

A094081


Smallest integral ladder whose ends slide over the respective distances 1 and m=2n+1 while slipping down along horizontal ground and vertical wall against which it leans.


0



5, 185, 1313, 4925, 13325, 29585, 57545, 101813, 167765, 261545, 390065, 561005, 782813, 1064705, 1416665, 1849445, 2374565, 3004313, 3751745, 4630685, 5655725, 6842225, 8206313, 9764885, 11535605, 13536905, 15787985, 18308813, 21120125
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OFFSET

0,1


COMMENTS

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*M3, 8*M5, 4*m)}.


LINKS

Table of n, a(n) for n=0..28.


FORMULA

a(n) = M(8*M  3), where M=2n^2 + 2n + 1=A001844(n).
a(n) = (2n^2 + 2n + 1)*(16n^2 + 16n + 5).
G.f.: (5+160*x+438*x^2+160*x^3+5*x^4)/(1x)^5.  Colin Barker, Jan 23 2012


EXAMPLE

The expressions associated with the first few entries are:
5^2=3^2 + 4^2=(3+1)^2 + (41)^2.
185^2=175^2 + 60^2=(175+1)^2 + (603)^2.
1313^2=1287^2 + 260^2=(1287+1)^2 + (2605)^2.
4925^2=4875^2 + 700^2=(4875+1)^2 + (7007)^2.
13325^2=13243^2 +1476^2=(13243+1)^2 + (14769)^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=(h1)^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(i1)  V(i2) + 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).


CROSSREFS

Sequence in context: A304277 A208403 A280797 * A189645 A228694 A268178
Adjacent sequences: A094078 A094079 A094080 * A094082 A094083 A094084


KEYWORD

nonn,easy


AUTHOR

Lekraj Beedassy, Apr 30 2004


STATUS

approved



