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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).

%I A053141 #178 Jan 05 2025 19:51:36
%S A053141 0,2,14,84,492,2870,16730,97512,568344,3312554,19306982,112529340,
%T A053141 655869060,3822685022,22280241074,129858761424,756872327472,
%U A053141 4411375203410,25711378892990,149856898154532,873430010034204,5090723162050694,29670908962269962,172934730611569080
%N 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).
%C A053141 Solution to b(b+1) = 2a(a+1) in natural numbers including 0; a = a(n), b = b(n) = A001652(n).
%C A053141 The solution of a special case of a binomial problem of H. Finner and K. Strassburger (strass(AT)godot.dfi.uni-duesseldorf.de).
%C A053141 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 A053141 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 A053141 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 A053141 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 A053141 Behera & Panda call these the balancers and A001109 are the balancing numbers. - _Michel Marcus_, Nov 07 2017
%H A053141 Reinhard Zumkeller, <a href="/A053141/b053141.txt">Table of n, a(n) for n = 0..1000</a>
%H A053141 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 A053141 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 A053141 A. Behera and G. K. Panda, <a href="https://web.archive.org/web/2024*/https://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 A053141 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 A053141 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 A053141 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 A053141 aBa Mbirika, Janee Schrader, and Jürgen Spilker, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL26/Mbirika/mbir5.html">Pell and Associated Pell Braid Sequences as GCDs of Sums of k Consecutive Pell, Balancing, and Related Numbers</a>, J. Int. Seq. (2023) Vol. 26, Art. 23.6.4.
%H A053141 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 A053141 G. K. Panda, <a href="https://web.archive.org/web/2024*/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 A053141 Michael Penn, <a href="https://www.youtube.com/watch?v=cMfFqYM04xk">(co) balancing numbers</a>, YouTube video, 2022.
%H A053141 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 A053141 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 A053141 Vladimir Pletser, <a href="https://arxiv.org/abs/2101.00998">Recurrent Relations for Multiple of Triangular Numbers being Triangular Numbers</a>, arXiv:2101.00998 [math.NT], 2021.
%H A053141 Vladimir Pletser, <a href="https://arxiv.org/abs/2102.12392">Closed Form Equations for Triangular Numbers Multiple of Other Triangular Numbers</a>, arXiv:2102.12392 [math.GM], 2021.
%H A053141 Vladimir Pletser, <a href="https://arxiv.org/abs/2102.13494">Triangular Numbers Multiple of Triangular Numbers and Solutions of Pell Equations</a>, arXiv:2102.13494 [math.NT], 2021.
%H A053141 Vladimir Pletser, <a href="https://arxiv.org/abs/2103.03019">Congruence Properties of Indices of Triangular Numbers Multiple of Other Triangular Numbers</a>, arXiv:2103.03019 [math.GM], 2021.
%H A053141 Vladimir Pletser, <a href="https://doi.org/10.13140/RG.2.2.35428.91527">Searching for multiple of triangular numbers being triangular numbers</a>, 2021.
%H A053141 Vladimir Pletser, <a href="https://www.researchgate.net/profile/Vladimir-Pletser/publication/359808848_USING_PELL_EQUATION_SOLUTIONS_TO_FIND_ALL_TRIANGULAR_NUMBERS_MULTIPLE_OF_OTHER_TRIANGULAR_NUMBERS/">Using Pell equation solutions to find all triangular numbers multiple of other triangular numbers</a>, 2021.
%H A053141 Burkard Polster, <a href="http://youtu.be/rjHFkx6eTL8">Nice merging together</a>, Mathologer video (2015).
%H A053141 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 A053141 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 A053141 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (7,-7,1).
%F A053141 a(n) = (A001653(n)-1)/2 = 2*A053142(n) = A011900(n)-1. [Corrected by _Pontus von Brömssen_, Sep 11 2024]
%F A053141 a(n) = 6*a(n-1) - a(n-2) + 2, a(0) = 0, a(1) = 2.
%F A053141 G.f.: 2*x/((1-x)*(1-6*x+x^2)).
%F A053141 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 A053141 a(n) = 2*Sum_{k = 0..n} A001109(k). - Mario Catalani (mario.catalani(AT)unito.it), Mar 22 2003
%F A053141 For n>=1, a(n) = 2*Sum_{k=0..n-1} (n-k)*A001653(k). - _Charlie Marion_, Jul 01 2003
%F A053141 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 A053141 From _Antonio Alberto Olivares_, Jan 13 2004: (Start)
%F A053141 a(n) = 7*a(n-1) - 7*a(n-2) + a(n-3).
%F A053141 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 A053141 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 A053141 a(n+1) + A055997(n+1) = A001541(n+1) + A001109(n+1). - _Creighton Dement_, Sep 16 2004
%F A053141 From _Charlie Marion_, Oct 18 2004: (Start)
%F A053141 For n>k, a(n-k-1) = A001541(n)*A001653(k)-A011900(n+k); e.g., 2 = 99*5 - 493.
%F A053141 For n<=k, a(k-n) = A001541(n)*A001653(k) - A011900(n+k); e.g., 2 = 3*29 - 85 + 2. (End)
%F A053141 a(n) = A084068(n)*A084068(n+1). - _Kenneth J Ramsey_, Aug 16 2007
%F A053141 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 A053141 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 A053141 a(n) = A001108(n+1) - A001109(n+1). - _Dylan Hamilton_, Nov 25 2010
%F A053141 a(n) = (a(n-1)*(a(n-1) - 2))/a(n-2) for n > 2. - _Vladimir Pletser_, Apr 08 2020
%F A053141 a(n) = (ChebyshevU(n, 3) - ChebyshevU(n-1, 3) - 1)/2 = (Pell(2*n+1) - 1)/2. - _G. C. Greubel_, Apr 27 2020
%F A053141 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
%F A053141 a(n) = A000194(A029549(n)) = A002024(A075528(n)). - _Pontus von Brömssen_, Sep 11 2024
%p A053141 A053141 := proc(n)
%p A053141     option remember;
%p A053141     if n <= 1 then
%p A053141         op(n+1,[0,2]) ;
%p A053141     else
%p A053141         6*procname(n-1)-procname(n-2)+2 ;
%p A053141     end if;
%p A053141 end proc: # _R. J. Mathar_, Feb 05 2016
%t A053141 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 A053141 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 *)
%t A053141 Table[(Fibonacci[2n + 1, 2] - 1)/2, {n, 0, 20}] (* _Vladimir Reshetnikov_, Sep 16 2016 *)
%o A053141 (Haskell)
%o A053141 a053141 n = a053141_list !! n
%o A053141 a053141_list = 0 : 2 : map (+ 2)
%o A053141    (zipWith (-) (map (* 6) (tail a053141_list)) a053141_list)
%o A053141 -- _Reinhard Zumkeller_, Jan 10 2012
%o A053141 (PARI) concat(0,Vec(2/(1-x)/(1-6*x+x^2)+O(x^30))) \\ _Charles R Greathouse IV_, May 14 2012
%o A053141 (PARI) {x=1+sqrt(2); y=1-sqrt(2); P(n) = (x^n - y^n)/(x-y)};
%o A053141 a(n) = round((P(2*n+1) - 1)/2);
%o A053141 for(n=0, 30, print1(a(n), ", ")) \\ _G. C. Greubel_, Jul 15 2018
%o A053141 (Magma) R<x>:=PowerSeriesRing(Integers(), 30); Coefficients(R!(2*x/((1-x)*(1-6*x+x^2)))); // _G. C. Greubel_, Jul 15 2018
%o A053141 (Sage) [(lucas_number1(2*n+1, 2, -1)-1)/2 for n in range(30)] # _G. C. Greubel_, Apr 27 2020
%Y A053141 Cf. A000129, A001108, A001109, A001652, A001653.
%Y A053141 Cf. A011900, A029549, A053142, A075528, A103200.
%Y A053141 Partial sums of A001542.
%Y A053141 Cf. A000194, A001541, A002024, A016278, A046090, A055997, A077259, A077288, A077398, A084068.
%K A053141 nonn,easy
%O A053141 0,2
%A A053141 _Wolfdieter Lang_
%E A053141 Name corrected by _Zak Seidov_, Apr 11 2011