This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
With[{b = DigitCount[Prime[Range[600]], 2, 1]}, Join[{2}, Select[Prime[Position[ 2*b - 1 - FoldList[Max, b], ?(# > 0 &)] // Flatten], Differences[DigitCount[#, 2]][[1]] >= 0 &]]] (* _Amiram Eldar, Jul 27 2023 *)
23 is in the sequence because 23 is a prime and 23_10 = 10111_2. '10111' has four 1's and one 0. - _Indranil Ghosh_, Jan 31 2017
Select[Prime[Range[70]], Plus@@IntegerDigits[#, 2] > Length[IntegerDigits[#, 2]]/2 &] (* Alonso del Arte, Jan 11 2011 *) Select[Prime[Range[100]], Differences[DigitCount[#, 2]][[1]] < 0 &] (* Amiram Eldar, Jul 25 2023 *)
B(x) = {nB = floor(log(x)/log(2)); b1 = 0; b0 = 0;
for(i = 0, nB, if(bittest(x,i), b1++;, b0++;); );
if(b1 > b0, return(1);, return(0););};
forprime(x = 3, 383, if(B(x), print1(x, ", "); ); ); \\ Washington Bomfim, Jan 11 2011
has(n)=hammingweight(n)>#binary(n)/2 select(has, primes(500)) \\ Charles R Greathouse IV, May 02 2013
# Program to generate the b-file
from sympy import isprime
i=1
j=1
while j<=200:
if isprime(i) and bin(i)[2:].count("1")>bin(i)[2:].count("0"):
print(str(j)+" "+str(i))
j+=1
i+=1 # Indranil Ghosh, Jan 31 2017
The binary expansion of 83 is (1,0,1,0,0,1,1), and 83 is the 23rd prime, so a(23) = 4 - 3 = 1.
Table[DigitCount[Prime[n],2,1]-DigitCount[Prime[n],2,0],{n,100}]
DigitCount[#,2,1]-DigitCount[#,2,0]&/@Prime[Range[100]] (* Harvey P. Dale, May 09 2025 *)
From _Indranil Ghosh_, Feb 03 2017: (Start) 5 is in the sequence because 5_10 = 101_2. '101' has two 1's and one 0. 17 is in the sequence because 17_10 = 10001_2. '10001' has two 1's and three 0's. (End)
Select[Prime[Range[200]],DigitCount[#,2,1]<=1+DigitCount[#,2,0]&] (* Harvey P. Dale, Apr 18 2023 *)
forprime(p=2,613,v=binary(p);s=0;for(k=1,#v,s+=if(v[k]==1,+1,-1));if(s<=1,print1(p,", "))) \\ Washington Bomfim, Jan 13 2011
from sympy import isprime
i=1
j=1
while j<=250:
if isprime(i) and bin(i)[2:].count("1")<=1+bin(i)[2:].count("0"):
print(str(j)+" "+str(i))
j+=1
i+=1 # Indranil Ghosh, Feb 03 2017
The binary expansion of 83 is (1,0,1,0,0,1,1) with ones minus zeros 4 - 3 = 1, and 83 is the 23rd prime, so 23 is in the sequence.
The primes A000040(a(n)) together with their binary expansions and binary indices begin:
5: 101 ~ {1,3}
19: 10011 ~ {1,2,5}
71: 1000111 ~ {1,2,3,7}
83: 1010011 ~ {1,2,5,7}
89: 1011001 ~ {1,4,5,7}
101: 1100101 ~ {1,3,6,7}
113: 1110001 ~ {1,5,6,7}
271: 100001111 ~ {1,2,3,4,9}
283: 100011011 ~ {1,2,4,5,9}
307: 100110011 ~ {1,2,5,6,9}
313: 100111001 ~ {1,4,5,6,9}
331: 101001011 ~ {1,2,4,7,9}
397: 110001101 ~ {1,3,4,8,9}
409: 110011001 ~ {1,4,5,8,9}
419: 110100011 ~ {1,2,6,8,9}
421: 110100101 ~ {1,3,6,8,9}
433: 110110001 ~ {1,5,6,8,9}
457: 111001001 ~ {1,4,7,8,9}
1103: 10001001111 ~ {1,2,3,4,7,11}
1117: 10001011101 ~ {1,3,4,5,7,11}
1181: 10010011101 ~ {1,3,4,5,8,11}
1223: 10011000111 ~ {1,2,3,7,8,11}
Select[Range[1000],DigitCount[Prime[#],2,1]-DigitCount[Prime[#],2,0]==1&]
read("transforms") ;A000120 := proc(n) wt(n) ; end proc:
isA177835 := proc(p) if isprime(p) then q := 2 ; while q < p do if A000120(q) >= 2*A000120(p)-1 then return true; end if; q := nextprime(q) ; end do: return false; else false; end if; end proc:
for i from 1 to 2000 do if isA177835(ithprime(i)) then printf("%d,",ithprime(i)) ; end if; end do: # R. J. Mathar, May 31 2010
With[{b = DigitCount[Prime[Range[200]], 2, 1]}, Rest@ Prime[Position[2*b - 1 - FoldList[Max, b], ?(# <= 0 &)] // Flatten]] (* _Amiram Eldar, Jul 25 2023 *)
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