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A001653
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Numbers n such that 2*n^2 - 1 is a square.
(Formerly M3955 N1630)
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149
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1, 5, 29, 169, 985, 5741, 33461, 195025, 1136689, 6625109, 38613965, 225058681, 1311738121, 7645370045, 44560482149, 259717522849, 1513744654945, 8822750406821, 51422757785981, 299713796309065, 1746860020068409
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OFFSET
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1,2
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COMMENTS
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Consider all Pythagorean triples (X,X+1,Z) ordered by increasing Z; sequence gives Z values.
The defining equation is X^2 + (X+1)^2 = Z^2, which when doubled gives 2Z^2 = (2X+1)^2 + 1. So the sequence gives Z's such that 2Z^2 = odd square + 1.
(x,y)=(a(n),a(n+1)) are the solutions with x<y of x/(yz)+y/(xz)+z/(xy)=3 with z=2. - Floor van Lamoen (fvlamoen(AT)hotmail.com), Nov 29 2001
Consequently the sum n^2*(2n^2 - 1) of the first n odd cubes (A002593) is also a square. - Lekraj Beedassy, Jun 05 2002
Numbers n such that 2*n^2=ceil(sqrt(2)*n*floor(sqrt(2)*n)) Benoit Cloitre, May 10 2003
Also, number of domino tilings in S_5 X P_2n. - R. Stephan, Mar 30 2004. Here S_5 is the star graph on 5 vertices with the edges {1,2}, {1,3}, {1,4}, {1,5}.
If x is in the sequence then so is x*(8*x^2-3). - James R. Buddenhagen, Jan 13 2005
In general, sum{k=0..n, binomial(2n-k,k)j^(n-k)}=(-1)^n*U(2n,I*sqrt(j)/2), I=sqrt(-1). - Paul Barry, Mar 13 2005
a(n) = L(n,6), where L is defined as in A108299; see also A002315 for L(n,-6). - Reinhard Zumkeller, Jun 01 2005
Define a T-circle to be a first-quadrant circle with integral radius that is tangent to the x- and y-axes. Such a circle has coordinates equal to its radius. Let C(0) be the T-circle with radius 1. Then for n>0, define C(n) to be the largest T-circle that intersects C(n-1). C(n) has radius a(n) and the coordinates of its points of intersection with C(n-1) are A001108(n) and A055997(n). Cf. A001109. - Charlie Marion, Sep 14 2005
Number of 01-avoiding words of length n on alphabet {0,1,2,3,4,5} which do not end in 0. - Tanya Khovanova, Jan 10 2007
The lower principal convergents to 2^(1/2), beginning with 1/1, 7/5, 41/29, 239/169, comprise a strictly increasing sequence; numerators=A002315 and denominators=A001653. - Clark Kimberling, Aug 26 2008
Apparently Ljunggren shows that 169 is the last square term.
The remainder of the division of a(n) by 5 is: 0, 1 or 4. [From Mohamed Bouhamida (bhmd95(AT)yahoo.fr), Aug 26 2009]
Number of units of a(n) belongs to a periodic sequence: 1, 5, 9, 9, 5, 1. The remainder of the division of a(n) by 5 belongs to a periodic sequence: 1, 0, 4, 4, 0, 1. [Mohamed Bouhamida (bhmd95(AT)yahoo.fr), Sep 01 2009]
If (p,q) is a solution of the Diophantine equation: X^2 + (X+1)^2 = Y^2 then (p+q) or (p+q+1) are perfect squares. If (p,q) is a solution of the Diophantine equation: X^2 + (X+1)^2 = Y^2 then (p+q) or (p+q)/8 are perfect squares. If (p,q) and (r,s) are two consecutive solutions of the Diophantine equation: X^2 + (X+1)^2 = Y^2 with p<r then s-r=p+q+1. [Mohamed Bouhamida (bhmd95(AT)yahoo.fr), Aug 29 2009]
If (p,q) and (r,s) are two consecutive solutions of the Diophantine equation: X^2 + (X + 1)^2 = Y^2 with p<r then r=3p+2q+1 and s=4p+3q+2. [From Mohamed Bouhamida (bhmd95(AT)yahoo.fr), Sep 02 2009]
Contribution from Gary W. Adamson, Jul 22 2010: (Start)
Equals INVERT transform of A005054: (1, 4, 20, 100, 500, 2500,...) and
INVERTi transform of A122074: (1, 6, 40, 268, 1796,...). (End)
a(n) is the number of compositions of n when there are 5 types of 1 and 4 types of other natural numbers. [Milan Janjic, Aug 13 2010]
The remainder after division of a(n) by a(k) appears to belong to
a periodic sequence: 1, 5,..., a(k-1), 0, a(k)-a(k-1),...,
a(k)-1, a(k)-1,..., a(k)-a(k-1), 0, a(k-1),..., 5, 1.
See Bouhamida’s Comment, Sep 01 2009. - Charlie Marion, May 02 2011
Apart from initial 1: subsequence of A198389, see also A198385. [Reinhard Zumkeller, Oct 25 2011]
(a(n+1),2*b(n+1)) and (a(n+2),2*b(n+1)), n>=0, with b(n):= A001109(n), give the (u(2*n),v(2*n)) and (u(2*n+1),v(2*n+1)) sequences, respectively, for Pythagorean triples (x,y,z), where x=|u^2-v^2|, y=2*u*v and z=u^2+v^2, with u odd and v even, which are generated from (u(0)=1,v(0)=2) by the substitution rule (u,v) -> (2*v+u,v) if u<v and (u,v) -> (u,2*u+v) if u>v. This leads to primitive triples because gcd(u,v) = 1 is respected. This corresponds to (primitive) Pythagorean triangles with |x-y|=1 (the catheti differ by one length unit). This (u,v) sequence starts with (1,2), (5,2), (5,12), (29,12), (29,70) ... - Wolfdieter Lang, 06 Mar 2012
a(n)^2 = 2*A182435(n)*(A182435(n)-1)+1. [Bruno Berselli, Oct 23 2012]
Area of the Fibonacci snowflake of order n. [José Luis Ramírez Ramírez, Dec 13 2012]
Area of the 3-generalized Fibonacci snowflake of order n, n>=3, [José Luis Ramírez Ramírez, Dec 13 2012]
For the o.g.f. given by Johannes W. Meijer, Aug 01 2010, in the formula section see a comment under A077445. - Wolfdieter Lang, Jan 18 2013.
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REFERENCES
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I. Adler, Three Diophantine equations - Part II, Fib. Quart., 7 (1969), 181-193.
A. H. Beiler, Recreations in the Theory of Numbers. New York: Dover, pp. 122-125, 1964.
A. Blondin-Massé, S. Brlek, S. Labbé, and M. Mendès France, Fibonacci snowflakes, Special Issue dedicated to Paulo Ribenboim, Annales des Sciences Mathématiques du Québec 35, No 2 (2011).
Daniel C. Fielder, Special integer sequences controlled by three parameters, Fibonacci Quarterly 6, 1968, 64-70.
T. W. Forget and T. A. Larkin, Pythagorean triads of the form X, X+1, Z described by recurrence sequences, Fib. Quart., 6 (No. 3, 1968), 94-104.
L. J. Gerstein, Pythagorean triples and inner products, Math. Mag., 78 (2005), 205-213.
W. Ljunggren, "Zur Theorie der Gleichung x^2+1=Dy^4", Avh. Norske Vid. Akad. Oslo I. 5, 27pp.
A. Martin, Table of prime rational right-angled triangles, Math. Mag., 2 (1910), 297-324.
Problem 47, Amer. Math. Monthly, 4 (1897), 25-28.
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).
Michael Z. Spivey and Laura L. Steil, The k-Binomial Transforms and the Hankel Transform, Journal of Integer Sequences, Vol. 9 (2006), Article 06.1.1.
David Wells, The Penguin Dictionary of Curious and Interesting Numbers (Rev. ed. 1997), p. 91.
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LINKS
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T. D. Noe, Table of n, a(n) for n=1..201
J.-P. Ehrmann et al., POLYA003: Integers of the form a/(bc) + b/(ca) + c/(ab).
Tanya Khovanova, Recursive Sequences
Ron Knott, Pythagorean Triples and Online Calculators
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 403
_Simon Plouffe_, Approximations de S\'{e}ries G\'{e}n\'{e}ratrices et Quelques Conjectures, Dissertation, Universit\'{e} du Qu\'{e}bec \`{a} Montr\'{e}al, 1992.
_Simon Plouffe_, 1031 Generating Functions and Conjectures, Universit\'{e} du Qu\'{e}bec \`{a} Montr\'{e}al, 1992.
José L. Ramírez, Gustavo N. Rubiano, and Rodrigo de Castro, A Generalization of the Fibonacci Word Fractal and the Fibonacci Snowflake, arXiv preprint arXiv:1212.1368, 2012
Eric Weisstein's World of Mathematics, NSW Number
Index entries for two-way infinite sequences
Index entries for sequences related to linear recurrences with constant coefficients
Index entries for sequences related to Chebyshev polynomials.
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FORMULA
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G.f.: (1-x)/(1-6*x+x^2).
a(n) = 6*a(n-1)-a(n-2) with a(0)=1, a(1)=5.
a(-1-n) = a(n).
a(n) = sqrt((A002315(n)^2 +1)/2). [Inserted by N. J. A. Sloane, May 08 2000]
a(n+1) = S(n, 6)-S(n-1, 6), n>=0, with S(n, 6) = A001109(n+1), S(-2, 6) := -1. S(n, x)=U(n, x/2) are Chebyshev's polynomials of the second kind. Cf. triangle A049310. a(n+1) = T(2*n+1, sqrt(2))/sqrt(2),n>=0, with T(n, x) Chebyshev's polynomials of the first kind. [Offset corrected by Wolfdieter Lang, Mar 06 2012]
a(n) ~ 1/4*sqrt(2)*(sqrt(2) + 1)^(2*n+1). - Joe Keane (jgk(AT)jgk.org), May 15 2002
a(n) = (((3 + 2*sqrt(2))^(n+1) - (3 - 2*sqrt(2))^(n+1)) - ((3 + 2*sqrt(2))^n - (3 - 2*sqrt(2))^n)) / (4*sqrt(2)). Lim. n->Inf. a(n)/a(n-1) = 3 + 2*sqrt(2). - Gregory V. Richardson, Oct 12 2002
Let q(n, x) = sum(i=0, n, x^(n-i)*binomial(2*n-i, i)); then q(n, 4) = a(n) - Benoit Cloitre, Nov 10 2002
For n and j >= 1, sum_{k=0..j}a(k)*a(n)-sum_{k=0..j-1}a(k)*a(n-1) = A001109(j+1)*a(n)-A001109(j)*a(n-1) = a(n+j); e.g. (1+5+29)*5-(1+5)*1=169. - Charlie Marion, Jul 07 2003
Contribution from Charlie Marion, Jul 16 2003: (Start)
For n>=k>=0, a(n)^2 = a(n+k)*a(n-k) - A084703(k)^2; e.g. 169^2 = 5741*5 - 144.
For n > 0, a(n) ^2 - a(n-1)^2 = 4*Sum_{k=0...2n-1}a(k) = 4*A001109(2n); e.g. 985^2 - 169^2 = 4*(1 + 5 + 29 + ... + 195025) = 4*235416.
Sum_{k=0...n}((-1)^(n-k)*a(k)) = A079291(n+1); e.g. -1 + 5 - 29 + 169 = 144.
A001652(n) + A046090(n) - a(n) = A001542(n); e.g. 119 + 120 - 169 = 70.
(End)
Sum_{k=0...n}((2k+1)*a(n-k)) = A001333(n+1)^2 - (1 + (-1)^(n+1))/2; e.g. 1*169+3*29+5*5+7*1=288=17^2-1; 1*29+3*5+5*1=49=7^2. - Charlie Marion, Jul 18 2003
Sum_{k = 0...n}a(k)*a(n) = sum_{k = 0...n}a(2k) and Sum_{k = 0...n}a(k)*a(n+1) = sum_{k = 0...n}a(2k+1); e.g. (1+5+29)*29 = 1+29+985 and (1+5+29)*169 = 5+169+5741. - Charlie Marion, Sep 22 2003
For n >= 3, a_{n} = 7(a_{n-1} - a_{n-2}) + a_{n-3}, with a_1 = 1, a_2 = 5 and a_3 = 29. a(n) = ((-1+2^(1/2))/2^(3/2))*(3 - 2^(3/2))^n + ((1+2^(1/2))/2^(3/2))*(3 + 2^(3/2))^n. - Antonio Alberto Olivares, Oct 13, 2003
Let a(n) = A001652(n), b(n) = A046090(n) and c(n) = this sequence. Then for k>j, c(i)*(c(k) - c(j)) = a(k+i)+...+a(i+j+1) + a(k-i-1)+...+a(j-i) + k - j. For n<0, a(n) = -b(-n-1) . Also a(n)*a(n+2k+1) + b(n)*b(n+2k+1) + c(n)*c(n+2k+1) = (a(n+k+1) - a(n+k))^2; a(n)*a(n+2k) + b(n)*b(n+2k) + c(n)*c(n+2k) = 2*c(n+k)^2. - Charlie Marion, Jul 01 2003
Let a(n) = A001652(n), b(n) = A046090(n) and c(n) = this sequence. Then for n>0, a(n)*b(n)*c(n)/(a(n)+b(n)+c(n))=sum_{k=0...n}c(2*k+1); e.g. 20*21*29/(20+21+29) = 5+169 = 174; a(n)*b(n)*c(n)/(a(n-1)+b(n-1)+c(n-1))=sum_{k=0...n}c(2*k); e.g. 119*120*169/(20+21+29)=1+29+985+33461=34476. - Charlie Marion, Dec 01 2003
Also solutions x>0 of the equation floor(x*r*floor(x/r))==floor(x/r*floor(x*r)) where r=1+sqrt(2) - Benoit Cloitre, Feb 15 2004
a(n)a(n+3) = 24 + a(n+1)a(n+2). - R. Stephan, May 29 2004
For n>=k, a(n)*a(n+2k+1)-a(n+k)*a(n+k+1)=a(k)^2-1; e.g. 29*195025-985*5741=840=29^2-1; 1*169-5*29=24=5^2-1; a(n)*a(n+2k)-a(n+k)^2=A001542(k)^2; e.g. 169*195025-5741^2=144=12^2; 1*29-5^2=4=2^2. - Charlie Marion Jun 02 2004
For all k, a(n) is a factor of a((2n+1)*k+n). a((2n+1)k+n)=a(n)*(sum_{j=0...k-1}(-1)^j*(a((2n+1)(k-j))+a((2n+1)(k-j)-1))+(-1)^k); e.g. 195025=5*(33461+5741-169-29+1); 7645370045=169*(6625109+1136689-1).- Charlie Marion, Jun 04 2004
a(n) = sum{k=0..n, binomial(n+k, 2k)4^k}. - Paul Barry, Aug 30 2004
a(n) = sum{k=0..n, binomial(2n+1, 2k+1)2^k}. - Paul Barry, Sep 30 2004
For n<k, a(n)*A001541(k)=A011900(n+k)+A053141(k-n-1); e.g. 5*99=495=493+2. For n>=k, a(n)*A001541(k)=A011900(n+k)+A053141(n-k); e.g. 29*3=87=85+2. - Charlie Marion, Oct 18 2004
a(n) = (-1)^n*U(2n, I*sqrt(4)/2)=(-1)^n*U(2n, I), U(n, x) Chebyshev polynomial of second kind, I=sqrt(-1). - Paul Barry, Mar 13 2005
a(n) = Pell(2n+1) = Pell(n)^2+Pell(n+1)^2. - Paul Barry, Jul 18 2005
a(n)*a(n+k) = A000129(k)^2+A000129(2n+k+1)^2; e.g., 29*5741=12^2+169^2. -Charlie Marion, Aug 02 2005
Let a(n)*a(n+k)=x. Then 2x^2-A001541(k)*x+A001109(k)^2=A001109(2n+k+1)^2; e.g., let x=29*985; then 2x^2-17x+6^2=40391^2; cf. A076218. - Charlie Marion, Aug 02 2005
With a=3+2sqrt(2), b=3-2sqrt(2): a(n) = (a^((2n+1)/2)+b^((2n+1)/2))/(2sqrt(2)). a(n)=A001109(n+1)-A001109(n). - Mario Catalani (mario.catalani(AT)unito.it), Mar 31 2003
If k is in the sequence, then the next term is floor[k*(3+2*sqrt(2))]. - Lekraj Beedassy, Jul 19 2005
a(n)=Jacobi_P(n,-1/2,1/2,3)/Jacobi_P(n,-1/2,1/2,1). - Paul Barry, Feb 03 2006
a(n)=sum{k=0..n, sum{j=0..n-k, C(n,j)C(n-j,k)Pell(n-j+1)}}, Pell(n)=A000129(n). - Paul Barry, May 19 2006
a(n) = round(sqrt(A002315(n)^2/2)). - Lekraj Beedassy, Jul 15 2006
a(n) = A079291(n) + A079291(n+1). - Lekraj Beedassy, Aug 14 2006
a(n+1) = 3*a(n)+(8*a(n)^2-4)^0.5, a(1)=1. - Richard Choulet (richard.choulet(AT)orange.fr), Sep 18 2007
6*a(n)*a(n+1) = a(n)^2+a(n+1)^2+4; e.g., 6*5*29=29^2+5^2+4; 6*169*985=169^2+985^2+4. - Charlie Marion, Oct 07 2007
2*A001541(k)*a(n)*a(n+k) = a(n)^2+a(n+k)^2+A001542(k)^2; e.g., 2*3*5*29=5^2+29^2+2^2; 2*99*29*5741=2*99*29*5741=29^2+5741^2+70^2. - Charlie Marion, Oct 12 2007
[a(n), A001109(n)] = [1,4; 1,5]^n * [1,0]. - Gary W. Adamson, Mar 21 2008
Contribution from Charlie Marion, Apr 10 2009: (Start)
In general, for n>=k, a(n+k)= 2*A001541(k)*a(n)-a(n-k);
e.g., a(n+0)=2*1*a(n)-a(n); a(n+1)=6*a(n)-a(n-1); a(6+0)=33461=2*33461-33461; a(5+1)=33461=6*5741-985; a(4+2)=33461=34*985-29; a(3+3)=33461=198*169-1.
(End)
G.f.: sqrt(x)*tan(4*arctan(sqrt(x)))/4. - Johannes W. Meijer, Aug 01 2010
Given k=(sqrt(2)+1)^2=3+2*sqrt(2); and a(0)=1; then a(n)=a(n-1)*k-((k-1)/(k^n)). - Charles L. Hohn, Mar 06 2011
Given k=(sqrt(2)+1)^2=3+2*sqrt(2); and a(0)=1; then a(n)=(k^n)+(k^(-n))-a(n-1) (which = a(n)=A003499(n)-a(n-1)) - Charles L. Hohn, Apr 04 2011
Let T(n) be the n-th triangular number; then, for n>0,
T(a(n))+A001109(n-1)=A046090(n)^2. See also A046090. - Charlie Marion, Apr 25 2011
For k>0, a(n+2k-1) - a(n) = 4*A001109(n+k-1)*A002315(k-1); a(n+2k) - a(n) = 4*A001109(k)*A002315(n+k-1). - Charlie Marion, Jan 06 2012
a(k+j+1) = (A001541(k)*A001541(j)+A002315(k)*A002315(j))/2. - Charlie Marion, Jun 25 2012
a(n) = A143608(n-1)*A143608(n) + 1. - Charlie Marion, Dec 11 2012
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MAPLE
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a[0]:=1: a[1]:=5: for n from 2 to 26 do a[n]:=6*a[n-1]-a[n-2] od: seq(a[n], n=0..20); - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Jul 26 2006
A001653:=-(-1+5*z)/(z**2-6*z+1); [Conjectured (correctly) by Simon Plouffe in his 1992 dissertation. Gives sequence except for one of the leading 1's.]
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MATHEMATICA
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a[n_] := (MatrixPower[{{1, 2, 2}, {2, 1, 2}, {2, 2, 3}}, n].{{1}, {0}, {1}})[[3, 1]]; Table[ a[n], {n, 0, 20}] (* Robert G. Wilson v, Jan 08 2005)
LinearRecurrence[{6, -1}, {1, 5}, 40] (* Harvey P. Dale, Jul 12 2011 *)
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PROG
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(PARI) a(n)=subst(poltchebi(abs(n+1))+poltchebi(abs(n)), x, 3)/4 /* Michael Somos */
(PARI) a(n)=([5, 2; 2, 1]^n)[1, 1] /* from Lambert Klasen */
(Haskell)
a001653 n = a001653_list !! n
a001653_list = 1 : 5 : zipWith (-) (map (* 6) $ tail a001653_list) a001653_list
-- Reinhard Zumkeller, May 07 2013
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CROSSREFS
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Other two sides are A001652, A046090.
Cf. A001519.
These numbers are the odd-indexed Pell numbers from A000129. The even-indexed Pell numbers are A001542. - Ira M. Gessel, Sep 27 2002.
Row 6 of array A094954.
Cf. A001109.
Cf. A005054, A122074.
Row 1 of array A188647.
Cf. A056220.
Cf. A077445(n) = 4*a(n).
Sequence in context: A015537 A182017 A141812 * A141814 A175883 A122370
Adjacent sequences: A001650 A001651 A001652 * A001654 A001655 A001656
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KEYWORD
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nonn,easy,nice,changed
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AUTHOR
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N. J. A. Sloane.
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EXTENSIONS
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Additional comments from Wolfdieter Lang, Feb 10 2000
Better description from Harvey P. Dale, Jan 15 2002
Edited by N. J. A. Sloane Nov 02 2002
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STATUS
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approved
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