A060462 Integers k such that k! is divisible by k*(k+1)/2.
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
Offset: 1
Keywords
Examples
5 is a term because 5*4*3*2*1 = 120 is divisible by 5 + 4 + 3 + 2 + 1 = 15.
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]
Programs
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Maple
for n from 1 to 300 do if n! mod (n*(n+1)/2) = 0 then printf(`%d,`,n) fi:od:
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Mathematica
Select[Range[94], Mod[#!, #*(# + 1)/2] == 0 &] (* Jayanta Basu, Apr 24 2013 *)
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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
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Python
from sympy import composite def A060462(n): return composite(n-1)-1 if n>1 else 1 # Chai Wah Wu, Aug 02 2024
Extensions
Corrected and extended by Henry Bottomley and James Sellers, Apr 11 2001
Offset corrected by Alois P. Heinz, Dec 11 2020
Comments