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A060462 Integers k such that k! is divisible by k*(k+1)/2. 18
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, 37, 38, 39, 41, 43, 44, 45, 47, 48, 49, 50, 51, 53, 54, 55, 56, 57, 59, 61, 62, 63, 64, 65, 67, 68, 69, 71, 73, 74, 75, 76, 77, 79, 80, 81, 83, 84, 85, 86, 87, 89, 90, 91, 92, 93, 94 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,2
COMMENTS
k! / (k-th triangular number) is an integer.
a(n) = A072668(n) for n>0.
From Bernard Schott, Dec 11 2020: (Start)
Numbers k such that Sum_{j=1..k} j divides Product_{j=1..k} j.
k is a term iff k != p-1 with p is an odd prime (see De Koninck & Mercier reference).
The ratios obtained a(n)!/T(a(n)) = A108552(n). (End)
REFERENCES
Jean-Marie De Koninck & Armel Mercier, 1001 Problèmes en Théorie Classique des Nombres, Problème 181 pp. 31 and 163, Ellipses, Paris, 2004.
Joseph D. E. Konhauser et al., Which Way Did The Bicycle Go?, Problem 98, pp. 29; 145-146, MAA Washington DC, 1996.
LINKS
Harry J. Smith, Table of n, a(n) for n = 1..2001 [offset adapted by Georg Fischer, Jan 04 2021]
EXAMPLE
5 is a term because 5*4*3*2*1 = 120 is divisible by 5 + 4 + 3 + 2 + 1 = 15.
MAPLE
for n from 1 to 300 do if n! mod (n*(n+1)/2) = 0 then printf(`%d, `, n) fi:od:
MATHEMATICA
Select[Range[94], Mod[#!, #*(# + 1)/2] == 0 &] (* Jayanta Basu, Apr 24 2013 *)
PROG
(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
CROSSREFS
Sequence in context: A299542 A144724 A196990 * A072668 A249433 A300737
KEYWORD
nonn
AUTHOR
Michel ten Voorde, Apr 09 2001
EXTENSIONS
Corrected and extended by Henry Bottomley and James A. Sellers, Apr 11 2001
Offset corrected by Alois P. Heinz, Dec 11 2020
STATUS
approved

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Last modified April 19 21:09 EDT 2024. Contains 371798 sequences. (Running on oeis4.)