%I A001109 M4217 N1760
%S A001109 0,1,6,35,204,1189,6930,40391,235416,1372105,7997214,46611179,
%T A001109 271669860,1583407981,9228778026,53789260175,313506783024,
%U A001109 1827251437969,10650001844790,62072759630771,361786555939836
%N A001109 a(n)^2 is a triangular number: a(n) = 6*a(n-1) - a(n-2) with a(0)=0,
a(1)=1.
%C A001109 8*a(n)^2 + 1 is a perfect square. - Gregory V. Richardson (omomom(AT)hotmail.com),
Oct 05 2002
%C A001109 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
%C A001109 (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
%C A001109 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
%C A001109 n such that 8*n^2=floor(sqrt(8)*n*ceil(sqrt(8)*n)) Benoit Cloitre (benoit7848c(AT)orange.fr),
May 10 2003
%C A001109 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
%C A001109 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
%C A001109 Kekule numbers for certain benzenoids (see the Cyvin-Gutman reference).
- Emeric Deutsch (deutsch(AT)duke.poly.edu), Jun 19 2005
%C A001109 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
%C A001109 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
%C A001109 Self convolution of central Delannoy numbers (A001850) - Benoit Cloitre
(benoit7848c(AT)orange.fr), Sep 28 2005
%C A001109 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
%C A001109 One half the bisection of the Pell numbers (A000129). - Frank Adams-Watters
(FrankTAW(AT)Netscape.net), Jan 08 2006
%C A001109 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
%C A001109 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
%C A001109 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
%C A001109 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
%C A001109 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
%C A001109 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
%C A001109 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]
%C A001109 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]
%C A001109 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]
%C A001109 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]
%D A001109 N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences,
Academic Press, 1995 (includes this sequence).
%D A001109 N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973
(includes this sequence).
%D A001109 I. Adler, Three Diophantine equations - Part II, Fib. Quart., 7 (1969),
181-193.
%D A001109 A. Auel, MSRI Emissary, Fall 2005, Jan 12 (2006), p. 1.
%D A001109 A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964,
p. 193.
%D A001109 Elwyn Berlekamp and Joe P. Buhler, Puzzle Column, Emissary, MSRI Newsletter,
Fall 2005. Problem 1.
%D A001109 D. M. Burton, The History of Mathematics, McGraw Hill, (1991), p. 213.
%D A001109 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}).
%D A001109 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 A001109 H. G. Forder, A Simple Proof of a Result on Diophantine Approximation,
Math. Gaz., 47 (1963), 237-238.
%D A001109 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?]
%D A001109 H. Harborth, Fermat-like binomial equations, Applications of Fibonacci
numbers, Proc. 2nd Int. Conf., San Jose/Ca., August 1986, 1-5 (1988).
%D A001109 B. Hayes, Calculemus!, American Scientist, 96 (Sep-Oct 2008), 362-366.
%D A001109 P. Lafer, Discovering the square-triangular numbers, Fib. Quart., 9 (1971),
93-105.
%D A001109 R. A. Sulanke, Bijective recurrences concerning Schroeder paths, Electron.
J. Combin. 5 (1998), Research Paper 47, 11 pp.
%H A001109 T. D. Noe, <a href="b001109.txt">Table of n, a(n) for n=0..200</a>
%H A001109 <a href="Sindx_Tu.html#2wis">Index entries for two-way infinite sequences</
a>
%H A001109 <a href="Sindx_Rea.html#recLCC">Index entries for sequences related to
linear recurrences with constant coefficients</a>
%H A001109 Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/
RecursiveSequences.html">Recursive Sequences</a>
%H A001109 A. Bogomolny, <a href="http://www.cut-the-knot.org/do_you_know/triSquare.shtml">
There exist triangular numbers that are also squares</a>
%H A001109 John C. Butcher, <a href="http://www.math.auckland.ac.nz/~butcher/miniature/
miniature2.html">On Ramanujan, continued Fractions and an interesting
number</a>
%H A001109 L. Euler, <a href="http://math.dartmouth.edu/~euler/pages/E029.html">
De solutione problematum diophanteorum per numeros integros</a>,
Par. 19
%H A001109 Madras College, St Andrews, <a href="http://www.madras.fife.sch.uk/maths/
amazingnofacts/fact017.html">Square Triangular Numbers</a>
%H A001109 MSRI newsletter, <a href="http://www.msri.org/communications/emissary/
index_html">Emissary</a>
%H A001109 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 A001109 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 A001109 Rajesh Ram, <a href="http://users.tellurian.net/hsejar/maths/triangle">
Triangle Numbers that are Perfect Squares</a>
%H A001109 K. J. Ramsey, <a href="http://groups.yahoo.com/group/Triangular_Numbers/
message/23">Relation of Mersenne Primes To Square Triangular Numbers</
a>
%H A001109 A. Sandhya, <a href="http://www.angelfire.com/ak/ashoksandhya/maths2.html">
Puzzle 4: A problem Srinivasa Ramanujan, the famous 20th century
Indian Mathematician Solved</a>
%H A001109 Sci.math Newsgroup, <a href="http://www.math.niu.edu/~rusin/known-math/
98/sq_tri">Square numbers which are triangular</a>
%H A001109 R. A. Sulanke, <a href="http://math.boisestate.edu/~sulanke/recentpapindex.html">
Moments, Narayana numbers and the cut and paste for lattice paths</
a>
%H A001109 Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/
BinomialCoefficient.html">Link to a section of The World of Mathematics.</
a>
%H A001109 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 A001109 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 A001109 Wikipedia, <a href="http://en.wikipedia.org/wiki/Triangular_square_number">
Triangular square number</a>
%H A001109 Rick Young, <a href="http://www.cob.ohio-state.edu/~young_53/Quote.ram.html">
Relevant quotation from biography of Ramanujan</a>
%H A001109 <a href="Sindx_Ch.html#Cheby">Index entries for sequences related to
Chebyshev polynomials.</a>
%F A001109 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).
%F A001109 a(n) = 3*a(n-1)+sqrt(8*a(n-1)^2+1) - R. J. Mathar (mathar(AT)strw.leidenuniv.nl),
Oct 09 2000
%F A001109 a(n) = A000129(n)*A001333(n) = A000129(n)*(A000129(n)+A000129(n-1)) =
ceiling(A001108(n)/sqrt(2)) - Henry Bottomley, Apr 19 2000.
%F A001109 a(n) ~ 1/8*sqrt(2)*(sqrt(2) + 1)^(2*n) - Joe Keane (jgk(AT)jgk.org),
May 15 2002
%F A001109 Lim n -> inf. a(n)/a(n-1) = 3 + 2*sqrt(2). - Gregory V. Richardson (omomom(AT)hotmail.com),
Oct 05 2002
%F A001109 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
%F A001109 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
%F A001109 a(n) = floor((3+2sqrt(2))^n/(4sqrt(2))). - Lekraj Beedassy (blekraj(AT)yahoo.com),
Apr 23 2003
%F A001109 G.f.: x/(1-6x+x^2). a(n)=6a(n-1)-a(n-2). a(-n)=-a(n). - Michael Somos,
Apr 07 2003
%F A001109 For n>=1, a(n) = Sum_{k=0...n-1}A001653(k) - Charlie Marion (charliem(AT)bestweb.net),
Jul 01 2003
%F A001109 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
%F A001109 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
%F A001109 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
%F A001109 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
%F A001109 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
%F A001109 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
%F A001109 E.g.f. : exp(3x)sinh(2sqrt(2)x)/(2sqrt(2)). - Paul Barry (pbarry(AT)wit.ie),
Apr 21 2004
%F A001109 A053141(n+1) + A055997(n+1) = A001541(n+1) + a(n+1). - Creighton Dement
(creighton.k.dement(AT)uni-oldenburg.de), Sep 16 2004
%F A001109 a(n)=sum{k=0..n, binomial(2n, 2k+1)2^(k-1)} - Paul Barry (pbarry(AT)wit.ie),
Oct 01 2004
%F A001109 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
%F A001109 a(n)=sum_{k=0...n}A001850(k)*A001850(n-k) - Benoit Cloitre (benoit7848c(AT)orange.fr),
Sep 28 2005
%F A001109 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
%F A001109 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
%F A001109 ((3 + 2Sqrt[2])^n - (3 - 2Sqrt[2])^n)/(4Sqrt[2]), - Artur Jasinski (grafix(AT)csl.pl),
Dec 10 2006
%F A001109 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
%F A001109 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
%F A001109 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
%F A001109 [A001653(n), a(n)] = [1,4; 1,5]^n * [1,0]. - Gary W. Adamson (qntmpkt(AT)yahoo.com),
Mar 21 2008
%F A001109 a(n)=sum{k=0..n-1, 4^k*C(n+k,2k+1)}. [From Paul Barry (pbarry(AT)wit.ie),
Apr 20 2009]
%p A001109 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)
%p A001109 A001109:=1/(z**2-6*z+1); [S. Plouffe in his 1992 dissertation.]
%p A001109 with (combinat):seq(fibonacci(2*n,2)/2, n=0..20); - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com),
Apr 20 2008
%t A001109 Expand[Table[((3 + 2Sqrt[2])^n - (3 - 2Sqrt[2])^n)/(4Sqrt[2]), {n, 0,
30}]] - Artur Jasinski (grafix(AT)csl.pl), Dec 10 2006
%t A001109 lst = {}; Do[AppendTo[lst, GegenbauerC[n, 1, 3]], {n, -1, 19}]; lst [From
Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Jul 14 2009]
%o A001109 (PARI) a(n)=imag((3+quadgen(32))^n)
%o A001109 (PARI) a(n)=subst(poltchebi(abs(n+1))-3*poltchebi(abs(n)),x,3)/8
%o A001109 sage: [lucas_number1(n,6,1) for n in range(27)] - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com),
Jun 25 2008
%Y A001109 sqrt(A001110). Cf. A001108, A002315. a(n)=sqrt((A001541(n)^2-1)/8) (cf.
Richardson comment).
%Y A001109 2*a(n) = A001542.
%Y A001109 Cf. A001653.
%Y A001109 Sequence in context: A081105 A161727 A121838 this_sequence A144638 A117671
A000399
%Y A001109 Adjacent sequences: A001106 A001107 A001108 this_sequence A001110 A001111
A001112
%K A001109 nonn,easy,nice
%O A001109 0,3
%A A001109 N. J. A. Sloane (njas(AT)research.att.com).
%E A001109 Additional comments from Wolfdieter Lang (wolfdieter.lang(AT)physik.uni-karlsruhe.de),
Feb 10 2000
%E A001109 More terms from Larry Reeves (larryr(AT)acm.org), Apr 19 2000.
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