A233836 -id:A233836 - cp's OEIS frontend

A004539 Expansion of sqrt(2) in base 2.

1, 0, 1, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 1

Offset: 1

N. J. A. Sloane

Bailey, Borwein, Crandall, & Pomerance prove a general result that the first n terms contain >> sqrt(n) 1's. Vandehey improves this to sqrt(2*n)(1 + o(1)). - Charles R Greathouse IV, Nov 07 2017

			1.0110101000001001111001...

T. D. Noe, Table of n, a(n) for n = 1..1000
B. Adamczewski and N. Rampersad, On patterns occurring in binary algebraic numbers, Proc. Amer. Math. Soc. 136 (2008), 3105-3109.
David H. Bailey, Jonathan M. Borwein, Richard E. Crandall, and Carl Pomerance, On the binary expansions of algebraic numbers, J. Théor. Nombres Bordeaux, 16 (2004), 487-518.
R. L. Graham and H. O. Pollak, Note on a nonlinear recurrence related to sqrt(2), Mathematics Magazine, Volume 43, Pages 143-145, 1970. Zbl 201.04705.
R. K. Guy, The strong law of small numbers. Amer. Math. Monthly 95 (1988), no. 8, 697-712. [Annotated scanned copy]
Richard Isaac, On the simple normality to base 2 of the square root of s, for s not a perfect square., arXiv:math/0512404 [math.NT], 2005-2006.
Jason Kimberley, Index of expansions of sqrt(d) in base b
Thomas Stoll, On families of nonlinear recurrences related to digits, Journal of Integer Sequences 8 (2005), 05.3.2.
Thomas Stoll, On a problem of Erdős and Graham concerning digits, Acta Arithmetica 125 (2006), pp. 89-100.
Thomas Stoll, A fancy way to obtain the binary digits of 759250125 sqrt{2}, (2009), Amer. Math. Monthly, 117 (2010), 611-617.
Joseph Vandehey, On the binary digits of sqrt(2), arXiv:1711.01722 [math.NT], 2017.
Eric Weisstein's World of Mathematics, Wolfram's Iteration
Eric Weisstein's World of Mathematics, Pythagoras's Constant

Cf. A002193 (decimal version), A233836 (run lengths of 0's and 1's).

a004539 n = a004539_list !! (n-1)
a004539_list = w 2 0 where
   w x r = bit : w (4 * (x - (4 * r + bit) * bit)) (2 * r + bit)
     where bit = head (dropWhile (\b -> (4 * r + b) * b < x) [0..]) - 1
-- Reinhard Zumkeller, Dec 16 2013

N[Sqrt[2], 200]; RealDigits[%, 2]
RealDigits[Sqrt[2],2,120][[1]] (* Harvey P. Dale, Aug 03 2024 *)

binary(sqrt(2)) \\ Michel Marcus, Nov 06 2017

a(n) = floor(quadgen(8)<<(n-1))%2; \\ Chittaranjan Pardeshi, Sep 09 2024

bc
```
obase=2 scale=200 sqrt(2)
```

a(k) = floor(Sum_{n>=1} A005875(n)/exp(Pi*n/(2^((2/3)*k+(1/3))))) mod 2. Will give the k-th binary digit of sqrt(2). A005875 : number of ways to write n as sum of 3 squares. - Simon Plouffe, Dec 30 2023

A084186 First occurrence of exactly n 1's in the binary expansion of sqrt(2).

Original entry on oeis.org

1, 3, 40, 17, 74, 265, 31, 336, 11937, 1403, 8894, 3524, 33223, 126903, 3067, 109312, 390536, 553171, 280266, 962560, 1747112, 1740081, 30793169, 13109551, 118101037, 1077718187, 44908294, 1528865059, 1647265647, 3913429742, 10501492774, 4702573600, 81557258556, 107498528405

Offset: 1

Ralf Stephan, May 18 2003

			The binary expansion of sqrt(2) is 1.0110101000001..(A004539) and at position 17, there are four 1's, framed by 0's, so a(4)=17.

Nick Hobson, C program

Cf. A084185, A084187.

Cf. A233836.

C
```
See Links section.
```

from itertools import count
from math import isqrt
def A084186(n):
    a, b = 2, (1<>1)^1
    for k in count(1-n):
        if isqrt(a)&b==c:
            return k
        a<<=2 # Chai Wah Wu, Jan 24 2024

a(21)-a(29) from Chai Wah Wu, Jan 25 2024

a(30)-a(34) from Nick Hobson, Feb 15 2024

A084187 First occurrence of exactly n 0's in the binary expansion of sqrt(2).

Original entry on oeis.org

2, 15, 63, 58, 9, 1003, 524, 454, 1303, 5335, 22472, 8882, 37469, 32279, 220311, 92988, 698343, 24002, 574131, 3333660, 5940559, 4079882, 8356569, 115885798, 76570753, 202460870, 1034477781, 457034356, 1005210009, 3753736439, 2204906858, 50747186116, 32242071604, 159423417084, 114244391078, 74632918239

Offset: 1

Ralf Stephan, May 18 2003

			The binary expansion of sqrt(2) is 1.0110101000001..(A004539) and at position 9, there are five 0's, framed by 1's, so a(5)=9.

Cf. A084185, A084186.

Cf. A233836.

C
```
See Links section of A084186.
```

With[{d=RealDigits[Sqrt[2],2,116*10^6][[1]]},Flatten[Table[SequencePosition[d,Join[ {1},PadRight[{},n,0],{1}],1][[All,1]],{n,25}]]]+1 (* Harvey P. Dale, Dec 12 2022 *)

from math import isqrt
from itertools import count
def A084187(n):
    a, b = 2, (1<>1)|1
    for k in count(1-n):
        if isqrt(a)&b==c:
            return k
        a<<=2 # Chai Wah Wu, Jan 25 2024

More terms from Ryan Propper, May 09 2006
a(26)-a(29) from Chai Wah Wu, Jan 25 2024
a(30)-a(36) from Nick Hobson, Feb 15 2024

A175911 Concatenate the run lengths of the runs of ones and zeros in the binary representation of n in the lowest possible base where it is possible to represent each run length as a single digit. Convert the result to base 10.

Original entry on oeis.org

1, 3, 2, 5, 7, 7, 3, 7, 16, 15, 14, 8, 22, 13, 4, 9, 29, 49, 17, 41, 31, 43, 23, 11, 25, 67, 23, 14, 53, 21, 5, 11, 46, 117, 30, 50, 148, 52, 27, 87, 124, 63, 122, 44, 130, 93, 34, 14, 45, 76, 26, 68, 202, 70, 39, 15, 57, 213, 54, 22, 106, 31, 6, 13, 67, 231, 47, 118, 469, 121

Offset: 1

Dylan Hamilton, Oct 14 2010

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Cf. A043276.

Cf. A101211, A233836.

a175911 n = foldl1 (\v d -> b * v + d) rls where
   b = maximum rls + 1
   rls = a101211_row n
-- Reinhard Zumkeller, Dec 16 2013

repcount[x_] := Length/@Split[x]
binrep[x_] := repcount[IntegerDigits[x, 2]]
Table[h = binrep[x]; FromDigits[h, Max[h] + 1], {x, 1, DESIRED_NUMBER_OF_DIGITS}]
f[n_] := Block[{a = Length /@ Split@ IntegerDigits[n, 2]}, FromDigits[a, Max@ a + 1]]; Array[f, 70] (* Robert G. Wilson v, Aug 17 2013 *)

A378472 Position of start of first run of exactly n zeros in the base-2 representation of Pi, or -1 if no such run exists.

Original entry on oeis.org

17, 1, 26, 7, 109, 135, 96, 189, 2610, 902, 4267, 36139, 17317, 8375, 479166, 11791, 112954, 436893, 1286743, 726844, 5572140, 27456324, 2005750, 42248747, 200643872, 547151636, 171498580, 469458286, 1222711767, 2151391703, 1407238214

Offset: 1

James S. DeArmon, Nov 27 2024

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.

No further terms <= 4*10^9. - Michael S. Branicky, Dec 04 2024

			The first run of a single "0" bit is at position 17, so a(1) = 17.
The first run of exactly 2 zeros is at position 1, so a(2) = 1.

Cf. A004601, A175945, A178708, A233836.

import gmpy2
gmpy2.get_context().precision = 2000000
pi = gmpy2.const_pi()
# Convert Pi to binary representation
binary_pi = gmpy2.digits(pi, 2)[0] # zero-th element is the string of bits
outVec = []
for lenRun in range(1,20):
  str0 = "".join( ["0" for _ in range (lenRun)])
  l1 = binary_pi.find("1"+str0+"1")
  outVec.append(l1)
print(outVec)

a(n) >= A178708(n). - Michael S. Branicky, Dec 13 2024

a(21)-a(31) from Michael S. Branicky, Dec 04 2024

Clarified definition, added escape clause - N. J. A. Sloane, Dec 23 2024

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.

Original entry on oeis.org

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

James S. DeArmon, Mar 21 2025

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

			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.

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

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)

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

A382667 Position of the first instance of prime(n), in base 2, in the binary representation of Pi after the binary point.

Original entry on oeis.org

3, 11, 16, 11, 16, 15, 25, 60, 91, 14, 11, 126, 58, 393, 207, 18, 14, 13, 6, 180, 141, 169, 58, 243, 47, 326, 168, 475, 15, 291, 451, 108, 64, 87, 327, 421, 358, 41, 356, 468, 343, 16, 618, 107, 80, 179, 57, 206, 291, 325, 361, 205, 427, 12, 95, 108, 436, 6, 996

Offset: 1

James S. DeArmon, Apr 02 2025

Positions are numbered starting from 1 for the first bit after the binary point in Pi.

			For n=19, the bits of Pi and their numbering, after the binary point, begin
          1 2 3 4 5 6 7 8 9 ...
   1 1 .  0 0 1 0 0 1 0 0 0 0 1 1 1 1 ...
                    \-----------/
                    prime(19) = 67
prime(19) = 1000011_2 begins at position a(19) = 6.
prime(58) = 271 = 100001111_2 also starts at 6 => a(58) = 6.

Cf. A000040, A004601, A178707, A233836, A378472, A382307.

p=Drop[RealDigits[Pi,2,1010][[1]],2](* increase for n>73 *);a[n_]:=First[SequencePosition[p,IntegerDigits[Prime[n],2]][[1]]] (* James C. McMahon, Apr 26 2025 *)

import gmpy2
from sympy import isprime
gmpy2.get_context().precision = 12000000
gmpy2.get_context().round = gmpy2.RoundDown
pi = gmpy2.const_pi()
binary_pi = gmpy2.digits(pi, 2)[0][2:] # Get binary digits and remove "11"
print([binary_pi.find(bin(cand)[2:])+1 for cand in range(2, 700) if isprime(cand)])

a(n) = A178707(A000040(n)). - Pontus von Brömssen, Apr 12 2025

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A004539 Expansion of sqrt(2) in base 2.

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A084186 First occurrence of exactly n 1's in the binary expansion of sqrt(2).

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A084187 First occurrence of exactly n 0's in the binary expansion of sqrt(2).

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A175911 Concatenate the run lengths of the runs of ones and zeros in the binary representation of n in the lowest possible base where it is possible to represent each run length as a single digit. Convert the result to base 10.

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A378472 Position of start of first run of exactly n zeros in the base-2 representation of Pi, or -1 if no such run exists.

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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.

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Python

Extensions

A382667 Position of the first instance of prime(n), in base 2, in the binary representation of Pi after the binary point.

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Keywords

Comments

Examples

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Programs

Mathematica

Python

Formula