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Search: id:A137502
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| A137502 |
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Reversal of sequence of powers in prime decomposition of n. |
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1
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| 2, 2, 4, 2, 6, 2, 8, 4, 10, 2, 18, 2, 14, 6, 16, 2, 12, 2, 50, 10, 22, 2, 54, 4, 26, 8, 98, 2, 30, 2, 32, 14, 34, 6, 36, 2, 38, 22, 250, 2, 70, 2, 242, 18, 46, 2, 162, 4, 4, 26, 338, 2, 24, 10, 686, 2, 58, 2, 150, 2, 150, 50, 64, 14, 154, 2, 578, 38, 42, 2, 108, 2, 74, 12, 1058, 6, 506
(list; graph; listen)
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OFFSET
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2,1
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COMMENT
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Using in the prime decomposition of n all the consecutive primes upto the higher prime divisor,null power admitted except for the last one
n=(p(1)^a1)*(p(2)^a2)*.......*(p(k)^ak) (ak>0, others ai>=0 and p(n) =n-th prime)
we reverse the sequence a1,a2,... ak to build c(n):
c(n)= (p(1)^ak)*(p(2)^a(k-1))* . . . . *(p(k)^a1)
As p(1)=2 and ak =/=0, c(n) is always even
If n is prime a(n) =2 and if n is a power of prime, a(n) is the same power of 2
If the sequence a1,a2,. . . . ak is palindromic, a(n)=n
For any given even number Q, we can by reversing the sequence of its powers define not only one but an infinity (by adding as many zeros as we want on the left end) of n such that c(n)=Q. Hence the sequence is a permutation of even integers where each even integer is infinitely repeated.
For example as Q=1224=(2^3)*(3^2)*(5^0)*(7^0)*(11^0)*(13^0)*(17^1),
Q =a((2^1)*(3^0)*(5^0)*(7^0)*(11^0)*(13^2)*(17^3))=a(1660594) but also of an infinity of other ones, the first one being a((2^0)*(3^1)*(5^0)*(7^0)*(11^0)*(13^0)*(17^2)*(19^1))=a(5946753)
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EXAMPLE
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n=9=(2^0)*(3^2), hence a(9) = (2^2)*(3^0)=4
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CROSSREFS
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Sequence in context: A067824 A107067 A046801 this_sequence A143112 A090624 A099735
Adjacent sequences: A137499 A137500 A137501 this_sequence A137503 A137504 A137505
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KEYWORD
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base,easy,nonn
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AUTHOR
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Philippe Lallouet (philip.lallouet(AT)orange.fr), Apr 22 2008
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