A382307 Position of start of first run of alternating bit values in the base-2 representation of Pi, or -1 if no such run exists.

1, 2, 2, 19, 19, 19, 19, 19, 1195, 1697, 1890, 1890, 1890, 1890, 15081, 63795, 206825, 206825, 206825, 470577, 470577, 557265, 557265, 557265, 557265, 557265, 447666572, 447666572, 699793337, 699793337, 2049646803, 2250772991

Offset: 1

Author

James S. DeArmon, Mar 21 2025

Keywords

• nonn
• base
• more

Comments

In base-2, Pi is: 11.00100100001111110110101010001... For this sequence, the integer part of Pi is ignored, and the first fractional bit is numbered one.

The bit substring may begin with either 0 or 1.

a(27) > 4*10^6. - Michael S. Branicky, Mar 23 2025

Conjecture: no term is -1 (would follow from a proof that Pi is normal in base 2). - Sean A. Irvine, Mar 30 2025

a(33) > 4*10^9. - Pontus von Brömssen, Apr 06 2025

Examples

			The first alternating bit run is "0", at position 1, so a(1) = 1. The second and third alternating bit runs are "01" and "010", starting at position 2, so a(2) and a(3) are both 2.
a(4)-a(8) = 19 since the binary digits of Pi are "10101010" starting at position 19.
		

Crossrefs

Cf. A004601, A175945, A178708, A233836, A378472.

Programs

Python

def binary_not(binary_string):
    return ''.join('1' if bit == '0' else '0' for bit in binary_string)
# !pip install gmpy2   # may be necessary
import gmpy2
gmpy2.get_context().precision = 12000000
pi = gmpy2.const_pi()
# Convert Pi to binary representation
binary_pi = gmpy2.digits(pi, 2)[0] # zero-th element is the string of bits
binary_pi = binary_pi[2:] # remove leading "11" left of decimal point
outVec = []
strSearch0 = "" # this substring starts with "0"
for lenRun in range(1,30):
  strSearch0 += "0" if lenRun%2==1 else "1"
  strSearch1 = binary_not(strSearch0)
  l0 = binary_pi.find(strSearch0)+1 # incr origin-0 result
  l1 = binary_pi.find(strSearch1)+1
  outVec.append(min(l0,l1))
print(outVec)

Extensions

a(26)-a(32) from Pontus von Brömssen, Apr 06 2025