OFFSET
0,3
COMMENTS
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). - Darin Stephenson (stephenson(AT)cs.hope.edu) and Alan Koch
If (X,X+1,Z) is a Pythagorean triple, then Z-X-1 and Z+X are 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_{j=1..k-1} a_j = Sum_{j=k+1..m} a_j. See Puzzle 1 in MSRI Emissary, Fall 2005. - Asher Auel, 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
If T_x = y^2 is a triangular number which is also a square, the least number which is both triangular and square and greater than T_x is T_(3*x + 4*y + 1) = (2*x + 3*y + 1)^2 (W. Sierpiński 1961). - Richard Choulet, Apr 28 2009
If (a,b) is a solution of the Diophantine equation 0 + 1 + 2 + ... + x = y^2, then a or (a+1) is a perfect square. If (a,b) is a solution of the Diophantine equation 0 + 1 + 2 + ... + x = y^2, then a or a/8 is a perfect square. 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, 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, Sep 02 2009
Also numbers k such that (ceiling(sqrt(k*(k+1)/2)))^2 - k*(k+1)/2 = 0. - Ctibor O. Zizka, Nov 10 2009
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 J 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
Conjecture: For n>1 a(n) is the index of the first occurrence of -n in sequence A123737. - Vaclav Kotesovec, Jun 02 2015
For n=2*k, k>0, a(n) is divisible by 8 (deficient), so since all proper divisors of deficient numbers are deficient, then a(n) is deficient. For n=2*k+1, k>0, a(n) is odd. If a(n) is a prime number, it is deficient; otherwise a(n) has one or two distinct prime factors and is therefore deficient again. sigma(a(5)) = 1723 < 3362 = 2*a(5). In either case, a(n) is deficient. - Muniru A Asiru, Apr 14 2016
The squares of NSW numbers (A008843) interleaved with twice squares from A084703, where A008843(n) = A002315(n)^2 and A084703(n) = A001542(n)^2. Conjecture: Also numbers n such that sigma(n) = A000203(n) and sigma(n-th triangular number) = A074285(n) are both odd numbers. - Jaroslav Krizek, Aug 05 2016
For n > 0, numbers for which the number of odd divisors of both n and of n + 1 is odd. - Gionata Neri, Apr 30 2018
For n >= 2, a(n) is the base for which A058331(A001109(n)) is a length-3 repunit. Example: for n=2, A001109(2)=6 and A058331(6)=73 and 73 in base a(2)=8 is 111. See Grantham and Graves. - Michel Marcus, Sep 11 2020
REFERENCES
A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 193.
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.
M. S. Klamkin, "International Mathematical Olympiads 1978-1985," (Supplementary problem N.T.6)
W. Sierpiński, 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).
LINKS
Indranil Ghosh, Table of n, a(n) for n = 0..1304 (terms 0..200 from T. D. Noe)
Marco Abrate, Stefano Barbero, Umberto Cerruti, and Nadir Murru, Polynomial sequences on quadratic curves, Integers, Vol. 15, 2015, #A38.
I. Adler, Three Diophantine equations - Part II, Fib. Quart., 7 (1969), pp. 181-193.
M. A. Asiru, All square chiliagonal numbers, Int J Math Edu Sci Technol, 47:7(2016), 1123-1134.
Elwyn Berlekamp and Joe P. Buhler, Puzzle Column, Emissary, MSRI Newsletter, Fall 2005. Problem 1, (6 MB).
Zongyun Chen, Steven J. Miller, and Chenghan Wu, Geometric Proof of the Irrationality of Square-Roots for Select Integers, arXiv:2410.14434 [math.HO], 2024. See pp. 10-11.
L. Euler, De solutione problematum diophanteorum per numeros integros, Par. 19.
H. G. Forder, A Simple Proof of a Result on Diophantine Approximation, Math. Gaz., 47 (1963), 237-238.
Jon Grantham and Hester Graves, The abc Conjecture Implies That Only Finitely Many Cullen Numbers Are Repunits, arXiv:2009.04052 [math.NT], 2020.
D. B. Hayes, Calculemus!, American Scientist, 96 (Sep-Oct 2008), 362-366.
Olcay Karaatli and Refik Keskin, On some diophantine equations related to square triangular and balancing numbers, J. Alg. Number Theory 4 (2) (2010) 71-89.
Refik Keskin and Olcay Karaatli, Some New Properties of Balancing Numbers and Square Triangular Numbers, Journal of Integer Sequences, Vol. 15 (2012), Article #12.1.4.
P. Lafer, Discovering the square-triangular numbers, Fib. Quart., 9 (1971), 93-105.
Ioana-Claudia Lazăr, Lucas sequences in t-uniform simplicial complexes, arXiv:1904.06555 [math.GR], 2019.
MSRI newsletter, Emissary, Fall 2005.
Vladimir Pletser, Recurrent Relations for Multiple of Triangular Numbers being Triangular Numbers, arXiv:2101.00998 [math.NT], 2020.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992.
B. Polster and M. Ross, Marching in squares, arXiv preprint arXiv:1503.04658 [math.HO], 2015.
K. Ramsey, Generalized Proof re Square Triangular Numbers.
K. Ramsey, Generalized Proof re Square Triangular Numbers, digest of 2 messages in Triangular_and_Fibonacci_Numbers Yahoo group, May 27, 2005 - Oct 10, 2011.
Eric Weisstein's World of Mathematics, Square Triangular Number.
Eric Weisstein's World of Mathematics, Triangular Number.
H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences, Integers, Volume 12A (2012) The John Selfridge Memorial Volume.
Index entries for linear recurrences with constant coefficients, signature (7,-7,1).
FORMULA
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 Alberto 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 a square or twice a square. - Labos Elemer, 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*(A000129(n)^2) - (n mod 2) = A001333(n)^2 - 1 + (n mod 2). - Henry Bottomley, Apr 19 2000, corrected by Eric Rowland, Jun 23 2017
A072221(n) = 3*a(n) + 1. - David Scheers, Dec 25 2006
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 J Ramsey, Oct 13 2011
a(n) = A048739(2n-2), n > 0. - Richard R. Forberg, Aug 31 2013
From Peter Bala, Jan 28 2014: (Start)
A divisibility sequence: that is, a(n) divides a(n*m) for all n and m. Case P1 = 8, P2 = 12, Q = 1 of the 3-parameter family of linear divisibility sequences found by Williams and Guy.
(2,1) entry of the 2 X 2 matrix T(n,M), where M = [0, -3; 1, 4] and T(n,x) is the Chebyshev polynomial of the first kind. (End)
E.g.f.: exp(x)*(exp(2*x)*cosh(2*sqrt(2)*x) - 1)/2. - Stefano Spezia, Oct 25 2024
EXAMPLE
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.
MAPLE
A001108:=-(1+z)/(z-1)/(z**2-6*z+1); # Simon Plouffe in his 1992 dissertation, without the leading 0
MATHEMATICA
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]] (* Harvey P. Dale, Sep 04 2011 *)
LinearRecurrence[{7, -7, 1}, {0, 1, 8}, 50] (* Vladimir Joseph Stephan Orlovsky, Feb 12 2012 *)
PROG
(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)
(PARI) x='x+O('x^99); concat(0, Vec(x*(1+x)/((1-x)*(1-6*x+x^2)))) \\ Altug Alkan, May 01 2018
(Haskell)
a001108 n = a001108_list !! n
a001108_list = 0 : 1 : map (+ 2)
(zipWith (-) (map (* 6) (tail a001108_list)) a001108_list)
-- Reinhard Zumkeller, Jan 10 2012
(Magma) m:=30; R<x>:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(x*(1+x)/((1-x)*(1-6*x+x^2)))); // G. C. Greubel, Jul 15 2018
CROSSREFS
a(n) = A115598(n), n > 0. - Hermann Stamm-Wilbrandt, Jul 27 2014
KEYWORD
nonn,easy,nice
AUTHOR
EXTENSIONS
More terms from Larry Reeves (larryr(AT)acm.org), Apr 19 2000
More terms from Lekraj Beedassy, Aug 21 2004
STATUS
approved