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%I #19 Jul 06 2019 01:50:26
%S 1,2,5,16,11,90,17,512,625,550,23,6480,31,1666,2695,65536,41,101250,
%T 47,110000,10285,5566,59,1866240,14641,10478,1953125,653072,67,
%U 1212750,73,33554432,19435,23698,31603,65610000,83,33934,44795,88000000,97,9071370,103
%N For any number n > 0, let f(n) be the polynomial of a single indeterminate x where the coefficient of x^e is the prime(1+e)-adic valuation of n (where prime(k) denotes the k-th prime); f establishes a bijection between the positive numbers and the polynomials of a single indeterminate x with nonnegative integer coefficients; let g be the inverse of f; a(n) = g(f(n)^2).
%C This sequence is the main diagonal of A297845.
%C This sequence has similarities with A296857.
%H Rémy Sigrist, <a href="/A297473/b297473.txt">Table of n, a(n) for n = 1..10000</a>
%H Rémy Sigrist, <a href="/A297473/a297473.png">Colored logarithmic scatterplot of the first 100000 terms</a> (where the color is function of A001222(n))
%F For any n > 0 and k > 0:
%F - A001221(a(n)) <= A001221(n)^2,
%F - A001222(a(n)) = A001222(n)^2,
%F - A055396(a(n)) = 2*A055396(n)-1 + [n=1],
%F - A061395(a(n)) = 2*A061395(n)-1 + [n=1],
%F - a(A000040(n)) = A031368(n),
%F - a(A000040(n)^k) = A031368(n)^(k^2).
%e For n = 12:
%e - 12 = 2^2 * 3 = prime(1+0)^2 * prime(1+1),
%e - f(12) = 2 + x,
%e - f(12)^2 = 4 + 4*x + x^2,
%e - a(12) = prime(1+0)^4 * prime(1+1)^4 * prime(1+2) = 2^4 * 3^4 * 5 = 6480.
%o (PARI) a(n) = my (f=factor(n), p=apply(primepi, f[,1]~)); prod (i=1, #p, prod(j=1, #p, prime(p[i]+p[j]-1)^(f[i,2]*f[j,2])))
%Y Cf. A000040, A001221, A001222, A031368, A055396, A061395, A296857, A297845.
%K nonn
%O 1,2
%A _Rémy Sigrist_, Dec 30 2017
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