0,3

8*a(n)^2 + 1 is a perfect square. - Gregory V. Richardson (omomom(AT)hotmail.com), Oct 05 2002

For n >= 2, A001108(n) gives exactly the positive integers m such that 1,2,...,m has a perfect median. The sequence of associated perfect medians is the present sequence. 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

(a(n),b(n)) where b(n)=A082291(n) are the integer solutions of the equation 2*binomial(b,a)=binomial(b+2,a). - Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de); comment revised by Michael Somos, Apr 07 2003

a(n) solves for y in x^2 - 8y^2 =1, or is the product xy, where (x,y) satisfies x^2 - 2y^2 = +-1, i.e. a(n)=A001333(n)*A000129(n). a(n) refers to inradius r of primitive Pythagorean triangles having consecutive legs, with corresponding semiperimeter s=a(n+1)={A001652(n)+A046090(n)+A001653(n)}/2 and area rs=A029549(n)=6*A029546(n). - Lekraj Beedassy (blekraj(AT)yahoo.com), Apr 23 2003

n such that 8*n^2=floor(sqrt(8)*n*ceil(sqrt(8)*n)) Benoit Cloitre (benoit7848c(AT)orange.fr), May 10 2003

For n>0, ratios a(n+1)/a(n) may be obtained as convergents to continued fraction expansion of 3+sqrt(8): either successive convergents of [6;-6] or odd convergents of [5;1, 4]. - Lekraj Beedassy (blekraj(AT)yahoo.com), Sep 09 2003

a(n+1) + A053141(n) = A001108(n+1). Generating floretion: - 2'i + 2'j - 'k + i' + j' - k' + 2'ii' - 'jj' - 2'kk' + 'ij' + 'ik' + 'ji' + 'jk' - 2'kj' + 2e ("jes" series) - Creighton Dement (creighton.k.dement(AT)uni-oldenburg.de), Dec 16 2004

Kekule numbers for certain benzenoids (see the Cyvin-Gutman reference). - Emeric Deutsch (deutsch(AT)duke.poly.edu), Jun 19 2005

Number of D steps on the line y=x in all Delannoy paths of length n (a Delannoy path of length n is a path from (0,0) to (n,n), consisting of steps E=(1,0), N=(0,1) and D=(1,1)). Example: a(2)=6 because in the 13 (=A001850(2)) Delannoy paths of length 2, namely (DD), (D)NE, (D)EN, NE(D), NENE, NEEN, NDE, NNEE, EN(D), ENNE, ENEN, EDN and EENN, we have alltogether six D steps on the line y=x (shown between parantheses). - Emeric Deutsch (deutsch(AT)duke.poly.edu), Jul 07 2005

Define a T-circle to be a first-quadrant circle with integral radius that is tangent to the x- and y-axes. Such a circle has coordinates equal to its radius. Let C(0) be the T-circle with radius 1. Then for n>0, define C(n) to be the smallest T-circle that does not intersect C(n-1). C(n) has radius a(n+1). Cf. A001653. - Charlie Marion (charliemath(AT)optonline.net), Sep 14 2005

Self convolution of central Delannoy numbers (A001850) - Benoit Cloitre (benoit7848c(AT)orange.fr), Sep 28 2005

Numbers such that there is an m with t(n+m)=2t(m), where t(n) are the triangular numbers A000217. For instance t(20)=2t(14)=210, so 6 is in the sequence. - Floor van Lamoen (fvlamoen(AT)hotmail.com), Oct 13 2005

One half the bisection of the Pell numbers (A000129). - Frank Adams-Watters (FrankTAW(AT)Netscape.net), Jan 08 2006

Pell trapezoids (cf. A084158); for n>0, a(n)=(A000129(n-1)+A000129(n+1))*A000129(n)/2; e.g. 204=(5+29)*12/2 - Charlie Marion (charliemath(AT)optonline.net), Apr 1 2006

Tested for 2<p<27: If and only if 2^p - 1 (the Mersenne number M(p)) is prime then M(p) divides a(2^(p-1)). - Kenneth J. Ramsey (RamseyKK2(AT)aol.com), May 16 2006

If 2^p - 1 is prime then M(p) divides a(2^(p-1)-1). Kenneth Ramsey (RamseyKK2(AT)aol.com), Jun 08 2006; comment corrected by Robert B. Israel (israel(AT)math.ubc.ca), Mar 18 2007

If 8n+5 and 8n+7 are twin primes then their product divides a(4n+3). - Kenneth Ramsey (RamseyKK2(AT)aol.com), Jun 08 2006

If p is an odd prime, then if p == 1 or 7 mod 8, then a((p-1)/2) == 0 mod p and a((p+1)/2) == 1 mod p; if p == 3 or 5 mod 8, then a((p-1)/2) == 1 mod p and a((p+1)/2) == 0 mod p. Kenneth Ramsey's comment about twin primes follows from this. - Robert B. Israel (israel(AT)math.ubc.ca), Mar 18 2007

a(n)*[a(n+b) - a(b-2)] = [a(n+1)+1]*[a(n+b-1) - a(b-1)] This identity also applies to any series a(0) = 0 a(1) = 1 a(n) = b*a(n-1) - a(n-2). - Kenneth J Ransey (Ramsey2879(AT)msn.com), Oct 17 2007

The remainder of the division of a(n) by 5 is: 0, 1 or 4. The remainder of the division of a(n) by 7 is: 0, 1 or 6. [From Mohamed Bouhamida (bhmd95(AT)yahoo.fr), Aug 26 2009]

Number of units of a(n) belongs to a periodic sequence: 0, 1, 6, 5, 4, 9. The remainder of the division of a(n) by 5 belongs to a periodic sequence: 0, 1, 1, 0, 4, 4. [From Mohamed Bouhamida (bhmd95(AT)yahoo.fr), Sep 01 2009]

Sequence gives y values of the Diophantine equation: 0+1+2+...+x=y^2. 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]

a(n)/A002315(n) converges to cos^2(Pi/8) = 1/2 + 2^(1/2)/4 [From Gary Detlefs (gdetlefs(AT)aol.com), Nov 25 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).

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

A. Auel, MSRI Emissary, Fall 2005, Jan 12 (2006), p. 1.

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.

D. M. Burton, The History of Mathematics, McGraw Hill, (1991), p. 213.

S. J. Cyvin and I. Gutman, Kekule structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (pp. 301, 302, P_{13}).

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.

P. Franklin, E. F. Beckenbach, H. S. M Coxeter, N. H. McCoy, K. Menger and J. L. Synge, The Carus Mathematical Monographs, The Mathematical Association of America, (1967), pp. 144-146 [Title of book?]

H. Harborth, Fermat-like binomial equations, Applications of Fibonacci numbers, Proc. 2nd Int. Conf., San Jose/Ca., August 1986, 1-5 (1988).

B. Hayes, Calculemus!, American Scientist, 96 (Sep-Oct 2008), 362-366.

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

R. A. Sulanke, Bijective recurrences concerning Schroeder paths, Electron. J. Combin. 5 (1998), Research Paper 47, 11 pp.

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

Tanya Khovanova, Recursive Sequences

A. Bogomolny, There exist triangular numbers that are also squares

John C. Butcher, On Ramanujan, continued Fractions and an interesting number

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

Madras College, St Andrews, Square Triangular Numbers

MSRI newsletter, Emissary

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.

Rajesh Ram, Triangle Numbers that are Perfect Squares

K. J. Ramsey, Relation of Mersenne Primes To Square Triangular Numbers

A. Sandhya, Puzzle 4: A problem Srinivasa Ramanujan, the famous 20th century Indian Mathematician Solved

Sci.math Newsgroup, Square numbers which are triangular

R. A. Sulanke, Moments, Narayana numbers and the cut and paste for lattice paths

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.

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

Wikipedia, Triangular square number

Rick Young, Relevant quotation from biography of Ramanujan

Index entries for sequences related to Chebyshev polynomials.

a(n) = S(n-1, 6) = U(n-1, 3) with U(n, x) Chebyshev's polynomials of the second kind. S(-1, x) := 0. Cf. triangle A049310 for S(n, x).

a(n) = 3*a(n-1)+sqrt(8*a(n-1)^2+1) - R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Oct 09 2000

a(n) = A000129(n)*A001333(n) = A000129(n)*(A000129(n)+A000129(n-1)) = ceiling(A001108(n)/sqrt(2)) - Henry Bottomley, Apr 19 2000.

a(n) ~ 1/8*sqrt(2)*(sqrt(2) + 1)^(2*n) - Joe Keane (jgk(AT)jgk.org), May 15 2002

Lim n -> inf. a(n)/a(n-1) = 3 + 2*sqrt(2). - Gregory V. Richardson (omomom(AT)hotmail.com), Oct 05 2002

a(n) = [(3 + sqrt(8))^(n-1) - [(3 - sqrt(8))^(n-1)] / (2*sqrt(8)). - Gregory V. Richardson (omomom(AT)hotmail.com), Oct 13 2002

a(n)=((3+2sqrt(2))^n-(3-2sqrt(2))^n)/(4sqrt(2)). a(2n)=a(n)*A003499(n). 4a(n)=A005319(n). - Mario Catalani (mario.catalani(AT)unito.it), Mar 21 2003

a(n) = floor((3+2sqrt(2))^n/(4sqrt(2))). - Lekraj Beedassy (blekraj(AT)yahoo.com), Apr 23 2003

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

For n>=1, a(n) = Sum_{k=0...n-1}A001653(k) - Charlie Marion (charliem(AT)bestweb.net), Jul 01 2003

For n > 0, 4*a(2n) = A001653(n)^2 - A001653(n-1)^2; e.g. 4*204 = 29^2 - 5^2 - Charlie Marion (charliem(AT)bestweb.net), Jul 16 2003

For n>0, a(n)=sum_{k = 0...n-1}((2k+1)*A001652(n-1-k))+A000217(n) e.g. 204=1*119+3*20+5*3+7*0+10 - Charlie Marion (charliem(AT)bestweb.net), Jul 18 2003

a(2n+1)=a(n+1)^2-a(n)^2; e.g. 40391=204^2-35^2 - Charlie Marion (charliemath(AT)verizon.net), Jan 12 2004

a(k)*a(2n+k)=a(n+k)^2-a(n)^2; e.g. 204*7997214=40391^2-35^2 - Charlie Marion (charliemath(AT)verizon.net), Jan 15 2004

For j<n+1, a(k+j)*a(2n+k-j)-sum_{i = 0...j-1}a(2n-(2i+1)) = a(n+k)^2-a(n)^2; e.g. 1189*40391-(1189+350) = 6930^2-35^2 - Charlie Marion (charliemath(AT)verizon.net), Jan 18 2004

a(n)=A000129(2n)/2; a(n) := ((1+sqrt(2))^(2n)-(1-sqrt(2))^(2n))sqrt(2)/8; a(n) := sum{i=0..n, sum{j=0..n, A000129(i+j)*n!/(i!j!(n-i-j)!)/2}}. - Paul Barry (pbarry(AT)wit.ie), Feb 06 2004

E.g.f. : exp(3x)sinh(2sqrt(2)x)/(2sqrt(2)). - Paul Barry (pbarry(AT)wit.ie), Apr 21 2004

A053141(n+1) + A055997(n+1) = A001541(n+1) + a(n+1). - Creighton Dement (creighton.k.dement(AT)uni-oldenburg.de), Sep 16 2004

a(n)=sum{k=0..n, binomial(2n, 2k+1)2^(k-1)} - Paul Barry (pbarry(AT)wit.ie), Oct 01 2004

a(n+1) = A001653(n+1) - A038723(n+1) (conjecture); (a(n)) = chuseq[J]( 'ii' + 'jj' + .5'kk' + 'ij' - 'ji' + 2.5e ), apart from initial term. - Creighton Dement (creighton.k.dement(AT)uni-oldenburg.de), Nov 19 2004

a(n)=sum_{k=0...n}A001850(k)*A001850(n-k) - Benoit Cloitre (benoit7848c(AT)orange.fr), Sep 28 2005

a_n = 7(a(n-1) - a(n-2)) + a(n-3), a(1) = 0, a(2) = 1, a(3) = 6, n > 3. Also a(n) = [ (1 + sqrt(2) )^2n - (1 - sqrt(2) )^2n ] / [4*sqrt(2)]. - Antonio Olivares, Oct 23 2003

a(n) = 5*(a(n-1)+a(n-2))-a(n-3). a(n) = 7*(a(n-1)-a(n-2))+a(n-3). - Mohamed Bouhamida (bhmd95(AT)yahoo.fr), Sep 20 2006

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

Define f[x,s] = s x + Sqrt[(s^2-1)x^2+1]; f[0,s]=0. a(n) = f[a(n-1),3]. - Marcos Carreira, Dec 27 2006

The Pell numbers A(000129) are defined by P(0)=0, P(1)=1; for n > 1, P(n) =2*P(n-1) +P(n-2). The perfect median m(n) can be expressed in terms of the Pell numbers by m(n) = P( n + 2) * ( P ( n + 2) + (P (n + 1)) for n >= 0. - Winston A. Richards (ugu(AT)psu.edu), Jun 11 2007

For k = 0,1,...,n, a(2n-k)-a(k)=2*a(n-k)*A001541(n); e.g., if n=5 and k=3, a(7)-a(3)= 40391-35=2*6*3363; also, a(2n+1-k)-a(k)=A002315(n-k)*A001653(n); e.g., if n=5 and k=3, a(8)-a(3)= 235416-35=41*5741 - Charlie Marion (charliemath(AT)optonline.net), Jul 18 2007

[A001653(n), a(n)] = [1,4; 1,5]^n * [1,0]. - Gary W. Adamson (qntmpkt(AT)yahoo.com), Mar 21 2008

a(n)=sum{k=0..n-1, 4^k*C(n+k,2k+1)}. [From Paul Barry (pbarry(AT)wit.ie), Apr 20 2009]

a[0]:=1: a[1]:=6: for n from 2 to 26 do a[n]:=6*a[n-1]-a[n-2] od: seq(a[n], n=0..26); (Deutsch)

A001109:=1/(z**2-6*z+1); [S. Plouffe in his 1992 dissertation.]

with (combinat):seq(fibonacci(2*n, 2)/2, n=0..20); - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Apr 20 2008

Expand[Table[((3 + 2Sqrt[2])^n - (3 - 2Sqrt[2])^n)/(4Sqrt[2]), {n, 0, 30}]] - Artur Jasinski (grafix(AT)csl.pl), Dec 10 2006

lst = {}; Do[AppendTo[lst, GegenbauerC[n, 1, 3]], {n, -1, 19}]; lst [From Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Jul 14 2009]

(PARI) a(n)=imag((3+quadgen(32))^n)

(PARI) a(n)=subst(poltchebi(abs(n+1))-3*poltchebi(abs(n)), x, 3)/8

sage: [lucas_number1(n, 6, 1) for n in range(27)] - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Jun 25 2008

sqrt(A001110). Cf. A001108, A002315. a(n)=sqrt((A001541(n)^2-1)/8) (cf. Richardson comment).

2*a(n) = A001542.

Cf. A001653.

Sequence in context: A081105 A161727 A121838 this_sequence A144638 A117671 A000399

Adjacent sequences: A001106 A001107 A001108 this_sequence A001110 A001111 A001112

nonn,easy,nice,new

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

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

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