OFFSET
1,7
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.
FORMULA
a(n) = a(n-1) + bitlength(prime(n)_2) - 2 * popcount(prime(n)_2) + 2, n > 1. - Sean A. Irvine, May 27 2019
PROG
(Python 3)
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
(PARI) s=0; forprime (p=2, 541, print1 (s += #binary(p\2)+1-2*hammingweight(p\2) ", ")) \\ Rémy Sigrist, Jul 13 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
CROSSREFS
KEYWORD
AUTHOR
Andrea Fornaciari, May 26 2019
STATUS
approved