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%I #21 Sep 09 2017 19:42:31
%S 1,1,1,1,1,1,1,1,1,1,0,1,1,1,1,1,1,1,1,1,1,0,0,1,1,1,1,1,0,1,0,1,0,1,
%T 1,1,1,1,1,1,0,1,0,0,1,0,0,1,1,1,1,1,0,1,0,1,1,0,0,1,0,0,1,1,1,0,0,1,
%U 0,1,0,1,1,1,1,1,0,1,0,1,1,0,0,1,1,0,0,0,0,1,1,0,0,0,1,1,1,1,0,1,0,1,0,1,1
%N Characteristic function of numbers having only factors that are Pierpont primes.
%H Antti Karttunen, <a href="/A122261/b122261.txt">Table of n, a(n) for n = 1..12289</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/PierpontPrime.html">Pierpont Prime</a>
%H <a href="/index/Ch#char_fns">Index entries for characteristic functions</a>
%F Multiplicative with a(p) = A065333(p-1), for p prime.
%F a(n) = if n=1 then 0 else A122262(n) - A122262(n-1).
%F a(A122260(n)) = 1.
%F a(n) = A122255(n) for n < 25.
%e For n = 11 = 11^1, 11 is not a Pierpoint prime because 11-1 = 10 = 2*5 has a prime factor larger than 3, thus a(11) = 0.
%e For n = 25 = 5^2, 5 is a Pierpoint prime as 5-1 = 4 = 2^2 does not have any prime factors larger than 3, thus a(25) = 1.
%t Block[{nn = 105, s}, s = Select[Sort@ Flatten@ Table[2^i*3^j + 1, {i, 0, Log2@ nn}, {j, 0, Log[3, nn/2^i]}] , PrimeQ]; Table[Boole[n == 1] + Boole@ AllTrue[FactorInteger[n][[All, 1]], MemberQ[s, #] &], {n, nn}]] (* _Michael De Vlieger_, Aug 23 2017, after _Robert G. Wilson v_ at A005109 *)
%o (PARI)
%o A065333(n) = ((3^valuation(n, 3)<<valuation(n, 2))==n); \\ This function from _Charles R Greathouse IV_, Aug 21 2011
%o A122261(n) = factorback(apply(p -> A065333(p-1), (factor(n)[, 1]))); \\ _Antti Karttunen_, Aug 22 2017
%Y Cf. A005109, A065333, A122255, A122262 (partial sums).
%Y Characteristic function of A122260.
%K nonn,mult
%O 1,1
%A _Reinhard Zumkeller_, Aug 29 2006
%E An unnecessary part removed from the formula and the Example section added by _Antti Karttunen_, Aug 22 2017
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