0,1

Ladder has upper end at height M(8*M - 5) and lower end distance 4*M*m off the wall (or vice versa), where 2M=m^2 + 1. {The Pythagorean triple is M times (8*M-3, 8*M-5, 4*m)}.

a(n)=M(8*M - 3), where M=2n^2 + 2n + 1=A001844(n).

(2n^2 - 2n + 1)*(16n^2 - 16n + 5).

The expressions associated with the first few entries are:

5^2=3^2 + 4^2=(3+1)^2 + (4-1)^2.

185^2=175^2 + 60^2=(175+1)^2 + (60-3)^2.

1313^2=1287^2 + 260^2=(1287+1)^2 + (260-5)^2.

4925^2=4875^2 + 700^2=(4875+1)^2 + (700-7)^2.

13325^2=13243^2 +1476^2=(13243+1)^2 + (1476-9)^2.

Consider the case n=2. For a ladder L with upper end at height h off ground and lower end at distance s off wall, we have relations L^2=h^2 + s^2=(h-1)^2 + (s+5)^2.....(*), which boil down to X^2 - 26*Y^2=-1 using the parameters X=2k+7, Y=L/13, h=5k+18, s=k+1, so that triples (L, h, s) are generated from the recurrence V(i)=102*V(i-1) - V(i-2) + W, where vectors V(i)=[L(i) h(i) s(i)], W=[0 -50 250], with V(-1)=[13 -12 -5], V(0)=[13 13 0], yielding solutions (1313, 1288, 255);(133913, 131313, 26260);(13657813, 13392588, 2678515);(1392963013, 1365912613, 273182520);...all satisfying relation (*) above with the smallest solution L(1) being 1313=a(2).

Sequence in context: A053363 A027572 A050239 this_sequence A096543 A138733 A015102

Adjacent sequences: A094078 A094079 A094080 this_sequence A094082 A094083 A094084

nonn

Lekraj Beedassy (blekraj(AT)yahoo.com), Apr 30 2004