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Search: id:A143792
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| A143792 |
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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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+0
3
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| 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; listen)
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OFFSET
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1,6
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COMMENT
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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]
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LINKS
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Leroy Quet, Home Page (listed in lieu of email address)
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EXAMPLE
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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.
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MATHEMATICA
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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]
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CROSSREFS
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Cf. A143791.
Sequence in context: A139146 A144032 A137686 this_sequence A029375 A071462 A101979
Adjacent sequences: A143789 A143790 A143791 this_sequence A143793 A143794 A143795
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KEYWORD
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base,nonn
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AUTHOR
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Leroy Quet, Sep 01 2008
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EXTENSIONS
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More terms from Robert G. Wilson v (rgwv(AT)rgwv.com), Sep 22 2008
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