|
| |
|
|
A143792
|
|
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.
|
|
2
| |
|
|
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, 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, 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
(list; graph; refs; listen; history; internal format)
|
|
|
|
OFFSET
| 1,6
|
|
|
COMMENTS
| a(2^k *p) = 2, where k = any positive integer and p = any odd prime.
a(p) = 1, where p = any prime.
a(2^k) = 1, where k = any positive integer.
a(n) <= A078826(n). [From Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Sep 08 2008]
|
|
|
EXAMPLE
| 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.
|
|
|
MATHEMATICA
| 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] [From Robert G. Wilson v (rgwv(AT)rgwv.com), Sep 22 2008]
|
|
|
CROSSREFS
| Cf. A143791.
Sequence in context: A139146 A144032 A137686 * A029375 A071462 A101979
Adjacent sequences: A143789 A143790 A143791 * A143793 A143794 A143795
|
|
|
KEYWORD
| base,nonn
|
|
|
AUTHOR
| Leroy Quet, Sep 01 2008
|
|
|
EXTENSIONS
| More terms from Robert G. Wilson v (rgwv(AT)rgwv.com), Sep 22 2008
|
| |
|
|
