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%I #38 Jan 04 2021 08:46:32
%S 1,3,5,7,8,9,11,13,14,15,17,19,20,21,23,24,25,26,27,29,31,32,33,34,35,
%T 37,38,39,41,43,44,45,47,48,49,50,51,53,54,55,56,57,59,61,62,63,64,65,
%U 67,68,69,71,73,74,75,76,77,79,80,81,83,84,85,86,87,89,90,91,92,93,94
%N Integers k such that k! is divisible by k*(k+1)/2.
%C k! / (k-th triangular number) is an integer.
%C a(n) = A072668(n) for n>0.
%C From _Bernard Schott_, Dec 11 2020: (Start)
%C Numbers k such that Sum_{j=1..k} j divides Product_{j=1..k} j.
%C k is a term iff k != p-1 with p is an odd prime (see De Koninck & Mercier reference).
%C The ratios obtained a(n)!/T(a(n)) = A108552(n). (End)
%D Jean-Marie De Koninck & Armel Mercier, 1001 Problèmes en Théorie Classique des Nombres, Problème 181 pp. 31 and 163, Ellipses, Paris, 2004.
%D Joseph D. E. Konhauser et al., Which Way Did The Bicycle Go?, Problem 98, pp. 29; 145-146, MAA Washington DC, 1996.
%H Harry J. Smith, <a href="/A060462/b060462.txt">Table of n, a(n) for n = 1..2001</a> [offset adapted by _Georg Fischer_, Jan 04 2021]
%e 5 is a term because 5*4*3*2*1 = 120 is divisible by 5 + 4 + 3 + 2 + 1 = 15.
%p for n from 1 to 300 do if n! mod (n*(n+1)/2) = 0 then printf(`%d,`,n) fi:od:
%t Select[Range[94], Mod[#!, #*(# + 1)/2] == 0 &] (* _Jayanta Basu_, Apr 24 2013 *)
%o (PARI) { f=1; t=0; n=-1; for (m=1, 4000, f*=m; t+=m; if (f%t==0, write("b060462.txt", n++, " ", m)); if (n==2000, break); ) } \\ _Harry J. Smith_, Jul 05 2009
%Y Cf. A000142, A000217, A072668, A108552.
%K nonn
%O 1,2
%A _Michel ten Voorde_, Apr 09 2001
%E Corrected and extended by _Henry Bottomley_ and _James A. Sellers_, Apr 11 2001
%E Offset corrected by _Alois P. Heinz_, Dec 11 2020
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