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A330687 Positions of records in A050377, number of ways to factor n into "Fermi-Dirac primes" (A050376).

%I A330687 #39 Nov 04 2023 21:52:23
%S A330687 1,4,16,64,144,256,576,1024,1296,2304,5184,9216,20736,82944,186624,
%T A330687 331776,746496,1327104,2073600,2985984,5308416,8294400,18662400,
%U A330687 21233664,26873856,33177600,47775744,51840000,74649600,107495424,132710400,207360000,429981696,530841600,671846400,829440000,1194393600,1719926784,1866240000,2687385600
%N A330687 Positions of records in A050377, number of ways to factor n into "Fermi-Dirac primes" (A050376).
%C A330687 From _David A. Corneth_, Dec 29 2019: (Start)
%C A330687 Each term is a perfect square. Proof: A050377(n) is multiplicative with a(p^e) = A018819(e) and A018819(2k) = A018819(2k+1) and this sequence considers just records so we only need exponents of the form 2k; i.e., terms are squares.
%C A330687 Furthermore, the exponent 2 occurs at most once in the prime factorization of a(n) as A018819(2)^2 = A018819(4) = 4. So if the last two exponents in the prime factorization of m are 2's then setting the first of those two exponents to 4 and the other to 0 gives the same A050377(m).
%C A330687 Example of an application of this proof: we have 3600 = 2^4 * 3^2 * 5^2. We see the last two exponents are 2's so we can set the first of those two to 4 and the second to 0. This gives 2^4 * 3^4 = 1296 and, indeed, A050377(1296) = A050377(3600) = 16.
%C A330687 It seems that most exponents of a(n) are divisible by 4.
%C A330687 More specifically: Let S(n) be the list, possibly with duplicates, of exponents occurring in the prime factorizations of terms with the sum of exponents in the prime factorization <= n.
%C A330687 Let R(n) = |{x : x==4, S(n)}| / |S(n)|.
%C A330687 For example, S(8) is found from the following terms: 4, 16, 64, 144, 256, 576 and 1296 as the exponents in the prime factorization are (2), (4), (6), (4, 2), (8), (6, 2), (4, 4). The sums of each of these exponents per term is <= 8. There are 10 exponents listed. Of these 10 there are 5 that are divisible by 4. Therefore R(8) = 5/10.
%C A330687 Then it seems that R(n) tends to some value > 0.8 as n grows. (End)
%H A330687 David A. Corneth, <a href="/A330687/b330687.txt">Table of n, a(n) for n = 1..1004</a> (terms <= 10^100; first 294 terms from Antti Karttunen)
%F A330687 A050377(a(n)) = A330688(n).
%F A330687 A329900(a(n)) = A330689(n).
%F A330687 a(n) = A330684(n)^2.
%t A330687 Block[{s = Rest@ Nest[Function[{a, n, b}, Append[a, {Times @@ Map[a[[# + 1, -1]] &, FactorInteger[n][[All, -1]] ], b}]] @@ {#1, #2, #1[[-1, -1]] + If[EvenQ@ #2, #1[[#2/2 + 1, -1]], 0 ]} & @@ {#, Length@ #} &, {{0, 1}, {1, 1}}, 10^5][[All, 1]], t}, t = Union@ FoldList[Max, s]; Map[FirstPosition[s, #][[1]] &, t]] (* _Michael De Vlieger_, Dec 29 2019 *)
%o A330687 (PARI)
%o A330687 upto_e = 101; \\ 101 --> 211 terms.
%o A330687 A018819(n) = if( n<1, n==0, if( n%2, A018819(n-1), A018819(n/2)+A018819(n-1))); \\ From A018819
%o A330687 v018819 = vector(upto_e,n,A018819(n)); \\ Precompute.
%o A330687 A050377(n) = factorback(apply(e -> v018819[e], factor(n)[, 2]));
%o A330687 A283980(n) = {my(f=factor(n)); prod(i=1, #f~, my(p=f[i, 1], e=f[i, 2]); if(p==2, 6, nextprime(p+1))^e)}; \\ From A283980
%o A330687 A330687list(e) = { my(lista = List([1, 2]), i=2, u = 2^e, t, m=0, v025487); while(lista[i] != u, if(2*lista[i] <= u, listput(lista,2*lista[i]); t = A283980(lista[i]); if(t <= u, listput(lista,t))); i++); v025487 = vecsort(Vec(lista)); lista = List([]); for(i=1,#v025487,if((t=A050377(v025487[i]))>m, listput(lista,v025487[i]); m=t)); Vec(lista); };
%o A330687 v330687 = A330687list(upto_e);
%o A330687 A330687(n) = v330687[n];
%o A330687 for(n=1,#v330687,print1(A330687(n),", "));
%Y A330687 Cf. A018819, A050376, A050377, A329900, A330684 (square roots), A330688 (the record values), A330689 (primorial deflation).
%Y A330687 Subsequence of A025487.
%K A330687 nonn
%O A330687 1,2
%A A330687 _Antti Karttunen_, Dec 28 2019