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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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125
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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
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 (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. - 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. - 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). - 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
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
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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.
A. Martin, Table of prime rational right-angled triangles, Math. Mag., 2 (1910), 297-324.
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, 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.
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 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 - 6*x + 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) = 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(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
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
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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] (* 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 *)
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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] ]; // 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, Jan 10 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)). - 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.
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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Additional comments from Wolfdieter Lang, Feb 10 2000
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STATUS
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approved
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