A095375 -id:A095375 - cp's OEIS frontend

A168161 Primes p which are equal to the sum of the binary digits in all primes <= p.

3, 5, 11, 19, 23, 47, 61

Offset: 1

M. F. Hasler, Nov 22 2009

A subsequence of A168162.

sbdQ[n_]:=Total[Flatten[IntegerDigits[#,2]&/@Prime[Range[PrimePi[ n]]]]] == n; Select[Prime[Range[20]],sbdQ] (* Harvey P. Dale, Feb 04 2015 *)

s=0; forprime(p=1, 9999, if(p==s+=norml2(binary(p)), print1(p, ", ")))

A168161 = { prime p | p=A095375(pi(p)) }, where pi(n) = A000720(n).

Cross reference added by Harvey P. Dale, Mar 26 2013

A095376 Values of k such that the total number of 1's in the binary expansions of the first k integers is a multiple of k.

Original entry on oeis.org

1, 2, 14, 62, 65, 77, 254, 322, 323, 327, 331, 332, 1022, 1281, 1341, 1348, 1349, 1350, 1352, 1353, 1354, 4094, 16382, 21505, 21757, 21762, 21820, 65534, 87299, 87355, 262142, 348161, 349181, 1048574, 1397762, 1398012, 1398020, 1398074, 4194302

Offset: 1

Labos Elemer, Jun 07 2004

All numbers of the form 4^k-2, with k>0, appear in this sequence. - Paul Tek, Sep 24 2013

			k=14: {1, 10, 11, 10, 101, 110, 111, 1000, 1001, 1010, 1011, 1100, 1101, 1110} includes 28 1's so A000788(14)/14 = 2 is an integer, thus 14 is here.

Donovan Johnson, Table of n, a(n) for n = 1..100

Cf. A000120, A000788, A014499, A095375.

lib[x_] := Count[IntegerDigits[x, 2], 1]; {s=0, ta=Table[0, {100}], tb=Table[0, {100}], u=1}; Do[s=s+lib[n]; w=n; If[IntegerQ[s/n], Print[{n, s/n}]; ta[[u]]=n; tb[[u]]=s/n; u=u+1], {n, 100000}]

Integer solutions to {A000788(x)/x is an integer}.

A328659 Partial sums of A035100: number of binary digits of the primes.

Original entry on oeis.org

0, 2, 4, 7, 10, 14, 18, 23, 28, 33, 38, 43, 49, 55, 61, 67, 73, 79, 85, 92, 99, 106, 113, 120, 127, 134, 141, 148, 155, 162, 169, 176, 184, 192, 200, 208, 216, 224, 232, 240, 248, 256, 264, 272, 280, 288, 296, 304, 312, 320, 328, 336, 344, 352, 360, 369, 378, 387, 396, 405, 414, 423, 432

Offset: 0

M. F. Hasler, Oct 25 2019

Useful to express the binary Copeland-Erdős constant, cf. formula.

Plotting a(n) against prime(n) might be a good tool for introducing students of mathematics, particularly those who are familiar with the use of binary representation, to the way the density of prime numbers decreases with increasing size. In essence, the graph of a(n) against prime(n) is approximately linear, and this becomes more obvious if we plot a(n)/prime(n): see the relevant plot in the links. - Peter Munn, Mar 03 2024

			Primes written in binary (A004676) read: 10, 11, 101, 111, 1011, 1101, 10001, ...
The length of the concatenation of the first n = 0, 1, 2, 3, .... terms is
  0, 2, 4, 7, 10, 14, 18, 23, ...: this sequence.

Robert Israel, Table of n, a(n) for n = 0..10000
Peter Munn, Plot2 graph of a(n)/prime(n)

Cf. A004676 (primes in binary), A035100 (their number of digits), A066747 & A191232: decimals and bits of the binary Copeland-Erdős constant.

Cf. A095375, A345867.

a := n -> add(ilog2(ithprime(k)), k=1..n) + n:
seq(a(n), n=0..62); # Peter Luschny, Oct 26 2019

Join[{0}, Accumulate[BitLength[Prime[Range[100]]]]] (* Paolo Xausa, Mar 20 2024 *)

s=0; A328659=vector(50,n,s+=logint(prime(n),2)+1)

from sympy import prime, primerange as primes
from itertools import accumulate
def f(n): return len(bin(n)[2:])
def aupton(nn): return [0]+list(accumulate(map(f, primes(2, prime(nn)+1))))
print(aupton(62)) # Michael S. Branicky, Jun 26 2021

a(n) = n + Sum_{k=1..n} floor(log_2(prime(k))).
A066747 = Sum_{n >= 1} prime(n)/2^a(n), the binary Copeland-Erdős constant.
a(n) = a(n-1) + A035100(n), n >= 1.
a(n) = A095375(n) + A345867(n) for n >= 1. - Alois P. Heinz, Jun 28 2021

A168157 Number of 0's in the matrix whose lines are the binary expansion of the first n primes.

Original entry on oeis.org

1, 1, 4, 4, 9, 10, 19, 21, 22, 23, 23, 37, 40, 42, 43, 45, 46, 47, 69, 72, 76, 78, 81, 84, 88, 91, 93, 95, 97, 100, 100, 136, 141, 145, 149, 152, 155, 159, 162, 165, 168, 171, 172, 177, 181, 184, 187, 188, 191, 194, 197, 198, 201, 202, 263, 268, 273, 277, 282, 287

Offset: 1

M. F. Hasler, Nov 21 2009

The matrix is to be taken of minimal size, i.e., have n lines and the number of columns needed to write the n-th prime in the last line, A035100(n). Otherwise said, there is no zero column except for n=1 (prime(1) = 2 = 10[2] in binary).
The number of zeros in the last line of the matrix is given by A035103(n).
One has a(n)=a(n-1) iff n = A059305(k) for some k, i.e. prime(n) is a Mersenne prime A000668(k) = A000225(A000043(k)).
If prime(n)=2^2^k+1 is a Fermat prime (A019434), n>2, then one has a(n)=a(n-1)+n-1+2^k-1.
More generally, the "big jumps" a(n+1) > a(n)+n happen whenever a column is added, i.e. when prime(n) = A014234(k) <=> prime(n+1) = A104080(k) for some k,n>1.

			a(4)=4 is the number of zeros in the matrix [010] /* = 2 in binary */ [011] /* = 3 in binary */ [101] /* = 5 in binary */ [111] /* = 7 in binary */

A168157(n)=n*#binary(prime(n))-sum(i=1,n,norml2(binary(prime(i))))

a(n)=n*A035100(n)-A095375(n).

A168162 Numbers n which do not exceed the sum of the binary digits in all primes <= n.

Original entry on oeis.org

3, 5, 7, 8, 11, 13, 14, 19, 23, 31, 32, 47, 61

Offset: 1

M. F. Hasler, Nov 22 2009

The sequence A168161 is a subsequence of the primes in this sequence.

			There is no prime <= 1 and 2 has only nonzero binary digit, therefore these numbers are not in the sequence.
However, a(1)=3 has two binary digits, so the total number of these equal 3.
Then, 4 is larger than this, but the prime p=5 again adds 2 nonzero binary digits adding to a total of 5=a(2).
Then 6 is larger than this, but the prime p=7 adds 3 more nonzero bits for a total of 8, such that a(3)=7 and a(4)=8 don't exceed this.

s=0; for(n=1,9999, isprime(n) && s+=norml2(binary(n)); n<=s & print1(n", "))

A168162 = { n | n <= A095375(pi(n)) }, where pi(n) = A000720(n).

A345867 Total number of 0's in the binary expansions of the first n primes.

Original entry on oeis.org

1, 1, 2, 2, 3, 4, 7, 9, 10, 11, 11, 14, 17, 19, 20, 22, 23, 24, 28, 31, 35, 37, 40, 43, 47, 50, 52, 54, 56, 59, 59, 64, 69, 73, 77, 80, 83, 87, 90, 93, 96, 99, 100, 105, 109, 112, 115, 116, 119, 122, 125, 126, 129, 130, 137, 142, 147, 151, 156, 161, 165, 170

Offset: 1

Alois P. Heinz, Jun 26 2021

			a(3) = 2: 2 = 10_2, 3 = 11_2, 5 = 101_2, so there are two 0's in the binary expansions of the first three primes.

Alois P. Heinz, Table of n, a(n) for n = 1..20000

Partial sums of A035103.

Cf. A000040, A059305, A095375, A328659.

a:= proc(n) option remember; `if`(n=0, 0, a(n-1)
      +add(1-i, i=Bits[Split](ithprime(n))))
    end:
seq(a(n), n=1..100);

Accumulate[DigitCount[Prime[Range[100]], 2, 0]] (* Paolo Xausa, Feb 26 2024 *)

from sympy import prime, primerange
from itertools import accumulate
def f(n): return (bin(n)[2:]).count('0')
def aupton(nn): return list(accumulate(map(f, primerange(2, prime(nn)+1))))
print(aupton(62)) # Michael S. Branicky, Jun 26 2021

a(n) = Sum_{i=1..n} A035103(i).
a(n) = a(n-1) for n in { A059305 }.
a(n) = A328659(n) - A095375(n).

A168154 Numbers n such that the sum of binary digits in prime(1), ..., prime(n) is prime.

Original entry on oeis.org

2, 3, 5, 8, 9, 15, 18, 21, 28, 34, 35, 45, 51, 53, 68, 77, 84, 86, 87, 90, 92, 102, 113, 116, 119, 121, 130, 131, 137, 149, 164, 174, 178, 192, 205, 210, 220, 221, 238, 240, 241, 250, 255, 275, 315, 318, 324, 344, 345, 363, 369, 375, 381, 386, 396, 397, 398, 404

Offset: 1

M. F. Hasler, Nov 20 2009

Indices of primes in A095375.

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000

Cf. A168153.

Module[{nn=500,prs},prs=Table[IntegerDigits[p,2],{p,Prime[Range[nn]]}];Select[Range[nn],PrimeQ[Total[Flatten[Take[prs,#]]]]&]] (* Harvey P. Dale, Aug 28 2021 *)

s=0; for(n=1,999, isprime(s+=norml2(binary(prime(n)))) & print1(n", "))
is_A168154(n)=isprime(A095375(n))

s=n=0;forprime(p=2,1e9,n++;if(isprime(s+=hammingweight(p)), print1(n", "))) \\ Charles R Greathouse IV, Mar 29 2013

Terms corrected by Charles R Greathouse IV, Mar 29 2013

A168155 Sum of binary digits of all primes < 2^n, i.e., with at most n binary digits.

Original entry on oeis.org

0, 3, 8, 14, 32, 61, 117, 230, 470, 922, 1807, 3597, 7071, 14022, 27693, 54876, 109077, 216301, 430183, 854696, 1700412, 3382868, 6733230, 13404811, 26704639, 53204936, 106034897, 211377718, 421466683, 840573072, 1676670824, 3345012214, 6674425203, 13319553281

Offset: 1

M. F. Hasler, Nov 20 2009

Partial sums of A168156.

			No prime can be written with only 1 binary digit, thus a(1)=0.
The primes that can be written with 2 binary digits are 2 = 10[2] and 3 = 11[2], they have 3 nonzero bits, so a(2)=3.
Primes with 3 binary digits are 5 = 101[2] and 7 = 111[3]. They add 5 more nonzero bits to yield a(3) = a(2)+5 = 8.

Cf. A168153.

s=0; L=p=2; while( L*=2, print1(s", "); until( L

a(n) = A095375( pi( 2^n-1 )), where pi = A000720.

a(25)-a(32) from Donovan Johnson, Jul 28 2010
a(33) from Chai Wah Wu, Apr 06 2020
a(34) from Chai Wah Wu, Apr 07 2020

A214344 Number of 1's in the first 10^n binary digits in the stream of prime numbers in base 2.

Original entry on oeis.org

1, 8, 69, 593, 5723, 56090, 541794, 5369528, 53803123, 527428642, 5249946808, 52800311682

Offset: 0

Tjandra Satria Gunawan, Jul 13 2012

Consider the stream (concatenation) of binary digits of primes in the MSB-first order featured in A191232. a(n) is the total count of 1's in the first 10^n of zeros and ones in this stream.

The complementary count of 0's is 10^n - a(n) = 0, 2, 31, 407, 4277, 43910, 458206, ... - R. J. Mathar, Jul 16 2012

Cf. A095375.

A214344 := proc()
    local stre,len,ct,p ;
    stre := [] ;
    len := 2 ;
    ct := 1 ;
    p := 2 ;
    while true do
        if nops(stre) = 0 then
            p := nextprime(p) ;
            stre := convert(p,base,2) ;
        end if;
        if op(-1,stre) = 1 then
            ct := ct+ 1;
        end if;
        stre := subsop(-1=NULL,stre) ;
        len := len+1 ;
        if ilog10(len-1) <> ilog10(len) then
            print(ct) ;
        end if;
    end do:
end proc: # R. J. Mathar, Jul 14 2012

pow = 1; sum1 = 0; sum2 = 0; p = 2;seq={}; k = 0; Do[d = IntegerDigits[p, 2]; sum1 += Count[d, 1]; sum2 += Length[d]; k++; If[sum2 >= pow, del = sum2 - pow; term = sum1 - Count[d[[-del ;; -1]], 1];   AppendTo[seq, term]; pow *= 10]; p = NextPrime[p], {10^4}]; seq (* Amiram Eldar, May 10 2019 *)

from sympy import nextprime
from itertools import islice
def bgen(p=2):
    while True: yield from (int(b) for b in bin(p)[2:]); p = nextprime(p)
def a(n): return sum(islice(bgen(), 10**n))
print([a(n) for n in range(7)]) # Michael S. Branicky, Jul 03 2022

a(9)-a(11) from Amiram Eldar, May 10 2019

A308430 Number of 0's minus number of 1's among the edge truncated binary representations of the first n prime numbers.

Original entry on oeis.org

0, 0, 1, 0, 0, 0, 3, 4, 3, 2, -1, 1, 3, 3, 1, 1, -1, -3, 0, 1, 4, 3, 4, 5, 8, 9, 8, 7, 6, 7, 2, 6, 10, 12, 14, 14, 14, 16, 16, 16, 16, 16, 12, 16, 18, 18, 18, 14, 14, 14, 14, 10, 10, 6, 13, 16, 19, 20, 23, 26, 27, 30, 31, 30, 31, 30, 31, 34, 33, 32, 35, 34, 31, 30, 27, 22, 25, 26, 29, 30, 31, 32, 29, 30, 27, 24, 27, 28, 27, 24, 23, 18, 15, 12, 9, 4, -1, 5, 9, 11

Offset: 1

Andrea Fornaciari, May 26 2019

By "edge truncated" we mean removing the first and last digit. For prime(3)=5 which has binary representation 101 edge truncating yields the string '0'. If there are 2 digits, then edge truncation yields the empty string ''. We count zero 1's and zero 0's in the empty string. The only cases of this are prime(1)=2 and prime(2)=3 which have binary representations 10 and 11.

Rémy Sigrist, Table of n, a(n) for n = 1..12251
Sean A. Irvine, Java program (github)
Jonas K. Sønsteby, Graph of 200 terms.
Jonas K. Sønsteby, Graph of 1000 terms.
Jonas K. Sønsteby, Graph of 5000 terms.
Jonas K. Sønsteby, Graph of 10000 terms.
Jonas K. Sønsteby, Graph of 100000 terms.

Cf. A004676, A095375, A014499, A177718, A296062.

s=0; forprime (p=2, 541, print1 (s += #binary(p\2)+1-2*hammingweight(p\2) ", ")) \\ Rémy Sigrist, Jul 13 2019

import gmpy2
def dec2bin(x):
    return str(bin(x))[2:]
def digitBalance(string):
    s = 0
    for char in string:
        if int(char) > 0:
            s -= 1
        else:
            s += 1
    return s
N = 100 # number of terms
seq = [0]
prime = 2
for i in range(N-1):
    prime = gmpy2.next_prime(prime)
    binary = dec2bin(prime)
    truncated = binary[1:-1]
    term = seq[-1] + digitBalance(truncated)
    seq.append(term)
print(seq) # Jonas K. Sønsteby, May 27 2019

def A308430list(b):
    L = []; s = 0
    for p in prime_range(2, b):
        q = (p//2).digits(2)
        s += 1 + len(q) - 2*sum(q)
        L.append(s)
    return L
print(A308430list(542)) # Peter Luschny, Jul 13 2019

a(n) = a(n-1) + bitlength(prime(n)2) - 2 * popcount(prime(n)_2) + 2, n > 1. - _Sean A. Irvine, May 27 2019

a(n) = Sum_{k=2..n} (A035100(k) - 2*A014499(k) + 2) = Sum_{k=2..n} (A070939(A000040(k)) - 2*A000120(A000040(k)) + 2). - Daniel Suteu, Jul 13 2019

cp's OEIS Frontend

A168161 Primes p which are equal to the sum of the binary digits in all primes <= p.

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A095376 Values of k such that the total number of 1's in the binary expansions of the first k integers is a multiple of k.

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A328659 Partial sums of A035100: number of binary digits of the primes.

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A168157 Number of 0's in the matrix whose lines are the binary expansion of the first n primes.

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A168162 Numbers n which do not exceed the sum of the binary digits in all primes <= n.

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A345867 Total number of 0's in the binary expansions of the first n primes.

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A168154 Numbers n such that the sum of binary digits in prime(1), ..., prime(n) is prime.

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A168155 Sum of binary digits of all primes < 2^n, i.e., with at most n binary digits.