A084188 -id:A084188 - cp's OEIS frontend

A017910 Powers of sqrt(2) rounded down.

1, 1, 2, 2, 4, 5, 8, 11, 16, 22, 32, 45, 64, 90, 128, 181, 256, 362, 512, 724, 1024, 1448, 2048, 2896, 4096, 5792, 8192, 11585, 16384, 23170, 32768, 46340, 65536, 92681, 131072, 185363, 262144, 370727, 524288

Offset: 0

N. J. A. Sloane

a(n) is the number of positive squares <= 2^n (cf. A136417). - Hans Havermann, Apr 05 2008
If expressed to two significant digits, these are the f-stop numbers in photography: 1, 1.4, 2, 2.8, 4, 5.6, 8, 11, 16, 22, ...
There are also "half stops" (sqrt(2)^(n/2)) and "third stops" (sqrt(2)^(n/3)): 1, 1.4, 1.6, 1.8, 2.0, 2.2, 2.5, 2.8, 3.2, 3.6, 4, 4.5, 5, 5.7, 6.3, 7.1, 8, 9, 10.
a(n) is also the ratio (rounded down) of the curvature of the circle inscribed in the n-th 45-45-90 triangle to that of the circle inscribed in the 1st triangle, with the triangles arranged in a spiral as shown in the illustration in the links section. - Kival Ngaokrajang, Aug 28 2013
a(n) is also the total length of Heighway dragon (rounded down) after n-iterations when L(0) = 1. See illustration in links. - Kival Ngaokrajang, Dec 15 2013

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Kival Ngaokrajang, Illustration of initial terms.
Kival Ngaokrajang, Illustration of Heighway dragon for n = 0..5.
Wikipedia, Dragon curve.

Cf. A136417, A017912. Bisections: A000079, A084188.

Partial sums of A190568.

[Floor(Sqrt(2^n)): n in [0..40]]; // Vincenzo Librandi, Nov 20 2011

[Isqrt(2^n):n in[0..40]]; // Jason Kimberley, Oct 25 2016

A017910 := n->floor(sqrt(2^n)); # Peter Luschny, Sep 20 2011

Floor[(Sqrt[2])^Range[0,40]] (* Vincenzo Librandi, Nov 20 2011 *)

a(n)=sqrtint(2^n) \\ Charles R Greathouse IV, Sep 22 2011

from math import isqrt
def A017910(n): return isqrt(1<>1) # Chai Wah Wu, Jul 26 2022

a(n) = A000196(A000079(n)). - Jason Kimberley, Oct 28 2016

a(n) = A017912(n)-1 if n is odd. a(n) = A017912(n) = 2^(n/2) if n is even. - Chai Wah Wu, Jul 26 2022

A219085 a(n) = floor((n + 1/2)^3).

Original entry on oeis.org

0, 3, 15, 42, 91, 166, 274, 421, 614, 857, 1157, 1520, 1953, 2460, 3048, 3723, 4492, 5359, 6331, 7414, 8615, 9938, 11390, 12977, 14706, 16581, 18609, 20796, 23149, 25672, 28372, 31255, 34328, 37595, 41063, 44738, 48627, 52734, 57066

Offset: 0

Clark Kimberling, Dec 20 2012

a(n) is the number k such that {k^p} < 1/2 < {(k+1)^p}, where p = 1/3 and { } = fractional part.  In general, suppose that f is a continuous strictly increasing downward concave function, with f(1)>=0 and f(k)+1/2 not an integer.  Let J(k) denote the inequality {f(k)} < 1/2 < {f(k+1)}, where {}= fractional part; equivalently, [{f(k)} + 1/2] = 0 and [{f(k+1)} + 1/2] = 1, where [ ] = floor.  Thus J(k) holds if the integer nearest f(k+1) exceeds the integer nearest f(k), so that k can be regarded as a "jump point for f".  The solutions of J(k) are the numbers [g(n)+1/2)] for n >= 0, where g = (inverse of f).
Conjecture:  if d is a positive integer and f(x) = x^(1/d), then the solutions of J(k) form a linearly recurrent sequence.
This conjecture was proved by David Moews; see Problem 21 in "Unsolved Problems and Rewards".  - Clark Kimberling, Feb 06 2013
Guide to related sequences:
f(x) ....... jump sequence ... linear recurrence order
x^(1/2) .... A002378 ......... 3
x^(1/3) .... A219085 ......... 7
x^(2/3) .... A203302 ......... (not linearly recurrent)
x^(1/4) .... A219086 ......... 5
x^(3/4) .... A219087 ......... (not linearly recurrent)
x^(1/5) .... A219088 ......... 21
x^(1/6) .... A219089 ......... 21
x^(1/7) .... A219090 ..........71
x^(1/8) .... A219091 ......... 23
log(x) ..... A219092 ......... (not linearly recurrent)
log_2(x) ... A084188 ......... (not linearly recurrent)

			Let p=1/3.  Then
3^p=1.44... and 4^p=1.58..., so 3 is a jump point.
15^p=2.46... and 16^p=2.51..., so 15 is a jump point.

Clark Kimberling, Table of n, a(n) for n = 0..10000
Clark Kimberling, Unsolved Problems and Rewards, Problem 21.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1,1,-3,3,-1).

Cf. A002378, A203302.

Mathematica
```
Table[Floor[(n + 1/2)^3], {n, 0, 100}]
```

a(n)=n^3 + (6*n^2 + 3*n)\4 \\ Charles R Greathouse IV, Oct 07 2015

a(n) = floor((n + 1/2)^3).
a(n) = 3*a(n-1) -3*a(n-2) +a(n-3) +a(n-4) -3*a(n-5) +3*a(n-6) -a(n-7).
G.f.: (3*x +6*x^2 +6*x^3 +7*x^4 +x^5 +x^6)/(u*v), where u = (1 - x)^4, v = 1 + x + x^2 + x^3.
a(n) = (n + 1/2)^3 + (2*i^(n*(n-1))+(-1)^n-4)/8, where i=sqrt(-1). - Bruno Berselli, Dec 21 2012

A084185 First occurrence of binary n in the binary expansion of sqrt(2).

Original entry on oeis.org

1, 1, 3, 8, 1, 3, 17, 8, 14, 4, 1, 19, 3, 18, 17, 8, 62, 51, 14, 6, 4, 1, 42, 80, 19, 3, 41, 18, 40, 17, 31, 8, 57, 62, 128, 51, 69, 20, 14, 6, 104, 4, 111, 66, 1, 96, 42, 146, 80, 50, 19, 118, 3, 77, 41, 126, 18, 98, 40, 17, 75, 33, 31, 453, 8, 57

Offset: 1

Ralf Stephan, May 18 2003

a(n)=1 iff n in A084188. a(2^(n+1)+1) = A084187(n)-1.

			The binary expansion of sqrt(2) is 1.0110101000001..(A004539) and binary(10)=1010 occurs first at index 4, so a(10)=4.

G. C. Greubel, Table of n, a(n) for n = 1..5000

Cf. A084186.

With[{s2=RealDigits[Sqrt[2],2,1000][[1]],n2=IntegerDigits[n,2]}, Flatten[ Table[First[ Position[Partition[s2,Length[n2],1],n2]],{n,70}]]] (* Harvey P. Dale, Oct 11 2012 *)

cp's OEIS Frontend

A017910 Powers of sqrt(2) rounded down.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Magma

Maple

Mathematica

PARI

Python

Formula

A219085 a(n) = floor((n + 1/2)^3).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A084185 First occurrence of binary n in the binary expansion of sqrt(2).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica