|
%I #18 Oct 30 2019 17:11:09
%S 2,6,4,12,20,3,6,12,30,42,8,24,56,18,72,10,15,40,90,110,4,6,12,24,36,
%T 60,132,156,14,35,84,182,10,30,60,210,16,48,112,240,272,9,18,36,63,90,
%U 144,306,342,5,20,30,60,80,180,380,28,42,126,420,22,99,220,462
%N Triangle read by rows: row n consists of the positive integers m where m+n divides m*n.
%C The maximum term of the n-th row, for n >= 2, is n*(n-1). The minimum term of row n is A063427(n). Row n contains A063647(n) terms (according to a comment by Benoit Cloitre). For p prime, row p^k has k terms. (Each term in row p^k is of the form p^k*(p^j -1), 1 <= j <= k.)
%H Nathaniel Johnston, <a href="/A127730/b127730.txt">Rows n = 2..500, flattened</a>
%F Let d_n be the sequence of divisors of n^2 that are less than n, in reverse order. Then T(n,k) = n*(n-d_n(k))/d_n(k). - _Franklin T. Adams-Watters_, Aug 07 2009
%e Row 6 is (3,6,12,30) because 6+3 = 9 divides 6*3 = 18, 6+6 = 12 divides 6*6 = 36, 6+12 = 18 divides 6*12 = 72 and 6+30 = 36 divides 6*30 = 180.
%p for n from 2 to 20 do for m from 1 to n*(n-1) do if(m*n mod (m+n) = 0)then printf("%d, ",m): fi: od: od: # _Nathaniel Johnston_, Jun 22 2011
%t f[n_] := Select[Range[n^2], Mod[n*#, n + # ] == 0 &];Table[f[n], {n, 2, 24}] // Flatten (* _Ray Chandler_, Feb 13 2007 *)
%o (PARI) arow(n)=local(d,m);d=divisors(n^2);vector(#d\2,k,m=d[ #d\2-k+1];n*(n-m)/m) \\ _Franklin T. Adams-Watters_, Aug 07 2009
%Y Cf. A063427, A063647, A127731, A191973.
%K nonn,easy,tabf
%O 2,1
%A _Leroy Quet_, Jan 26 2007
%E Extended by _Ray Chandler_, Feb 13 2007
|
