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%I #18 Feb 02 2020 17:52:51
%S 1,2,2,3,1,3,4,3,3,4,5,4,1,4,5,6,5,2,2,5,6,7,6,5,1,5,6,7,8,7,6,3,3,6,
%T 7,8,9,8,7,3,1,3,7,8,9,10,9,8,7,2,2,7,8,9,10,11,10,9,8,4,1,4,8,9,10,
%U 11,12,11,10,9,4,4,4,4,9,10,11,12,13,12,11,10
%N T(n, k) is the least positive m such that floor(n/m) = floor(k/m). Square array T(n, k) read by antidiagonals, n >= 0 and k >= 0.
%H Rémy Sigrist, <a href="/A331886/b331886.txt">Table of n, a(n) for n = 0..10010</a> (antidiagonals 0..140)
%H Rémy Sigrist, <a href="/A331886/a331886.png">Colored representation of T(n, k) for n, k = 0..1000</a> (where the hue is function of T(n, k))
%H Rémy Sigrist, <a href="/A331886/a331886_1.png">Colored representation of floor(n/T(n, k)) for n, k = 0..1000</a> (where the hue is function of floor(n/T(n, k)), red pixels correspond to 0's)
%F T(n, k) = T(k, n).
%F T(n, k) = 1 iff n = k.
%F T(n, k) <= 1 + max(n, k) with equality iff max(n, k) >= 2*min(n, k).
%F T(n, n+1) = A007978(n+1).
%e Array T(n, k) begins:
%e n\k| 0 1 2 3 4 5 6 7 8 9 10 11 12
%e ---+----------------------------------------------------
%e 0| 1 2 3 4 5 6 7 8 9 10 11 12 13
%e 1| 2 1 3 4 5 6 7 8 9 10 11 12 13
%e 2| 3 3 1 2 5 6 7 8 9 10 11 12 13
%e 3| 4 4 2 1 3 3 7 8 9 10 11 12 13
%e 4| 5 5 5 3 1 2 4 4 9 10 11 12 13
%e 5| 6 6 6 3 2 1 4 4 5 5 11 12 13
%e 6| 7 7 7 7 4 4 1 2 3 5 6 6 13
%e 7| 8 8 8 8 4 4 2 1 3 5 6 6 7
%e 8| 9 9 9 9 9 5 3 3 1 2 4 4 7
%e 9| 10 10 10 10 10 5 5 5 2 1 3 3 7
%e 10| 11 11 11 11 11 11 6 6 4 3 1 2 5
%e 11| 12 12 12 12 12 12 6 6 4 3 2 1 5
%e 12| 13 13 13 13 13 13 13 7 7 7 5 5 1
%o (PARI) T(n,k) = for (m=1, oo, if (n\m==k\m, return (m)))
%Y Cf. A007978, A331902.
%K nonn,tabl,look
%O 0,2
%A _Rémy Sigrist_, Jan 30 2020
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