A242268 Squares not ending in 00 that remain squares if prefixed with the digit 1.
225, 5625, 5405625, 23765625, 2127515625, 58503515625, 51921031640625, 250727431640625, 20090404775390625, 608180644775390625, 498431438615478515625, 2642208974615478515625, 189450791534674072265625, 6319494849134674072265625, 9981411957966851806640625
Offset: 1
Examples
225 = 15*15 and 1225 = 35*35.
Links
- Reiner Moewald, Table of n, a(n) for n = 1..102
Crossrefs
Cf. A167035.
Programs
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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.')
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