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First integral ladder to be largest perpendicular-corner-bending for exactly n distinct pairs of integral corridor widths.
0

%I #5 Mar 31 2012 10:26:03

%S 125,15625,1953125,274625,30517578125,3814697265625,34328125,

%T 59604644775390625,7450580596923828125,4291015625,

%U 116415321826934814453125,75418890625,1349232625

%N First integral ladder to be largest perpendicular-corner-bending for exactly n distinct pairs of integral corridor widths.

%C In general the largest-bending ladder L across perpendicular corner where corridors of widths M and N meet,is given by L^(2/3)=M^(2/3)+ N^(2/3).

%F a(n)=d^3, where d=A006339(n).

%e a(4)=274625 because this is the smallest largest-integral-bending-ladder in 4 distinct stances, viz. with corridor width pairs (4096, 250047), (15625, 216000), (35937, 175616), (59319, 140608).

%Y Cf. A088896.

%K nonn

%O 1,1

%A _Lekraj Beedassy_, May 25 2004