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.

Previous Showing 11-14 of 14 results.

A048562 a(n+1)^2 is next smallest nontrivial square beginning with a(n)^2, initial square is 9.

Original entry on oeis.org

3, 31, 3101, 31010001, 3101000100000001, 9806223340409913487644162249555, 310100010000000100000000000000014337285304591730671503423653032
Offset: 0

Views

Author

Patrick De Geest, May 15 1999

Keywords

Comments

Next terms are 126, 251 and 502 digits long.

Crossrefs

Cf. A048561.

A050632 a(n+1) is next smallest nontrivial square containing a(n) as a substring, initial term is 9.

Original entry on oeis.org

9, 49, 1849, 18496, 103184964, 7383508103184964, 544680602158273835081031849649
Offset: 1

Views

Author

Patrick De Geest, Jul 15 1999

Keywords

Crossrefs

Starting with a(2)=49, the sequence is identical to A050630.

Formula

a(n) = A050633(n)^2.

Extensions

a(7) from Max Alekseyev, Feb 15 2012

A249834 a(n+1) is the next smallest nontrivial cube beginning with a(n), initial term is 7.

Original entry on oeis.org

7, 729, 7290099019, 72900990191475181426079596544
Offset: 1

Views

Author

Derek Orr, Dec 03 2014

Keywords

Comments

a(5) is an 87-digit number. - Jon E. Schoenfield, Dec 04 2014

Crossrefs

Programs

  • PARI
    a(n)=k=n; s=1; while(s<10^7, if(s%10, if(s^3\(10^(#Str(s^3)-#Str(k)))==k, print1(s^3, ", "); k=s^3)); s++)
    a(7)
  • Python
    def f(x):
      n = x
      s = 1
      while s < 10**7:
        if s % 10:
          S = str(s**3)
          if S.startswith(str(n)):
            print(s**3, end=', ')
            n = s**3
        s += 1
    f(7)
    

Extensions

a(4) from Jon E. Schoenfield, Dec 04 2014

A249837 a(n+1) is the next smallest nontrivial cube beginning with a(n), initial term is 9.

Original entry on oeis.org

9, 9261, 92615351886784, 9261535188678457128255779014690172977343833, 926153518867845712825577901469017297734383369607525414854584903918819898290730346512973206455943454340951813592133138664220381927
Offset: 1

Views

Author

Derek Orr, Dec 03 2014

Keywords

Comments

a(7) has 1163 digits. - Robert Israel, Dec 04 2014

Crossrefs

Programs

  • Maple
    nextterm:= proc(x) local d,s,t;
       for d from 1 do
         s:= traperror(ceil((x*10^d+1)^(1/3)));
         while not type(s,integer) do
            Digits:= Digits *2;
            s:= traperror(ceil((x*10^d+1)^(1/3)));
         od:
         t:= traperror(floor(((x+1)*10^d-1)^(1/3)));
         while not type(t,integer) do
            Digits:= Digits *2;
            t:= traperror(floor(((x+1)*10^d-1)^(1/3)));
         od:
         if s <= t then return s^3 fi;
       od:
    end proc:
    a[1]:= 9;
    for n from 2 to 6 do a[n]:= nextterm(a[n-1]) od; # Robert Israel, Dec 04 2014
  • PARI
    a(n)=k=n; s=1; while(s<10^7, if(s%10, if(s^3\(10^(#Str(s^3)-#Str(k)))==k, print1(s^3, ", "); k=s^3)); s++)
    a(9)
  • Python
    def f(x):
      n = x
      s = 1
      while s < 10**7:
        if s % 10:
          S = str(s**3)
          if S.startswith(str(n)):
            print(s**3, end=', ')
            n = s**3
        s += 1
    f(9)
    

Extensions

a(4) and a(5) from Robert Israel, Dec 04 2014
Previous Showing 11-14 of 14 results.