cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-7 of 7 results.

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

Original entry on oeis.org

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

Views

Author

Erich Friedman, Jan 29 2000

Keywords

Comments

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

Examples

			32^3=32768, which starts with 32.
		

Crossrefs

Programs

  • Mathematica
    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[ # ]& ] ] ]
  • PARI
    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

Extensions

0 inserted by Juhani Heino, Aug 31 2015

A052211 Numbers k such that k^4 starts with k itself (in base 10).

Original entry on oeis.org

0, 1, 10, 100, 1000, 10000, 46416, 100000, 464159, 1000000, 2154435, 4641589, 10000000, 21544347, 100000000, 1000000000, 10000000000, 100000000000, 1000000000000, 2154434690032, 4641588833613, 10000000000000, 21544346900319, 46415888336128
Offset: 1

Views

Author

Erich Friedman, Jan 29 2000

Keywords

Examples

			46416^4 starts with 46416.
		

Crossrefs

Programs

  • Mathematica
    Join[{0}, Sort[ Table[ 10^i,{i,0,22} ]~Join~Select[ Table[ Ceiling[ 10.^(1/3)*10^i ],{i,0,22} ], Take[ IntegerDigits[ #^4 ],Length[ IntegerDigits[ # ] ] ]==IntegerDigits[ # ]& ]~Join~Select[ Table[ Ceiling[ 10.^(2/3)*10^i ],{i,0,22} ],Take[ IntegerDigits[ #^4 ],Length[ IntegerDigits[ # ] ] ]==IntegerDigits[ # ]& ] ] ]
  • PARI
    r=41; print1(1, ", "); e=4; 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

Extensions

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

Views

Author

Arkadiusz Wesolowski, Dec 10 2013

Keywords

Examples

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

Crossrefs

Programs

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

Extensions

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

Views

Author

Arkadiusz Wesolowski, Dec 10 2013

Keywords

Examples

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

Crossrefs

Programs

  • Mathematica
    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 *)
  • PARI
    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, ", ")));

Extensions

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

Views

Author

Arkadiusz Wesolowski, Dec 10 2013

Keywords

Examples

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

Crossrefs

Programs

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

Extensions

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

Views

Author

Arkadiusz Wesolowski, Dec 10 2013

Keywords

Examples

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

Crossrefs

Programs

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

Extensions

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

Views

Author

Arkadiusz Wesolowski, Dec 10 2013

Keywords

Examples

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

Crossrefs

Programs

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

Extensions

0 inserted by Juhani Heino, Aug 31 2015
Showing 1-7 of 7 results.