
Cf. A000332.

Programs

Formula

a(n) = (1/6)*(n^4 - 2*n^3 + 2*n^2 + 5*n). - James A. Record (james.record(AT)gmail.com), Nov 07 2004
a(n) = binomial(n+3,4) - 2*binomial(n+2,4) + 5*binomial(n+1,4). - Johannes W. Meijer, Apr 29 2011

A002309 Sum of fourth powers of first n odd numbers.

Original entry on oeis.org

1, 82, 707, 3108, 9669, 24310, 52871, 103496, 187017, 317338, 511819, 791660, 1182285, 1713726, 2421007, 3344528, 4530449, 6031074, 7905235, 10218676, 13044437, 16463238, 20563863, 25443544, 31208345, 37973546, 45864027, 55014652, 65570653, 77688014
Offset: 1

Views

Author

Keywords

Examples

			a(1) = 1^4 = 1.
a(2) = 1^4 + 3^4 = 82.
a(3) = 1^4 + 3^4 + 5^4 = 707.
		

References

  • F. E. Croxton and D. J. Cowden, Applied General Statistics. 2nd ed., Prentice-Hall, Englewood Cliffs, NJ, 1955, p. 742.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • Maple
    A002309:=(1+76*z+230*z**2+76*z**3+z**4)/(z-1)**6; # conjectured by Simon Plouffe in his 1992 dissertation
  • Mathematica
    Table[(48*n^5 - 40*n^3 + 7*n)/15, {n, 0, 40}] (* Stefan Steinerberger, Apr 10 2006 *)
    s = 0; lst = {s}; Do[s += n^4; AppendTo[lst, s], {n, 1, 60, 2}]; lst (* Zerinvary Lajos, Jul 12 2009 *)
    Accumulate[Range[1,63,2]^4] (* Harvey P. Dale, Oct 24 2011 *)
    LinearRecurrence[{6,-15,20,-15,6,-1},{1,82,707,3108,9669,24310},20] (* Harvey P. Dale, Sep 29 2015 *)
  • PARI
    a(n)=(48*n^5-40*n^3+7*n)/15 \\ Charles R Greathouse IV, Apr 07 2016
    
  • Python
    def A002309(n): return n*(n**2*(6*n**2-5<<3)+7)//15 # Chai Wah Wu, Oct 02 2024

Formula

a(n) = (48*n^5 - 40*n^3 + 7*n)/15. - Ralf Stephan, Jan 29 2003
a(1)=1, a(2)=82, a(3)=707, a(4)=3108, a(5)=9669, a(6)=24310, a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6). - Harvey P. Dale, Oct 24 2011
a(n) = v(n,n-2) - v(n,n-1)*V(n,n-1), where v(n,k) and V(n,k) are the central factorial numbers of the first kind and the second kind, respectively, with odd indices. - Mircea Merca, Jan 25 2014
From Wolfdieter Lang, Mar 11 2017: (Start)
G.f.: x*(1 + 76*x + 230*x^2 + 76*x^3 + 1*x^4)/(1-x)^6.
E.g.f. (with offset 0): exp(x)*(1 + 81*x + 544*x^2/2! + 1232*x^3/3! + 1152*x^4/4! + 384*x^5/5!). (End)

Extensions

Definition changed by David A. Corneth, Mar 11 2017
Name clarified by Mohammed Yaseen, Jul 24 2023

A063522 a(n) = n*(5*n^2 - 3)/2.

Original entry on oeis.org

0, 1, 17, 63, 154, 305, 531, 847, 1268, 1809, 2485, 3311, 4302, 5473, 6839, 8415, 10216, 12257, 14553, 17119, 19970, 23121, 26587, 30383, 34524, 39025, 43901, 49167, 54838, 60929, 67455, 74431, 81872, 89793, 98209, 107135, 116586, 126577, 137123, 148239, 159940
Offset: 0

Views

Author

N. J. A. Sloane, Aug 02 2001

Keywords

Crossrefs

(1/12)*t*(n^3 - n) + n for t = 2, 4, 6, ... gives A004006, A006527, A006003, A005900, A004068, A000578, A004126, A000447, A004188, A004466, A004467, A007588, A062025, A063521, A063522, A063523.
Bisections: A160674, A160699.

Programs

  • Magma
    [n*(5*n^2 -3)/2: n in [0..30]]; // G. C. Greubel, May 02 2018
  • Mathematica
    lst={};Do[AppendTo[lst, LegendreP[3, n]], {n, 10^2}];lst (* Vladimir Joseph Stephan Orlovsky, Sep 11 2008 *)
    CoefficientList[Series[x*(1 + 13*x + x^2)/(1-x)^4, {x, 0, 50}], x] (* G. C. Greubel, Sep 01 2017 *)
    LinearRecurrence[{4,-6,4,-1},{0,1,17,63},40] (* Harvey P. Dale, Sep 06 2023 *)
  • PARI
    a(n) = { n*(5*n^2 - 3)/2 } \\ Harry J. Smith, Aug 25 2009
    

Formula

G.f.: x*(1 + 13*x + x^2)/(1-x)^4. - Colin Barker, Jan 10 2012
E.g.f.: (x/2)*(2 + 15*x + 5*x^2)*exp(x). - G. C. Greubel, Sep 01 2017

A004467 a(n) = n*(11*n^2 - 5)/6.

Original entry on oeis.org

0, 1, 13, 47, 114, 225, 391, 623, 932, 1329, 1825, 2431, 3158, 4017, 5019, 6175, 7496, 8993, 10677, 12559, 14650, 16961, 19503, 22287, 25324, 28625, 32201, 36063, 40222, 44689, 49475, 54591, 60048
Offset: 0

Views

Author

Albert D. Rich (Albert_Rich(AT)msn.com)

Keywords

Comments

3-dimensional analog of centered polygonal numbers, that is: centered hendecagonal pyramidal numbers (see Deza paper in References).

References

  • E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 140.

Crossrefs

1/12*t*(n^3-n)+n for t = 2, 4, 6, ... gives A004006, A006527, A006003, A005900, A004068, A000578, A004126, A000447, A004188, A004466, A004467, A007588, A062025, A063521, A063522, A063523.

Programs

Formula

G.f.: x*(1+9*x+x^2)/(1-x)^4. - Colin Barker, Jan 08 2012
a(0)=0, a(1)=1, a(2)=13, a(3)=47; for n>3, a(n) = 4*a(n-1)-6*a(n-2)+4*a(n-3)-a(n-4). - Harvey P. Dale, Sep 22 2013
E.g.f.: (x/6)*(6 + 33*x + 11*x^2)*exp(x). - G. C. Greubel, Sep 01 2017

A062025 a(n) = n*(13*n^2 - 7)/6.

Original entry on oeis.org

0, 1, 15, 55, 134, 265, 461, 735, 1100, 1569, 2155, 2871, 3730, 4745, 5929, 7295, 8856, 10625, 12615, 14839, 17310, 20041, 23045, 26335, 29924, 33825, 38051, 42615, 47530, 52809, 58465, 64511, 70960, 77825, 85119, 92855, 101046, 109705, 118845, 128479, 138620, 149281
Offset: 0

Views

Author

N. J. A. Sloane, Aug 02 2001

Keywords

Crossrefs

1/12*t*(n^3-n)+n for t = 2, 4, 6, ... gives A004006, A006527, A006003, A005900, A004068, A000578, A004126, A000447, A004188, A004466, A004467, A007588, A062025, A063521, A063522, A063523.

Programs

Formula

From G. C. Greubel, Sep 01 2017: (Start)
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4).
G.f.: (x + 11*x^2 + x^3)/(1 - x)^4.
E.g.f.: (x/6)*(6 + 39*x + 13*x^2)*exp(x). (End)

A063523 a(n) = n*(8*n^2 - 5)/3.

Original entry on oeis.org

0, 1, 18, 67, 164, 325, 566, 903, 1352, 1929, 2650, 3531, 4588, 5837, 7294, 8975, 10896, 13073, 15522, 18259, 21300, 24661, 28358, 32407, 36824, 41625, 46826, 52443, 58492, 64989, 71950, 79391, 87328, 95777, 104754, 114275, 124356, 135013, 146262, 158119, 170600
Offset: 0

Views

Author

N. J. A. Sloane, Aug 02 2001

Keywords

Comments

Also as a(n)=(1/6)*(16*n^3-10*n), n>0: structured octagonal anti-diamond numbers (vertex structure 17) (Cf. A100187 = alternate vertex; A100188 = structured anti-diamonds; A100145 for more on structured numbers). - James A. Record (james.record(AT)gmail.com), Nov 07 2004

Crossrefs

1/12*t*(n^3-n)+n for t = 2, 4, 6, ... gives A004006, A006527, A006003, A005900, A004068, A000578, A004126, A000447, A004188, A004466, A004467, A007588, A062025, A063521, A063522, A063523.

Programs

  • Mathematica
    Table[n(8n^2-5)/3,{n,0,80}] (* Vladimir Joseph Stephan Orlovsky, Apr 18 2011 *)
    LinearRecurrence[{4,-6,4,-1},{0,1,18,67},81] (* or *) CoefficientList[ Series[ (x+14 x^2+x^3)/(x-1)^4,{x,0,80}],x] (* Harvey P. Dale, Jul 11 2011 *)
  • PARI
    a(n) = n*(8*n^2 - 5)/3 \\ Harry J. Smith, Aug 25 2009

Formula

a(0)=0, a(1)=1, a(2)=18, a(3)=67, a(n)=4*a(n-1)-6*a(n-2)+4*a(n-3)- a(n-4). - Harvey P. Dale, Jul 11 2011
G.f.: (x+14*x^2+x^3)/(x-1)^4. - Harvey P. Dale, Jul 11 2011
E.g.f.: (x/3)*(3 + 24*x + 8*x^2)*exp(x). - G. C. Greubel, Sep 01 2017

A002593 a(n) = n^2*(2*n^2 - 1); also Sum_{k=0..n-1} (2k+1)^3.

Original entry on oeis.org

0, 1, 28, 153, 496, 1225, 2556, 4753, 8128, 13041, 19900, 29161, 41328, 56953, 76636, 101025, 130816, 166753, 209628, 260281, 319600, 388521, 468028, 559153, 662976, 780625, 913276, 1062153, 1228528, 1413721, 1619100, 1846081
Offset: 0

Views

Author

Keywords

Comments

The m-th term, for m = A065549(n), is perfect (A000396). - Lekraj Beedassy, Jun 04 2002
Partial sums of A016755. - Lekraj Beedassy, Jan 06 2004
Also, the k-th triangular number, where k = 2n^2 - 1 = A056220(n), i.e., a(n) = A000217(A056220(n)). - Lekraj Beedassy, Jun 11 2004
Also, the j-th hexagonal number, where j = n^2 = A000290(n), i.e., a(n) = A000384(A000290(n)) and a(n) = A056220(n) * A000290(n) or j * k. This sequence is a subsequence of the hexagonal number sequence and retains the aspect intrinsic to the hexagonal number sequence that each number in this sequence can be found by multiplying its triangular number by its hexagonal number. - Bruce J. Nicholson, Aug 22 2017
Odd numbers and their squares both having the form 2x-+1, we may write (2r+1)^3 = (2r+1)*(2s-1), where s = centered squares = (r+1)^2 + r^2. Since 2r+1 = (r+1)^2 - r^2, it follows immediately from summing telescopingly over n-1, the product 2*{(r+1)^4 - r^4} - {(r+1)^2 - r^2}, that Sum_{r=0..n-1} (2r+1)^3 = 2*n^4 - n^2 = n^2*(2n^2 - 1). - Lekraj Beedassy, Jun 16 2004
a(n) is also the starting term in the sum of a number M(n) of consecutive cubed integers equaling a squared integer (A253724) for M(n) equal to twice a squared integer (A001105). Numbers a(n) such that a^3 + (a+1)^3 + ... + (a+M-1)^3 = c^2 has nontrivial solutions over the integers for M equal to twice a squared integer (A001105). If M is twice a squared integer, there always exists at least one nontrivial solution for the sum of M consecutive cubed integers starting from a^3 and equaling a squared integer c^2. For n >= 1, M(n) = 2n^2 (A001105), a(n) = M(M-1)/2 = n^2(2n^2 - 1), and c(n) = sqrt(M/2) (M(M^2-1)/2) = n^3(4n^4 - 1). The trivial solutions with M < 1 and a < 2 are not considered. - Vladimir Pletser, Jan 10 2015
Binomial transform of the sequence with offset 1 is (1, 27, 98, 120, 48, 0, 0, 0, ...). - Gary W. Adamson, Jul 23 2015

References

  • Louis Comtet, Advanced Combinatorics, Reidel, 1974, p. 169, #31.
  • F. E. Croxton and D. J. Cowden, Applied General Statistics. 2nd ed., Prentice-Hall, Englewood Cliffs, NJ, 1955, p. 742.
  • L. B. W. Jolley, Summation of Series. 2nd ed., Dover, NY, 1961, p. 7.
  • Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, page 47.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • Magma
    [n^2*(2*n^2 - 1): n in [0..40]]; // Vincenzo Librandi, Sep 07 2011
    
  • Maple
    A002593:=-z*(z+1)*(z**2+22*z+1)/(z-1)**5; # conjectured by Simon Plouffe in his 1992 dissertation
    a:= n-> n^2*(2*n^2-1): seq(a(n), n=0..50);  # Vladimir Pletser, Jan 10 2015
  • Mathematica
    CoefficientList[Series[(-x^4-23x^3-23x^2-x)/(x-1)^5,{x,0, 80}],x]  (* or *)
    Table[ n^2 (2n^2-1),{n,0,80}]  (* Harvey P. Dale, Mar 28 2011 *)
    Join[{0},Accumulate[Range[1,91,2]^3]] (* or *) LinearRecurrence[{5,-10,10,-5,1},{0,1,28,153,496},40] (* Harvey P. Dale, Mar 22 2017 *)
  • PARI
    a(n) = n^2*(2*n^2 - 1) \\ Charles R Greathouse IV, Feb 07 2017

Formula

a(n) = A000217(A056220(n)). - Lekraj Beedassy, Jun 11 2004
G.f.: (-x^4 - 23*x^3 - 23*x^2 - x)/(x - 1)^5. - Harvey P. Dale, Mar 28 2011
a(n) = n^2*(2n^2 - 1). - Vladimir Pletser, Jan 10 2015
E.g.f.: exp(x)*x*(1 + 13*x + 24*x^2/2! + 12*x^3/3!). - Wolfdieter Lang, Mar 11 2017
a(n) = A000384(A000290(n)) = A056220(n) * A000290(n). - Bruce J. Nicholson, Aug 22 2017
From Amiram Eldar, Aug 25 2022: (Start)
Sum_{n>=1} 1/a(n) = 1 - Pi^2/6 - cot(Pi/sqrt(2))*Pi/sqrt(2).
Sum_{n>=1} (-1)^(n+1)/a(n) = cosec(Pi/sqrt(2))*Pi/sqrt(2) - Pi^2/12 - 1. (End)

A059722 a(n) = n*(2*n^2 - 2*n + 1).

Original entry on oeis.org

0, 1, 10, 39, 100, 205, 366, 595, 904, 1305, 1810, 2431, 3180, 4069, 5110, 6315, 7696, 9265, 11034, 13015, 15220, 17661, 20350, 23299, 26520, 30025, 33826, 37935, 42364, 47125, 52230, 57691, 63520, 69729, 76330, 83335, 90756, 98605, 106894, 115635, 124840
Offset: 0

Views

Author

Henry Bottomley, Feb 07 2001

Keywords

Comments

Mean of the first four nonnegative powers of 2n+1, i.e., ((2n+1)^0 + (2n+1)^1 + (2n+1)^2 + (2n+1)^3)/4. E.g., a(2) = (1 + 3 + 9 + 27)/4 = 10.
Equatorial structured meta-diamond numbers, the n-th number from an equatorial structured n-gonal diamond number sequence. There are no 1- or 2-gonal diamonds, so 1 and (n+2) are used as the first and second terms since all the sequences begin as such. - James A. Record (james.record(AT)gmail.com), Nov 07 2004
Starting with offset 1 = row sums of triangle A143803. - Gary W. Adamson, Sep 01 2008
Form an array from the antidiagonals containing the terms in A002061 to give antidiagonals 1; 3,3; 7,4,7; 13,8,8,13; 21,14,9,14,21; and so on. The difference between the sum of the terms in n+1 X n+1 matrices and those in n X n matrices is a(n) for n>0. - J. M. Bergot, Jul 08 2013
Sum of the numbers from (n-1)^2 to n^2. - Wesley Ivan Hurt, Sep 08 2014

Crossrefs

Cf. A000330, A005900, A081436, A100178, A100179, A059722: "equatorial" structured diamonds; A000447: "polar" structured meta-diamond; A006484 for other structured meta numbers; and A100145 for more on structured numbers. - James A. Record (james.record(AT)gmail.com), Nov 07 2004

Programs

Formula

a(n) = A053698(2*n-1)/4.
a(n) = Sum_{j=1..n} ((n+j-1)^2-j^2+1). - Zerinvary Lajos, Sep 13 2006
From R. J. Mathar, Sep 02 2008: (Start)
G.f.: x*(1 + x)*(1 + 5*x)/(1 - x)^4.
a(n) = A002414(n-1) + A002414(n).
a(n+1) - a(n) = A136392(n+1). (End)
a(n) = (A000290(n) + A000290(n+1)) * (A000217(n+1) - A000217(n)). - J. M. Bergot, Nov 02 2012
a(n) = n * A001844(n-1). - Doug Bell, Aug 18 2015
a(n) = A000217(n^2) - A000217(n^2-2*n). - Bruno Berselli, Jun 26 2018
E.g.f.: exp(x)*x*(1 + 4*x + 2*x^2). - Stefano Spezia, Jun 20 2021

Extensions

Edited with new definition by N. J. A. Sloane, Aug 29 2008

A100188 Polar structured meta-anti-diamond numbers, the n-th number from a polar structured n-gonal anti-diamond number sequence.

Original entry on oeis.org

1, 6, 27, 84, 205, 426, 791, 1352, 2169, 3310, 4851, 6876, 9477, 12754, 16815, 21776, 27761, 34902, 43339, 53220, 64701, 77946, 93127, 110424, 130025, 152126, 176931, 204652, 235509, 269730, 307551, 349216
Offset: 1

Views

Author

James A. Record (james.record(AT)gmail.com), Nov 07 2004

Keywords

Examples

			There are no 1- or 2-gonal anti-diamonds, so 1 and (2n+2) are the first and second terms since all the sequences begin as such.
		

Crossrefs

Cf. A000578, A000447, A004466, A007588, A063521, A062523 - "polar" structured anti-diamonds; A100189 - "equatorial" structured meta-anti-diamond numbers; A006484 for other structured meta numbers; and A100145 for more on structured numbers.

Programs

  • Magma
    [(1/6)*(2*n^4-2*n^2+6*n): n in [1..40]]; // Vincenzo Librandi, Aug 18 2011
    
  • Mathematica
    Table[(2n^4-2n^2+6n)/6,{n,40}] (* or *) LinearRecurrence[{5,-10,10,-5,1}, {1,6,27,84,205},40] (* Harvey P. Dale, May 11 2016 *)
  • PARI
    vector(40, n, (n^4 -n^2 +3*n)/3) \\ G. C. Greubel, Nov 08 2018

Formula

a(n) = (1/6)*(2*n^4 - 2*n^2 + 6*n).
G.f.: x*(1 + x + 7*x^2 - x^3)/(1-x)^5. - Colin Barker, Apr 16 2012
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5); a(1)=1, a(2)=6, a(3)=27, a(4)=84, a(5)=205. - Harvey P. Dale, May 11 2016
E.g.f.: (3*x + 6*x^2 + 6*x^3 + x^4)*exp(x)/3. - G. C. Greubel, Nov 08 2018
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