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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). 35
0, 2, 14, 84, 492, 2870, 16730, 97512, 568344, 3312554, 19306982, 112529340, 655869060, 3822685022, 22280241074, 129858761424, 756872327472, 4411375203410, 25711378892990, 149856898154532, 873430010034204 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,2

COMMENTS

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

REFERENCES

P Catarino, H Campos, P Vasco, On some identities for balancing and cobalancing numbers, Annales Mathematicae et Informaticae, 45 (2015) pp. 11-24; http://ami.ektf.hu.~

LINKS

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

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.

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

Burkard Polster, Nice merging together, YouTube, 2015.

B. Polster, M. Ross, Marching in squares, arXiv preprint arXiv:1503.04658, 2015

A. Tekcan, M. Tayat, M. E. Ozbek, 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).

FORMULA

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

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

a(n) = 7a(n-1)-7a(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. - Antonio Alberto Olivares, Jan 13 2004

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 [corrected a typo by Vaclav Kotesovec, Feb 05 2016]

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

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. - Charlie Marion, Oct 18 2004

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

Let G(n,m) = (2m+1)*a(n)+ m and H(n,m) = (2m+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) = 2T(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

MAPLE

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

MATHEMATICA

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-Fran├žois Alcover, Nov 28 2013 *)

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

PROG

(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^98))) \\ Charles R Greathouse IV, May 14 2012

CROSSREFS

Cf. A001653, A001652, A053142, A011900, A001108, A001109, A075528, A029549, A001652, A103200. Partial sums of A001542.

Sequence in context: A077444 A138126 A268881 * A036692 A075140 A037563

Adjacent sequences:  A053138 A053139 A053140 * A053142 A053143 A053144

KEYWORD

nonn,easy

AUTHOR

Wolfdieter Lang

EXTENSIONS

Name corrected by Zak Seidov, Apr 11 2011

STATUS

approved

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Last modified August 19 01:39 EDT 2017. Contains 290789 sequences.