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A001108
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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.
(Formerly M4536 N1924)
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41
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0, 1, 8, 49, 288, 1681, 9800, 57121, 332928, 1940449, 11309768, 65918161, 384199200, 2239277041, 13051463048, 76069501249, 443365544448, 2584123765441, 15061377048200, 87784138523761, 511643454094368, 2982076586042449, 17380816062160328, 101302819786919521
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OFFSET
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0,3
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COMMENTS
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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, 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, Nov 28 2007
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). [Richard Choulet, 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. [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. [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. [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. [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. [Ctibor O. Zizka, Nov 10 2009]
Contribution from Lekraj Beedassy, Mar 04 2011: (Start)
Let x=a(n) be the index of the associated triangular number T_x=1+2+3+...+x and y=A001109(n) be the base of the associated perfect square S_y=y^2. Now using the identity S_y = T_y + T_{y-1}, the defining T_x = S_y may be rewritten as T_y = T_x - T_{y-1}, or 1+2+3+...+y = y+(y+1)+...+x. This solves the Strand Magazine House Number problem mentioned in A001109 in references from Poo-Sung Park and John C. Butcher. In a variant of the problem, solving the equation 1+3+5+...+(2*x+1) = (2*x+1)+(2*x+3)+...+(2*y-1) implies S_(x+1) = S_y - S_x, i.e., with (x,x+1,y) forming a Pythagorean triple, the solutions are given by pairs of x=A001652(n), y=A001653(n). (End)
If P = 8*n +/- 1 is a prime, then P divides a((P-1)/2); e.g. 7 divides a(3) and 41 divides a(20). Also, if P = 8*n +/- 3 is prime, then 4*P divides (a((P-1)/2) + a((P+1)/2 +3). [Kenneth Ramsey, Mar 05 2012]
Starting at a(2), a(n) gives all the dimensions of Euclidean k-space in which the ratio of outer to inner Soddy hyperspheres' radii for k+1 identical kissing hyperspheres is rational. The formula for this ratio is (1+3k+2*Sqrt(2k*(k+1)))/(k-1) where k is the dimension. So for a(3) = 49, the ratio is 6 in the 49th dimension. See comment for A010502. - Frank M Jackson, Feb 09 2013
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REFERENCES
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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
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
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LINKS
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T. D. Noe, Table of n, a(n) for n=0..200
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?]
_Simon 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.
_Simon Plouffe_, 1031 Generating Functions and Conjectures, Universit\'{e} du Qu\'{e}bec \`{a} Montr\'{e}al, 1992.
D. L. Vestal, Review of "Pythagorean Triangles" (Chapter 4) by W. Sierpinski
Eric Weisstein's World of Mathematics, Square Triangular Number.
Eric Weisstein's World of Mathematics, Triangular Number.
Index entries for sequences related to Chebyshev polynomials.
Index entries for sequences related to linear recurrences with constant coefficients, signature (7,-7,1).
Index entries for two-way infinite sequences
K. Ramsey, Generalized Proof re Square Triangular Numbers
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FORMULA
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a(0) = 0, a(n+1) = 3*a(n) + 1 + 2*sqrt(2*a(n)*(a(n)+1)). - Jim Nastos, Jun 18 2002
a(n) = floor( (1/4) * (3+2*sqrt(2))^n ) - Benoit Cloitre, Sep 04 2002
a(n) = A001653(k)*A001653(k+n) - A001652(k)*A001652(k+n) -A046090(k)*A046090(k+n) - Charlie Marion, 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_n = 7*a(n-1) -7*a(n-2) +a(n-3). - Antonio Olivares, Oct 23 2003
a(n)=sum_(r=1, ..., n) 2^(r-1)*binomial(2n, 2r). - Lekraj Beedassy, 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). Wolfdieter Lang, Oct 18 2004
G.f.: x*(1+x)/((1-x)*(1-6*x+x^2)). Binet form: a(n)=((3+2*sqrt(2))^n +(3-2*sqrt(2))^n-2)/4. - Bruce Corrigan (scentman(AT)myfamily.com), Oct 26 2002
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
A072221(n) = 3*a(n) + 1. - David Scheers, Dec 25 2006
A028982(a(n))+1=A028982(a(n)+1). [From Juri-Stepan Gerasimov, Mar 28 2011]
a(n+1)^2+a(n)^2+1=6*a(n+1)*a(n)+2*a(n+1)+2*a(n). - Charlie Marion, Sep 28 2011
a(n) = 2*A001653(m)*A053141(n-m-1) + A002315(m)*A046090(n-m-1) + a(m) with m<n; otherwise, a(n) = 2*A001653(m)*A053141(m-n) - A002315(m)*A001652(m-n) + a(m). See Link to Generalized Proof re Square Triangular Numbers. - _Kenneth Ramsey_, Oct 13 2011
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EXAMPLE
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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.
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MAPLE
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A001108:=-(1+z)/(z-1)/(z**2-6*z+1); [Simon Plouffe in his 1992 dissertation without the leading 0.]
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MATHEMATICA
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Table[(1/2)(-1 + Sqrt[1 + Expand[8(((3 + 2Sqrt[2])^n - (3 - 2Sqrt[2])^n)/(4Sqrt[2]))^2]]), {n, 0, 100}] - Artur Jasinski, Dec 10 2006
Transpose[NestList[{#[[2]], #[[3]], 6#[[3]]-#[[2]]+2}&, {0, 1, 8}, 20]][[1]] (* From Harvey P. Dale, Sep 04 2011 *)
LinearRecurrence[{7, -7, 1}, {0, 1, 8}, 50] (* From Vladimir Joseph Stephan Orlovsky, Feb 12 2012 *)
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PROG
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(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)
(Haskell)
a001108 n = a001108_list !! n
a001108_list = 0 : 1 : map (+ 2)
(zipWith (-) (map (* 6) (tail a001108_list)) a001108_list)
-- Reinhard Zumkeller, 10 Jan 2012
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CROSSREFS
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Cf. A001109, A001110, A007913, A000203, A084301, A001652, A072221.
Partial sums of A002315.
Sequence in context: A089383 A200660 A028443 * A097204 A037539 A037483
Adjacent sequences: A001105 A001106 A001107 * A001109 A001110 A001111
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KEYWORD
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nonn,easy,nice
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AUTHOR
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N. J. A. Sloane.
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EXTENSIONS
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More terms from Larry Reeves (larryr(AT)acm.org), Apr 19 2000
More terms from Lekraj Beedassy, Aug 21 2004
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STATUS
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approved
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