OFFSET
1,3
COMMENTS
Previous name: There exists an isomorphism from the positive rationals under multiplication to Z[x] under addition, defined by f(q) = e1 + (e2)x + (e3)(x^2) +...+ (ek)(x^(k-1)) + ... (where e_i is the exponent of the i-th prime in q's prime factorization) The a(n) above are calculated by a(n) = f^(-1)[d/dx f(n)] (In other words: differentiate n's image in Z[x] and return to Q).
With primes noted p_0 = 2, p_1 = 3, etc., let f be the function that maps n = Product_{i=0..d} p_i^e_i to P = Sum_{i=0..d} e_i*X^i; and let g be the inverse function of f. a(n) is by definition g(P') = g((f(n))'). - Luc Rousseau, Aug 06 2018
REFERENCES
Joseph J. Rotman, The Theory of Groups: An Introduction, 2nd ed. Boston: Allyn and Bacon, Inc. 1973. Page 9, problem 1.26.
LINKS
Andrew Howroyd, Table of n, a(n) for n = 1..500
Sam Alexander, Post to sci.math.
Wikipedia, Transport of structure
EXAMPLE
504 = 2^3 * 3^2 * 7 is mapped to polynomial 3+2X+X^3, whose derivative is 2+3X^2, which is mapped to 2^2 * 5^3 = 500. Then, a(504) = 500. - Luc Rousseau, Aug 06 2018
PROG
(PARI) a(n)={my(f=factor(n)); prod(i=1, #f~, my([p, e]=f[i, ]); if(p==2, 1, precprime(p-1)^(e*primepi(p-1))))} \\ Andrew Howroyd, Jul 31 2018
CROSSREFS
KEYWORD
easy,nonn,mult
AUTHOR
Sam Alexander, Dec 12 2003
EXTENSIONS
More terms from Ray Chandler, Dec 20 2003
New name from Peter Munn, Aug 10 2022 using existing formula (Andrew Howroyd, Jul 31 2018) and introductory comment.
STATUS
approved