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%I #24 Sep 08 2022 08:46:02
%S 2,3,2,2,4,2,2,3,2,3,2,5,2,2,3,2,2,4,2,2,2,3,3,2,2,6,2,3,2,2,2,4,4,2,
%T 3,2,2,5,2,2,2,2,3,3,2,4,2,2,3,2,2,3,2,7,2,3,3,2,4,2,2,2,2,3,2,2,5,4,
%U 2,3,2,2,2,2,4,2,2,3,2,3,6,2,2,2,3,2,2,2
%N Row n of table represents second signature of A013929(n): list of exponents >= 2 in canonical prime factorization of A013929(n), in nonincreasing order.
%C Length of row n equals A212177(n).
%D M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 844.
%H Jason Kimberley, <a href="/A212174/b212174.txt">Table of i, a(i) for i = 1..4492 (n = 1..3917)</a>
%H M. Abramowitz and I. A. Stegun, eds., <a href="http://www.convertit.com/Go/ConvertIt/Reference/AMS55.ASP">Handbook of Mathematical Functions</a>, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
%H Primefan, <a href="http://primefan.tripod.com/500factored.html">The First 2500 Integers Factored</a> (1st of 5 pages)
%F a(n) = A212172(A013929(n)).
%F This sequence is both the subsequence of A212171 formed by omitting all 1s and the subsequence of A212172 formed by omitting all 0's. - _Jason Kimberley_, Jun 13 2012
%e First rows of table read: 2; 3; 2; 2; 4; 2; 2; 3;...
%e 12 = 2^2*3 has positive exponents 2 and 1 in its prime factorization, but only exponents that are 2 or greater appear in a number's second signature. Hence, 12's second signature is {2}. Since 12 = A013929(4), row 4 of the table represents the second signature {2}.
%o (Magma) &cat[Reverse(Sort([pe[2]:pe in Factorisation(n)|pe[2]gt 1])):n in[1..247]]; // _Jason Kimberley_, Jun 13 2012
%Y Cf. A013929, A212172, A212175, A212176, A212177.
%K nonn,tabf
%O 1,1
%A _Matthew Vandermast_, Jun 03 2012
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