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This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A364169 Smallest integer m = b*c which satisfies (b + c)*n = m - 1.

Original entry on oeis.org

6, 21, 40, 105, 126, 301, 204, 273, 550, 1221, 936, 697, 690, 3165, 2176, 4641, 1242, 1333, 4200, 8841, 1786, 3213, 2508, 15025, 9126, 18981, 3700, 6105, 13950, 3901, 3876, 4161, 6106, 5781, 23976, 49321, 8178, 6765, 32800, 67281, 6930, 18565, 7440, 11001, 49726, 8925, 9072, 26977
Offset: 1

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Author

Jose Aranda, Jul 12 2023

Keywords

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)

Examples

			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.

Links

Crossrefs

Cf. A009112 ("1-pqrs" numbers), A364171 (increasing m).

Programs

  • 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 *)
  • 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

Extensions

More terms from Michel Marcus, Jul 12 2023
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