0,2

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 (olivares14031(AT)yahoo.com), 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 (pierrecami(AT)tele2.fr), 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 (blekraj(AT)yahoo.com), Aug 14 2006

Let b(n) = A053141 then a(n)*b(n+1) = b(n)*a(n+1) + b(n) - Kenneth J Ramsey (Ramsey2879(AT)msn.com), 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 (charliemath(AT)optonline.net), Nov 19 2007

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.

Zhang Zaiming, Problem #502, Pell's Equation - Once Again, Two-Year College Math. Jnl., 25 (1994), 241-243.

T. D. Noe, Table of n, a(n) for n=0..200

Ron Knott, Pythagorean Triples and Online Calculators

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.

a(n) = [ (1+ sqrt(2))^(2n+1) + (1-sqrt(2))^(2n+1) - 2 ]/4.

G.f.: x(3-x)/((1-6x+x^2)(1-x)). - Michael Somos, Apr 07 2003

a_{n} = 7(a_{n-1} - 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 Olivares (olivares14031(AT)yahoo.com), Oct 13, 2003

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 (charliem(AT)bestweb.net), Jul 01 2003

a(n)*a(n+1) + A046090(n)*A046090(n+1) = A001542(n+1)^2 = A084703(n+1). - Charlie Marion (charliem(AT)bestweb.net), 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 (charliem(AT)bestweb.net), 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 (charliem(AT)bestweb.net), 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 (charliem(AT)bestweb.net), Jul 21 2003

a(n)={A002315(n) - 1}/2. - Lekraj Beedassy (blekraj(AT)yahoo.com), Nov 25 2003

Sum_{k=0...n}a(k)=A089950(n); e.g., 0+3+20+119=142 - Charlie Marion (charliemath(AT)verizon.net), 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 (charliemath(AT)verizon.net), Mar 17 2003

a(n)=(A001653(n+1) - 3*A001653(n) - 2)/4. - Lekraj Beedassy (blekraj(AT)yahoo.com), Jul 13 2004

a(n)={2*A084159(n) - 1 + (-1)^(n+1)}/2. - Lekraj Beedassy (blekraj(AT)yahoo.com), Jul 21 2004

a(n)=5*(a(n-1)+a(n-2))-a(n-3)+4; a(n)=7*(a(n-1)-a(n-2))+a(n-3). - 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 (richardchoulet(AT)yahoo.fr), 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 (charliemath(AT)optonline.net), 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 (charliemath(AT)optonline.net), Jan 21 2008

The first few triples are (0,1,1), (3,4,5), (20,21,29), (119,120,169), ...

A001652:=z*(-3+z)/(z-1)/(z**2-6*z+1); [Conjectured by S. Plouffe in his 1992 dissertation.]

pythList[ n_Integer?Positive ] := With[ {p = 3 + 2 Sqrt[ 2 ]}, NestList[ Floor[ p # ] + 3 &, 3, n - 1 ] ]

(PARI) {a(n)=subst(poltchebi(n+1)-poltchebi(n)-2, x, 3)/4} /* Michael Somos Aug 11 2006 */

Cf. A001653. A046090(n)=-a(-1-n).

Cf. A001108.

Adjacent sequences: A001649 A001650 A001651 this_sequence A001653 A001654 A001655

Sequence in context: A139471 A108911 A005096 this_sequence A128910 A037788 A037669

nonn,easy,nice

njas

Mathematica formula from Harvey P. Dale (hpd1(AT)is2.nyu.edu).

Additional comments from Wolfdieter Lang (wolfdieter.lang(AT)physik.uni-karlsruhe.de), Feb 10 2000