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First number of divisor symmetry n: d(n-k) = d(n+k) for 1 <= k <= n, but d(n-k-1) != d(n+k+1).
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%I #14 May 01 2013 20:58:19

%S 4,9,216,30,20376,432,18000,13338864,15194736,866452464,5175273600,

%T 35399473200

%N First number of divisor symmetry n: d(n-k) = d(n+k) for 1 <= k <= n, but d(n-k-1) != d(n+k+1).

%C 3*10^11 < a(13) <= 1245273287760. a(14) = 72462882816. - _Donovan Johnson_, Dec 25 2011

%e 8 and 10 have 4 divisors each, 7 and 11 have 2 divisors each, but 6 and 12 have different numbers of divisors; thus 9 has divisor symmetry 2. Since no smaller number has this, a(2) = 9.

%o (PARI) a(n)=my(k=n);while(k++, for(i=1,n,if(numdiv(k-i)!=numdiv(k+i),next(2))); if(numdiv(k-n-1)==numdiv(k+n+1), next); return(k))

%Y Cf. A093492, A006558.

%K nonn,hard

%O 1,1

%A _Charles R Greathouse IV_, Dec 19 2011

%E a(11) from _Donovan Johnson_, Dec 20 2011

%E a(12) from _Donovan Johnson_, Dec 25 2011