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a(n) = the number of distinct prime divisors, p, of n that, when p is represented in binary, each p occurs at least once in the binary representation of n.
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%I #13 Aug 01 2015 13:41:58

%S 0,1,1,1,1,2,1,1,0,2,1,2,1,2,1,1,1,1,1,2,0,2,1,2,0,2,1,2,1,2,1,1,0,2,

%T 0,1,1,2,1,2,1,1,1,2,2,2,1,2,0,1,1,2,1,2,2,2,1,2,1,2,1,2,2,1,0,1,1,2,

%U 0,1,1,1,1,2,2,2,0,2,1,2,0,2,1,1,1,2,1,2,1,3,1,2,1,2,1,2,1,1,1,1,1,2,1,2,2

%N a(n) = the number of distinct prime divisors, p, of n that, when p is represented in binary, each p occurs at least once in the binary representation of n.

%C a(2^k * p) = 2, where k = any positive integer and p = any odd prime.

%C a(p) = 1, where p = any prime.

%C a(2^k) = 1, where k = any positive integer.

%C a(n) <= A078826(n). - _Reinhard Zumkeller_, Sep 08 2008

%C Size of intersection of n-th rows of tables A225243 and A027748. - _Reinhard Zumkeller_, Aug 14 2013

%H Reinhard Zumkeller, <a href="/A143792/b143792.txt">Table of n, a(n) for n = 1..10000</a>

%e 60 in binary is 111100. The distinct primes dividing 60 are 2 (which is 10 in binary), 3 (11 in binary) and 5 (101) in binary. The string 10 does occur within 111100 like so: 111(10)0. The string 11 also occurs (multiple times) within 111100, in one way like so: (11)1100. But the string 101 does not occur in 111100. Since 2 and 3 occur within 60 (when each of these numbers is written in binary), but 5 does not, then a(60) = 2.

%t f[n_] := Block[{nb = ToString@ FromDigits@ IntegerDigits[n, 2], psb = ToString@ FromDigits@ IntegerDigits[ #, 2] & /@ First@ Transpose@ FactorInteger@ n, c = 0, k = 1}, lmt = 1 + Length@ psb; While[k < lmt, If[ StringCount[nb, psb[[k]]] > 0, c++ ]; k++ ]; c]; f[1] = 0; Array[f, 105] (* _Robert G. Wilson v_, Sep 22 2008 *)

%o (Haskell)

%o import Data.List (intersect)

%o a143792 n = length $ a225243_row n `intersect` a027748_row (fromIntegral n)

%o -- _Reinhard Zumkeller_, Aug 14 2013

%Y Cf. A143791.

%K base,nonn

%O 1,6

%A _Leroy Quet_, Sep 01 2008

%E More terms from _Robert G. Wilson v_, Sep 22 2008