|
|
A060462
|
|
Integers k such that k! is divisible by k*(k+1)/2.
|
|
17
|
|
|
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
|
Cf. A000142, A000217, A072668, A108552.
Sequence in context: A299542 A144724 A196990 * A072668 A249433 A300737
Adjacent sequences: A060459 A060460 A060461 * A060463 A060464 A060465
|
|
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
|
|
|
|