A052331 Inverse of A052330; A binary encoding of Fermi-Dirac factorization of n, shown in decimal.

0, 1, 2, 4, 8, 3, 16, 5, 32, 9, 64, 6, 128, 17, 10, 256, 512, 33, 1024, 12, 18, 65, 2048, 7, 4096, 129, 34, 20, 8192, 11, 16384, 257, 66, 513, 24, 36, 32768, 1025, 130, 13, 65536, 19, 131072, 68, 40, 2049, 262144, 258, 524288, 4097, 514, 132, 1048576, 35

Offset: 1

Author

Christian G. Bower, Dec 15 1999

Keywords

• nonn

Comments

Every number can be represented uniquely as a product of numbers of the form p^(2^k), sequence A050376. This sequence is a binary representation of this factorization, with a(p^(2^k)) = 2^(i-1), where i is the index (A302778) of p^(2^k) in A050376. Additive with a(p^e) = sum a(p^(2^e_k)) where e = sum(2^e_k) is the binary representation of e and a(p^(2^k)) is as described above. - Franklin T. Adams-Watters, Oct 25 2005 - Index offset corrected by Antti Karttunen, Apr 17 2018

Examples

			n = 84 has Fermi-Dirac factorization A050376(5) * A050376(3) * A050376(2) = 7*4*3. Thus a(84) = 2^(5-1) + 2^(3-1) + 2^(2-1) = 16 + 4 + 2 = 22 ("10110" in binary = A182979(84)). - _Antti Karttunen_, Apr 17 2018
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..4096
• Index entries for sequences that are permutations of the natural numbers

Crossrefs

Cf. A050376, A052330, A064547, A302024, A302029, A302776, A302778, A302785, A302786, A302787, A302790, A302784.

Cf. A182979 (same sequence shown in binary).

One less than A064358.

Cf. also A156552.

Programs

PARI

A052331=a(n)={for(i=1,#n=factor(n)~,n[2,i]>1||next; m=binary(n[2,i]); n=concat(n,Mat(vector(#m-1,j,[n[1,i]^2^(#m-j),m[j]]~)));n[2,i]%=2); n||return(0); m=vecsort(n[1,]); forprime(p=1,m[#m],my(j=0);while(p^2^j>1} \\ M. F. Hasler, Apr 08 2015

PARI

up_to_e = 8192;
v050376 = vector(up_to_e);
ispow2(n) = (n && !bitand(n,n-1));
i = 0; for(n=1,oo,if(ispow2(isprimepower(n)), i++; v050376[i] = n); if(i == up_to_e,break));
A052331(n) = { my(s=0,e); while(n > 1, fordiv(n, d, if(((n/d)>1)&&ispow2(isprimepower(n/d)), e = vecsearch(v050376, n/d); if(!e, print("v050376 too short!"); return(1/0)); s += 2^(e-1); n = d; break))); (s); }; \\ Antti Karttunen, Apr 12 2018

Formula

a(1)=0; a(n*A050376(k)) = a(n) + 2^k for a(n) < 2^k, k=0, 1, ... - Thomas Ordowski, Mar 23 2005

From Antti Karttunen, Apr 13 2018: (Start)

a(1) = 0; for n > 1, a(n) = A000079(A302785(n)-1) + a(A302776(n)).

For n > 1, a(n) = A000079(A302786(n)-1) * A302787(n).

a(n) = A064358(n)-1.

A000120(a(n)) = A064547(n).

A069010(a(n)) = A302790(n).

(End)