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)=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

Contribution from Charlie Marion (charliemath(AT)optonline.net), Jun 12 2009: (Start)

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.

(End)

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.

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 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

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 (charliem(AT)bestweb.net), Jul 01 2003

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

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

a(n+1) + A055997(n+1) = A001541(n+1) + A001109(n+1). - Creighton Dement (creighton.k.dement(AT)uni-oldenburg.de), 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 (charliemath(AT)optonline.net), Oct 18 2004

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

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

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

Cf. A001653, A001652, A053142, A011900. Partial sums of A001542 - Barry E. Williams, May 07 2000.

Cf. A001109, A075528, A029549, A001652.

Cf. A103200.

Sequence in context: A077461 A077444 A138126 this_sequence A036692 A075140 A037563

Adjacent sequences: A053138 A053139 A053140 this_sequence A053142 A053143 A053144

nonn,easy

Wolfdieter Lang (wolfdieter.lang(AT)physik.uni-karlsruhe.de)