A212710 Smallest number k such that the difference between the greatest prime divisor of k^2+1 and the sum of the other prime distinct divisors equals n.
411, 1, 3, 447, 2, 57, 212, 8, 307, 13, 5, 38, 319, 99, 3310, 70, 4, 242, 132, 50, 73, 17, 192, 12, 133, 3532, 41, 22231, 999, 43, 172, 68, 83, 11878, 294, 30, 6, 111, 9, 776, 2059, 922, 818, 46, 1183, 23, 216, 182, 557, 2010, 1818, 3323, 945, 512, 568, 76
Offset: 1
Keywords
Examples
a(1) = 411 because 411^2+1 = 2 * 13 * 73 * 89 and 89 - (2 + 13 + 73) = 89 - 88 = 1.
Programs
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Maple
A212710 := proc(n) local fs,gpf,opf,k ; for k from 1 do fs := numtheory[factorset](k^2+1) ; gpf := max(op(fs)) ; opf := add( f,f=fs)-gpf ; if gpf-opf = n then return k; end if; end do: end proc: seq(A212710(n),n=1..50) ; # R. J. Mathar, Nov 14 2014
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Mathematica
lst={};Do[k=1;[While[!2*FactorInteger[k^2+1][[-1,1]]-Total[Transpose[FactorInteger[k^2+1]][[1]]]==n,k++]];AppendTo[lst,k],{n,0,60}];lst (* Michel Lagneau, Oct 28 2014 *)
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PARI
a(n) = {k = 1; ok = 0; while (!ok, f = factor(k^2+1); nbp = #f~; ok = (f[nbp, 1] - sum(i=1, nbp-1, f[i,1]) == n); if (!ok, k++);); k;} \\ Michel Marcus, Nov 09 2014