This site is supported by donations to The OEIS Foundation.

 Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
 A067634 a(1) = 1; string of digits of a(n)^2 is a substring of the string of digits of a(n+1)^2. 2
 1, 4, 13, 130, 1300, 13000, 130000, 1300000, 13000000, 130000000, 1300000000, 13000000000, 130000000000, 1300000000000, 13000000000000, 130000000000000, 1300000000000000, 13000000000000000, 130000000000000000 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS Provably infinite. The obvious pattern continues. Proof: By induction, assume that 13*10^k is the (k+2)nd element in the sequence for some k >= 1. Clearly 13*10^{k+1} satisfies the required condition; we need to show that no other number works. Equivalently, we need to show that 169*10^{2k+2} is the smallest square of one of the forms: 169*10^{2k+1}+a, a*10^{2k+3}+169*10^{2k}, 169*10^{2k+2}+a*10+b, a*10^{2k+4}+169*10^{2k+1}+b, a*10^{2k+4}+b*10^{2k+3}+169*10^{2k}, where 0 <= a,b <= 9. Insisting that the number be less than 169*10^{2k+2} and checking that it is a 2-adic, 3-adic and 5-adic square eliminates all but 169*10^{2k+1}+9 and 1169*10^{2k+1}+1. To eliminate these, reduce modulo the primes 101, 137=(10^4+1)/173 and 5882353=(10^8+1)/17; these all divide 10^16+1, so it suffices to check k=0,1,2,3,4,5,6,7. QED. Eric Rains, Jan 29, 2002. LINKS CROSSREFS Cf. A014563, A066825. Sequence in context: A197897 A203380 A006104 * A203215 A198159 A042537 Adjacent sequences:  A067631 A067632 A067633 * A067635 A067636 A067637 KEYWORD nonn,base AUTHOR David W. Wilson, Feb 05 2002 EXTENSIONS More terms from David W. Wilson, Feb 05 2002 STATUS approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Transforms | Puzzles | Hot | Classics
Recent Additions | More pages | Superseeker | Maintained by The OEIS Foundation Inc.

Content is available under The OEIS End-User License Agreement .