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Suppose n=(p1^e1)(p2^e2)... where p1,p2,... are the prime numbers and e1,e2,... are nonnegative integers. Then a(n) = e1 + (e2)*4 + (e3)*16 + (e4)*64 + ... + (ek)*(4^(k-1)) + ...
6

%I #6 Oct 09 2013 14:20:16

%S 0,1,4,2,16,5,64,3,8,17,256,6,1024,65,20,4,4096,9,16384,18,68,257,

%T 65536,7,32,1025,12,66,262144,21,1048576,5,260,4097,80,10,4194304,

%U 16385,1028,19,16777216,69,67108864,258,24,65537,268435456,8,128,33,4100,1026

%N Suppose n=(p1^e1)(p2^e2)... where p1,p2,... are the prime numbers and e1,e2,... are nonnegative integers. Then a(n) = e1 + (e2)*4 + (e3)*16 + (e4)*64 + ... + (ek)*(4^(k-1)) + ...

%C Replace "4" with "x" and extend the definition of a to positive rationals and a becomes an isomorphism between positive rationals under multiplication and polynomials over Z under addition. This remark generalizes A001222, A048675 and A054841: evaluate said polynomial at x=1, x=2 and x=10, respectively.

%D Joseph J. Rotman, The Theory of Groups: An Introduction, 2nd ed. Boston: Allyn and Bacon, Inc. 1973. Page 9, problem 1.26.

%H Sam Alexander, <a href="http://tinyurl.com/yzjw">Post to sci.math</a>.

%Y Cf. A001222, A048675, A054841, A090880, A090882, A090883, A090884.

%K easy,nonn

%O 1,3

%A _Sam Alexander_, Dec 12 2003

%E More terms from _Ray Chandler_, Dec 20 2003