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A001652
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a(n) = 6*a(n-1) - a(n-2) + 2 with a(0) = 0, a(1) = 3.
(Formerly M3074 N1247)
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123
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0, 3, 20, 119, 696, 4059, 23660, 137903, 803760, 4684659, 27304196, 159140519, 927538920, 5406093003, 31509019100, 183648021599, 1070379110496, 6238626641379, 36361380737780, 211929657785303, 1235216565974040
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OFFSET
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0,2
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COMMENTS
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Consider all Pythagorean triples (X,X+1,Z) ordered by increasing Z; sequence gives X values.
Numbers n such that triangular number t(n) (see A000217) = n(n+1)/2 is a product of two consecutive integers (cf. A097571).
Members of Diophantine pairs. Solution to a(a+1)=2b(b+1) in natural numbers including 0; a=a(n), b=b(n)= A053141(n) - The solution of a special case of a binomial problem of H. Finner and K. Strassburger (strass(AT)godot.dfi.uni-duesseldorf.de).
The three sequences x (A001652), y (A046090) and z (A001653) may be obtained by setting u and v equal to the Pell numbers (A000129) in the formulae x = 2uv, y = u^2 - v^2, z = u^2 + v^2. [Joseph Wiener and Donald Skow]. - Antonio Alberto Olivares, Dec 22 2003
Define a(1)=0 a(2)=3 such that 2*(a(1)^2)+2*a(1)+1=j(1)^2=1^2 and 2*(a(2)^2)+2*a(2)+1=j(2)^2=5^2=25. Then a(n)=a(n-2)+4*sqrt(2*(a(n-1)^2)+2*a(n-1)+1). Another definition: a(n) such that 2*(a(n)^2)+2*a(n)+1 = j(n)^2. - Pierre CAMI, Mar 30 2005
The complete Pythagorean triple {X(n),Y(n)=X(n)+1,Z(n)} with X<Y<Z,may be recursively generated through the mapping W(n) -> M*W(n), where W(n)=transpose of vector [X(n) Y(n) Z(n)] and M a 3 X 3 matrix given by [2 1 2 / 1 2 2 / 2 2 3]. - Lekraj Beedassy, Aug 14 2006
Let b(n) = A053141 then a(n)*b(n+1) = b(n)*a(n+1) + b(n) - Kenneth J Ramsey, Sep 22 2007
In general, if b(n)= A053141(n), then a(n)*b(n+k) = a(n+k)*b(n)+b(k); e.g. 3*84 = 119*2+14; 3*2870 = 4059*2+492; 20*2870=5741*14+84 - Charlie Marion, Nov 19 2007
lim_{n -> infinity} a(n)/a(n-1) = 3+2*sqrt(2). [From Klaus Brockhaus, Feb 17 2009]
The remainder of the division of a(n) by 5 is: 0, 1, 3 or 4. [From Mohamed Bouhamida (bhmd95(AT)yahoo.fr), Aug 26 2009]
Number of units of a(n) belongs to a periodic sequence: 0, 3, 0, 9, 6, 9. The remainder of the division of a(n) by 5 belongs to a periodic sequence: 0, 3, 0, 4, 1, 4. [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 consecutive 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]
a(n+k) = A001541(k)*a(n)+A001542(k)*A001653(n+1)+A001108(k). - Charlie Marion, Dec 10 2010
The numbers 3*A001652=(0,9,60,357,2088,12177,70980,...) are all the nonnegative values of X such that X^2+(X+3)^2=Z^2 (Z is in A075841). [From Bruno Berselli, Aug 26 2010]
Let T(n) = n*(n+1)/2 (the n-th triangular number). For n>0,
T(a(n)+2k*A001653(n+1)) = 2*T(A053141(n-1)+k*A002315(n))+k^2 and
T(a(n)+(2k+1)*A001653(n+1)) = (A001109(n+1)+k*A002315(n))^2+k(k+1).
Also (a(n)+k*A001653(n))^2+(a(n)+k*A001653(n)+1)^2 = (A001653(n+1)+k*A002315(n))^2+k^2. - Charlie Marion, Dec 09 2010
For n>0, A143608(n) divides a(n). - _Kenneth J. Ramsey_, Jun 28 2012
Set a(n)=p; a(n)+1=q; the generated triplet x=p^2+pq; y=q^2+pq; k=p^2+q^2 satisfies x^2+y^2=k(x+y). - Carmine Suriano, Dec 17 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.
Martin V. Bonsangue, Gerald E. Gannon and Laura J. Pheifer, Misinterpretations can sometimes be a good thing, Math. Teacher, vol. 95, No. 6 (2002) pp. 446-449.
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.
A. Martin, Table of prime rational right-angled triangles, Math. Mag., 2 (1910), 297-324.
S. P. Mohanty, Which triangular numbers are products of three consecutive integers, Acta Math. Hungar., 58 (1991), 31-36.
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).
Zhang Zaiming, Problem #502, Pell's Equation - Once Again, Two-Year College Math. Jnl., 25 (1994), 241-243.
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LINKS
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T. D. Noe, Table of n, a(n) for n=0..200
Ron Knott, Pythagorean Triples and Online Calculators
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992.
Simon Plouffe, 1031 Generating Functions and Conjectures, Université du Québec à Montréal, 1992.
Index entries for two-way infinite sequences
Index entries for sequences related to linear recurrences with constant coefficients, signature (7,-7,1)
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FORMULA
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G.f.: x(3-x)/((1-6x+x^2)(1-x)). - Michael Somos, Apr 07 2003
a(n) = 7*a(n-1) - 7*a(n-2) + a(n-3). a_{n} = -1/2 + ((1-2^{1/2})/4)*(3 - 2^{3/2})^n + ((1+2^{1/2})/4)*(3 + 2^{3/2})^n . - Antonio Alberto Olivares, Oct 13 2003
a(n) = (sinh((2*n+1)*log(1+sqrt(2)))-1)/2 = (sqrt(1+8*A029549)-1)/2. - Bill Gosper, Feb 07 2010
binomial(A001652+1,2) = 2*binomial(A053141+1,2) = A029549. See A053141. - Bill Gosper, Feb 07 2010
Let b(n) = A046090(n) and c(n) = A001653(n). 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
a(n)*a(n+1) + A046090(n)*A046090(n+1) = A001542(n+1)^2 = A084703(n+1). - Charlie Marion, Jul 01 2003
For n and j >= 1, sum_{k=0..j}A001653(k)*a(n) - sum_{k=0...j-1}A001653(k)*a(n-1) + A053141(j) = A001109(j+1)*a(n) - A001109(j)*a(n-1) + A053141(j)= a(n+j); e.g. (1+5+29)*119-(1+5)*20+14=4059. - Charlie Marion, Jul 07 2003
Sum_{k=0...n} (2k+1)*a(n-k) = A001109(n+1)-A000217(n+1); e.g. 1*119+3*20+5*3+7*0=194=204-10. - Charlie Marion, Jul 18 2003
a(n)=A055997(n)-1+(2*A055997(n)*A001108(n))^.5; e.g. 119=50-1+(2*50*49)^.5. - Charlie Marion, Jul 21 2003
a(n)={A002315(n) - 1}/2. - Lekraj Beedassy, Nov 25 2003
Sum_{k=0...n}a(k)=A089950(n); e.g., 0+3+20+119=142. - Charlie Marion, Jan 21 2004
a(2n+k)+a(k)+1=A001541(n)*A002315(n+k); e.g., 23660+696+1=3*8119; for k>0, a(2n+k)-a(k-1)=A001541(n+k)*A002315(n);e.g. 803760-119=19601*41. - Charlie Marion, Mar 17 2003
a(n)=(A001653(n+1) - 3*A001653(n) - 2)/4. - Lekraj Beedassy, Jul 13 2004
a(n)={2*A084159(n) - 1 + (-1)^(n+1)}/2. - Lekraj Beedassy, Jul 21 2004
a(n)=5*(a(n-1)+a(n-2))-a(n-3)+4. - Mohamed Bouhamida (bhmd95(AT)yahoo.fr), Sep 02 2006
a(n+1)=3*a(n)+(8*a(n)^2+8*a(n)+4)^0.5+1, a(1)=0. - Richard Choulet, Sep 18 2007
As noted (Sep 20 2006), a(n) = 5*(a(n-1)+a(n-2))-a(n-3)+4. In general, for n>2k, a(n)=A001653(k)*(a(n-k)+a(n-k-1)+1)-a(n-2k-1)-1; e.g. 696=5*(119+20+1)-3-1; 137903=29(4059+696+1)-20-1. Also a(n) = 7*(a(n-1)-a(n-2))+a(n-3). In general, for n>2k, A002378(k)*(a(n-k)-a(n-k-1))+a(n-2k-1); e.g. 696=7(119-20)+3; 137903=41(4059-696)+20. - Charlie Marion, Dec 26 2007
In general, for n>=k>0, a(n)= (A001653(n+k)-A001541(k)*A001653(n)-2*A001109(k-1))/(4*A001109(k-1)); e.g., 4059=(33461-3*5741-2*1)/(4*1); 4059=(195025-17*5741-2*6)/(4*6). - Charlie Marion, Jan 21 2008
a(n) = ((1+ sqrt(2))^(2n+1) + (1-sqrt(2))^(2n+1) - 2)/4 = (A001333(2n+1)-1)/2.
a(2n+k-1) = Pell(2n-1)*Pell(2n+2k) + Pell(2n-2)*Pell(2n+2k+1) + A001108(k+1);
a(2n+k) = Pell(2n)*Pell(2n+2k+1) + Pell(2n-1)*Pell(2n+2k+2) - A055997(k+2).
- Charlie Marion, Jan 04 2010
a(n) = A048739(2n-1) for n > 0. - Richard R. Forberg, Aug 31 2013
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EXAMPLE
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The first few triples are (0,1,1), (3,4,5), (20,21,29), (119,120,169), ...
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MAPLE
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A001652:=z*(-3+z)/(z-1)/(z**2-6*z+1); [Simon Plouffe in his 1992 dissertation.]
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MATHEMATICA
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LinearRecurrence[{7, -7, 1}, {0, 3, 20}, 30] (* From Harvey P. Dale, Aug 19 2011 *)
With[{c=3+2*Sqrt[2]}, NestList[Floor[c*#]+3&, 3, 30]] (* Harvey P. Dale, Oct 22 2012 *)
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PROG
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(PARI) {a(n)=subst(poltchebi(n+1)-poltchebi(n)-2, x, 3)/4} /* Michael Somos, Aug 11 2006 */
(MAGMA) Z<x>:=PolynomialRing(Integers()); N<r2>:=NumberField(x^2-2); S:=[ (-2+(r2+1)*(3+2*r2)^n-(r2-1)*(3-2*r2)^n)/4: n in [1..20] ]; [ Integers()!S[j]: j in [1..#S] ]; [From Klaus Brockhaus, Feb 17 2009]
(Haskell)
a001652 n = a001652_list !! n
a001652_list = 0 : 3 : map (+ 2)
(zipWith (-) (map (* 6) (tail a001652_list)) a001652_list)
-- Reinhard Zumkeller, 10 Jan 2012
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CROSSREFS
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Cf. A001653. A046090(n)=-a(-1-n).
Cf. A001108, A143608.
Cf. A156035 (decimal expansion of 3+2*sqrt(2)). [From Klaus Brockhaus, Feb 17 2009]
Cf. numbers m such that k*A000217(m)+1 is a square: A006451 for k=1; m=0 for k=2; A233450 for k=3; this sequence for k=4; A129556 for k=5; A001921 for k=6. [Bruno Berselli, Dec 16 2013]
Sequence in context: A108911 A005096 A164535 * A128910 A037788 A037669
Adjacent sequences: A001649 A001650 A001651 * A001653 A001654 A001655
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KEYWORD
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nonn,easy,nice
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AUTHOR
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N. J. A. Sloane.
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EXTENSIONS
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Mathematica formula from Harvey P. Dale.
Additional comments from Wolfdieter Lang, Feb 10 2000
Additional comments from Carmine Suriano
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STATUS
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approved
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