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

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

Author

Andrea Fornaciari, May 26 2019

Keywords

• sign
• base
• look

Comments

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.

Links

• 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.

Crossrefs

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

Programs

PARI

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

Python

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

Sage

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

Formula

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