OFFSET
1,1
COMMENTS
All terms have b,c > 1.
a(n) is the smallest of a certain n-class. The 1-class would be related to the numbers denoted as "pqrs" in A009112.
From David A. Corneth, Jul 19 2023: (Start)
n < min(b, c) <= 2*n.
Proof of n < min(b, c):
As m = b*c and (b + c)*n = m - 1 we have m - (m - 1) = 1 = b*c - (b + c)*n.
Solving 1 = b*c = (b + c)*n for c gives c = (n*b + 1) / (b - n) > 0.
As 0 < b, c, n we have b - n > 0 so b > n.
Similarily as b = (n*c + 1)/(c - n) we have c > n.
Proof of min(b, c) <= 2*n by contradiction.
Suppose 2*n < min(b, c). Then 2*n + 1 <= min(b, c) as both b and c are integers.
Let b = 2n + b' and c = 2n + c' where 1 <= b', c', n. Then
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)
LINKS
David A. Corneth, Table of n, a(n) for n = 1..10000
EXAMPLE
a(1) = 6 as 6 = 2*3, with (2 + 3)*1 = 6 - 1.
a(2) = 21 as 21 = 3*7, with (3 + 7)*2 = 21 - 1.
a(6) = 301 as 301 = 7*43, with (7 + 43)*6 = 301 - 1.
MAPLE
f:= proc(n) local t, d, b, c, m;
d:= max(select(`<=`, numtheory:-divisors(n^2+1), n));
b:= d+n;
c:= (n^2+1)/d + n;
b*c
end proc:
map(f, [$1..100]); # Robert Israel, Jul 19 2023
MATHEMATICA
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 *)
PROG
(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
(PARI) a(n) = {forstep(i = 1, oo, n, if(iscan(i, n), return(i)))}
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
(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
(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
CROSSREFS
KEYWORD
AUTHOR
Jose Aranda, Jul 12 2023
EXTENSIONS
More terms from Michel Marcus, Jul 12 2023
STATUS
approved