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%I #68 Oct 25 2023 20:21:36
%S 6,21,40,105,126,301,204,273,550,1221,936,697,690,3165,2176,4641,1242,
%T 1333,4200,8841,1786,3213,2508,15025,9126,18981,3700,6105,13950,3901,
%U 3876,4161,6106,5781,23976,49321,8178,6765,32800,67281,6930,18565,7440,11001,49726,8925,9072,26977
%N Smallest integer m = b*c which satisfies (b + c)*n = m - 1.
%C All terms have b,c > 1.
%C a(n) is the smallest of a certain n-class. The 1-class would be related to the numbers denoted as "pqrs" in A009112.
%C From _David A. Corneth_, Jul 19 2023: (Start)
%C n < min(b, c) <= 2*n.
%C Proof of n < min(b, c):
%C As m = b*c and (b + c)*n = m - 1 we have m - (m - 1) = 1 = b*c - (b + c)*n.
%C Solving 1 = b*c = (b + c)*n for c gives c = (n*b + 1) / (b - n) > 0.
%C As 0 < b, c, n we have b - n > 0 so b > n.
%C Similarily as b = (n*c + 1)/(c - n) we have c > n.
%C Proof of min(b, c) <= 2*n by contradiction.
%C Suppose 2*n < min(b, c). Then 2*n + 1 <= min(b, c) as both b and c are integers.
%C Let b = 2n + b' and c = 2n + c' where 1 <= b', c', n. Then
%C 1 = b*c - (b + c)*n = (2n + c') * (2n + b') - (2n + c' + 2n + b') * n = (b' + c')*n + c'*b' > 1. A contradiction. Q.e.d. (End)
%H David A. Corneth, <a href="/A364169/b364169.txt">Table of n, a(n) for n = 1..10000</a>
%e a(1) = 6 as 6 = 2*3, with (2 + 3)*1 = 6 - 1.
%e a(2) = 21 as 21 = 3*7, with (3 + 7)*2 = 21 - 1.
%e a(6) = 301 as 301 = 7*43, with (7 + 43)*6 = 301 - 1.
%p f:= proc(n) local t,d,b,c,m;
%p d:= max(select(`<=`,numtheory:-divisors(n^2+1),n));
%p b:= d+n;
%p c:= (n^2+1)/d + n;
%p b*c
%p end proc:
%p map(f, [$1..100]); # _Robert Israel_, Jul 19 2023
%t seq[max_] := Module[{len = Floor[Sqrt[max]/2], s, r}, s = Table[max + 1, {len}]; Do[r = (b*c - 1)/(b + c); If[IntegerQ[r] && r <= len && b*c < s[[r]], s[[r]] = b*c], {b, 2, max}, {c, 2, max/b}]; TakeWhile[s, # <= max &]]; seq[70000] (* _Amiram Eldar_, Jul 12 2023 *)
%o (PARI) a(n) = for (x=1, oo, my(d=divisors(x)); for (i=1, #d\2, b = d[i]; c = x/d[i]; if ((b+c)*n == (x-1), return(x)););); \\ _Michel Marcus_, Jul 12 2023
%o (PARI) a(n) = {forstep(i = 1, oo, n, if(iscan(i, n), return(i)))}
%o iscan(c, n) = {D = (1 - c)^2 - 4*n^2*c; if(!issquare(D), return(0)); b = ((c - 1) + sqrtint((1-c)^2 - 4*n^2*c)) / (2*n); if(denominator(b) == 1, return(1))} \\ _David A. Corneth_, Jul 12 2023
%o (PARI) a(n) = {res = oo; for(b = n+1, 2*n, c = (n*b + 1)/(b - n); if(denominator(c) == 1, res = min(res, b*c))); res} \\ _David A. Corneth_, Jul 19 2023
%o (PARI) a(n) = my(d = divisors(n^2 + 1), t = d[#d \ 2], b = t+n, c = (n^2 + 1)/t + n); return(b*c) \\ _David A. Corneth_, Jul 20 2023, adapted from _Robert Israel_, Jul 19 2023
%Y Cf. A009112 ("1-pqrs" numbers), A364171 (increasing m).
%K nonn,easy,look
%O 1,1
%A _Jose Aranda_, Jul 12 2023
%E More terms from _Michel Marcus_, Jul 12 2023
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