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%I #14 May 19 2019 22:05:41
%S 0,1,1,2,1,11,1,3,2,11,1,21,1,11,11,4,1,21,1,21,11,11,1,31,2,11,3,21,
%T 1,111,1,5,11,11,11,22,1,11,11,31,1,111,1,21,21,11,1,41,2,21,11,21,1,
%U 31,11,31,11,11,1,211,1,11,21,6,11,111,1,21,11,111,1
%N Prime signature of n (sorted in decreasing order), concatenated.
%C In the variant A037916, the exponents of the prime factorization are concatenated without being sorted first (i.e., rows of A124010).
%F a(n) = concatenation of row n of A212171.
%F a(n) = a(A046523(n)). - _David A. Corneth_, Oct 13 2018
%e For n = 1, the prime signature is the empty sequence, so the concatenation of its terms yields 0 by convention.
%e For n = 2 = 2^1, n = 3 = 3^1 and any prime p = p^1, the prime signature is (1), and concatenation yields a(n) = 1.
%e For n = 4 = 2^2, the prime signature is (2), and concatenation yields a(n) = 2.
%e For n = 6 = 2^1 * 3^1, the prime signature is (1,1), and concatenation yields a(n) = 11.
%e For n = 12 = 2^2 * 3^1 but also n = 18 = 2^1 * 3^2, the prime signature is (2,1) since exponents are sorted in decreasing order; concatenation yields a(n) = 21.
%e For n = 30 = 2^1 * 3^1 * 5^1, the prime signature is (1,1,1), and concatenation yields a(n) = 111.
%e For n = 3072 = 2^10 * 3^1, the prime signature is (10,1), and concatenation yields a(n) = 101. This is the first term with nondecreasing digits.
%t {0}~Join~Array[FromDigits@ Flatten[IntegerDigits /@ FactorInteger[#][[All, -1]] ] &, 78, 2] (* _Michael De Vlieger_, Oct 13 2018 *)
%o (PARI) a(n)=fromdigits(vecsort(factor(n)[,2]~,,4)) \\ Except for multiples of 2^10, 3^10, etc.
%o (PARI) a(n)=eval(concat(apply(t->Str(t),vecsort(factor(n)[,2]~,,4)))) \\ Slower but correct for all n.
%Y Cf. A037916, A118914, A124010, A212171.
%K nonn,easy,base
%O 1,4
%A _M. F. Hasler_, Oct 12 2018
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