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A094081
Smallest integral ladder whose ends slide over the respective distances 1 and m=2*n+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
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 2*M = m^2 + 1. {The Pythagorean triple is M times (8*M-3, 8*M-5, 4*m)}.
FORMULA
a(n) = M*(8*M - 3), where M = 2*n^2 + 2*n + 1 = A001844(n).
a(n) = (2*n^2 + 2*n + 1)*(16*n^2 + 16*n + 5).
G.f.: (5+160*x+438*x^2+160*x^3+5*x^4)/(1-x)^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 + (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).
CROSSREFS
Sequence in context: A304277 A208403 A280797 * A189645 A228694 A268178
KEYWORD
nonn,easy
AUTHOR
Lekraj Beedassy, Apr 30 2004
STATUS
approved