A178776 - OEIS

%I #15 Oct 05 2013 00:30:48

%S 4,4,12,12,24,36,40,40,72,72,84,120,144,144,180,180,240,252,252,252,

%T 360,400,400,432,504,504,600,600,672,672,672,840,900,900,900,936,1120,

%U 1120,1260,1260,1320,1440,1440,1440,1680,1764,1800,1800,1872,1872

%N a(n) is the largest number that appears twice in an n x n multiplication table of positive integers, disregarding the pairs that are obviously produced by the commutative property.

%C It appears that all terms of this sequence are multiple of 4 (checked up to 10^7). - _Michel Marcus_, Aug 26 2013

%F Let xy = n be the factorization of n such that x+y is a minimum. We then have a(4)= 4 and a(n)= max{a(n-1), xy(x-1)(y-1)} for all n>4.

%e The first term is 4, which appears twice in a 4 x 4 multiplication table. a(4) = 2x2 = 4x1 = 4. For all n = prime a(n) = a(n-1) so a(5) = 4. a(6) = 12 = 2x6 = 3x4, a(7) = 12, a(8) = 24 = 3x8 = 4x6, a(9) = 36 = 4x9 = 6x6.

%o (PARI) a(n) = {skeep = Set(); mmax = 0; for (i = 1, n, for (j = i, n, v = i*j; if (! setsearch(skeep, v), skeep = setunion(skeep, Set(v)), mmax = max(mmax, v)););); mmax;} \\ _Michel Marcus_, Aug 26 2013

%o (PARI) findxy(n, d) = {mins = 2*n; if (#d % 2, nd = #d\2 +1, nd = #d/2); for (i = 1, nd, if ((s = d[i]+n/d[i]) < mins, mins = s; mini = i);); x = d[mini]; y = n/x; return (x*y*(x-1)*(y-1));}

%o lista(nn) = {lmmx = 4; print1(lmmx, ", "); for (n=5, nn, d = divisors(n); nmmx = findxy(n, d); lmmx = max(lmmx, nmmx); print1(lmmx, ", "););} \\ _Michel Marcus_, Aug 26 2013

%K nonn

%O 4,1

%A _Gary Yane_, Jun 11 2010

%E Edited by _N. J. A. Sloane_, Jun 12 2010

%E Corrected by _Michel Marcus_, Aug 26 2013