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 A053141 a(0)=0, a(1)=2 then a(n) = a(n-2) + 2*sqrt(8*a(n-1)^2 + 8*a(n-1) + 1). 40

%I

%S 0,2,14,84,492,2870,16730,97512,568344,3312554,19306982,112529340,

%T 655869060,3822685022,22280241074,129858761424,756872327472,

%U 4411375203410,25711378892990,149856898154532,873430010034204

%N a(0)=0, a(1)=2 then a(n) = a(n-2) + 2*sqrt(8*a(n-1)^2 + 8*a(n-1) + 1).

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

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

%C 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

%C 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

%C 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

%C 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

%C Behera & Panda call these the balancers and A001109 are the balancing numbers. - _Michel Marcus_, Nov 07 2017

%H Reinhard Zumkeller, <a href="/A053141/b053141.txt">Table of n, a(n) for n = 0..1000</a>

%H Jeremiah Bartz, Bruce Dearden, and Joel Iiams, <a href="https://arxiv.org/abs/1810.07895">Classes of Gap Balancing Numbers</a>, arXiv:1810.07895 [math.NT], 2018.

%H Jeremiah Bartz, Bruce Dearden, and Joel Iiams, <a href="https://ajc.maths.uq.edu.au/pdf/77/ajc_v77_p318.pdf">Counting families of generalized balancing numbers</a>, The Australasian Journal of Combinatorics (2020) Vol. 77, Part 3, 318-325.

%H A. Behera and G. K. Panda, <a href="http://www.fq.math.ca/Scanned/37-2/behera.pdf">On the Square Roots of Triangular Numbers</a>, Fib. Quart., 37 (1999), pp. 98-105.

%H Martin V. Bonsangue, Gerald E. Gannon and Laura J. Pheifer, <a href="https://www.jstor.org/stable/20871097">Misinterpretations can sometimes be a good thing</a>, Math. Teacher, vol. 95, No. 6 (2002) pp. 446-449.

%H P. Catarino, H. Campos, and P. Vasco, <a href="http://ami.ektf.hu/uploads/papers/finalpdf/AMI_45_from11to24.pdf">On some identities for balancing and cobalancing numbers</a>, Annales Mathematicae et Informaticae, 45 (2015) pp. 11-24.

%H Refik Keskin and Olcay Karaatli, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL15/Karaatli/karaatli5.html">Some New Properties of Balancing Numbers and Square Triangular Numbers</a>, Journal of Integer Sequences, Vol. 15 (2012), Article #12.1.4.

%H J. S. Myers, R. Schroeppel, S. R. Shannon, N. J. A. Sloane, and P. Zimmermann, <a href="http://arxiv.org/abs/2004.14000">Three Cousins of Recaman's Sequence</a>, arXiv:2004:14000 [math.NT], April 2020.

%H G. K. Panda, <a href="https://www.fq.math.ca/Papers1/45-3/panda.pdf">Sequence balancing and cobalancing numbers</a>, Fib. Q., Vol. 45, No. 3 (2007), 265-271. See p. 266.

%H Robert Phillips, <a href="https://web.archive.org/web/20100713033314/http://www.usca.edu/math/~mathdept/bobp/pdf/polgonal.pdf">Polynomials of the form 1+4ke+4ke^2</a>, 2008.

%H Robert Phillips, <a href="https://web.archive.org/web/20100713033404/http://www.usca.edu/math/~mathdept/bobp/pdf/result.pdf">A triangular number result</a>, 2009.

%H Burkard Polster, <a href="http://youtu.be/rjHFkx6eTL8">Nice merging together</a>, Mathologer video (2015).

%H B. Polster and M. Ross, <a href="https://arxiv.org/abs/1503.04658">Marching in squares</a>, arXiv preprint arXiv:1503.04658 [math.HO], 2015.

%H A. Tekcan, M. Tayat, and M. E. Ozbek, <a href="https://doi.org/10.1155/2014/897834">The diophantine equation 8x^2-y^2+8x(1+t)+(2t+1)^2=0 and t-balancing numbers</a>, ISRN Combinatorics, Volume 2014, Article ID 897834, 5 pages.

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (7,-7,1).

%F a(n) = (A001653(n)-1)/2 = 2*A053142(n) = A011900(n-1).

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

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

%F 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

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

%F For n>=1, a(n) = 2*Sum_{k=0..n-1} (n-k)*A001653(k). - _Charlie Marion_, Jul 01 2003

%F 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

%F From _Antonio Alberto Olivares_, Jan 13 2004: (Start)

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

%F 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)

%F 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]

%F a(n+1) + A055997(n+1) = A001541(n+1) + A001109(n+1). - _Creighton Dement_, Sep 16 2004

%F From _Charlie Marion_, Oct 18 2004: (Start)

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

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

%F a(n) = A084068(n)*A084068(n+1). - _Kenneth J Ramsey_, Aug 16 2007

%F 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

%F 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

%F a(n) = A001108(n+1) - A001109(n+1). - _Dylan Hamilton_, Nov 25 2010

%F a(n) = (a(n-1)*(a(n-1) - 2))/a(n-2) for n > 2. - _Vladimir Pletser_, Apr 08 2020

%F a(n) = (ChebyshevU(n, 3) - ChebyshevU(n-1, 3) - 1)/2 = (Pell(2*n+1) - 1)/2. - _G. C. Greubel_, Apr 27 2020

%p A053141 := proc(n)

%p option remember;

%p if n <= 1 then

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

%p else

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

%p end if;

%p end proc: # _R. J. Mathar_, Feb 05 2016

%t 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 *)

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

%t Table[(Fibonacci[2n + 1, 2] - 1)/2, {n, 0, 20}] (* _Vladimir Reshetnikov_, Sep 16 2016 *)

%o a053141 n = a053141_list !! n

%o a053141_list = 0 : 2 : map (+ 2)

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

%o -- _Reinhard Zumkeller_, Jan 10 2012

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

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

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

%o for(n=0, 30, print1(a(n), ", ")) \\ _G. C. Greubel_, Jul 15 2018

%o (MAGMA) R<x>:=PowerSeriesRing(Integers(), 30); Coefficients(R!(2*x/((1-x)*(1-6*x+x^2)))); // _G. C. Greubel_, Jul 15 2018

%o (Sage) [(lucas_number1(2*n+1, 2, -1)-1)/2 for n in range(30)] # _G. C. Greubel_, Apr 27 2020

%Y Cf. A000129, A001108, A001109, A001652, A001653.

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

%Y Partial sums of A001542.

%K nonn,easy,changed

%O 0,2

%A _Wolfdieter Lang_

%E Name corrected by _Zak Seidov_, Apr 11 2011

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Last modified September 23 22:32 EDT 2020. Contains 337315 sequences. (Running on oeis4.)