A158118 -id:A158118 - cp's OEIS frontend

A063865 Number of solutions to +- 1 +- 2 +- 3 +- ... +- n = 0.

1, 0, 0, 2, 2, 0, 0, 8, 14, 0, 0, 70, 124, 0, 0, 722, 1314, 0, 0, 8220, 15272, 0, 0, 99820, 187692, 0, 0, 1265204, 2399784, 0, 0, 16547220, 31592878, 0, 0, 221653776, 425363952, 0, 0, 3025553180, 5830034720, 0, 0, 41931984034, 81072032060, 0, 0

Offset: 0

N. J. A. Sloane, suggested by J. H. Conway, Aug 27 2001

Number of sum partitions of the half of the n-th-triangular number by distinct numbers in the range 1 to n. Example: a(7)=8 since triangular(7)=28 and 14 = 2+3+4+5 = 1+3+4+6 = 1+2+5+6 = 3+5+6 = 7+1+2+4 = 7+3+4 = 7+2+5 = 7+1+6. - Hieronymus Fischer, Oct 20 2010
The asymptotic formula below was stated as a conjecture by Andrica & Tomescu in 2002 and proved by B. D. Sullivan in 2013. See his paper and H.-K. Hwang's review MR 2003j:05005 of the JIS paper. - Jonathan Sondow, Nov 11 2013
a(n) is the number of subsets of {1..n} whose sum is equal to the sum of their complement. See example below. - Gus Wiseman, Jul 04 2019

			From _Gus Wiseman_, Jul 04 2019: (Start)
For example, the a(0) = 1 through a(8) = 14 subsets (empty columns not shown) are:
  {}  {3}    {1,4}  {1,6,7}    {3,7,8}
      {1,2}  {2,3}  {2,5,7}    {4,6,8}
                    {3,4,7}    {5,6,7}
                    {3,5,6}    {1,2,7,8}
                    {1,2,4,7}  {1,3,6,8}
                    {1,2,5,6}  {1,4,5,8}
                    {1,3,4,6}  {1,4,6,7}
                    {2,3,4,5}  {2,3,5,8}
                               {2,3,6,7}
                               {2,4,5,7}
                               {3,4,5,6}
                               {1,2,3,4,8}
                               {1,2,3,5,7}
                               {1,2,4,5,6}
(End)

T. D. Noe, N. J. A. Sloane and Ray Chandler, Table of n, a(n) for n = 0..3339 (terms < 10^1000, first 101 terms from T. D. Noe, next 300 terms from N. J. A. Sloane)
Dorin Andrica and Ovidiu Bagdasar, On k-partitions of multisets with equal sums, The Ramanujan J. (2021) Vol. 55, 421-435.
Dorin Andrica, Ovidiu Bagdasar, and George Cătălin Ţurcaş, The Number of Partitions of a Set and Superelliptic Diophantine Equations, Disc. Math. and Applications, Springer, Cham (2020), 35-55.
D. Andrica and E. J. Ionascu, Variations on a result of Erdős and Surányi, INTEGERS 2013 slides.
D. Andrica and I. Tomescu, On an Integer Sequence Related to a Product of Trigonometric Functions, and Its Combinatorial Relevance, J. Integer Seq., 5 (2002), Article 02.2.4
Ovidiu Bagdasar and Dorin Andrica, New results and conjectures on 2-partitions of multisets, 2017 7th International Conference on Modeling, Simulation, and Applied Optimization (ICMSAO).
Steven R. Finch, Signum equations and extremal coefficients, February 7, 2009. [Cached copy, with permission of the author]
B. D. Sullivan, On a Conjecture of Andrica and Tomescu, J. Int. Sequences, 16 (2013), Article 13.3.1.
zbMATH, Review of Andrica and Tomescu

Cf. A000980, A025591, A058377, A063866, A063867, A113036, A113037, A141000.
"Decimations": A060468 = 2*A060005, A123117 = 2*A104456.
Analogous sequences for sums of squares and cubes are A158092, A158118, see also A019568. - Pietro Majer, Mar 15 2009
Cf. A053632, A059529, A326173, A326174, A326175.

M:=400; t1:=1; lprint(0,1); for n from 1 to M do t1:=expand(t1*(x^n+1/x^n)); lprint(n, coeff(t1,x,0)); od: # N. J. A. Sloane, Jul 07 2008

f[n_, s_] := f[n, s]=Which[n==0, If[s==0, 1, 0], Abs[s]>(n*(n+1))/2, 0, True, f[ n-1, s-n]+f[n-1, s+n]]; a[n_] := f[n, 0]
nmax = 50; d = {1}; a1 = {};
Do[
  i = Ceiling[Length[d]/2];
  AppendTo[a1, If[i > Length[d], 0, d[[i]]]];
  d = PadLeft[d, Length[d] + 2 n] + PadRight[d, Length[d] + 2 n];
  , {n, nmax}];
a1 (* Ray Chandler, Mar 13 2014 *)

a(n)=my(x='x); polcoeff(prod(k=1,n,x^k+x^-k)+O(x),0) \\ Charles R Greathouse IV, May 18 2015

a(n)=0^n+floor(prod(k=1,n,2^(n*k)+2^(-n*k)))%(2^n) \\ Tani Akinari, Mar 09 2016

Asymptotic formula: a(n) ~ sqrt(6/Pi)*n^(-3/2)*2^n for n = 0 or 3 (mod 4) as n approaches infinity.
a(n) = 0 unless n == 0 or 3 (mod 4).
a(n) = constant term in expansion of Product_{ k = 1..n } (x^k + 1/x^k). - N. J. A. Sloane, Jul 07 2008
If n = 0 or 3 (mod 4) then a(n) = coefficient of x^(n(n+1)/4) in Product_{k=1..n} (1+x^k). - D. Andrica and I. Tomescu.
a(n) = 2*A058377(n) for any n > 0. - Rémy Sigrist, Oct 11 2017

More terms from Dean Hickerson, Aug 28 2001

Corrected and edited by Steven Finch, Feb 01 2009

A158092 Number of solutions to +- 1 +- 2^2 +- 3^2 +- 4^2 +- ... +- n^2 = 0.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 2, 2, 0, 0, 2, 10, 0, 0, 86, 114, 0, 0, 478, 860, 0, 0, 5808, 10838, 0, 0, 55626, 100426, 0, 0, 696164, 1298600, 0, 0, 7826992, 14574366, 0, 0, 100061106, 187392994, 0, 0, 1223587084, 2322159814, 0, 0, 16019866270, 30353305134, 0, 0

Offset: 1

Pietro Majer, Mar 12 2009

nonn

Twice A083527.
Number of partitions of the half of the n-th-square-pyramidal number into parts that are distinct square numbers in the range 1 to n^2. Example: a(7)=2 since, squarePyramidal(7)=140 and 70=1+4+16+49=9+25+36. - Hieronymus Fischer, Oct 20 2010
Erdős & Surányi prove that this sequence is unbounded. More generally, there are infinitely many ways to write a given number k as such a sum. - Charles R Greathouse IV, Nov 05 2012
The expansion and integral representation formulas below are due to Andrica & Tomescu. The asymptotic formula is a conjecture; see Andrica & Ionascu. - Jonathan Sondow, Nov 11 2013

			For n=8 the a(8)=2 solutions are: +1-4-9+16-25+36+49-64=0 and -1+4+9-16+25-36-49+64=0.

Alois P. Heinz and Ray Chandler, Table of n, a(n) for n = 1..500 (first 240 terms from Alois P. Heinz)
D. Andrica and E. J. Ionascu, Variations on a result of Erdős and Surányi, INTEGERS 2013 slides.
Dorin Andrica and Ioan Tomescu, On an Integer Sequence Related to a Product of Trigonometric Functions, and Its Combinatorial Relevance, J. Integer Sequences, 5 (2002), Article 02.2.4.
P. Erdős and J. Surányi, Egy additív számelméleti probléma (in Hungarian; Russian and German summaries), Mat. Lapok 10 (1959), pp. 284-290.

Cf. A063865, A158118, A019568, A083527, A231015.

From Pietro Majer, Mar 15 2009: (Start)
N:=60: p:=1: a:=[]: for n from 1 to N do p:=expand(p*(x^(n^2)+x^(-n^2))):
a:=[op(a), coeff(p, x, 0)]: od:a; (End)
# second Maple program:
b:= proc(n, i) option remember; local m; m:= (1+(3+2*i)*i)*i/6;
      `if`(n>m, 0, `if`(n=m, 1, b(abs(n-i^2), i-1) +b(n+i^2, i-1)))
    end:
a:= n-> `if`(irem(n-1, 4)<2, 0, 2*b(n^2, n-1)):
seq(a(n), n=1..60);  # Alois P. Heinz, Nov 05 2012

b[n_, i_] := b[n, i] = With[{m = (1+(3+2*i)*i)*i/6}, If[n>m, 0, If[n == m, 1, b[ Abs[n-i^2], i-1] + b[n+i^2, i-1]]]]; a[n_] := If[Mod[n-1, 4]<2, 0, 2*b[n^2, n-1]]; Table[a[n], {n, 1, 60}] (* Jean-François Alcover, Mar 13 2015, after Alois P. Heinz *)

a(n)=2*sum(i=0,2^(n-1)-1,sum(j=1,n-1,(-1)^bittest(i,j-1)*j^2)==n^2) \\ Charles R Greathouse IV, Nov 05 2012

from itertools import count, islice
from collections import Counter
def A158092_gen(): # generator of terms
    ccount = Counter({0:1})
    for i in count(1):
        bcount = Counter()
        for a in ccount:
            bcount[a+(j:=i**2)] += ccount[a]
            bcount[a-j] += ccount[a]
        ccount = bcount
        yield(ccount[0])
A158092_list = list(islice(A158092_gen(),20)) # Chai Wah Wu, Jan 29 2024

Constant term in the expansion of (x + 1/x)(x^4 + 1/x^4)..(x^n^2 + 1/x^n^2).
a(n)=0 for any n == 1 or 2 (mod 4).
Integral representation: a(n)=((2^n)/pi)*int_0^pi prod_{k=1}^n cos(x*k^2) dx
Asymptotic formula: a(n) = (2^n)*sqrt(10/(pi*n^5))*(1+o(1)) as n-->infty; n == -1 or 0 (mod 4).
a(n) = 2 * A083527(n). - T. D. Noe, Mar 12 2009
min{n : a(n) > 0} = A231015(0) = 7. - Jonathan Sondow, Nov 06 2013

a(51)-a(56) from R. H. Hardin, Mar 12 2009

Edited by N. J. A. Sloane, Sep 15 2009

A158380 Number of solutions to +-1 +- 3 +- 6 +- ... +- n(n+1)/2 = 0.

Original entry on oeis.org

1, 0, 0, 0, 2, 0, 2, 2, 4, 0, 12, 16, 26, 0, 66, 104, 210, 0, 620, 970, 1748, 0, 5948, 10480, 18976, 0, 60836, 111430, 209460, 0, 704934, 1284836, 2387758, 0, 8331820, 15525814, 28987902, 0, 101242982, 190267598, 358969426, 0, 1275032260, 2404124188, 4547419694

Offset: 0

Pietro Majer, Mar 17 2009

Equivalently, number of partitions of the set of the first n triangular numbers {t(1),...,t(n)} into two classes with equal sums.
Constant term in the expansion of (x + 1/x)(x^3 + 1/x^3)...(x^t(n) + 1/x^t(n)).
a(n) = 0 for all n == 1 (mod 4).
Andrica & Tomescu give a more general integral formula than the one below. - Jonathan Sondow, Nov 11 2013

			For n=6 the 2 solutions are +1-3+6-10-15+21 = 0 and -1+3-6+10+15-21 = 0.

Ray Chandler, Table of n, a(n) for n = 0..633
Dorin Andrica and Ioan Tomescu, On an Integer Sequence Related to a Product of Trigonometric Functions, and Its Combinatorial Relevance, J. Integer Sequences, 5 (2002), Article 02.2.4.

Cf. A058498, A063865, A158092, A158118.

N:=70: p:=1: a:=[]: for n from 0 to N do
p:=expand(p*(x^(n*(n+1)/2)+x^(-n*(n+1)/2))):
a:=[op(a), coeff(p, x, 0)]: od:a;
# second Maple program:
b:= proc(n, i) option remember; (m-> `if`(n>m, 0,
      `if`(n=m, 1, b(abs(n-i*(i+1)/2), i-1)+
      b(n+i*(i+1)/2, i-1))))((2+(3+i)*i)*i/6)
    end:
a:= n-> `if`(irem(n, 4)=1, 0, b(0, n)):
seq(a(n), n=0..50);  # Alois P. Heinz, Sep 17 2017

a[n_] := With[{t = Table[k(k+1)/2, {k, 1, n}]}, Coefficient[Times @@ (x^t + 1/x^t), x, 0]];
Table[an = a[n]; Print["a(", n, ") = ", an]; an, {n, 0, 50}] (* Jean-François Alcover, Jun 16 2018 *)

t(k) = k*(k+1)/2;
a(n) = polcoeff(prod(k=1, n, (x^t(k)+ 1/x^t(k))), 0); \\ Michel Marcus, May 19 2015

a(n) = (2^n/Pi) * Integral_{x=0..Pi} cos(x)*cos(3x)*...*cos(n(n+1)x/2) dx.
a(n) ~ 2^(n+1)*sqrt(10/Pi)*n^(-5/2)*(1+o(1)) as n --> infinity, n !== 1 (mod 4).
a(n) = 2 * A058498(n) for n > 0. - Alois P. Heinz, Nov 01 2011

a(0) = 1 prepended by Joerg Arndt, Sep 17 2017

Example corrected by Ilya Gutkovskiy, Feb 02 2022

A158465 Number of solutions to +-1+-2^4+-3^4+-4^4...+-n^4=0.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 16, 18, 0, 0, 32, 100, 0, 0, 424, 510, 0, 0, 2792, 5988, 0, 0, 29058, 45106, 0, 0, 276828, 473854, 0, 0, 2455340, 4777436, 0, 0, 27466324, 46429640, 0, 0, 280395282, 526489336, 0, 0, 3193589950, 5661226928, 0, 0

Offset: 1

Pietro Majer, Mar 19 2009

nonn

Constant term in the expansion of (x + 1/x)(x^16 + 1/x^16)..(x^n^4 + 1/x^n^4).
a(n)=0 for any n=1 (mod 4) or n=2 (mod 4).
Andrica & Tomescu give a more general integral formula than the one below. The asymptotic formula below is a conjecture by Andrica & Ionascu; it remains unproven. - Jonathan Sondow, Nov 11 2013

			For n=16 the a(16) = 2 solutions are +1 +16 +81 +256 -625 -1296 -2401 +4096 +6561 +10000 +14641 +20736 -28561 -38416 -50625 +65536 = 0 and the opposite.

D. Andrica and E. J. Ionascu, Variations on a result of Erdős and SurányiINTEGERS 2013 slides.
Dorin Andrica and Ioan Tomescu, On an Integer Sequence Related to a Product of Trigonometric Functions, and Its Combinatorial Relevance, J. Integer Sequences, 5 (2002), Article 02.2.4.

Cf. A063865, A158092, A158118, A158380, A019568.

A111253(n) = a(n)/2. - Alois P. Heinz, Oct 31 2011

N:=32: p:=1 a:=[]: for n from 32 to N do p:=expand
(p*(x^(n^4)+x^(-n^4))): a:=[op(a), coeff(p,x,0)]: od:a;

Integral representation: a(n) = ((2^n)/Pi)*int_0^pi prod_{k=1}^n cos(x*k^4) dx.

Asymptotic formula: a(n) = (2^n)*sqrt(18/(Pi*n^9))*(1+o(1)) as n->infinity; n=-1 or 0 (mod 4).

a(35)-a(58) from Alois P. Heinz, Oct 31 2011

A348892 Number of solutions to +-1^3 +- 2^3 +- 3^3 +- ... +- n^3 = n.

Original entry on oeis.org

1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 2, 4, 0, 0, 83, 69, 0, 0, 353, 414, 0, 0, 7800, 12496, 0, 0, 48162, 56870, 0, 0, 733392, 1253467, 0, 0, 4892337, 10022277, 0, 0, 45859303, 149422926, 0, 0, 623257759, 1339056922, 0, 0, 7453502893, 13446831198

Offset: 0

Ilya Gutkovskiy, Jan 28 2022

nonn

Cf. A063890, A158118, A348165.

from functools import lru_cache
@lru_cache(maxsize=None)
def b(n, i):
    if n > (i*(i+1)//2)**2: return 0
    if i == 0: return 1
    return b(n+i**3, i-1) + b(abs(n-i**3), i-1)
def a(n): return b(n, n)
print([a(n) for n in range(54)]) # Michael S. Branicky, Jan 28 2022

a(n) = [x^n] Product_{k=1..n} (x^(k^3) + 1/x^(k^3)).

A350861 Number of solutions to +-1^3 +- 2^3 +- 3^3 +- ... +- n^3 = 0 or 1.

Original entry on oeis.org

1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 2, 4, 1, 4, 2, 6, 1, 4, 124, 12, 344, 536, 712, 1140, 713, 4574, 2260, 4384, 5956, 10634, 73758, 48774, 197767, 406032, 638830, 1147500, 1097442, 4249160, 3263500, 6499466, 11844316, 21907736, 82561050, 85185855, 261696060

Offset: 0

Ilya Gutkovskiy, Jan 19 2022

nonn

			a(12) = 2: +1^3 + 2^3 - 3^3 + 4^3 - 5^3 - 6^3 - 7^3 + 8^3 + 9^3 - 10^3 - 11^3 + 12^3 = -1^3 - 2^3 + 3^3 - 4^3 + 5^3 + 6^3 + 7^3 - 8^3 - 9^3 + 10^3 + 11^3 - 12^3 = 0.

Robert Israel, Table of n, a(n) for n = 0..99

Cf. A000578, A025591, A113263, A158118.

f:= proc(n) local S,k,x,s;
  S:= mul(1 + x^(2*k^3),k=1..n);
  s:= sum(k^3,k=1..n);
  coeff(S,x,s) + coeff(S,x,s+1)
end proc:
map(f, [$0..50]); # Robert Israel, Mar 15 2023

from functools import lru_cache
@lru_cache(maxsize=None)
def b(n, i):
    if n > (i*(i+1)//2)**2: return 0
    if i == 0: return 1
    return b(n+i**3, i-1) + b(abs(n-i**3), i-1)
def a(n): return b(0, n) + b(1, n)
print([a(n) for n in range(46)]) # Michael S. Branicky, Jan 19 2022

A368845 Number of solutions to +- 1^3 +- 2^3 +- 3^3 +- ... +- n^3 = n^3.

Original entry on oeis.org

1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 6, 4, 0, 0, 8, 187, 0, 0, 831, 1086, 0, 0, 7127, 3983, 0, 0, 20086, 120445, 0, 0, 674006, 1056938, 0, 0, 6983613, 5964500, 0, 0, 40031490, 142694311, 0, 0, 853687222, 1622335105, 0, 0, 10288998770, 12509111104

Offset: 0

Ilya Gutkovskiy, Jan 22 2024

nonn

Cf. A063890, A158118, A348892.

b:= proc(n, i) option remember; (m-> `if`(n>m, 0, `if`(n=m, 1,
      b(abs(n-i^3), i-1) +b(n+i^3, i-1))))((i*(i+1)/2)^2)
    end:
a:= n-> `if`(irem(n, 4)>1, 0, b(n^3, n)):
seq(a(n), n=0..53);  # Alois P. Heinz, Jan 22 2024

a(n) = [x^(n^3)] Product_{k=1..n} (x^(k^3) + 1/x^(k^3)).

A369345 a(n) is the constant term in expansion of Product_{k=1..n} (x^(k^3) + 1 + 1/x^(k^3)).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 3, 3, 5, 17, 31, 61, 139, 309, 701, 1651, 3849, 8929, 22295, 53777, 131025, 335619, 837999, 2107947, 5484373, 14071891, 36275323, 95881995, 250956301, 659257445, 1763642977, 4685724391, 12496708267, 33766814039, 90846586161, 245197523769

Offset: 0

Ilya Gutkovskiy, Jan 20 2024

nonn

All terms are odd.

Cf. A000578, A007576, A158118, A350249.

b:= proc(n, i) option remember; `if`(n>(i*(i+1)/2)^2, 0,
     `if`(i=0, 1, b(n, i-1)+b(n+i^3, i-1)+b(abs(n-i^3), i-1)))
    end:
a:= n-> b(0, n):
seq(a(n), n=0..35);  # Alois P. Heinz, Jan 21 2024

Table[Coefficient[Product[x^(k^3) + 1 + 1/x^(k^3), {k, 1, n}], x, 0], {n, 0, 33}]

A369731 Number of solutions to +- 1^3 +- 2^3 +- 3^3 +- ... +- n^3 = 1.

Original entry on oeis.org

0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 4, 1, 0, 0, 6, 1, 0, 0, 12, 344, 0, 0, 1140, 713, 0, 0, 4384, 5956, 0, 0, 48774, 197767, 0, 0, 1147500, 1097442, 0, 0, 6499466, 11844316, 0, 0, 85185855, 261696060, 0, 0, 1649383741, 2039067290, 0, 0, 13301106607, 25603704324

Offset: 0

Ilya Gutkovskiy, Jan 30 2024

nonn

Cf. A000578, A063866, A158118, A348892, A350861, A369730, A369732.

Table[Coefficient[Product[(x^(k^3) + 1/x^(k^3)), {k, 1, n}], x, 1], {n, 0, 50}]

a(n) = [x^1] Product_{k=1..n} (x^(k^3) + 1/x^(k^3)).

A292522 Number of solutions to +- 1^3 +- 3^3 +- 5^3 +- 7^3 +- ... +- (4*n-1)^3 = 0.

Original entry on oeis.org

1, 0, 0, 0, 0, 0, 0, 0, 2, 6, 2, 10, 118, 88, 254, 3308, 2558, 9578, 84568, 121804, 496396, 3312400, 5755724, 19021024, 116780256, 241754350, 883730786, 4923089216, 11668601596, 42357336066, 205859270250, 538878582526, 1974181071852, 9194146886086, 26277093562150

Offset: 0

Seiichi Manyama, Sep 18 2017

nonn

			For n=8 the 2 solutions are
+1^3-3^3-5^3+7^3-9^3+11^3+13^3-15^3-17^3+19^3+21^3-23^3+25^3-27^3-29^3+31^3 = 0 and
-1^3+3^3+5^3-7^3+9^3-11^3-13^3+15^3+17^3-19^3-21^3+23^3-25^3+27^3+29^3-31^3 = 0.

Cf. A158118, A292476, A292496.

Constant term in the expansion of Product_{k=1..2*n} (x^((2*k-1)^3)+1/x^((2*k-1)^3)).

a(29)-a(34) from Alois P. Heinz, Sep 18 2017

cp's OEIS Frontend

A063865 Number of solutions to +- 1 +- 2 +- 3 +- ... +- n = 0.

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A158092 Number of solutions to +- 1 +- 2^2 +- 3^2 +- 4^2 +- ... +- n^2 = 0.

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A158380 Number of solutions to +-1 +- 3 +- 6 +- ... +- n(n+1)/2 = 0.

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A158465 Number of solutions to +-1+-2^4+-3^4+-4^4...+-n^4=0.

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A348892 Number of solutions to +-1^3 +- 2^3 +- 3^3 +- ... +- n^3 = n.

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A350861 Number of solutions to +-1^3 +- 2^3 +- 3^3 +- ... +- n^3 = 0 or 1.

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A368845 Number of solutions to +- 1^3 +- 2^3 +- 3^3 +- ... +- n^3 = n^3.

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