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%I #52 Nov 03 2023 16:07:58
%S 1,2,3,4,5,6,8,9,7,10,12,15,16,18,14,20,24,27,21,25,30,32,36,11,28,40,
%T 45,48,54,35,42,50,60,64,72,81,13,22,56,63,75,80,90,96,108,33,49,70,
%U 84,100,120,128,135,144,162,26,44,105,112,125,126,150,160,180,192,216,243
%N Positive integers sorted by A001414(n), the sum of their prime divisors, as the major key and n as the minor key.
%C This is a permutation of the positive integers.
%C a(1) could be taken as 0 because 1 is not a member of A001414 and one could start with a(0)=1 (see the W. Lang link).
%C The row length sequence of this array is A000607(n), n>=2.
%C If the array is [1,0,2,3,4,5,6,6,...] with offset 0 then the row length sequence is A000607(n), n>=0.
%C From _David James Sycamore_, May 11 2018: (Start)
%C For n > 1, a(n) is the smallest number not yet seen such that sopfr(a(n)) is the least possible integer. The sequence lists in increasing order elements of the finite sets S(k) = {x: sopfr(x)=k}, k >= 0, where sopfr(x) = 0 iff x = 1. When a(n) = A056240(k) for some k >= 2, then sopfr(a(n)) = k and a(n) is the first of A000607(k) terms, all of which have sopfr = k. (A000607(k) is the number of partitions of k into prime parts.) Consequently the sequence follows a sawtooth profile, rising from a(n) = A056240(k) to A000792(k), the greatest number with sopfr = k, then starting over with A056240(k+1) for the next larger value of sopfr. (End) [Edited by _M. F. Hasler_, Jan 19 2019]
%H Alois P. Heinz, <a href="/A064364/b064364.txt">Rows n = 1..60, flattened</a> (first 32 rows from Reinhard Zumkeller)
%H H. Havermann: <a href="http://chesswanks.com/seq/sopfr/000.txt">The first 100 sums (complete, a 6 MB file)</a>
%H H. Havermann: <a href="http://chesswanks.com/seq/sopfr/">Tables of sum-of-prime-factors sequences (overview with links to the first 50000 sums)</a>
%H Wolfdieter Lang, <a href="/A064364/a064364.txt">First 16 rows.</a>
%H <a href="/index/Per#IntegerPermutation">Index entries for sequences that are permutations of the natural numbers</a>
%F If a(n) = A056240(k) for some k then a(n+A000607(k)-1) = A000792(k). - _David James Sycamore_, May 11 2018
%e The triangle reads:
%e 1,
%e (0,) (see comment in link to "first 16 rows" by W. Lang)
%e 2,
%e 3,
%e 4,
%e 5, 6,
%e 8, 9,
%e 7, 10, 12,
%e 15, 16, 18,
%e 14, 20, 24, 27,
%e 21, 25, 30, 32, 36,
%e 11, 28, 40, 45, 48, 54,
%e 35, 42, 50, 60, 64, 72, 81,
%e 13, 22, 56, 63, 75, 80, 90, 96, 108,
%e ...
%t terms = 1000; nmax0 = 100000 (* a rough estimate of max sopfr *);
%t sopfr[n_] := sopfr[n] = Total[Times @@@ FactorInteger[n]];
%t f[n1_, n2_] := Which[t1 = sopfr[n1]; t2 = sopfr[n2]; t1 < t2, True, t1 == t2, n1 <= n2, True, False];
%t Clear[g];
%t g[nmax_] := g[nmax] = Sort[Range[nmax], f][[1 ;; terms]];
%t g[nmax = nmax0];
%t g[nmax += nmax0];
%t While[g[nmax] != g[nmax - nmax0], Print[nmax]; nmax += nmax0];
%t A064364 = g[nmax] (* _Jean-François Alcover_, Mar 13 2019 *)
%o (Haskell)
%o import Data.List (partition, union)
%o a064364 n k = a064364_tabf !! (n-1) !! (k-1)
%o a064364_row n = a064364_tabf !! (n-1)
%o a064364_tabf = [1] : tail (f 1 [] 1 (map a000792 [2..])) where
%o f k pqs v (w:ws) = (map snd pqs') :
%o f (k + 1) (union pqs'' (zip (map a001414 us) us )) w ws where
%o us = [v + 1 .. w]
%o (pqs', pqs'') = partition ((== k) . fst) pqs
%o a064364_list = concat a064364_tabf
%o -- _Reinhard Zumkeller_, Jun 11 2015
%o (PARI) lista(nn) = {nmax = A000792(nn); v = vector(nmax, k, A001414(k)); for (n=1, nn, vn = select(x->x==n, v, 1); for (k = 1, #vn, print1(vn[k], ", ")))} \\ _Michel Marcus_, May 01 2018
%o (PARI) A064364_vec(N, k=6, L=9)={vector(N, i, if(i<7, N=i, until(A001414(N+=1)==k, ); N<L, N, k++; L=3^((k-2)\3)*(2+(k-2)%3); N+0*N=A056240(k)-1))} \\ To compute terms up to a given value of k=sopfr(n) and/or for large N >> 1000, it is more efficient to use code similar to lista() above, with "for(k...)" replaced by "a=concat(a, vn)". - _M. F. Hasler_, Jan 19 2019
%Y Cf. A001414.
%Y Cf. A000607 (row lengths), A002098 (row sums), A056240 (least = first term in the n-th row), A000792 (greatest term in the n-th row).
%Y Cf. A257815 (inverse).
%K easy,nonn,look,tabf
%O 1,2
%A _Howard A. Landman_, Sep 25 2001
%E More terms from _Vladeta Jovovic_, Sep 27 2005
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