0,3

b(0)=0, c(0)=1, b(i+1)=b(i)+c(i), c(i+1)=b(i+1)+b(i); then a(i) (the number in the sequence) is 2b(i)^2 if i is even, c(i)^2 if i is odd and b(n)=A000129(n) and c(n)=A001333(n) - from stephenson(AT)cs.hope.edu (Darin Stephenson and Alan Koch)

For n>1 gives solutions to A007913(2x)=A007913(x+1) - Benoit Cloitre (benoit7848c(AT)orange.fr), Apr 07 2002

If (X,X+1,Z) is a Pythagorean triple, then Z-X-1 and Z+X is in the sequence.

For n >= 2, a(n) gives exactly the positive integers m such that 1,2,...,m has a perfect median. The sequence of associated perfect medians is A001109. Let a_1,...,a_m be an (ordered) sequence of real numbers, then a term a_k is a perfect median if sum_{1<=j<k} a_j = sum_{k<j<=m} a_j. See Puzzle 1 in MSRI Emissary, Fall 2005. - Asher Auel (auela(AT)math.upenn.edu), Jan 12 2006

This is the r=8 member of the r-family of sequences S_r(n) defined in A092184 where more information can be found.

Also, 1^3+2^3+3^3+...+a(n)^3 = k(n)^4 where k(n) is A001109 - Anton Vrba (antonvrba(AT)yahoo.com), Nov 18 2006

The sequence lists the numbers n for which Sum_{i=0..n}{i} is a perfect square. - Paolo P. Lava (ppl(AT)spl.at), Nov 28 2007

Example: 8*9/2 = 36 = 6^2; 49*50/2 = 1225 = 35^2 [From Vincenzo Librandi (vincenzo.librandi(AT)tin.it), Jan 26 2009]

If T_x=y^2 is a triangular number which is also a square, the least both triangular and square number which is greater as T_x is T_(3*x+4*y+1)=(2*x+3*y+1)^2 (W. Sierpinski 1961). [From Richard Choulet (richardchoulet(AT)yahoo.fr), Apr 28 2009]

The remainder of the division of a(n) by 5 is: 0, 1, 3 or 4. The remainder of the division of a(n) by 7 is: 0 or 1. [From Mohamed Bouhamida (bhmd95(AT)yahoo.fr), Aug 26 2009]

Number of units of a(n) belongs to a periodic sequence: 0, 1, 8, 9, 8, 1. The remainder of the division of a(n) by 5 belongs to a periodic sequence: 0, 1, 3, 4, 3, 1. [From Mohamed Bouhamida (bhmd95(AT)yahoo.fr), Sep 01 2009]

If (a,b) is a solution of the Diophantine equation: 0+1+2+...+x=y^2 then a or (a+1) are perfect squares. If (a,b) is a solution of the Diophantine equation: 0+1+2+...+x=y^2 then a or a/8 are perfect squares. If (a,b) and (c,d) are two consecutive solutions of the Diophantine equation: 0+1+2+...+x=y^2 with a<c then a+b=c-d and ((d+b)^2,d^2-b^2) is a solution too. If (a,b), (c,d) and (e,f) are three consecutive solutions of the Diophantine equation: 0+1+2+...+x=y^2 with a<c<e then (8*d^2,d*(f-b)) is a solution too. [From Mohamed Bouhamida (bhmd95(AT)yahoo.fr), Aug 29 2009]

If (p,q) and (r,s) are two consecutive solutions of the Diophantine equation: 0+1+2+...+x=y^2 with p<r then r=3p+4q+1 and s=2p+3q+1. [From Mohamed Bouhamida (bhmd95(AT)yahoo.fr), Sep 02 2009]

Also numbers n such that (ceiling(sqrt(n*(n+1)/2)))^2 - n*(n+1)/2 = 0. [From Ctibor O. Zizka (c.zizka(AT)email.cz), Nov 10 2009]

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).

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.

I. Adler, Three Diophantine equations - Part II, Fib. Quart., 7 (1969), 181-193.

A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 193.

Elwyn Berlekamp and Joe P. Buhler, Puzzle Column, Emissary, MSRI Newsletter, Fall 2005. Problem 1.

L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 2, p. 10.

H. G. Forder, A Simple Proof of a Result on Diophantine Approximation, Math. Gaz., 47 (1963), 237-238.

M. S. Klamkin, "International Mathematical Olympiads 1978-1985," (Supplementary problem N.T.6)

P. Lafer, Discovering the square-triangular numbers, Fib. Quart., 9 (1971), 93-105.

W. Sierpinski, Pythagorean triangles, Dover Publications, Inc., Mineola, NY, 2003, pp. 21-22 MR2002669

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

Index entries for two-way infinite sequences

Index entries for sequences related to linear recurrences with constant coefficients

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.

L. Euler, De solutione problematum diophanteorum per numeros integros, Par. 19

MSRI newsletter, Emissary [This must to refer to a puzzle in one of the issues of the Emissary magazine - but which issue?]

D. L. Vestal, Review of "Pythagorean Triangles" (Chapter 4) by W. Sierpinski

Eric Weisstein's World of Mathematics, Link to a section of The World of Mathematics.

Eric Weisstein's World of Mathematics, Link to a section of The World of Mathematics.

Index entries for sequences related to Chebyshev polynomials.

a(0) = 0, a(n+1) = 3*a(n) + 1 + 2*sqrt(2*a(n)*(a(n)+1)). - Jim Nastos (nastos(AT)gmail.com), Jun 18 2002

a(n) = floor( (1/4) * (3+2*sqrt(2))^n ) - Benoit Cloitre (benoit7848c(AT)orange.fr), Sep 04 2002

a(n) = A001653(k)*A001653(k+n) - A001652(k)*A001652(k+n) -A046090(k)*A046090(k+n) - Charlie Marion (charliem(AT)bestweb.net), Jul 01 2003

a(n)=A001652(n-1)+A001653(n-1)=A001653(n)-A046090(n)=(A001541(n)-1)/2=a(-n). - Michael Somos Mar 03 2004

a_0 = 0, a_1 = 1, a_2 = 8, a_n = 7(a_{n-1} - a_{n-2}) + a_{n-3}. - Antonio Olivares, Oct 23 2003

a(n)=sum_(r=1, ..., n) 2^(r-1)*C(2n, 2r). - Lekraj Beedassy (blekraj(AT)yahoo.com), Aug 21 2004

If n>1, then both A000203[n] and A000203[n+1] are odd numbers: n is either square or twice square. - Labos E. (labos(AT)ana.sote.hu), Aug 23 2004

a(n)= (T(n, 3)-1)/2 with Chebyshev's polynomials of the first kind evaluated at x=3: T(n, 3)= A001541(n). W. Lang (wolfdieter.lang_AT_physik_DOT_uni-karlsruhe_DOT_de), Oct 18 2004

G.f.: A(x)=x*(1+x)/((1-x)*(1-6*x+x^2)); closed form: a(n)=((3+2*sqrt(2))^n +(3-2*sqrt(2))^n-2)/4 - Bruce Corrigan (scentman(AT)myfamily.com), Oct 26 2002

G.f.: x*(1+x)/(1-7*x+7*x^2-x^3).

a(n) = floor(sqrt(2*A001110(n)))=floor(A001109(n)*sqrt(2))=2*(A00012 9(n)^2)+[n mod 2]=A001333^2+1-[n mod 2] - Henry Bottomley, Apr 19, 2000

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

(1/2)(-1 + Sqrt[1 + 8(((3 + 2Sqrt[2])^n - (3 - 2Sqrt[2])^n)/(4Sqrt[2]))^2]]) - Artur Jasinski (grafix(AT)csl.pl), Dec 10 2006

A072221(n) = 3*a(n) + 1. - David Scheers, Dec 25 2006

a(1)=((3+2*sqrt(2))+(3-2*sqrt(2))-2)/4=(3+3-2)/4=4/4=1 a(2)=((3+2*sqrt(2))^2+(3-2*sqrt(2))^2-2)/4=(9+4*sqrt(2)+8+9-4*sqrt(2)+8-2)/4= (18+16-2)/4=(34-2)/4=32/4=8 etc.

A001108:=-(1+z)/(z-1)/(z**2-6*z+1); [Conjectured by S. Plouffe in his 1992 dissertation.]

Table[(1/2)(-1 + Sqrt[1 + Expand[8(((3 + 2Sqrt[2])^n - (3 - 2Sqrt[2])^n)/(4Sqrt[2]))^2]]), {n, 0, 100}] - Artur Jasinski (grafix(AT)csl.pl), Dec 10 2006

(PARI) a(n)=(real((3+quadgen(32))^n)-1)/2

(PARI) a(n)=(subst(poltchebi(abs(n)), x, 3)-1)/2

(PARI) a(n)=if(n<0, a(-n), (polsym(1-6*x+x^2, n)[n+1]-2)/4)

Cf. A001109, A001110, A007913, A000203, A084301, A001652, A072221.

Partial sums of A002315.

Sequence in context: A026774 A089383 A028443 this_sequence A097204 A037539 A037483

Adjacent sequences: A001105 A001106 A001107 this_sequence A001109 A001110 A001111

nonn,easy,nice

N. J. A. Sloane (njas(AT)research.att.com).

More terms from Larry Reeves (larryr(AT)acm.org), Apr 19 2000

More terms from Lekraj Beedassy (blekraj(AT)yahoo.com), Aug 21 2004