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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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138
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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
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,2
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COMMENTS
| 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 (blekraj(AT)yahoo.com), Jun 05 2002
Numbers n such that 2*n^2=ceil(sqrt(2)*n*floor(sqrt(2)*n)) Benoit Cloitre (benoit7848c(AT)orange.fr), 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 Buddenhagen (jbuddenh(AT)gmail.com), 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 (pbarry(AT)wit.ie), Mar 13 2005
a(n) = L(n,6), where L is defined as in A108299; see also A002315 for L(n,-6). - Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), 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 (tanyakh(AT)yahoo.com), 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 (ck6(AT)evansville.edu), 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. [From 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 cosecutive solutions of the Diophantine equation: X^2 + (X+1)^2 = Y^2 with p<r then s-r=p+q+1. [From 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 (qntmpkt(AT)yahoo.com), 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. [From Milan R. Janjic (agnus(AT)blic.net), 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 2 2011.
Apart from initial 1: subsequence of A198389, see also A198385. [Reinhard Zumkeller, Oct 25 2011]
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REFERENCES
| 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.
Fielder, Daniel C.; Special integer sequences controlled by three parameters. Fibonacci Quart 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
| 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
S. 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.
S. Plouffe, 1031 Generating Functions and Conjectures, Universit\'{e} du Qu\'{e}bec \`{a} Montr\'{e}al, 1992.
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
| G.f.: (1-x)/(1-6*x+x^2). a(n)=6a(n-1)-a(n-2) with a(0)=1, a(1)=5. a(-1-n)=a(n).
a(n) = S(n, 6)-S(n-1, 6) 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) = T(2*n+1, sqrt(2))/sqrt(2), with T(n, x) Chebyshev's polynomials of the first kind.
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 (omomom(AT)hotmail.com), 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 (benoit7848c(AT)orange.fr), 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
For n>=k>=0, a(n)^2 = a(n+k)*a(n-k) - A084703(k)^2; e.g. 169^2 = 5741*5 - 144. - Charlie Marion, Jul 16 2003
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 - Charlie Marion, Jul 16 2003
Sum_{k=0...n}((-1)^(n-k)*a(k)) = A079291(n+1); e.g. -1 + 5 - 29 + 169 = 144 - Charlie Marion, Jul 16 2003
A001652(n) + A046090(n) - a(n) = A001542(n); e.g. 119 + 120 - 169 = 70 - Charlie Marion, Jul 16 2003
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 Olivares (olivares14031(AT)yahoo.com), 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 (benoit7848c(AT)orange.fr), 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 (pbarry(AT)wit.ie), Aug 30 2004
a(n)=sum{k=0..n, binomial(2n+1, 2k+1)2^k}. - Paul Barry (pbarry(AT)wit.ie), 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 (pbarry(AT)wit.ie), Mar 13 2005
a(n)=Pell(2n+1)=Pell(n)^2+Pell(n+1)^2; - Paul Barry (pbarry(AT)wit.ie), Jul 18 2005
a(n)*a(n+k)=Pell(2n+k+1)^2+Pell(k)^2;e.g., 5*29=12^2+1^2; 29*985=169^2+2^2 - 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
a(n)*a(n+k)=A000129(k)^2+A000129(2n+k+1)^2;e.g., 29*5741=12^2+169^2; - Charlie Marion, Aug 03 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 (blekraj(AT)yahoo.com), Jul 19 2005
a(n)=Jacobi_P(n,-1/2,1/2,3)/Jacobi_P(n,-1/2,1/2,1); - Paul Barry (pbarry(AT)wit.ie), 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 (pbarry(AT)wit.ie), May 19 2006
a(n)=round[sqrt({A002315(n)}^2/2)]. - Lekraj Beedassy (blekraj(AT)yahoo.com), Jul 15 2006
a(n) = A079291(n) + A079291(n+1). - Lekraj Beedassy (blekraj(AT)yahoo.com), Aug 14 2006
a(n+1)=3*a(n)+(8*a(n)^2-4)^0.5, a(1)=1. - Richard Choulet (richardchoulet(AT)yahoo.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 (qntmpkt(AT)yahoo.com), Mar 21 2008
a(n)=6*a(n-1)-a(n-2), a(0)=1, a(1)=5. [From Ctibor O. Zizka (c.zizka(AT)email.cz), Mar 31 2009]
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)
Contribution from Johannes W. Meijer (meijgia(AT)hotmail.com), Aug 01 2010: (Start)
G.f.: sqrt(x)*tan(4*arctan(sqrt(x)))/4
(End)
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 (ch+oeis(AT)1111-internet.com), 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 (ch+oeis(AT)1111-internet.com), 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
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MAPLE
| 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 S. Plouffe in his 1992 dissertation. Gives sequence except for one of the leading 1's.]
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MATHEMATICA
| a[n_] := (MatrixPower[{{1, 2, 2}, {2, 1, 2}, {2, 2, 3}}, n].{{1}, {0}, {1}})[[3, 1]]; Table[ a[n], {n, 0, 20}] (from Robert G. Wilson v Jan 08 2005)
LinearRecurrence[{6, -1}, {1, 5}, 40] (* From Harvey P. Dale, Jul 12 2011 *)
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PROG
| (PARI) a(n)=subst(poltchebi(abs(n+1))+poltchebi(abs(n)), x, 3)/4 (from Michael Somos)
(PARI) a(n)=([5, 2; 2, 1]^n)[1, 1] (from Lambert Klasen)
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CROSSREFS
| Other two sides are A001652, A046090. a(n)=sqrt{(A002315(n)^2 +1)/2}.
Cf. A001519.
These numbers are the odd-indexed Pell numbers from A000129. The even-indexed Pell numbers are A001542. - Ira Gessel (gessel(AT)brandeis.edu), Sep 27 2002.
Row 6 of array A094954.
Cf. A001109.
Cf. A005054, A122074 [From Gary W. Adamson (qntmpkt(AT)yahoo.com), Jul 22 2010]
Row 1 of array A188647.
Cf. A056220.
Sequence in context: A146178 A015537 A141812 * A141814 A175883 A122370
Adjacent sequences: A001650 A001651 A001652 * A001654 A001655 A001656
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KEYWORD
| nonn,easy,nice
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AUTHOR
| N. J. A. Sloane (njas(AT)research.att.com).
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EXTENSIONS
| Additional comments from Wolfdieter Lang (wolfdieter.lang(AT)physik.uni-karlsruhe.de), Feb 10 2000
Better description from Harvey P. Dale (hpd1(AT)nyu.edu), Jan 15 2002
Edited by N. J. A. Sloane (njas(AT)research.att.com) Nov 02 2002
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