|
%I #74 Apr 07 2020 22:55:31
%S 0,1,10,100,1000,11,10000,101,100000,1001,1000000,110,10000000,10001,
%T 1010,100000000,1000000000,100001,10000000000,1100,10010,1000001,
%U 100000000000,111,1000000000000,10000001,100010,10100,10000000000000,1011,100000000000000,100000001,1000010,1000000001,11000,100100
%N Fermi-Dirac representation of n. Let n have factorization p1^(2^e1) * p2^(2^e2) * ... * pr^(2^er), where each factor is in A050376. The number n is represented by a binary string that indicates which terms of A050376 appear in the factorization of n.
%C Every number has a unique representation as a product of terms from A050376. - _N. J. A. Sloane_, Feb 11 2011
%C The "Fermi-Dirac factorization" of n, i.e., the factorization of n into prime powers of the form p_k^(2^e_k), e_k >= 0, (A050376) allows each of those prime powers to be used at most once, since this corresponds to the binary representation of the exponents of the prime powers p^a of the "Bose-Einstein factorization" of n, i.e., the classic prime factorization of n. (Cf. A050376 comments.)
%C The prime powers of the form p_k^(2^e_k), e_k >= 0 (A050376) might be called "Fermi-Dirac primes" since they may appear at most once (thus raised to powers 0 or 1) in the "Fermi-Dirac factorization" of n. Compare with the classic prime factorization of n, which might be called the "Bose-Einstein factorization" of n, where the primes (which might be called "Bose-Einstein primes") may appear any number of times >= 0.
%C In the "Fermi-Dirac representation" of n, if a given prime power with powers of two as exponents does not appear in the factorization of n into prime powers with powers of two as exponents, we use 0 as a placeholder; otherwise, we use 1 to indicate that the given prime power with powers of two as exponents does appear in the "Fermi-Dirac factorization" of n.
%C In the base-b representation of n, we do not show the leading 0's, except for 0 where it is more convenient to show it than to show nothing. Similarly, for the "Fermi-Dirac representation" of n, we do not show the leading 0's, except for 0, which is the representation of 1, where it is more convenient to show it than to show nothing.
%C The limit of the supremum of the number of "binary digits" of the representation of n is asymptotic to the number of primes up to n, i.e., n/log(n), making this representation absolutely impractical!
%C See A052330 for the numbers having representation as 0, 1, 10, 11, 100, 101, 110, 111, ... which is an ordering of the positive integers. (Cf. OEIS Wiki page.)
%C Let n have factorization (f_r)^g_r * ... * (f_2)^g_2 * (f_1)^g_1, where f_i is the i-th prime power of the form p_k^(2^e_k), e_k >= 0 (A050376, A302778); then a(n) = Sum_{i=1..r} g_i * 2^(i-1).
%C The number of 1's in a(n) is the number of terms of A050376 dividing n with odd maximal exponent. For example, if n=96, then the maximal exponent of 2 that divides 96 is 5, for 3 it is 1, for 4 it is 2, for 16 it is 1. Thus only 2, 3 and 16 divide n with odd maximal exponents. Therefore, the number of 1's in a(96) is 3. Moreover, since 2=A050376(1), 3=A050376(2) and 16=A050376(9), then 1's appear in positions 1,2,9 from the right. - _Vladimir Shevelev_, Nov 02 2013
%H T. D. Noe, <a href="/A182979/b182979.txt">Table of n, a(n) for n = 1..1000</a>
%H OEIS Wiki, <a href="/wiki/%22Fermi-Dirac_representation%22_of_n">"Fermi-Dirac representation" of n</a>
%H OEIS Wiki, <a href="/wiki/Ordering_of_positive_integers_by_increasing_%22Fermi-Dirac_representation%22">Ordering of positive integers by increasing "Fermi-Dirac representation"</a>
%F Let q_1,q_2,q_3,... be consecutive terms of A050376 and n = q_1^a_1 * q_2^a_2 *...* q_r^a_r, where a_i = 0 or 1. Then a(n) = a_1 + 10*a_2 + ... +10^(r-1)*a_r. For example, since 30 = 2^1 * 3^1 * 4^0 * 5^1, then a(30)= 1 + 10 + 1000 = 1011. - _Vladimir Shevelev_, Nov 02 2013
%F a(n) = A007088(A052331(n)). - _Antti Karttunen_, Apr 17 2018
%e "Fermi-Dirac factorizations" (cf. A050376 examples, here with exponents of "Fermi-Dirac primes" being either 0 or 1):
%e 6 = 3*2 = 3^1*2^1, so a(6) = 11;
%e 8 = 4*2 = 4^1*3^0*2^1, so a(8) = 101;
%e 20 = 5*4 = 5^1*4^1*3^0*2^0, so a(20) = 1100;
%e 24 = 4*3*2 = 4^1*3^1*2^1, so a(24) = 111;
%e 27 = 9*3 = 9^1*7^0*5^0*4^0*3^1*2^0, so a(27) = 100010;
%e 32 = 16*2 = 16^1*13^0*11^0*9^0*7^0*5^0*4^0*3^0*2^1, so a(32) = 100000001;
%e 64 = 16*4 = 16^1*13^0*11^0*9^0*7^0*5^0*4^1*3^0*2^0, so a(64) = 100000100;
%e 108 = 9*4*3 = 9^1*7^0*5^0*4^1*3^1*2^0, so a(108) = 100110;
%e 120 = 5*4*3*2 = 5^1*4^1*3^1*2^1, so a(120) = 1111;
%e ...
%t nn=24; p=Select[Range[nn], PrimeQ]; Do[p=Select[Union[p,p^2], #<=nn&], {Floor[Log[2,Log[2,nn]]]}]; Table[m=n; FromDigits[Table[If[Mod[m,i]==0, m=m/i; 1, 0], {i,Reverse[p]}]],{n,nn}]
%Y Cf. A050376, A052330, A052331, A302778.
%K nonn
%O 1,3
%A _Daniel Forgues_, Feb 10 2011, Feb 13 2011
%E Clearer definition from _T. D. Noe_, Feb 11 2011
%E Edited by _N. J. A. Sloane_, Jul 21 2018
|
