0,2

Solution to b(b+1) = 2a(a+1) in natural numbers including 0; a = a(n), b = b(n) = A001652(n).

The solution of a special case of a binomial problem of H. Finner and K. Strassburger (strass(AT)godot.dfi.uni-duesseldorf.de).

Also the indices of triangular numbers that are half other triangular numbers [a of T(a) such that 2T(a)=T(b)]. The T(a)'s are in A075528, the T(b)'s are in A029549 and the b's are in A001652. - Bruce Corrigan (scentman(AT)myfamily.com), Oct 30 2002

Sequences A053141 (this entry), A016278, A077259, A077288 and A077398 are part of an infinite series of sequences. Each depends upon the polynomial p(n) = 4k*n^2 + 4k*n + 1, when 4k is not a perfect square. Equivalently, they each depend on the equation k*t(x)=t(z) where t(n) is the triangular number formula n(n+1)/2. The dependencies are these: they are the sequences of positive integers n such that p(n) is a perfect square and there exists a positive integer m such that k*t(n)=t(m). A053141 is for k=2, A016278 is for k=3, A077259 is for k=5. - Robert Phillips (bobanne(AT)bellsouth.net), Oct 11 2007, Nov 27 2007

Jason Holt observes that a pair drawn from a drawer with A053141(n)+1 red socks and A001652(n) - A053141(n) blue socks will as likely as not be matching reds: (A053141+1)*A053141/((A001652+1)*A001652) = 1/2, n>0. - Bill Gosper, Feb 07 2010

The values x(n)=A001652(n), y(n)=A046090(n) and z(n)=A001653(n) form a nearly isosceles Pythagorean triple since y(n)=x(n)+1 and x(n)^2 + y(n)^2 = z(n)^2; e.g., for n=2, 20^2 + 21^2 = 29^2. In a similar fashion, if we define b(n)=A011900(n) and c(n)=A001652(n), a(n), b(n) and c(n) form a nearly isosceles anti-Pythagorean triple since b(n)=a(n)+1 and a(n)^2 + b(n)^2 = c(n)^2 + c(n) + 1; i.e., the value a(n)^2 + b(n)^2 lies almost exactly between two perfect squares; e.g., 2^2 + 3^2 = 13 = 4^2 - 3 = 3^2 + 4; 14^2 + 15^2 = 421 = 21^2 - 20 = 20^2 + 21. - Charlie Marion, Jun 12 2009

Behera & Panda call these the balancers and A001109 are the balancing numbers. - Michel Marcus, Nov 07 2017

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000

Max A. Alekseyev, Joseph Samuel Myers, Richard Schroeppel, S. R. Shannon, N. J. A. Sloane, and Paul Zimmermann, Three Cousins of Recamán's Sequence, The Fibonacci Quarterly, Vol. 60, No. 3 (2022), pp. 201-219; arXiv preprint, arXiv:2004.14000 [math.NT], 2020-2021.

Jeremiah Bartz, Bruce Dearden, and Joel Iiams, Classes of Gap Balancing Numbers, arXiv:1810.07895 [math.NT], 2018.

Jeremiah Bartz, Bruce Dearden, and Joel Iiams, Counting families of generalized balancing numbers, The Australasian Journal of Combinatorics, Vol. 77, Part 3 (2020), 318-325.

A. Behera and G. K. Panda, On the Square Roots of Triangular Numbers, Fib. Quart., Vol. 37, No. 2 (1999), pp. 98-105.

Martin V. Bonsangue, Gerald E. Gannon and Laura J. Pheifer, Misinterpretations can sometimes be a good thing, Math. Teacher, Vol. 95, No. 6 (2002) pp. 446-449.

Paula Catarino, Helena Campos, and Paulo Vasco, On some identities for balancing and cobalancing numbers, Annales Mathematicae et Informaticae, 45 (2015) pp. 11-24.

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.

aBa Mbirika, Janee Schrader, and Jürgen Spilker, Pell and Associated Pell Braid Sequences as GCDs of Sums of k Consecutive Pell, Balancing, and Related Numbers, J. Int. Seq., Vol. 26 (2023), Article 23.6.4.

G. K. Panda, Sequence balancing and cobalancing numbers, Fib. Q., Vol. 45, No. 3 (2007), 265-271. See p. 266.

Michael Penn, (co) balancing numbers, YouTube video, 2022.

Robert Phillips, Polynomials of the form 1+4ke+4ke^2, 2008.

Robert Phillips, A triangular number result, 2009.

Vladimir Pletser, Recurrent Relations for Multiple of Triangular Numbers being Triangular Numbers, arXiv:2101.00998 [math.NT], 2020.

Vladimir Pletser, Closed Form Equations for Triangular Numbers Multiple of Other Triangular Numbers, arXiv:2102.12392 [math.GM], 2021.

Vladimir Pletser, Triangular Numbers Multiple of Triangular Numbers and Solutions of Pell Equations, arXiv:2102.13494 [math.NT], 2021.

Vladimir Pletser, Congruence Properties of Indices of Triangular Numbers Multiple of Other Triangular Numbers, arXiv:2103.03019 [math.GM], 2021.

Vladimir Pletser, Searching for multiple of triangular numbers being triangular numbers, 2021.

Vladimir Pletser, Using Pell equation solutions to find all triangular numbers multiple of other triangular numbers, 2021.

Burkard Polster, Nice merging together, Mathologer video (2015).

Burkard Polster and Marty Ross, Marching in squares, arXiv preprint arXiv:1503.04658 [math.HO], 2015.

Ahmet Tekcan, Merve Tayat, and Meltem E. Özbek, The diophantine equation 8x^2-y^2+8x(1+t)+(2t+1)^2=0 and t-balancing numbers, ISRN Combinatorics, Volume 2014, Article ID 897834, 5 pages.

Index entries for linear recurrences with constant coefficients, signature (7,-7,1).

a(n) = (A001653(n)-1)/2 = 2*A053142(n) = A011900(n)-1. [Corrected by Pontus von Brömssen, Sep 11 2024]

a(n) = 6*a(n-1) - a(n-2) + 2, a(0) = 0, a(1) = 2.

G.f.: 2*x/((1-x)*(1-6*x+x^2)).

Let c(n) = A001109(n). Then a(n+1) = a(n)+2*c(n+1), a(0)=0. This gives a generating function (same as existing g.f.) leading to a closed form: a(n) = (1/8)*(-4+(2+sqrt(2))*(3+2*sqrt(2))^n + (2-sqrt(2))*(3-2*sqrt(2))^n). - Bruce Corrigan (scentman(AT)myfamily.com), Oct 30 2002

a(n) = 2*Sum_{k = 0..n} A001109(k). - Mario Catalani (mario.catalani(AT)unito.it), Mar 22 2003

For n>=1, a(n) = 2*Sum_{k=0..n-1} (n-k)*A001653(k). - Charlie Marion, Jul 01 2003

For n and j >= 1, A001109(j+1)*A001652(n) - A001109(j)*A001652(n-1) + a(j) = A001652(n+j). - Charlie Marion, Jul 07 2003

From Antonio Alberto Olivares, Jan 13 2004: (Start)

a(n) = 7*a(n-1) - 7*a(n-2) + a(n-3).

a(n) = -(1/2) - (1-sqrt(2))/(4*sqrt(2))*(3-2*sqrt(2))^n + (1+sqrt(2))/(4*sqrt(2))*(3+2*sqrt(2))^n. (End)

a(n) = sqrt(2)*cosh((2*n+1)*log(1+sqrt(2)))/4 - 1/2 = (sqrt(1+4*A029549)-1)/2. - Bill Gosper, Feb 07 2010 [typo corrected by Vaclav Kotesovec, Feb 05 2016]

a(n+1) + A055997(n+1) = A001541(n+1) + A001109(n+1). - Creighton Dement, Sep 16 2004

From Charlie Marion, Oct 18 2004: (Start)

For n>k, a(n-k-1) = A001541(n)*A001653(k)-A011900(n+k); e.g., 2 = 99*5 - 493.

For n<=k, a(k-n) = A001541(n)*A001653(k) - A011900(n+k); e.g., 2 = 3*29 - 85 + 2. (End)

a(n) = A084068(n)*A084068(n+1). - Kenneth J Ramsey, Aug 16 2007

Let G(n,m) = (2*m+1)*a(n)+ m and H(n,m) = (2*m+1)*b(n)+m where b(n) is from the sequence A001652 and let T(a) = a*(a+1)/2. Then T(G(n,m)) + T(m) = 2*T(H(n,m)). - Kenneth J Ramsey, Aug 16 2007

Let S(n) equal the average of two adjacent terms of G(n,m) as defined immediately above and B(n) be one half the difference of the same adjacent terms. Then for T(i) = triangular number i*(i+1)/2, T(S(n)) - T(m) = B(n)^2 (setting m = 0 gives the square triangular numbers). - Kenneth J Ramsey, Aug 16 2007

a(n) = A001108(n+1) - A001109(n+1). - Dylan Hamilton, Nov 25 2010

a(n) = (a(n-1)*(a(n-1) - 2))/a(n-2) for n > 2. - Vladimir Pletser, Apr 08 2020

a(n) = (ChebyshevU(n, 3) - ChebyshevU(n-1, 3) - 1)/2 = (Pell(2*n+1) - 1)/2. - G. C. Greubel, Apr 27 2020

E.g.f.: (exp(3*x)*(2*cosh(2*sqrt(2)*x) + sqrt(2)*sinh(2*sqrt(2)*x)) - 2*exp(x))/4. - Stefano Spezia, Mar 16 2024

a(n) = A000194(A029549(n)) = A002024(A075528(n)). - Pontus von Brömssen, Sep 11 2024

Sum_{n>=1} 1/a(n) = 2 - sqrt(2) (A101465). - Amiram Eldar, Jan 30 2026

A053141 := proc(n)

option remember;

if n <= 1 then

op(n+1, [0, 2]) ;

else

6*procname(n-1)-procname(n-2)+2 ;

end if;

end proc: # R. J. Mathar, Feb 05 2016

Join[{a=0, b=1}, Table[c=6*b-a+1; a=b; b=c, {n, 60}]]*2 (* Vladimir Joseph Stephan Orlovsky, Jan 18 2011 *)

a[n_] := Floor[1/8*(2+Sqrt[2])*(3+2*Sqrt[2])^n]; Table[a[n], {n, 0, 20}] (* Jean-François Alcover, Nov 28 2013 *)

Table[(Fibonacci[2n + 1, 2] - 1)/2, {n, 0, 20}] (* Vladimir Reshetnikov, Sep 16 2016 *)

(Haskell)

a053141 n = a053141_list !! n

a053141_list = 0 : 2 : map (+ 2)

(zipWith (-) (map (* 6) (tail a053141_list)) a053141_list)

-- Reinhard Zumkeller, Jan 10 2012

(PARI) concat(0, Vec(2/(1-x)/(1-6*x+x^2)+O(x^30))) \\ Charles R Greathouse IV, May 14 2012

(PARI) {x=1+sqrt(2); y=1-sqrt(2); P(n) = (x^n - y^n)/(x-y)};

a(n) = round((P(2*n+1) - 1)/2);

for(n=0, 30, print1(a(n), ", ")) \\ G. C. Greubel, Jul 15 2018

(Magma) R<x>:=PowerSeriesRing(Integers(), 30); Coefficients(R!(2*x/((1-x)*(1-6*x+x^2)))); // G. C. Greubel, Jul 15 2018

(SageMath) [(lucas_number1(2*n+1, 2, -1)-1)/2 for n in range(30)] # G. C. Greubel, Apr 27 2020

Cf. A000129, A001108, A001109, A001652, A001653, A101465.

Cf. A011900, A029549, A053142, A075528, A103200.

Partial sums of A001542.

Cf. A000194, A001541, A002024, A016278, A046090, A055997, A077259, A077288, A077398, A084068.

Sequence in context: A335693 A138126 A268881 * A339281 A361552 A036692

Adjacent sequences: A053138 A053139 A053140 * A053142 A053143 A053144

nonn,easy

Name corrected by Zak Seidov, Apr 11 2011

approved