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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)*3 + (e3)*9 + (e4)*27 + ... + (ek)*(3^(k-1)) + ...
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%I #18 Sep 09 2017 19:43:21

%S 0,1,3,2,9,4,27,3,6,10,81,5,243,28,12,4,729,7,2187,11,30,82,6561,6,18,

%T 244,9,29,19683,13,59049,5,84,730,36,8,177147,2188,246,12,531441,31,

%U 1594323,83,15,6562,4782969,7,54,19,732,245,14348907,10,90,30,2190

%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)*3 + (e3)*9 + (e4)*27 + ... + (ek)*(3^(k-1)) + ...

%C Replace "3" 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.

%C For examples of such evaluations at x=3, see "Other identities" in the Formula section. - _Antti Karttunen_, Jul 31 2015

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

%H Antti Karttunen, <a href="/A090880/b090880.txt">Table of n, a(n) for n = 1..512</a>

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

%F a(1) = 0; for n > 1, a(n) = 3^(A055396(n)-1) + a(A032742(n)). [Where A055396(n) gives the index of the smallest prime dividing n and A032742(n) gives the largest proper divisor of n.] - _Antti Karttunen_, Jul 29 2015

%F Other identities. For all n >= 0:

%F a(A206296(n)) = A006190(n).

%F a(A260443(n)) = A178590(n).

%o (Scheme) (define (A090880 n) (if (= 1 n) (- n 1) (+ (A000244 (- (A055396 n) 1)) (A090880 (A032742 n))))) ;; _Antti Karttunen_, Jul 29 2015

%Y Row 3 of A104244.

%Y Cf. A001222, A006190, A048675, A054841, A090881, A090882, A090883, A090884, A178590, A206296, A260443.

%K easy,nonn

%O 1,3

%A _Sam Alexander_, Dec 12 2003

%E More terms from _Ray Chandler_, Dec 20 2003