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Numbers N such that N=(a+b)*c=a*b+c for some a,b,c>1.
3

%I #18 Dec 23 2024 14:53:42

%S 14,33,39,60,64,84,95,110,138,150,155,174,189,217,248,258,259,272,315,

%T 324,360,368,390,399,405,410,430,473,504,530,539,564,584,624,663,670,

%U 732,754,770,819,852,854,869,885,897,915,1005,1008,1024,1053,1056,1065,1104,1110,1120,1139,1155,1248,1278,1292,1360,1378,1422

%N Numbers N such that N=(a+b)*c=a*b+c for some a,b,c>1.

%C Without imposing c>1, there would be the trivial decomposition a=c=1, b=N-1, for any N.

%C One has a>c, b>c, since, e.g., a<=c would imply N = ab+c <= c(b+1) < c(b+a) = N. Therefore one can impose the restriction 1 < c < b <= a.

%H Robert Israel, <a href="/A191676/b191676.txt">Table of n, a(n) for n = 1..10000</a>

%H Claudio Meller, <a href="https://web.archive.org/web/*/http://list.seqfan.eu/oldermail/seqfan/2011-June/014955.html">posting on SeqFan mailing list</a>, June 9, 2011.

%p N:= 10^4: # for terms <= N

%p Res:= NULL:

%p for a from 3 to N/3 do

%p for b from 3 to a while a*b < N do

%p c:= a*b/(a+b-1);

%p if c::posint and c>1 then

%p v:= (a+b)*c;

%p if v<=N then Res:= Res, v fi

%p fi

%p od od:

%p sort(convert({Res},list)); # _Robert Israel_, Nov 06 2019

%t mx = 1424; lmt = Floor[9 Sqrt[mx]/2]; lst = {}; Do[ If[a*b + c == (a + b) c < mx, AppendTo[lst, a*b + c]], {a, 2, lmt}, {b, a + 1, lmt}, {c, 2, a - 1}]; Sort@ lst (* _Robert G. Wilson v_, Jun 17 2011 *)

%o (PARI) is_A191676(N)={fordiv(N,c,c*c>N & return; c>1 & fordiv(N-c,a,a*a>N-c & break; a>c & (a+(N-c)/a)*c==N & return(1)))}

%K nonn

%O 1,1

%A _M. F. Hasler_, based on suggestion of _Claudio Meller_ (cf. link), Jun 10 2011