|
|
A089631
|
|
a(n) = (Product_{p is a prime factor of n} p) mod (Product_{p is a prime factor of n} p-1).
|
|
1
|
|
|
0, 0, 1, 0, 1, 0, 1, 0, 1, 2, 1, 0, 1, 2, 7, 0, 1, 0, 1, 2, 9, 2, 1, 0, 1, 2, 1, 2, 1, 6, 1, 0, 13, 2, 11, 0, 1, 2, 15, 2, 1, 6, 1, 2, 7, 2, 1, 0, 1, 2, 19, 2, 1, 0, 15, 2, 21, 2, 1, 6, 1, 2, 9, 0, 17, 6, 1, 2, 25, 22, 1, 0, 1, 2, 7, 2, 17, 6, 1, 2, 1, 2, 1, 6, 21, 2, 31, 2, 1, 6, 19, 2, 33, 2, 23, 0, 1, 2, 13
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,10
|
|
COMMENTS
|
It can be shown that the unsolved problem of finding a composite solution n of the congruence n-1 = 1 mod phi(n) is equivalent to finding a squarefree composite number n such that a(n) = 1.
|
|
LINKS
|
C. Rivera, Puzzle 248, The Prime Puzzles and Problems Connection.
|
|
FORMULA
|
|
|
EXAMPLE
|
Let n = 10; then Product_{p is a prime factor of 10} p = 2*5 = 10, Product_{p is a prime factor of 10} p-1 = (2-1) * (5-1) = 1*4 = 4, so a(10) = 10 mod 4 = 2.
|
|
MATHEMATICA
|
p[n_] := Transpose[FactorInteger[n]][[1]]; t = Table[p[i], {i, 2, 100}]; r = {}; s = {}; For[i = 1, i <= 99, i++, r = Append[r, Product[t[[i]][[j]], {j, 1, Length[t[[i]]]}]]; s = Append[s, Product[t[[i]][[j]] - 1, {j, 1, Length[t[[i]]]}]]]; Mod[r, s]
|
|
PROG
|
(PARI)
A173557(n) = my(f=factor(n)[, 1]); prod(k=1, #f, f[k]-1); \\ This function from Michel Marcus, Oct 31 2017
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|