A115598 -id:A115598 - cp's OEIS frontend

A001542 a(n) = 6*a(n-1) - a(n-2) for n > 1, a(0)=0 and a(1)=2.

0, 2, 12, 70, 408, 2378, 13860, 80782, 470832, 2744210, 15994428, 93222358, 543339720, 3166815962, 18457556052, 107578520350, 627013566048, 3654502875938, 21300003689580, 124145519261542, 723573111879672

Offset: 0

N. J. A. Sloane

Consider the equation core(x) = core(2x+1) where core(x) is the smallest number such that x*core(x) is a square: solutions are given by a(n)^2, n > 0. - Benoit Cloitre, Apr 06 2002
Terms > 0 give numbers k which are solutions to the inequality |round(sqrt(2)*k)/k - sqrt(2)| < 1/(2*sqrt(2)*k^2). - Benoit Cloitre, Feb 06 2006
Also numbers m such that A125650(6*m^2) is an even perfect square, where A124650(m) is a numerator of m*(m+3)/(4*(m+1)*(m+2)) = Sum_{k=1..m} 1/(k*(k+1)*(k+2)). Sequence A033581 is a bisection of A125651. - Alexander Adamchuk, Nov 30 2006
The upper principal convergents to 2^(1/2), beginning with 3/2, 17/12, 99/70, 577/408, comprise a strictly decreasing sequence; essentially, numerators = A001541 and denominators = {a(n)}. - Clark Kimberling, Aug 26 2008
Even Pell numbers. - Omar E. Pol, Dec 10 2008
Numbers k such that 2*k^2+1 is a square. - Vladimir Joseph Stephan Orlovsky, Feb 19 2010
These are the integer square roots of the Half-Squares, A007590(k), which occur at values of k given by A001541. Also the numbers produced by adding m + sqrt(floor(m^2/2) + 1) when m is in A002315. See array in A227972. - Richard R. Forberg, Aug 31 2013
A001541(n)/a(n) is the closest rational approximation of sqrt(2) with a denominator not larger than a(n), and 2*a(n)/A001541(n) is the closest rational approximation of sqrt(2) with a numerator not larger than 2*a(n). These rational approximations together with those obtained from the sequences A001653 and A002315 give a complete set of closest rational approximations of sqrt(2) with restricted numerator as well as denominator. - A.H.M. Smeets, May 28 2017
Conjecture: Numbers k such that c/m < k for all natural a^2 + b^2 = c^2 (Pythagorean triples), a < b < c and a+b+c = m. Numbers which correspondingly minimize c/m are A002939. - Lorraine Lee, Jan 31 2020
All of the positive integer solutions of a*b + 1 = x^2, a*c + 1 = y^2, b*c + 1 = z^2, x + z = 2*y, 0 < a < b < c are given by a=a(n), b=A005319(n), c=a(n+1), x=A001541(n), y=A001653(n+1), z=A002315(n) with 0 < n. - Michael Somos, Jun 26 2022

			G.f. = 2*x + 12*x^2 + 70*x^3 + 408*x^4 + 2378*x^5 + 13860*x^6 + ...

Jay Kappraff, Beyond Measure, A Guided Tour Through Nature, Myth and Number, World Scientific, 2002; pp. 480-481.
Thomas Koshy, Fibonacci and Lucas Numbers with Applications, 2001, Wiley, pp. 77-79.
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).
James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, pages 257-258.
P.-F. Teilhet, Query 2376, L'Intermédiaire des Mathématiciens, 11 (1904), 138-139. - N. J. A. Sloane, Mar 08 2022

T. D. Noe, Table of n, a(n) for n = 0..100
I. Adler, Three Diophantine equations - Part II, Fib. Quart., 7 (1969), 181-193.
Christian Aebi and Grant Cairns, Lattice Equable Parallelograms, arXiv:2006.07566 [math.NT], 2020.
Hacène Belbachir, Soumeya Merwa Tebtoub, László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
H. Brocard, Notes élémentaires sur le problème de Peel, Nouvelle Correspondance Mathématique, 4 (1878), 161-169.
A. J. C. Cunningham, Binomial Factorisations, Vols. 1-9, Hodgson, London, 1923-1929. See Vol. 1, page xxxv.
S. Falcon, Relationships between Some k-Fibonacci Sequences, Applied Mathematics, 2014, 5, 2226-2234.
R. J. Hetherington, Letter to N. J. A. Sloane, Oct 26 1974.
J. M. Katri and D. R. Byrkit, Problem E1976, Amer. Math. Monthly, 75 (1968), 683-684.
Tanya Khovanova, Recursive Sequences.
E. Kilic, Y. T. Ulutas, and N. Omur, A Formula for the Generating Functions of Powers of Horadam's Sequence with Two Additional Parameters, J. Int. Seq. 14 (2011) #11.5.6, Table 3, k=1.
D. H. Lehmer, On the multiple solutions of the Pell equation, Annals Math., 30 (1928), 66-72.
D. H. Lehmer, On the multiple solutions of the Pell equation (annotated scanned copy).
Mathematical Reflections, Solution to Problem O271, Issue 5, 2013, p 22.
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.
Mark A. Shattuck, Tiling proofs of some formulas for the Pell numbers of odd index, Integers, 9 (2009), 53-64.
R. A. Sulanke, Moments, Narayana numbers and the cut and paste for lattice paths.
Soumeya M. Tebtoub, Hacène Belbachir, and László Németh, Integer sequences and ellipse chains inside a hyperbola, Proceedings of the 1st International Conference on Algebras, Graphs and Ordered Sets (ALGOS 2020), hal-02918958 [math.cs], 17-18.
Index entries for linear recurrences with constant coefficients, signature (6,-1).

Bisection of Pell numbers A000129: {a(n)} and A001653(n+1), n >= 0.

Cf. A001108, A001353, A001541, A001835, A003499, A007805, A007913, A115598, A125650, A125651, A125652.

a:=[0,2];; for n in [3..20] do a[n]:=6*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Dec 23 2019

a001542 n = a001542_list !! n
a001542_list =
   0 : 2 : zipWith (-) (map (6 *) $ tail a001542_list) a001542_list
-- Reinhard Zumkeller, Aug 14 2011

I:=[0,2]; [n le 2 select I[n] else 6*Self(n-1) -Self(n-2): n in [1..20]]; // G. C. Greubel, Dec 23 2019

A001542:=2*z/(1-6*z+z**2); # conjectured by Simon Plouffe in his 1992 dissertation
seq(combinat:-fibonacci(2*n, 2), n = 0..20); # Peter Luschny, Jun 28 2018

LinearRecurrence[{6, -1}, {0, 2}, 30] (* Harvey P. Dale, Jun 11 2011 *)
Fibonacci[2*Range[0,20], 2] (* G. C. Greubel, Dec 23 2019 *)
Table[2 ChebyshevU[-1 + n, 3], {n, 0, 20}] (* Herbert Kociemba, Jun 05 2022 *)

a[0]:0$
a[1]:2$
a[n]:=6*a[n-1]-a[n-2]$
A001542(n):=a[n]$
makelist(A001542(x),x,0,30); /* Martin Ettl, Nov 03 2012 */

{a(n) = imag( (3 + 2*quadgen(8))^n )}; /* Michael Somos, Jan 20 2017 */

vector(21, n, 2*polchebyshev(n-1, 2, 33) ) \\ G. C. Greubel, Dec 23 2019

l=[0, 2]
for n in range(2, 51): l+=[6*l[n - 1] - l[n - 2], ]
print(l) # Indranil Ghosh, Jun 06 2017

[2*chebyshev_U(n-1,3) for n in (0..20)] # G. C. Greubel, Dec 23 2019

a(n) = 2*A001109(n).
a(n) = ((3+2*sqrt(2))^n - (3-2*sqrt(2))^n) / (2*sqrt(2)).
G.f.: 2*x/(1-6*x+x^2).
a(n) = sqrt(2*(A001541(n))^2 - 2)/2. - Barry E. Williams, May 07 2000
a(n) = (C^(2n) - C^(-2n))/sqrt(8) where C = sqrt(2) + 1. - Gary W. Adamson, May 11 2003
For all terms x of the sequence, 2*x^2 + 1 is a square. Limit_{n->oo} a(n)/a(n-1) = 3 + 2*sqrt(2). - Gregory V. Richardson, Oct 10 2002
For n > 0: a(n) = A001652(n) + A046090(n) - A001653(n); e.g., 70 = 119 + 120 - 169. Also a(n) = A001652(n - 1) + A046090(n - 1) + A001653(n - 1); e.g., 70 = 20 + 21 + 29. Also a(n)^2 + 1 = A001653(n - 1)*A001653(n); e.g., 12^2 + 1 = 145 = 5*29. Also a(n + 1)^2 = A084703(n + 1) = A001652(n)*A001652(n + 1) + A046090(n)*A046090(n + 1). - Charlie Marion, Jul 01 2003
a(n) = ((1+sqrt(2))^(2*n) - (1-sqrt(2))^(2*n))/(2*sqrt(2)). - Antonio Alberto Olivares, Dec 24 2003
2*A001541(k)*A001653(n)*A001653(n+k) = A001653(n)^2 + A001653(n+k)^2 + a(k)^2; e.g., 2*3*5*29 = 5^2 + 29^2 + 2^2; 2*99*29*5741 = 29^2 + 5741^2 + 70^2. - Charlie Marion, Oct 12 2007
a(n) = sinh(2*n*arcsinh(1))/sqrt(2). - Herbert Kociemba, Apr 24 2008
For n > 0, a(n) = A001653(n) + A002315(n-1). - Richard R. Forberg, Aug 31 2013
a(n) = 3*a(n-1) + 2*A001541(n-1); e.g., a(4) = 70 = 3*12 + 2*17. - Zak Seidov, Dec 19 2013
a(n)^2 + 1^2 = A115598(n)^2 + (A115598(n)+1)^2. - Hermann Stamm-Wilbrandt, Jul 27 2014
Sum {n >= 1} 1/( a(n) + 1/a(n) ) = 1/2. - _Peter Bala, Mar 25 2015
E.g.f.: exp(3*x)*sinh(2*sqrt(2)*x)/sqrt(2). - Ilya Gutkovskiy, Dec 07 2016
A007814(a(n)) = A001511(n). See Mathematical Reflections link. - Michel Marcus, Jan 06 2017
a(n) = -a(-n) for all n in Z. - Michael Somos, Jan 20 2017
From A.H.M. Smeets, May 28 2017: (Start)
A051009(n) = a(2^(n-2)).
a(2n) = 2*a(2)*A001541(n).
A001541(n)/a(n) > sqrt(2) > 2*a(n)/A001541(n). (End)
a(A298210(n)) = A002349(2*n^2). - A.H.M. Smeets, Jan 25 2018
a(n) = A000129(n)*A002203(n). - Adam Mohamed, Jul 20 2024

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.

Original entry on oeis.org

0, 1, 8, 49, 288, 1681, 9800, 57121, 332928, 1940449, 11309768, 65918161, 384199200, 2239277041, 13051463048, 76069501249, 443365544448, 2584123765441, 15061377048200, 87784138523761, 511643454094368, 2982076586042449, 17380816062160328, 101302819786919521

Offset: 0

N. J. A. Sloane

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
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 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
a(n) will be solutions to some (A000217(k) + A000217(k+1))/2. - Art Baker, Jul 16 2019
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

			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.

A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 193.
John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 204.
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).
James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, pages 257-258.

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).
Henk Bruin and Robbert Fokkink, On the records and zeros of a deterministic random walk, arXiv:2503.11734 [math.DS], 2025. See p. 5.
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.
Leonhard 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.
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D. L. Vestal, Review of "Pythagorean Triangles" (Chapter 4) by W. Sierpiński.
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Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (7,-7,1).
Index entries for two-way infinite sequences

Cf. A000217, A000290, A001109, A001110, A007913, A000203, A084301, A001652, A072221.
Partial sums of A002315. A000129, A005319.
a(n) = A115598(n), n > 0. - Hermann Stamm-Wilbrandt, Jul 27 2014

a001108 n = a001108_list !! n
a001108_list = 0 : 1 : map (+ 2)
   (zipWith (-) (map (* 6) (tail a001108_list)) a001108_list)
-- Reinhard Zumkeller, Jan 10 2012

m:=30; R:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(x*(1+x)/((1-x)*(1-6*x+x^2)))); // G. C. Greubel, Jul 15 2018

A001108:=-(1+z)/(z-1)/(z**2-6*z+1); # Simon Plouffe in his 1992 dissertation, without the leading 0

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 *)

PARI
```
a(n)=(real((3+quadgen(32))^n)-1)/2
```

a(n)=(subst(poltchebi(abs(n)),x,3)-1)/2

a(n)=if(n<0,a(-n),(polsym(1-6*x+x^2,n)[n+1]-2)/4)

x='x+O('x^99); concat(0, Vec(x*(1+x)/((1-x)*(1-6*x+x^2)))) \\ Altug Alkan, May 01 2018

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
A028982(a(n)) + 1 = A028982(a(n) + 1). - 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 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.
a(2*n+1) = A002315(n)^2 = Sum_{k = 0..4*n + 1} Pell(n), where Pell(n) = A000129(n).
a(2*n) = (1/2)*A005319(n)^2 = 8*A001109(n)^2.
(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

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

More terms from Lekraj Beedassy, Aug 21 2004

A012132 Numbers z such that x(x+1) + y(y+1) = z*(z+1) is solvable in positive integers x,y.

Original entry on oeis.org

3, 6, 8, 10, 11, 13, 15, 16, 18, 20, 21, 23, 26, 27, 28, 31, 33, 36, 37, 38, 40, 41, 43, 44, 45, 46, 48, 49, 51, 52, 53, 54, 55, 56, 57, 58, 59, 61, 62, 63, 64, 66, 67, 68, 71, 73, 74, 75, 76, 77, 78, 80, 81, 83, 86, 88, 89, 91, 92, 93

Offset: 1

Sander van Rijnswou (sander(AT)win.tue.nl)

nonn

Theorem (Sierpinski, 1963): n is a term iff n^2+(n+1)^2 is a composite number. - N. J. A. Sloane, Feb 29 2020
For n > 1, A047219 is a subset of this sequence. This is because n^2 + (n+1)^2 is divisible by 5 if n is (1 or 3) mod 5 (also see A027861). - Dmitry Kamenetsky, Sep 02 2008
From Hermann Stamm-Wilbrandt, Sep 10 2014: (Start)
For n > 0, A212160 is a subset of this sequence (n^2 + (n+1)^2 is divisible by 13 if n == (2 or 10) (mod 13)).
For n >= 0, A212161 is a subset of this sequence (n^2 + (n+1)^2 is divisible by 17 if n == (6 or 10) (mod 17)).
The above are for divisibility by 5, 13, 17; notation (1,3,5), (2,10,13), (6,10,17). Divisibility by p for a and p-a-1; notation (a,p-a-1,p). These are the next tuples: (8,20,29), (15,21,37), (4,36,41), (11,41,53), ... . The corresponding sequences are a subset of this sequence (8,20,37,49,66,78,... for (8,20,29)). These sequences have no entries in the OEIS yet. For any prime of the form 4*k+1 there is exactly one of these tuples/sequences.
For n > 1, A000217 (triangular numbers) is a subset of this sequence (3,6,10,15,...); z=A000217(n), y=z-1, x=n.
For n > 0, A001652 is a subset of this sequence; z=A001652(n), x=y=A053141(n).
For n > 1, A001108(=A115598) is a subset of this sequence; z=A001108(n), x=A076708(n), y=x+1.
For n > 0, A124124(2*n+1)(=A098790(2*n)) is a subset of this sequence (6,37,218,...); z=A124124(2*n+1), x=a(n)-1, y=a(n)+1, a(m) = 6*a(m-1) - a(m-2) + 2, a(0)=0, a(1)=4.
(End)

Aviezri S. Fraenkel, Diophantine equations involving generalized triangular and tetrahedral numbers, pp. 99-114 of A. O. L. Atkin and B. J. Birch, editors, Computers in Number Theory. Academic Press, NY, 1971.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
H. Finner and K. Strassburger, Increasing sample sizes do not necessarily increase the power of UMPU-tests for 2 X 2-tables, Metrika, 54, 77-91, (2001).
Heiko Harborth, Fermat-like binomial equations, Applications of Fibonacci numbers, Proc. 2nd Int. Conf., San Jose/Ca., August 1986, 1-5 (1988).
W. Sierpinski, On triangular numbers which are sums of two smaller triangular numbers, (Polish), Wiadom. Mat. (2) 7 (1963), 27-28. See MR0182602.

Complement of A027861. - Michael Somos, Jun 08 2000
Cf. A000217, A047219, A027861, A166080.
Cf. also A212160, A212161, A001652, A001108, A115598, A124124, A098790.

Select[Range[100], !PrimeQ[#^2 + (#+1)^2]& ] (* Jean-François Alcover, Jan 17 2013, after Michael Somos *)

More terms and references from Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de), Feb 09 2000

cp's OEIS Frontend

A001542 a(n) = 6*a(n-1) - a(n-2) for n > 1, a(0)=0 and a(1)=2.

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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.

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A012132 Numbers z such that x*(x+1) + y*(y+1) = z*(z+1) is solvable in positive integers x,y.

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A012132 Numbers z such that x(x+1) + y(y+1) = z*(z+1) is solvable in positive integers x,y.