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A326952
a(n) = A001222(A028905(n)).
2
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 2, 1, 4, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 4, 1, 1, 1, 2, 2, 1, 1, 1, 5, 1, 1, 1, 1, 1, 3, 3, 1, 3, 1, 1, 1, 7, 3, 1, 1, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 1, 1, 3, 1, 1, 2, 2, 1, 3, 2, 2, 1, 4, 1, 1, 2, 2, 1, 1, 2, 1, 1, 2, 1, 3, 2, 2
OFFSET
1,13
COMMENTS
Multiplicity of prime divisors of n, where n is a number composed of the sorted digits of a prime number.
Conjecture: the sum of the first n terms of A326952 (smallest to largest sorting) is <= the sum of the first n terms of A326953 (largest to smallest sorting). This is true for the first 9592 terms.
LINKS
Joshua Michael McAteer, Table of n, a(n) for n = 1..9592
EXAMPLE
The 13th prime number is 41. Sorting the digits gives 14. 14 has 2 factors, 2 and 7. The 13th term of this sequence is 2.
PROG
(MATLAB)
nmax= 100;
p = primes(nmax);
lp = length(p);
sfac = zeros(1, lp);
for i = 1:lp
digp=str2double(regexp(num2str(p(i)), '\d', 'match'));
sdigp = sort(digp);
l=length(digp);
conv = 10.^flip(0:(l-1));
snum = sum(conv.*sdigp);
sfac(i) = numel(factor(snum));
end
CROSSREFS
Cf. A001222 (bigomega), A028905, A326953 (for reverse sorting).
Sequence in context: A187201 A262403 A343491 * A109909 A254573 A144387
KEYWORD
nonn,base
AUTHOR
STATUS
approved