A010467 -id:A010467 - cp's OEIS frontend

A176441 Decimal expansion of sqrt(210).

1, 4, 4, 9, 1, 3, 7, 6, 7, 4, 6, 1, 8, 9, 4, 3, 8, 5, 7, 3, 7, 1, 8, 6, 6, 4, 1, 5, 7, 1, 6, 9, 7, 7, 1, 7, 2, 3, 1, 4, 0, 1, 3, 2, 8, 7, 4, 7, 5, 8, 9, 7, 3, 0, 8, 8, 6, 9, 5, 9, 2, 4, 8, 0, 7, 1, 1, 8, 1, 4, 4, 3, 7, 2, 6, 5, 3, 6, 8, 0, 4, 2, 1, 7, 1, 2, 5, 6, 3, 1, 9, 2, 0, 0, 3, 6, 1, 7, 4, 9, 7, 7, 5, 3, 0

Offset: 2

Klaus Brockhaus, Apr 19 2010

Continued fraction expansion of sqrt(210) is A040195.

			sqrt(210) = 14.49137674618943857371...

Index entries for algebraic numbers, degree 2.

Cf. A010467 (decimal expansion of sqrt(10)), A010477 (decimal expansion of sqrt(21)), A176440 (decimal expansion of (14+sqrt(210))/4), A040195 (14 followed by (repeat 2, 28)).

RealDigits[Sqrt[210],10,120][[1]]  (* Harvey P. Dale, Apr 21 2011 *)

A018072 Powers of fourth root of 10 rounded down.

Original entry on oeis.org

1, 1, 3, 5, 10, 17, 31, 56, 100, 177, 316, 562, 1000, 1778, 3162, 5623, 10000, 17782, 31622, 56234, 100000, 177827, 316227, 562341, 1000000, 1778279, 3162277, 5623413, 10000000, 17782794, 31622776, 56234132, 100000000, 177827941, 316227766, 562341325, 1000000000, 1778279410

Offset: 0

N. J. A. Sloane

			a(2) = 3 because 10^(2/4) = 10^(1/2) = sqrt(10) = 3.16228...
a(3) = 5 because 10^(3/4) = 5.62341...
a(4) = 10 because 10^(4/4) = 10^1 = 10.
a(5) = 17 because 10^(5/4) = 17.78279...

Vincenzo Librandi, Table of n, a(n) for n = 0..2000

Cf. A017934, A011007, A010467.

[Floor(10^(n/4)): n in [0..37]]; // Vincenzo Librandi, May 30 2025

Table[Floor[10^(n/4)], {n, 0, 39}] (* Alonso del Arte, Jan 26 2013 *)
Floor[Surd[10,4]^Range[0,40]] (* Harvey P. Dale, Oct 26 2019 *)

a(n)=sqrtint(sqrtint(10^n)) \\ Charles R Greathouse IV, Jan 27 2013

A few additional terms from Alonso del Arte, Jan 26 2013

A042937 Denominators of continued fraction convergents to sqrt(1000).

Original entry on oeis.org

1, 1, 2, 3, 5, 8, 53, 114, 281, 4329, 8939, 22207, 142181, 164388, 306569, 470957, 777526, 1248483, 78183472, 79431955, 157615427, 237047382, 394662809, 631710191, 4184923955, 9001558101, 22188040157, 341822160456, 705832361069, 1753486882594

Offset: 0

N. J. A. Sloane

			sqrt(1000) = 31.62... = 31 + 1/(1 + 1/(1 + ...)) with convergents 31/1, 32/1, 63/2, 95/3, 158/5, ... - _M. F. Hasler_, Nov 02 2019

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 78960998, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1).

Cf. A042936 (numerators), A040968 (continued fraction), A010467 (decimals).

Analog for sqrt(m): A000129 (m=2), A002530 (m=3), A001076 (m=5), A041007 (m=6), A041009 (m=7), A041011 (m=8), A005663 (m=10), A041015 (m=11), A041017 (m=12), ..., A042933 (m=998), A042935 (m=999).

Denominator[Convergents[Sqrt[1000], 30]] (* Vincenzo Librandi, Feb 01 2014 *)

A42937=contfracpnqn(c=contfrac(sqrt(1000)),#c-1)[2,] \\ Possibly incorrect last term ignored. NB: a(n) = A42937[n+1]. For more terms use e.g. \p999, or compute any a(n) from this as in A042936. - M. F. Hasler, Nov 01 2019

More terms from Vincenzo Librandi, Feb 01 2014

A081168 Differences of Beatty sequence for square root of 10.

Original entry on oeis.org

3, 3, 3, 3, 3, 3, 4, 3, 3, 3, 3, 3, 4, 3, 3, 3, 3, 3, 4, 3, 3, 3, 3, 3, 4, 3, 3, 3, 3, 3, 4, 3, 3, 3, 3, 3, 4, 3, 3, 3, 3, 3, 3, 4, 3, 3, 3, 3, 3, 4, 3, 3, 3, 3, 3, 4, 3, 3, 3, 3, 3, 4, 3, 3, 3, 3, 3, 4, 3, 3, 3, 3, 3, 4, 3, 3, 3, 3, 3, 3, 4, 3, 3, 3, 3, 3, 4, 3, 3, 3, 3, 3, 4, 3, 3, 3, 3, 3, 4, 3, 3, 3, 3, 3, 4

Offset: 0

Benoit Cloitre, Apr 16 2003

nonn

Let S(0) = 3; obtain S(k) from S(k-1) by applying the morphism 3 -> 333334, 4 -> 3333334; sequence is S(0), S(1), S(2), ...

More generally, for a(n,m) = floor((n+1)*sqrt(m^2+ 1)) - floor(n*sqrt(m^2+1)) start with m and apply the morphism: m -> m^(2m-1), m+1; m+1 -> m^(2m), m+1.

G. C. Greubel, Table of n, a(n) for n = 0..10000

Cf. A010467, A006337, A177102.

A081168:= func< n | Floor((n+1)*Sqrt(10)) - Floor(n*Sqrt(10)) >;
[A081168(n): n in [0..120]]; // G. C. Greubel, Jan 15 2024

Differences[Floor[Sqrt[10]*Range[0, 120]]] (* G. C. Greubel, Jan 15 2024 *)

a(n)=floor((n+1)*sqrt(10))-floor(n*sqrt(10))

def A081168(n): return floor((n+1)*sqrt(10)) - floor(n*sqrt(10))
[A081168(n) for n in range(121)] # G. C. Greubel, Jan 15 2024

a(n) = floor((n+1)*sqrt(10)) - floor(n*sqrt(10)).

A088995 Least k > 0 such that the first n digits of 2^k and 5^k are identical.

Original entry on oeis.org

5, 98, 1068, 1068, 127185, 2728361, 15917834, 73482154, 961700165, 961700165, 83322853582, 1404948108914, 7603192018819, 167022179253602, 3550275020220728, 5729166542536373, 106675272785875442, 331484151442699072, 2330288635428230258177, 2330288635428230258177

Offset: 1

Lekraj Beedassy, Dec 01 2003

The number of matching first digits of 2^n and 5^n increases with n and forms the sequence 3,1,6,2,2,7,7,6,6,... which approaches sqrt(10).
Numbers are half of the denominator of some convergent to log_10(2). - J. Mulder (jasper.mulder(AT)planet.nl), Feb 03 2010 [WARNING: This holds only for n < 6, it is wrong from a(6) = 2728361 on: The denominators are ..., 325147, 6107016, ... which would yield the larger solution 3053508 for n = 6. - M. F. Hasler, Mar 20 2025]
Xianwen Wang guesses that if the length of the continued fraction of m/k is h (where m is the difference between the numbers of digits of 2^k and 5^k), the first h-1 items of the continued fractions of m/k and log_10(2.5) agree. But this guess is not true for the similar sequence A359698. - Zhao Hui Du, Jun 06 2023
The terms grow like 10^n, oscillating by a few orders of magnitude around this value. I conjecture that log(a(n)/10^n) = O(log log n). - M. F. Hasler, Mar 22 2025

			a(2) = 98: 2^98 = 316912650057057350374175801344 and 5^98 = 315544362088404722164691426113114491869282574043609201908111572265625.

Zhao Hui Du, Table of n, a(n) for n = 1..1000
Zhao Hui Du, Chinese page to introduce algorithm for a(n) to prove the guess of _Xianwen Wang_
Robert Munafo, Sequence A88995: The 32 and 3125 Property
T. Sillke, Powers of 2 and 5 Puzzle

Cf. A088935, A010467, A359698, A073733.

L2 = N[ Log[ 10, 2 ], 50 ]; L5 = N[ Log[ 10, 5 ], 50 ]; k = 1; Do[ While[ Take[ RealDigits[ 10^FractionalPart[ L2*k ] ][[ 1 ] ], n ] != Take[ RealDigits[ 10^FractionalPart[ L5*k ] ][[ 1 ] ], n ], k++ ]; Print[ k ], {n, 1, 10} ]
L2 = N[ Log[ 10, 2 ], 50 ]; L5 = N[ Log[ 10, 5 ], 50 ]; k = 1; Do[ While[ Take[ RealDigits[ 10^FractionalPart[ L2*k ]][[ 1 ]], n ] != Take[ RealDigits[ 10^FractionalPart[ L5*k ]][[ 1 ]], n ], k++ ]; Print[ k ], {n, 1, 7} ]
f[n_, k_] := {Floor[ 10^(k - 1 + N[FractionalPart[n Log[5]/Log[10]], 20])], Floor[10^(k - 1 + N[FractionalPart[n Log[2]/Log[10]], 20])]} Flatten@Block[{$MaxExtraPrecision = \[Infinity]}, Block[{l = Denominator /@ Convergents[Log10[2], 1000]}, Array[k \[Function] l[[Flatten@Position[f[ #/2, k] & /@ l, {x_, x_}, {1}, 1]]]/2, 20]]] (* J. Mulder (jasper.mulder(AT)planet.nl), Feb 03 2010 *)
(* alternate program *)
n = 100; $MaxExtraPrecision = n; ans =
 ContinuedFraction[Log10[5/2], n]; data =
 Denominator /@
  Flatten[Table[
    FromContinuedFraction[Join[ans[[1 ;; p - 1]], {#}]] & /@
     Range[1, ans[[p]]], {p, 2, n}]]; sol =
 Select[Table[{k, a = N[FractionalPart[{k Log10[2], k Log10[5]}], n];
    10^a, b = RealDigits[10^a][[All, 1]];
    LengthWhile[Range[Length[b[[1]]]], b[[1, #]] == b[[2, #]] &],
    10^a . {-1, 1}, RealDigits[10^a . {-1, 1}][[-1]]}, {k, data}],
  Abs[#[[-2]]] < 1 &];
acc = Association[{}]; s = sol[[All, {1, 3}]]; For[i = 1,
 i < Length[s], i++,
 If[Lookup[acc, s[[i, 2]], 0] == 0,
  acc[s[[i, 2]]] = s[[i, 1]]]]; final =
 Rest[Sort[Normal[acc]]] /. Rule -> List;
bcc = Association[{}]; For[i = Max[Keys[acc]], i >= Min[Keys[acc]], i--,
  j = i; While[Lookup[acc, j, 0] == 0 && j < Max[Keys[acc]], j++];
 bcc[i] = acc[j]; j = i; While[bcc[j] >= bcc[j + 1], j++];
 bcc[i] = Min[bcc[i], bcc[j]]]; bb =
 Rest[Sort[Normal[Reverse[bcc]]]] /. Rule -> List (* Xianwen Wang, Jun 02 2023 *)

apply( {A088995(n) = localprec(max(n/.4,38)); my(L1=log(10), L2=log(2)/L1, L5=1-L2, c=contfrac(L5-L2), T=sqrtint(10^(2*n-1)), a=log(T)/L1%1, b=log(T+1)/L1%1, d(x)=a3, while(t(c-cv[i]), c-=cv[i]))+return(c))}, [1..15]) \\ - M. F. Hasler, Mar 22 2025

# NOTE: Although sympy's frac() may give incorrect results in some cases, care has been taken to ensure there should be no issue here. - M. F. Hasler, Mar 26 2025
from sympy import S, sqrt, log, frac
def A088995(n):
    T = 10**n//sqrt(10); prec = (n+9)/.4; cf = 'continued_fraction'
    L2, L5, a, b = (frac(log(x,10).n(prec)) for x in (2, 5, T, T+1))
    L = []; test = lambda k: a < frac(k*L2) < b > frac(k*L5) > a
    for K in S(f"{cf}_convergents({cf}_iterator(log(5/2,10)))"):
        if test(k := K.q):    # = K.denominator but K.denominator() in old versions
            for step in L[::-1]:
              while test(k - step): k -= step
            return k
        if k > 99: L.append(k) # M. F. Hasler, Mar 22 2025, edited Mar 26 2025
print(first30 := [A088995(n) for n in range(1,30)]) # M. F. Hasler, Mar 18 2025, edited Mar 24 2025, Mar 28 2025

Edited by Robert G. Wilson v, Dec 02 2003
More terms from J. Mulder (jasper.mulder(AT)planet.nl), Feb 03 2010
a(6) and a(7) corrected by Keith F. Lynch, May 25 2023
a(11), a(13)-a(15), a(17) corrected by Zhao Hui Du, Jun 07 2023

A194386 Numbers m such that Sum_{k=1..m} (<1/2 + k*r> - ) > 0, where r=sqrt(10) and < > denotes fractional part.

Original entry on oeis.org

1, 2, 3, 4, 5, 7, 8, 9, 10, 11, 13, 14, 15, 16, 17, 19, 20, 21, 22, 23, 25, 26, 27, 28, 29, 31, 32, 33, 34, 35, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75

Offset: 1

Clark Kimberling, Aug 23 2011

nonn

See A194368.

Cf. A010467, A194368, A194385.

r = Sqrt[10]; c = 1/2;
x[n_] := Sum[FractionalPart[k*r], {k, 1, n}]
y[n_] := Sum[FractionalPart[c + k*r], {k, 1, n}]
t1 = Table[If[y[n] < x[n], 1, 0], {n, 1, 300}];
Flatten[Position[t1, 1]]   (* empty *)
t2 = Table[If[y[n] == x[n], 1, 0], {n, 1, 400}];
Flatten[Position[t2, 1]]     (* A194385 *)
t3 = Table[If[y[n] > x[n], 1, 0], {n, 1, 100}];
Flatten[Position[t3, 1]]     (* A194386 *)

A136583 n such that floor(sqrt(10^(2*n-1))) is (probably) prime.

Original entry on oeis.org

1, 2, 7, 18, 33, 206, 468, 1061, 6831, 40377

Offset: 1

Lekraj Beedassy, Jan 09 2008

Number of digits of sqrt(10)-primes (A136582).

The n such that A017934(2*n-1) is (probably) prime.

Cf. A010467, A017934, A017934, A131581, A132153, A136582, A175733, A175734.

for n in [1..10^6] do if IsPrime(Isqrt(10^(2*n-1))) then printf "%o, ", n; end if; end for; // Jason Kimberley, Sep 03 2011

rd = RealDigits[Sqrt[10], 10, 10^5][[1]]; Do[ If[ PrimeQ@ FromDigits@ Take[rd, n], Print@n], {n, 10^5}] (* Robert G. Wilson v, Jan 20 2008 *)

a(6) - a(8) from Robert G. Wilson v, Jan 20 2008

Probable terms a(9) and a(10) from Jason Kimberley, Aug 19 and Sep 03 2011

A176219 Decimal expansion of (6+2*sqrt(10))/3.

Original entry on oeis.org

4, 1, 0, 8, 1, 8, 5, 1, 0, 6, 7, 7, 8, 9, 1, 9, 5, 5, 4, 6, 6, 5, 9, 2, 9, 0, 2, 9, 6, 2, 1, 8, 1, 2, 3, 5, 5, 8, 1, 3, 0, 3, 6, 7, 5, 9, 5, 5, 0, 1, 4, 4, 5, 5, 1, 2, 3, 8, 3, 3, 6, 5, 6, 8, 5, 2, 8, 3, 9, 6, 2, 9, 2, 4, 2, 6, 1, 5, 8, 8, 1, 4, 2, 2, 9, 4, 9, 8, 7, 3, 8, 9, 1, 9, 5, 3, 3, 5, 3, 0, 1, 2, 4, 8, 9

Offset: 1

Klaus Brockhaus, Apr 12 2010

Continued fraction expansion of (6+2*sqrt(10))/3 is A010714.

			(6+2*sqrt(10))/3 = 4.10818510677891955466...

Daniel Starodubtsev, Table of n, a(n) for n = 1..10000

Cf. A010467 (decimal expansion of sqrt(10)), A010714 (repeat 4, 9).

A194385 Numbers m such that Sum_{k=1..m} (<1/2 + k*r> - ) = 0, where r=sqrt(10) and < > denotes fractional part.

Original entry on oeis.org

6, 12, 18, 24, 30, 36, 228, 234, 240, 246, 252, 258, 264, 456, 462, 468, 474, 480, 486, 492, 684, 690, 696, 702, 708, 714, 720

Offset: 1

Clark Kimberling, Aug 23 2011

nonn

See A194368.

Cf. A010467, A194368, A194386.

r = Sqrt[10]; c = 1/2;
x[n_] := Sum[FractionalPart[k*r], {k, 1, n}]
y[n_] := Sum[FractionalPart[c + k*r], {k, 1, n}]
t1 = Table[If[y[n] < x[n], 1, 0], {n, 1, 300}];
Flatten[Position[t1, 1]]  (* empty *)
t2 = Table[If[y[n] == x[n], 1, 0], {n, 1, 800}];
Flatten[Position[t2, 1]]     (* A194385 *)
t3 = Table[If[y[n] > x[n], 1, 0], {n, 1, 100}];
Flatten[Position[t3, 1]]     (* A194386 *)

A335320 Decimal expansion of a prime generating constant c for which round(c^k) is prime for 2 <= k <= 388.

Original entry on oeis.org

3, 1, 6, 2, 2, 7, 7, 6, 7, 1, 8, 5, 5, 9, 5, 6, 9, 3, 4, 1, 1, 8, 1, 9, 7, 8, 7, 0, 6, 1, 4, 2, 8, 8, 0, 9, 2, 1, 1, 9, 6, 0, 5, 0, 1, 9, 8, 7, 0, 6, 6, 9, 1, 6, 8, 5, 0, 9, 2, 6, 0, 4, 3, 9, 7, 2, 3, 1, 7, 8, 5, 0, 3, 6, 2, 0, 7, 4, 0, 8, 6, 4, 7, 5, 4

Offset: 5

Hugo Pfoertner, Jun 30 2020

See A332308 for more information and references.

Apart from the offset, first differs from A010467 at the 9th digit. - Omar E. Pol, Jun 30 2020

			31622.7767185595693411819787061428809211960501987066916850926...

Hugo Pfoertner, Table of n, a(n) for n = 5..4100, provides all 4096 digits.

Cf. A332308, A010467.

cp's OEIS Frontend

A176441 Decimal expansion of sqrt(210).

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A018072 Powers of fourth root of 10 rounded down.

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A042937 Denominators of continued fraction convergents to sqrt(1000).

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A081168 Differences of Beatty sequence for square root of 10.

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A088995 Least k > 0 such that the first n digits of 2^k and 5^k are identical.

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A194386 Numbers m such that Sum_{k=1..m} (<1/2 + k*r> - ) > 0, where r=sqrt(10) and < > denotes fractional part.

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Mathematica

A136583 n such that floor(sqrt(10^(2*n-1))) is (probably) prime.

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Magma

Mathematica

Extensions

A176219 Decimal expansion of (6+2*sqrt(10))/3.

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