A052211 -id:A052211 - cp's OEIS frontend

A052210 Numbers k such that k^3 starts with k itself (in base 10).

0, 1, 10, 32, 100, 1000, 10000, 31623, 100000, 316228, 1000000, 3162278, 10000000, 31622777, 100000000, 1000000000, 10000000000, 31622776602, 100000000000, 316227766017, 1000000000000, 10000000000000, 31622776601684, 100000000000000, 316227766016838

Offset: 1

Erich Friedman, Jan 29 2000

Replace the first term with 4, then add 1 to all the others to find numbers k where k^3 starts with k+2. Similar processes can be used for any k+2m. (conjectured) - Dhilan Lahoti, Aug 30 2015

10^k is in the sequence for all k. For odd k, m = ceil(10^(k/2)) is in the sequence if and only if m^3/(m+1) < 10^k. A necessary condition for this is that m - 1/2 > 10^(k/2), i.e. the first digit after the decimal point in 10^(k/2) is at least 5. Is this sufficient as well as necessary? - Robert Israel, Aug 31 2015

			32^3=32768, which starts with 32.

Andrew Howroyd, Table of n, a(n) for n = 1..400

Cf. A052211, A233451, A233452, A233453, A233454, A233455, A233456, A261751.

Join[{0}, Sort[ Table[ 10^i,{i,0,22} ]~Join~Select[ Table[ Ceiling[ Sqrt[ 10 ]*10^i ],{i,0,22} ], Take[ IntegerDigits[ #^3 ],Length[ IntegerDigits[ # ] ] ]==IntegerDigits[ # ]& ] ] ]

r=29; print1(1, ", "); e=3; for(n=2, r, p=round((10^(1/(e-1)))^n); f=p^e; b=10^(#Str(f)-#Str(p)); if((f-lift(Mod(f, b)))/b==p, print1(p, ", "))); \\ Arkadiusz Wesolowski, Dec 10 2013

0 inserted by Juhani Heino, Aug 31 2015

A307600 Numbers k such that the digits of k^(1/4) begin with k.

Original entry on oeis.org

0, 1, 21, 463, 464, 9999, 10000, 215443, 4641588, 99999999, 100000000, 2154434689, 2154434690, 46415888335, 46415888336, 999999999999, 1000000000000, 21544346900318, 464158883361277, 9999999999999999, 10000000000000000, 215443469003188371, 215443469003188372

Offset: 1

Dmitry Kamenetsky, Apr 17 2019

Program is in A307371.
From Bernard Schott, May 01 2019: (Start)
There are two nontrivial families in this sequence:
1st: 21, 215443, 2154434689, 2154434690, 21544346900318, ...
2nd: 463, 464, 4641588, 46415888335, 46415888336, ... (End)
From Jon E. Schoenfield, May 04 2019: (Start)
For each number k such that the digits of k^(1/m) begin with k, we have, for each m >= 2, floor(k^(1/m) * 10^d) = k for some integer d, so k^(1/m) * 10^d ~= k; solving for k gives k ~= 10^(d*m/(m-1)).
In the m=4 case (this sequence), this gives k ~= 10^(d*4/3) so, as d is incremented by 1, 10^(d*4/3) increases by a factor of 10^(4/3) = 10000^1/3 = 21.5443469...:
.
   d |     10^(d*4/3)
  ---+---------------------
   0 |             1
   1 |            21.544...
   2 |           464.158...
   3 |         10000
   4 |        215443.469...
   5 |       4641588.833...
   6 |     100000000
   7 |    2154434690.031...
   8 |   46415888336.127...
   9 | 1000000000000
.
Each nonnegative integer d corresponds to one or two terms in the sequence. Letting j = floor(10000^(d/3)), j is necessarily a term; j-1 is also a term iff (j-1)^(1/4)*10^d < j. This inequality is satisified
   for d == 1 (mod 3) at d = 7, 13, 16, 34, 37, ...;
   for d == 2 (mod 3) at d = 2, 8, 20, 29, 32, 35, ...;
   and at every d == 0 (mod 3).
(The sequence contains no other terms than numbers k of the form j or j-1 where j = floor(10000^(d/3)) for some nonnegative integer d.)
(End)

			215443^(1/4) = 21.544335..., which begins with "215443", so 215443 is in the sequence.

Chai Wah Wu, Table of n, a(n) for n = 1..1172

Cf. A307371, A307588.

Cf. A052211 (analog for 4th power instead of 1/4).

a(12)-a(23) from Jon E. Schoenfield, May 01 2019

A233451 Numbers k such that k^5 starts with k itself (in base 10).

Original entry on oeis.org

0, 1, 10, 18, 100, 178, 1000, 10000, 17783, 31623, 100000, 177828, 316228, 1000000, 10000000, 100000000, 1000000000, 5623413252, 10000000000, 100000000000, 177827941004, 316227766017, 1000000000000, 1778279410039, 10000000000000, 31622776601684

Offset: 1

Arkadiusz Wesolowski, Dec 10 2013

			18^5 = 1889568 begins with 18, so 18 is in the sequence.

Robert Israel, Table of n, a(n) for n = 1..1756

Cf. A052210, A052211, A233452, A233453, A233454, A233455, A233456.

R:= 0,1:
for d from 1 to 100 do
  k:= ceil(10^(d/4));
  if k^5 - 10^d * k < 10^d then
    R:= R, k
  fi
od:
R; # Robert Israel, Sep 11 2024

kmax=10^6; Select[Range[0,kmax],FromDigits[Drop[IntegerDigits[#^5],-(IntegerLength[#^5]-IntegerLength[#])]]==# &] (* Stefano Spezia, Aug 27 2023 *)

r=54; print1(1, ", "); e=5; for(n=2, r, p=round((10^(1/(e-1)))^n); f=p^e; b=10^(#Str(f)-#Str(p)); if((f-lift(Mod(f, b)))/b==p, print1(p, ", ")));

0 inserted by Juhani Heino, Aug 31 2015

A233452 Numbers k such that k^6 starts with k itself (in base 10).

Original entry on oeis.org

0, 1, 4, 10, 16, 40, 100, 631, 1000, 1585, 2512, 10000, 15849, 25119, 100000, 1000000, 10000000, 15848932, 100000000, 1000000000, 6309573445, 10000000000, 100000000000, 251188643151, 1000000000000, 3981071705535, 6309573444802, 10000000000000

Offset: 1

Arkadiusz Wesolowski, Dec 10 2013

			4^6 = 4096 begins with 4, so 4 is in the sequence.

Cf. A052210, A052211, A233451, A233453, A233454, A233455, A233456.

kmax=10^6; Select[Range[0,kmax],FromDigits[Drop[IntegerDigits[#^6],-(IntegerLength[#^6]-IntegerLength[#])]]==# &] (* Stefano Spezia, Aug 27 2023 *)

r=65; print1(1, ", "); e=6; for(n=2, r, p=round((10^(1/(e-1)))^n); f=p^e; b=10^(#Str(f)-#Str(p)); if((f-lift(Mod(f, b)))/b==p, print1(p, ", ")));

0 inserted by Juhani Heino, Aug 31 2015

A233453 Numbers k such that k^7 starts with k itself (in base 10).

Original entry on oeis.org

0, 1, 10, 100, 1000, 6813, 10000, 14678, 46416, 100000, 146780, 464159, 1000000, 4641589, 10000000, 21544347, 68129207, 100000000, 1000000000, 10000000000, 100000000000, 316227766017, 681292069058, 1000000000000, 2154434690032, 10000000000000, 21544346900319

Offset: 1

Arkadiusz Wesolowski, Dec 10 2013

			6813^7 = 681347587591081074493576917 begins with 6813, so 6813 is in the sequence.

Cf. A052210, A052211, A233451, A233452, A233454, A233455, A233456.

Join[{0},Select[Range[0,5*10^6],IntegerDigits[#]==Take[IntegerDigits[ #^7],IntegerLength[ #]]&]] (* The program generates the first 14 terms of the sequence. To generate more, increase the Range constant. *) (* Harvey P. Dale, Mar 31 2022 *)

r=80; print1(1, ", "); e=7; for(n=2, r, p=round((10^(1/(e-1)))^n); f=p^e; b=10^(#Str(f)-#Str(p)); if((f-lift(Mod(f, b)))/b==p, print1(p, ", ")));

0 inserted by Juhani Heino, Aug 31 2015

A233454 Numbers k such that k^8 starts with k itself (in base 10).

Original entry on oeis.org

0, 1, 2, 10, 14, 72, 100, 139, 518, 1000, 10000, 13895, 19307, 26827, 37276, 100000, 1000000, 10000000, 13894955, 26826958, 100000000, 193069773, 517947468, 1000000000, 1930697729, 10000000000, 100000000000, 268269579528, 1000000000000, 3727593720315

Offset: 1

Arkadiusz Wesolowski, Dec 10 2013

			2^8 = 256 begins with 2, so 2 is in the sequence.

Cf. A052210, A052211, A233451, A233452, A233453, A233455, A233456.

Join[{0}, Select[Range@ 1000000, Take[IntegerDigits[#^8], IntegerLength@ #] == IntegerDigits@ # &]] (* Michael De Vlieger, Aug 31 2015 *)

r=88; print1(1, ", "); e=8; for(n=2, r, p=round((10^(1/(e-1)))^n); f=p^e; b=10^(#Str(f)-#Str(p)); if((f-lift(Mod(f, b)))/b==p, print1(p, ", ")));

0 inserted by Juhani Heino, Aug 31 2015

A233455 Numbers k such that k^9 starts with k itself (in base 10).

Original entry on oeis.org

0, 1, 10, 75, 100, 750, 1000, 4217, 7499, 10000, 100000, 177828, 1000000, 10000000, 74989421, 100000000, 1000000000, 5623413252, 10000000000, 100000000000, 177827941004, 1000000000000, 1778279410039, 10000000000000, 56234132519035, 100000000000000

Offset: 1

Arkadiusz Wesolowski, Dec 10 2013

			750^9 = 75084686279296875000000000 begins with 750, so 750 is in the sequence.

G. L. Honaker, Jr. and Chris Caldwell, Prime Curios! 7499

Cf. A052210, A052211, A233451, A233452, A233453, A233454, A233456.

Join[{0}, Select[Range@ 1000000, Take[IntegerDigits[#^9], IntegerLength@ #] == IntegerDigits@ # &]] (* Michael De Vlieger, Aug 31 2015 *)

r=112; print1(1, ", "); e=9; for(n=2, r, p=round((10^(1/(e-1)))^n); f=p^e; b=10^(#Str(f)-#Str(p)); if((f-lift(Mod(f, b)))/b==p, print1(p, ", ")));

0 inserted by Juhani Heino, Aug 31 2015

A233456 Numbers k such that k^10 starts with k itself (in base 10).

Original entry on oeis.org

0, 1, 6, 10, 13, 36, 60, 100, 1000, 10000, 100000, 129155, 278256, 1000000, 10000000, 21544347, 100000000, 1000000000, 10000000000, 59948425032, 100000000000, 599484250319, 1000000000000, 10000000000000, 100000000000000, 464158883361278, 1000000000000000

Offset: 1

Arkadiusz Wesolowski, Dec 10 2013

			6^10 = 60466176 begins with 6, so 6 is in the sequence.

Cf. A052210, A052211, A233451, A233452, A233453, A233454, A233455.

kmax=10^6; Select[Range[0,kmax],FromDigits[Drop[IntegerDigits[#^10],-(IntegerLength[#^10]-IntegerLength[#])]]==# &] (* Stefano Spezia, Aug 27 2023 *)

r=135; print1(1, ", "); e=10; for(n=2, r, p=round((10^(1/(e-1)))^n); f=p^e; b=10^(#Str(f)-#Str(p)); if((f-lift(Mod(f, b)))/b==p, print1(p, ", ")));

0 inserted by Juhani Heino, Aug 31 2015

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A052210 Numbers k such that k^3 starts with k itself (in base 10).

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A307600 Numbers k such that the digits of k^(1/4) begin with k.

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A233451 Numbers k such that k^5 starts with k itself (in base 10).

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A233452 Numbers k such that k^6 starts with k itself (in base 10).

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A233453 Numbers k such that k^7 starts with k itself (in base 10).

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A233454 Numbers k such that k^8 starts with k itself (in base 10).

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A233455 Numbers k such that k^9 starts with k itself (in base 10).

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A233456 Numbers k such that k^10 starts with k itself (in base 10).

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