%I A001108 M4536 N1924
%S A001108 0,1,8,49,288,1681,9800,57121,332928,1940449,11309768,65918161,
%T A001108 384199200,2239277041,13051463048,76069501249,443365544448,
%U A001108 2584123765441,15061377048200,87784138523761,511643454094368
%N A001108 a(n)-th triangular number is a square: a(n+1) = 6*a(n)-a(n-1)+2, with 
               a(0) = 0, a(1) = 1.
%C A001108 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)
%C A001108 For n>1 gives solutions to A007913(2x)=A007913(x+1) - Benoit Cloitre 
               (benoit7848c(AT)orange.fr), Apr 07 2002
%C A001108 If (X,X+1,Z) is a Pythagorean triple, then Z-X-1 and Z+X is in the sequence.
%C A001108 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
%C A001108 This is the r=8 member of the r-family of sequences S_r(n) defined in 
               A092184 where more information can be found.
%C A001108 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
%C A001108 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
%C A001108 Example: 8*9/2 = 36 = 6^2; 49*50/2 = 1225 = 35^2 [From Vincenzo Librandi 
               (vincenzo.librandi(AT)tin.it), Jan 26 2009]
%C A001108 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]
%C A001108 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]
%C A001108 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]
%C A001108 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]
%C A001108 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]
%C A001108 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]
%D A001108 N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, 
               Academic Press, 1995 (includes this sequence).
%D A001108 N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 
               (includes this sequence).
%D A001108 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.
%D A001108 I. Adler, Three Diophantine equations - Part II, Fib. Quart., 7 (1969), 
               181-193.
%D A001108 A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, 
               p. 193.
%D A001108 Elwyn Berlekamp and Joe P. Buhler, Puzzle Column, Emissary, MSRI Newsletter, 
               Fall 2005. Problem 1.
%D A001108 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.
%D A001108 H. G. Forder, A Simple Proof of a Result on Diophantine Approximation, 
               Math. Gaz., 47 (1963), 237-238.
%D A001108 M. S. Klamkin, "International Mathematical Olympiads 1978-1985," (Supplementary 
               problem N.T.6)
%D A001108 P. Lafer, Discovering the square-triangular numbers, Fib. Quart., 9 (1971), 
               93-105.
%D A001108 W. Sierpinski, Pythagorean triangles, Dover Publications, Inc., Mineola, 
               NY, 2003, pp. 21-22 MR2002669
%H A001108 T. D. Noe, <a href="b001108.txt">Table of n, a(n) for n=0..200</a>
%H A001108 <a href="Sindx_Tu.html#2wis">Index entries for two-way infinite sequences</
               a>
%H A001108 <a href="Sindx_Rea.html#recLCC">Index entries for sequences related to 
               linear recurrences with constant coefficients</a>
%H A001108 S. Plouffe, <a href="http://www.lacim.uqam.ca/%7Eplouffe/articles/MasterThesis.pdf">
               Approximations de S\'{e}ries G\'{e}n\'{e}ratrices et Quelques Conjectures</
               a>, Dissertation, Universit\'{e} du Qu\'{e}bec \`{a} Montr\'{e}al, 
               1992.
%H A001108 S. Plouffe, <a href="http://www.lacim.uqam.ca/%7Eplouffe/articles/FonctionsGeneratrices.pdf">
               1031 Generating Functions and Conjectures</a>, Universit\'{e} du 
               Qu\'{e}bec \`{a} Montr\'{e}al, 1992.
%H A001108 L. Euler, <a href="http://math.dartmouth.edu/~euler/pages/E029.html">
               De solutione problematum diophanteorum per numeros integros</a>, 
               Par. 19
%H A001108 MSRI newsletter, <a href="http://www.msri.org/communications/emissary/
               index_html">Emissary</a> [This must to refer to a puzzle in one of 
               the issues of the Emissary magazine - but which issue?]
%H A001108 D. L. Vestal, <a href="http://www.maa.org/reviews/pythtriangles.html">
               Review of "Pythagorean Triangles" (Chapter 4) by W. Sierpinski</a>
%H A001108 Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/
               SquareTriangularNumber.html">Link to a section of The World of Mathematics.</
               a>
%H A001108 Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/
               TriangularNumber.html">Link to a section of The World of Mathematics.</
               a>
%H A001108 <a href="Sindx_Ch.html#Cheby">Index entries for sequences related to 
               Chebyshev polynomials.</a>
%F A001108 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
%F A001108 a(n) = floor( (1/4) * (3+2*sqrt(2))^n ) - Benoit Cloitre (benoit7848c(AT)orange.fr), 
               Sep 04 2002
%F A001108 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
%F A001108 a(n)=A001652(n-1)+A001653(n-1)=A001653(n)-A046090(n)=(A001541(n)-1)/2=a(-n). 
               - Michael Somos Mar 03 2004
%F A001108 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
%F A001108 a(n)=sum_(r=1, ..., n) 2^(r-1)*C(2n, 2r). - Lekraj Beedassy (blekraj(AT)yahoo.com), 
               Aug 21 2004
%F A001108 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
%F A001108 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
%F A001108 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
%F A001108 G.f.: x*(1+x)/(1-7*x+7*x^2-x^3).
%F A001108 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
%F A001108 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
%F A001108 (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
%F A001108 A072221(n) = 3*a(n) + 1. - David Scheers, Dec 25 2006
%e A001108 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.
%p A001108 A001108:=-(1+z)/(z-1)/(z**2-6*z+1); [Conjectured by S. Plouffe in his 
               1992 dissertation.]
%t A001108 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
%o A001108 (PARI) a(n)=(real((3+quadgen(32))^n)-1)/2
%o A001108 (PARI) a(n)=(subst(poltchebi(abs(n)),x,3)-1)/2
%o A001108 (PARI) a(n)=if(n<0,a(-n),(polsym(1-6*x+x^2,n)[n+1]-2)/4)
%Y A001108 Cf. A001109, A001110, A007913, A000203, A084301, A001652, A072221.
%Y A001108 Partial sums of A002315.
%Y A001108 Sequence in context: A026774 A089383 A028443 this_sequence A097204 A037539 
               A037483
%Y A001108 Adjacent sequences: A001105 A001106 A001107 this_sequence A001109 A001110 
               A001111
%K A001108 nonn,easy,nice,new
%O A001108 0,3
%A A001108 N. J. A. Sloane (njas(AT)research.att.com).
%E A001108 More terms from Larry Reeves (larryr(AT)acm.org), Apr 19 2000
%E A001108 More terms from Lekraj Beedassy (blekraj(AT)yahoo.com), Aug 21 2004