|
| |
|
|
A087207
|
|
A binary representation of the primes that divide a number, shown in decimal.
|
|
50
|
|
|
|
0, 1, 2, 1, 4, 3, 8, 1, 2, 5, 16, 3, 32, 9, 6, 1, 64, 3, 128, 5, 10, 17, 256, 3, 4, 33, 2, 9, 512, 7, 1024, 1, 18, 65, 12, 3, 2048, 129, 34, 5, 4096, 11, 8192, 17, 6, 257, 16384, 3, 8, 5, 66, 33, 32768, 3, 20, 9, 130, 513, 65536, 7, 131072, 1025, 10, 1, 36, 19, 262144, 65, 258
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,3
|
|
|
COMMENTS
|
The binary representation of a(n) shows which prime numbers divide n, but not the multiplicities. a(2)=1, a(3)=10, a(4)=1, a(5)=100, a(6)=11, a(10)=101, a(30)=111, etc.
For n > 1, a(n) gives the (one-based) index of the column where n is located in array A285321. A008479 gives the other index. - Antti Karttunen, Apr 17 2017
For all n > 1 for which the value of A285331(n) is well-defined, we have A285331(a(n)) <= floor(A285331(n)/2), because then n is included in the binary tree A285332 and a(n) is one of its ancestors (in that tree), and thus must be at least one step nearer to its root than n itself.
Conjecture: Starting at any n and iterating the map n -> a(n), we will always reach 0 (see A288569). This conjecture is equivalent to the conjecture that at any n that is neither a prime nor a power of two, we will eventually hit a prime number (which then becomes a power of two in the next iteration). If this conjecture is false then sequence A285332 cannot be a permutation of natural numbers. On the other hand, if the conjecture is true, then A285332 must be a permutation of natural numbers, because all primes and powers of 2 occur in definite positions in that tree. This conjecture also implies the conjectures made in A019565 and A285320 that essentially claim that there are neither finite nor infinite cycles in A019565.
If there are any 2-cycles in this sequence, then both terms of the cycle should be present in A286611 and the larger one should be present in A286612.
(End)
Binary rank of the distinct prime indices of n, where the binary rank of an integer partition y is given by Sum_i 2^(y_i-1). For all prime indices (with multiplicity) we have A048675. - Gus Wiseman, May 25 2024
|
|
|
LINKS
|
|
|
|
FORMULA
|
Additive with a(p^e) = 2^(i-1) where p is the i-th prime. - Vladeta Jovovic, Oct 29 2003
(End)
a(n^2) = a(n).
(End)
|
|
|
EXAMPLE
|
a(38) = 129 because 38 = 2*19 = prime(1)*prime(8) and 129 = 2^0 + 2^7 (in binary 10000001).
a(140) = 13, binary 1101 because 140 is divisible by the first, third and fourth primes and 2^(1-1) + 2^(3-1) + 2^(4-1) = 13.
|
|
|
MATHEMATICA
|
a[n_] := Total[ 2^(PrimePi /@ FactorInteger[n][[All, 1]] - 1)]; a[1] = 0; Table[a[n], {n, 1, 69}] (* Jean-François Alcover, Dec 12 2011 *)
|
|
|
PROG
|
(Haskell)
a087207 = sum . map ((2 ^) . (subtract 1) . a049084) . a027748_row
(PARI) a(n) = {if (n==1, 0, my(f=factor(n), v = []); forprime(p=2, vecmax(f[, 1]), v = concat(v, vecsearch(f[, 1], p)!=0); ); fromdigits(Vecrev(v), 2)); } \\ Michel Marcus, Jun 05 2017
(PARI) A087207(n)=vecsum(apply(p->1<<primepi(p-1), factor(n)[, 1])) \\ Significantly faster than using sum(...). - M. F. Hasler, Jun 23 2017
(Python)
from sympy import factorint, primepi
def a(n):
return sum(2**primepi(i - 1) for i in factorint(n))
(Scheme)
|
|
|
CROSSREFS
|
Cf. A000040, A000120, A001221, A005117, A008479, A019565, A055396, A285320, A285321, A285329, A285330, A285332.
Positions of particular values are: A000079\{1} (1), A000244\{1} (2), A033845 (3), A000351\{1} (4), A033846 (5), A033849 (6), A143207 (7), A000420\{1} (8), A033847 (9), A033850 (10), A033851 (12), A147576 (14), A147571 (15), A001020\{1} (16), A033848 (17).
A048675 gives binary rank of prime indices.
Cf. A000720, A005940, A018819, A023506, A071814, A225620, A277319, A277905, A304818, A372689, A372890.
|
|
|
KEYWORD
|
nonn,base,nice
|
|
|
AUTHOR
|
Mitch Cervinka (puritan(AT)planetkc.com), Oct 26 2003
|
|
|
EXTENSIONS
|
|
|
|
STATUS
|
approved
|
| |
|
|
