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.

A242268 Squares not ending in 00 that remain squares if prefixed with the digit 1.

Original entry on oeis.org

225, 5625, 5405625, 23765625, 2127515625, 58503515625, 51921031640625, 250727431640625, 20090404775390625, 608180644775390625, 498431438615478515625, 2642208974615478515625, 189450791534674072265625, 6319494849134674072265625, 9981411957966851806640625
Offset: 1

Views

Author

Reiner Moewald, Aug 16 2014

Keywords

Comments

It can easily be shown that all squares that remain squares if prefixed with the digit 1 end in 00 or 25 and, moreover, that all squares ending in 00 are multiples of the squares ending in 5 (factor: 10^(2*n)).
Subsequence of A167035. - Michel Marcus, Sep 08 2014

Examples

			225 = 15*15 and 1225 = 35*35.
		

Crossrefs

Cf. A167035.

Programs

  • Maple
    A:= {}:
    for m from 3 to 100 do
      cand1:= floor(log[5](1/2*(1+sqrt(2))*10^(m/2)));
      cand2:= floor(log[5](2*(1+sqrt(2))*(5/2)^(m/2)));
      s1:= 5^cand1 - 10^m/4/5^cand1;
      s2:=  2^m/4*5^cand2 - 5^(m-cand2);
      if s1^2 >= 10^(m-1) then A:= A union {s1^2} fi;
      if s2^2 >= 10^(m-1) then A:= A union {s2^2} fi;
    od:
    A; # Robert Israel, Sep 08 2014
  • PARI
    for(n=1,10^20,p=n^2;if(p%100,s=concat("1",Str(p));if(issquare(eval(s)),print1(p,", ")))) \\ Derek Orr, Aug 23 2014
  • Python
    import math
    def power(a, n):
       pow = 1
       for i in range(0, n):
          pow = pow * a
       return pow
    end = 50
    for n in range(1, end):
       l1 = 1/math.log(5)*(math.log(math.sqrt(2)-1)+(n-2)/2*math.log(2))+ n/2
       u1 = 1/math.log(5)*(math.log(math.sqrt(11)-1)+(n-3)/2*math.log(2))+ (n-1)/2
       if math.ceil(l1) == math.floor(u1) and math.ceil(l1)>0:
          p = math.ceil(l1)
          x = power(5, p)*(-1)+power(2, n-2)*power(5, n-p)
          print(x*x)
       l2 = 1/math.log(5)*(math.log(math.sqrt(11)+1)+(n-3)/2*math.log(2))+ (n-1)/2
       u2 = 1/math.log(5)*(math.log(math.sqrt(2)+1)+(n-2)/2*math.log(2))+ n/2
       if math.ceil(l2) == math.floor(u2) and math.ceil(l2)>0:
          p = math.ceil(l2)
          x = power(5, p)-power(2, n-2)*power(5, n-p)
          print(x*x)
    print('End.')