|
|
A342930
|
|
Least positive number k such that n^n divides k*(k+1)/2.
|
|
2
|
|
|
1, 7, 26, 511, 3124, 16767, 823542, 33554431, 387420488, 1787109375, 285311670610, 6737830608896, 302875106592252, 10190301669556224, 12913848876953124, 36893488147419103231, 827240261886336764176, 22831345258932427292672, 1978419655660313589123978, 35357007743740081787109375
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,2
|
|
LINKS
|
|
|
FORMULA
|
|
|
EXAMPLE
|
n | a(n) | T(a(n)) = n^n * A342931(n).
----+----------+------------------------------------
1 | 1 | 1 = 1^1 * 1.
2 | 7 | 28 = 2^2 * 7.
3 | 26 | 351 = 3^3 * 13.
4 | 511 | 130816 = 4^4 * 511.
5 | 3124 | 4881250 = 5^5 * 1562.
6 | 16767 | 140574528 = 6^6 * 3013.
7 | 823542 | 339111124653 = 7^7 * 411771.
8 | 33554431 | 562949936644096 = 8^8 * 33554431.
|
|
PROG
|
(PARI) a(n) = my(k=1, m=n^n); while(k*(k+1)/2%m!=0, k++); k;
(PARI) a(n) = { my(p = 2*n^n, f = factor(p), res = oo); for(i = 2^(#f~-1), 2^#f~-1, b = binary(i); pr = prod(j = 1, #f~, f[j, 1]^(b[j]*f[j, 2])); ipr = p/pr; for(j = -1, 0, c = lift(chinese(Mod(-1-j, ipr), Mod(j, pr))); if(c > 0, res = min(res, c)))); res } \\ David A. Corneth, Mar 29 2021
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,easy
|
|
AUTHOR
|
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|