|
%I #19 Jan 23 2018 07:03:48
%S 3,6,4,9,18,5,12,8,36,6,15,30,45,60,7,18,12,10,24,90,8,21,42,63,84,
%T 105,126,9,24,16,72,12,120,48,168,10,27,54,15,108,135,30,189,216,11,
%U 30,20,90,40,14,60,210,80,270,12,33,66,99,132,165,198,231,264,297,330,13
%N Triangular array T read by rows, T(n, k) = n*k*(gcd(n,k)+2)/gcd(n,k)^2.
%C Described in the CNRS link as a puzzle where op(n,k) is defined by: op(n,n)=n+2, op(n,k)=op(k,n) and op(n,n+k)/op(n,k)=(n+k)/k.
%C If gcd(n,k)+2 is replaced by gcd(n,k), then triangle A051173 is obtained.
%H Indranil Ghosh, <a href="/A242224/b242224.txt">Rows 1..100 of triangle, flattened</a>
%H Ana Rechtman, <a href="http://images.math.cnrs.fr/Mai-1er-defi.html">Mai, 1er défi</a>, Images des Mathématiques, CNRS, 2014 (in French).
%H See also <a href="http://images.math.cnrs.fr/spip.php?page=forum&id_article=2583">Comments</a>, Images des Mathématiques, CNRS, 2014.
%e Triangle begins:
%e 3;
%e 6, 4;
%e 9, 18, 5;
%e 12, 8, 36, 6;
%e 15, 30, 45, 60, 7;
%e 18, 12, 10, 24, 90, 8;
%e ...
%t t[n_, k_] := n*k*(GCD[n, k] +2)/GCD[n, k]^2; Table[ t[n, k], {n, 12}, {k, n}] // Flatten (* _Robert G. Wilson v_, Jan 21 2018 *)
%o (PARI) tabl(nn) = {for (n=1, nn, for (k=1, n, print1(n*k*(gcd(n,k)+2)/gcd(n,k)^2, ", ");); print(););}
%K nonn,tabl
%O 1,1
%A _Michel Marcus_, May 08 2014
|
