%I A053141
%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
%N A053141 Define a(1)=0, a(2)=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)=4kn^2+4kn+1, when 4k is not a perfect square. Equivantly, 
               they each depend on the equation kt(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 kt(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 Contribution from Charlie Marion (charliemath(AT)optonline.net), Jun 
               12 2009: (Start)
%C A053141 The values x(n)=A001652(n), y(n)=A046090(n) and z(n)=A001653(n) form 
               a nearly
%C A053141 isosceles Pythagorean triple since y(n)=x(n)+1 and x(n)^2+y(n)^2=z(n)^2; 
               e.g.,
%C A053141 for n=2, 20^2+21^2=29^2.In a similar fashion, if we define b(n)=A011900(n) 
               and
%C A053141 c(n)=A001652(n), a(n), b(n) and c(n) form a nearly isosceles anti-Pythagorean
%C A053141 triple since b(n)=a(n)+1 and a(n)^2+b(n)^2=c(n)^2+c(n)+1; i.e, the value
%C A053141 a(n)^2+b(n)^2 lies almost exactly between two perfect squares; e.g.,
%C A053141 2^2+3^2=13=4^2-3=3^2+4; 14^2+15^2=421=21^2-20=20^2+21.
%C A053141 (End)
%D A053141 Martin V. Bonsangue, Gerald E. Gannon & Laura J. Pheifer, Misinterpretations 
               can sometimes be a good thing, Math. Teacher, vol. 95, No. 6 (2002) 
               pp. 446-449.
%F A053141 a(n) = (A001653(n)-1)/2 = 2*A053142(n) = A011900(n-1).
%F A053141 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)).
%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 of 2x/((1-x)*(1-6x+x^2)) (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 (charliem(AT)bestweb.net), 
               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); e.g. 204*119-35*20+84=23660 - Charlie Marion 
               (charliem(AT)bestweb.net), Jul 07 2003
%F A053141 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 (tonioolivares(AT)todito.com), 
               Jan 13 2004
%F A053141 a(n+1) + A055997(n+1) = A001541(n+1) + A001109(n+1). - Creighton Dement 
               (creighton.k.dement(AT)uni-oldenburg.de), Sep 16 2004
%F A053141 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 (charliemath(AT)optonline.net), Oct 18 2004
%F A053141 a(n) is the product of adjacent terms g(n) and g(n+1) of the sequence 
               0,1,2,7,12,41... (A0084068) which has the recursive formula g(n) 
               = 2g(n-1)-g(n-2) for n even and g(n) = 4g(n-1)-g(n-2) for n odd. 
               - Kenneth J Ramsey (Ramsey2879(AT)msn.com), Aug 16 2007
%F A053141 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 (Ramsey2879(AT)msn.com), 
               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 (Ramsey2879(AT)msn.com), Aug 16 2007
%Y A053141 Cf. A001653, A001652, A053142, A011900. Partial sums of A001542 - Barry 
               E. Williams, May 07 2000.
%Y A053141 Cf. A001109, A075528, A029549, A001652.
%Y A053141 Cf. A103200.
%Y A053141 Sequence in context: A077461 A077444 A138126 this_sequence A036692 A075140 
               A037563
%Y A053141 Adjacent sequences: A053138 A053139 A053140 this_sequence A053142 A053143 
               A053144
%K A053141 nonn,easy
%O A053141 0,2
%A A053141 Wolfdieter Lang (wolfdieter.lang(AT)physik.uni-karlsruhe.de)