|
| |
|
|
A289629
|
|
Smallest positive k such that (k+1)^n + (-k)^n is divisible by a square greater than 1.
|
|
6
|
|
|
|
3, 7, 113, 14, 3, 23, 19, 7, 1, 2, 113, 75, 3, 7, 765, 36, 3, 2476, 87, 1, 3, 165, 19, 14, 2, 7, 28, 149, 1, 2972, 151, 2, 3, 14, 113, 977, 3, 5, 19, 34, 3, 135, 113, 7, 3, 136, 335, 23, 1, 7, 113, 11, 3, 2, 19, 2, 3
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
2,1
|
|
|
COMMENTS
|
Most of the factorizations to prove known terms of this sequence have been uploaded to factordb.com.
For known terms up to n = 100, the square that divides (k+1)^n + (-k)^n is very small, less than 1500^2, with only one example greater than 1000^2.
a(20m+10) = 1 for m >= 0. With k = 1 and starting at n = 20*0 + 10 = 10, (k+1)^n + (-k)^n = 2^10 + 1 = 1024 + 1 = 1025 which is divisible by 5^2. Since the last two digits of 2^n repeat in a cycle of length 20, (k+1)^n + (-k)^n will always be divisible by 5^2 for n = 20m + 10.
Conjecture: (k+1)^n + (-k)^n is not squarefree for the following (n, k) patterns, with m >= 1: (22m-11, 2), (20m-6, 3), (20m-2, 3), (3^m, 7), (15m, 7), (20m-15, 14), (16m-8, 19), (42m-35, 23), and (8m-4, 113). In each case, the value of a(n) in this sequence is usually equal to the value specified for k, but sometimes this value is not the smallest such k. For example, in the (n, k) = (20m-2, 3) case, a(20m-2) = 3 for m = 1..3, but at m = 4, a(20*4-2) = a(78) = 1.
(End)
|
|
|
LINKS
|
|
|
|
FORMULA
|
|
|
|
EXAMPLE
|
a(2) = 3 because (1+1)^2 + (-1)^2 = 5 is squarefree, (2+1)^2 + (-2)^2 = 13 is squarefree, and (3+1)^2 + (-3)^2 = 25 is divisible by 5^2.
|
|
|
MATHEMATICA
|
|
|
|
PROG
|
(PARI) a(n) = my(k=1); while (issquarefree((k+1)^n + (-k)^n), k++); k; \\ Michel Marcus, Dec 04 2021
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn,more
|
|
|
AUTHOR
|
|
|
|
EXTENSIONS
|
Offset corrected; a(16), a(32), a(36), a(44), and a(48) corrected; and a(50)-a(58) added by Kevin P. Thompson, Dec 05 2021.
|
|
|
STATUS
|
approved
|
| |
|
|
