A278909 Binary Smith numbers: composite numbers n such that sum of bits of n = sum of bits of prime factors of n (counted with multiplicity).

15, 51, 55, 85, 125, 159, 185, 190, 205, 215, 222, 238, 246, 249, 253, 287, 303, 319, 374, 407, 438, 442, 469, 471, 475, 489, 494, 501, 507, 591, 623, 639, 670, 679, 687, 699, 730, 745, 755, 763, 765, 771, 799, 807, 822, 830, 843, 867, 890, 893, 917, 923, 925, 935, 939, 951, 970, 973, 979, 986, 989, 995, 1010, 1015, 1017, 1020, 1023, 1135, 1167, 1203, 1243

Offset

1,1

Comments

Binary equivalent of A006753 as well as A176670. (Since bits can only be 0 or 1, having equal sums of bits is logically equivalent to having the same nonzero bits.)

There are 615 terms up to 10^4, 6412 up to 10^5, 66369 up to 10^6, 630106 up to 10^7, 6268949 up to 10^8, 62159262 up to 10^9, and 596587090 up to 10^10. - Charles R Greathouse IV, Dec 09 2016

Links

Ely Golden, Table of n, a(n) for n = 1..10000

Example

a(1) = 15, as 15 (1111) in binary has the same number of 1 bits as its prime factors (11 and 101).

Mathematica

Select[Range@ 1250, And[CompositeQ@ #, DigitCount[#, 2, 1] = Total@ Flatten@ Apply[DigitCount[#, 2, 1] & /@ ConstantArray[#1, #2] &, FactorInteger@ #, 1]] &] (* Michael De Vlieger, Dec 02 2016 *)

Prog

(SageMath)
def numfactorbits(x):
    if(x<2):
        return 0;
    s=0;
    f=list(factor(x));
    #ensures inequality of numfactorbits(x) and bin(x).count("1") if x is prime
    if((len(f)==1)&(f[0][1]==1)):
        return 0;
    for c in range(len(f)):
        s+=bin(f[c][0]).count("1")*f[c][1]
    return s;
counter=2
index=1
while(index<=10000):
    if(numfactorbits(counter)==bin(counter).count("1")):
        print(str(index)+" "+str(counter))
        index+=1;
    counter+=1;
(PARI) is(n) = my(f=factor(n)[, 1]~, expo=factor(n)[, 2]~, v=[], s=0); for(k=1, #f, while(expo[k] > 0, expo[k]--; v=concat(v, f[k]))); for(k=1, #v, v[k]=binary(v[k])); my(w=[]); for(y=1, #v, w=concat(w, v[y])); if(vecsum(w)==vecsum(binary(n)), return(1), return(0))
terms(n) = my(i=0); forcomposite(c=1, , if(is(c), print1(c, ", "); i++; if(i==n, break)))
/* Print initial 70 terms as follows: */
terms(70) \\ Felix Fröhlich, Dec 01 2016
(PARI) is(n)=my(f=factor(n), t=#f~); (t>1 || (t==1 && f[1, 2]>1)) && hammingweight(n)==sum(i=1, t, hammingweight(f[i, 1])*f[i, 2]) \\ Charles R Greathouse IV, Dec 02 2016
(Python)
from sympy import factorint
def sbd(n): return bin(n).count('1')
def ok(n):
  f = factorint(n)
  return sum(f[p] for p in f) > 1 and sbd(n) == sum(sbd(p)*f[p] for p in f)
print(list(filter(ok, range(1244)))) # Michael S. Branicky, Apr 22 2021

Crossrefs

Cf. A006753, A176670.

Sequence in context: A191746 A029941 A349817 * A194851 A075928 A020214

Adjacent sequences: A278906 A278907 A278908 * A278910 A278911 A278912

Keyword

nonn,base,easy

Author

Ely Golden, Nov 30 2016

Status

approved