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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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145
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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
(list;
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refs;
listen;
history;
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internal format)
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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 formulas x = 2uv, y = u^2 - v^2, z = u^2 + v^2. [Joseph Wiener and Donald Skow]. - Antonio Alberto Olivares, Dec 22 2003
All Pythagorean triples {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). - Klaus Brockhaus, Feb 17 2009
The remainder of the division of a(n) by 5 is: 0, 1, 3 or 4. - Mohamed Bouhamida, 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. - Mohamed Bouhamida, 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, 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. - Mohamed Bouhamida, 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). - 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 triple 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
The arms of the triangle are found with (b(n),c(n)) for 2*b(n)*c(n) and c(n)^2 - b(n)^2. Let b(1)=1 and c(1)=2, then b(n)=c(n-1) and c(n)= 2*c(n-1) + b(n-1). Alternatively, b(n)=c(n-1) and c(n) equals the nearest integer to b(n)*(1+sqrt(2)). - J. M. Bergot, Oct 09 2014
Conjecture: For n>1 a(n) is the index of the first occurrence of n in sequence A123737. - Vaclav Kotesovec, Jun 02 2015
Numbers n such that Product_{k=1..n} (4*k^4+1) is a square (see A274307). - Chai Wah Wu, Jun 21 2016
Numbers n such that n^2+(n+1)^2 is a square. - César Aguilera, Aug 14 2017
For integers a and d, let P(a,d,1) = a, P(a,d,2) = a+d, and, for n>2, P(a,d,n) = 2*P(a,d,n-1) + P(a,d,n-1). Further, let p(n) = P(a,d,1) + ... + P(a,d,2n). Then p(n)^2 + (p(n)+d)^2 + a^2 = P(a,d,2n+1)^2 + d^2. When a = 1 and d = 1, p(n) = a(n) and P(a,d,n) = A000129(n), the n-th Pell number. - Charlie Marion, Dec 08 2018
The terms of this sequence satisfy the Diophantine equation k^2 + (k+1)^2 = m^2, which is equivalent to (2k+1)^2 - 2*m^2 = -1. Now, with x=2k+1 and y=m, we get the Pell-Fermat equation x^2 - 2*y^2 = -1. The solutions (x,y) of this equation are respectively in A002315 and A001653. The relation k = (x-1)/2 explains Lekraj Beedassy's Nov 25 2003 formula. Thus, the corresponding numbers m = y, which express the length of the hypotenuse of these right triangles (k,k+1,m) are in A001653. - Bernard Schott, Mar 10 2019
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REFERENCES
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A. H. Beiler, Recreations in the Theory of Numbers. New York: Dover, pp. 122-125, 1964.
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).
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LINKS
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T. D. Noe, Table of n, a(n) for n = 0..200
I. Adler, Three Diophantine equations - Part II, Fib. Quart., 7 (1969), 181-193.
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.
Ron Knott, Pythagorean Triples and Online Calculators
A. Martin, Table of prime rational right-angled triangles, The Mathematical Magazine, 2 (1910), 297-324.
A. Martin, Table of prime rational right-angled triangles (annotated scans of a few pages)
A. Martin, On rational right-angled triangles , Proceedings of the Fifth International Congress of Mathematicians (Cambridge, 22-28 August 1912)
S. P. Mohanty, Which triangular numbers are products of three consecutive integers, Acta Math. Hungar., 58 (1991), 31-36.
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.
Burkard Polster, Nice merging together, Mathologer video (2015).
B. Polster, M. Ross, Marching in squares, arXiv:1503.04658 [math.HO], 2015.
Zhang Zaiming, Problem #502, Pell's Equation - Once Again, The College Mathematics Journal, 25 (1994), 241-243.
Index entries for two-way infinite sequences
Index entries for linear recurrences with constant coefficients, signature (7,-7,1).
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FORMULA
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G.f.: x *(3 - x) / ((1 - 6*x + x^2) * (1 - x)). - Simon Plouffe in his 1992 dissertation
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) = a(n-2) + 4*sqrt(2*(a(n-1)^2)+2*a(n-1)+1). - Pierre CAMI, Mar 30 2005
a(n) = (sinh((2*n+1)*log(1+sqrt(2)))-1)/2 = (sqrt(1+8*A029549)-1)/2. - Bill Gosper, Feb 07 2010
binomial(a(n)+1,2) = 2*binomial(A053141(n)+1,2) = A029549(n). 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). - Charlie Marion, Jul 07 2003
Sum_{k=0...n} (2k+1)*a(n-k) = A001109(n+1) - A000217(n+1). - Charlie Marion, Jul 18 2003
a(n) = A055997(n) - 1 + sqrt(2*A055997(n)*A001108(n)). - Charlie Marion, Jul 21 2003
a(n) = {A002315(n) - 1}/2. - Lekraj Beedassy, Nov 25 2003
a(2n+k) + a(k) + 1 = A001541(n)*A002315(n+k). 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, Sep 02 2006
a(n+1) = 3*a(n) + sqrt(8*a(n)^2 +8*a(n) +4) + 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. 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). - 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
From Charlie Marion, Jan 04 2010: (Start)
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). (End)
a(n) = A048739(2n-1) for n > 0. - Richard R. Forberg, Aug 31 2013
a(n+1) = 3*a(n) + 2*A001653(n) + 1 [Mohamed Bouhamida's 2009 (p,q)(r,s) comment above rewritten]. - Hermann Stamm-Wilbrandt, Jul 27 2014
a(n)^2 + (a(n)+1)^2 = A001653(n+1)^2. - Pierre CAMI, Mar 30 2005; clarified by Hermann Stamm-Wilbrandt, Aug 31 2014
a(n+1) = 3*A001541(n) + 10*A001109(n) + A001108(n). - Hermann Stamm-Wilbrandt, Sep 09 2014
For n>0, a(n) = Sum_{k=1..2n) A000129(k). - Charlie Marion, Nov 07 2015
a(n) = 3*A053142(n) - A053142(n-1). - R. J. Mathar, Feb 05 2016
E.g.f.: (1/4)*(-2*exp(x) - (sqrt(2) - 1)*exp((3-2*sqrt(2))*x) + (1 + sqrt(2))*exp((3+2*sqrt(2))*x)). - Ilya Gutkovskiy, Apr 11 2016
a(n) = A001108(n) + 2*sqrt(A000217(A001108(n))). - Dimitri Papadopoulos, Jul 06 2017
a(A000217(n-1)) = ((A001653(n)+1)/2) * ((A001653(n)-1)/2), n > 1. - Ezhilarasu Velayutham, Mar 10 2019
a(n) = ((a(n-1)+1)*(a(n-1)-3))/a(n-2) for n>2. - Vladimir Pletser, Apr 08 2020
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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 := proc(n)
option remember;
if n <= 1 then
op(n+1, [0, 3]) ;
else
6*procname(n-1)-procname(n-2)+2 ;
end if;
end proc: # R. J. Mathar, Feb 05 2016
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MATHEMATICA
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LinearRecurrence[{7, -7, 1}, {0, 3, 20}, 30] (* Harvey P. Dale, Aug 19 2011 *)
With[{c=3+2*Sqrt[2]}, NestList[Floor[c*#]+3&, 3, 30]] (* Harvey P. Dale, Oct 22 2012 *)
CoefficientList[Series[x (3 - x)/((1 - 6 x + x^2) (1 - x)), {x, 0, 30}], x] (* Vincenzo Librandi, Oct 21 2014 *)
Table[(LucasL[2*n + 1, 2] - 2)/4, {n, 0, 30}] (* G. C. Greubel, Jul 15 2018 *)
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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 */
(PARI) concat(0, Vec(x*(3-x)/((1-6*x+x^2)*(1-x)) + O(x^50))) \\ Altug Alkan, Nov 08 2015
(PARI) {a=1+sqrt(2); b=1-sqrt(2); Q(n) = a^n + b^n};
for(n=0, 30, print1(round((Q(2*n+1) - 2)/4), ", ")) \\ G. C. Greubel, Jul 15 2018
(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] ]; // Klaus Brockhaus, Feb 17 2009
(MAGMA) m:=30; R<x>:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(x*(3-x)/((1-6*x+x^2)*(1-x)))); // G. C. Greubel, Jul 15 2018
(Haskell)
a001652 n = a001652_list !! n
a001652_list = 0 : 3 : map (+ 2)
(zipWith (-) (map (* 6) (tail a001652_list)) a001652_list)
-- Reinhard Zumkeller, Jan 10 2012
(GAP) a:=[0, 3];; for n in [3..25] do a[n]:=6*a[n-1]-a[n-2]+2; od; a; # Muniru A Asiru, Dec 08 2018
(Sage) (x*(3-x)/((1-6*x+x^2)*(1-x))).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Mar 08 2019
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CROSSREFS
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Cf. A046090(n) = -a(-1-n).
Cf. A001108, A143608, A089950 (partial sums).
Cf. A156035 (decimal expansion of 3+2*sqrt(2)). - 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
Cf. A001541, A001109, A274307.
Cf. A002315, A001653 (solutions of x^2 - 2*y^2 = -1).
Sequence in context: A005096 A275796 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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Additional comments from Wolfdieter Lang, Feb 10 2000
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STATUS
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approved
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