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 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 Antti Karttunen, Table of n, a(n) for n = 1..20000 C. Rivera, Puzzle 248, The Prime Puzzles and Problems Connection. FORMULA a(n) = A007947(n) mod A173557(n). - Antti Karttunen, Feb 25 2018 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) A007947(n) = factorback(factorint(n)[, 1]); \\ This function from Andrew Lelechenko, May 09 2014 A173557(n) = my(f=factor(n)[, 1]); prod(k=1, #f, f[k]-1); \\ This function from Michel Marcus, Oct 31 2017 A089631(n) = (A007947(n)%A173557(n)); \\ Antti Karttunen, Feb 25 2018 CROSSREFS Cf. A007947, A173557. Sequence in context: A129620 A074766 A138107 * A298878 A195982 A102761 Adjacent sequences:  A089628 A089629 A089630 * A089632 A089633 A089634 KEYWORD nonn AUTHOR Joseph L. Pe, Jan 04 2004 EXTENSIONS Term a(1) = 0 prepended by Antti Karttunen, Feb 25 2018 STATUS approved

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Last modified November 19 00:12 EST 2018. Contains 317332 sequences. (Running on oeis4.)