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A326953
a(n) = A001222(A028906(n)).
2
1, 1, 1, 1, 1, 1, 1, 2, 5, 3, 1, 1, 1, 1, 2, 1, 2, 1, 3, 1, 1, 1, 1, 3, 1, 3, 3, 3, 4, 1, 2, 1, 2, 3, 1, 2, 1, 1, 1, 2, 1, 1, 1, 3, 1, 1, 1, 3, 3, 2, 3, 3, 1, 1, 5, 4, 3, 2, 3, 1, 7, 3, 3, 1, 1, 2, 1, 1, 1, 2, 2, 1, 2, 1, 2, 3, 1, 2, 3, 4, 1, 1, 1, 1, 2, 1, 5, 3, 1, 1, 3, 2, 3, 1, 3, 3, 4, 1, 4, 1
OFFSET
1,8
COMMENTS
Multiplicity of prime divisors of n, where n is a number composed of the reverse sorted digits of a prime number.
Conjecture: the sum of the first n terms of A326953 (largest to smallest sorting) is >= the sum of the first n terms of A326952 (smallest to largest sorting). This is true for the first 9592 terms.
LINKS
Joshua Michael McAteer, Table of n, a(n) for n = 1..9592
EXAMPLE
The 28th prime number is 107. The reverse sorted digits are 710. The factorization of 710 is 2, 5, 71, therefore the 28th term in this sequence is 3.
PROG
(MATLAB)
nmax= 100;
p = primes(nmax);
lp = length(p);
lfac = zeros(1, lp);
for i = 1:lp
digp=str2double(regexp(num2str(p(i)), '\d', 'match'));
ldigp = flip(sort(digp));
l=length(digp);
conv = 10.^flip(0:(l-1));
lnum = sum(conv.*ldigp);
lfac(i) = numel(factor(lnum));
end
CROSSREFS
Cf. A001222 (bigomega), A028906, A326952 (for ascending sorted version).
Sequence in context: A265318 A279536 A269954 * A234255 A062706 A059217
KEYWORD
nonn,base
AUTHOR
STATUS
approved