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

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

1,1

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

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

Sequence in context: A363621 A225150 A056237 * A364171 A199194 A268223

Adjacent sequences: A364166 A364167 A364168 * A364170 A364171 A364172

Keyword

nonn,easy,look

Author

Jose Aranda, Jul 12 2023

Extensions

More terms from Michel Marcus, Jul 12 2023

Status

approved