A068802 Smaller of two consecutive squares which have no common digits.

0, 1, 4, 9, 16, 25, 36, 64, 196, 361, 484, 841, 1936, 5929, 8836, 69696, 1999396, 29997529

Offset: 0

Author

Amarnath Murthy, Mar 06 2002

Keywords

• base
• fini
• full
• nonn

Comments

There are no more terms. Sketch of proof: Suppose n^2 and (n+1)^2 have no common digits. Then their first digits differ, so n+1 = ceiling(sqrt(d*10^r)) with 1<=d<=9 and r>=0. In other words,

n+1 = ceiling(sqrt(e*100^s)) with e in {1,2,...,9,10,20,...,90} and s>=0. The cases e=1, 4 and 9 are easy. Otherwise note that about half the digits of (n+1)^2 equal 0, so n^2 has no 0's. We have

(n+1)^2 - n^2 = 2n+1 ~ 10^s*2*sqrt(e). For e=20, this is about 10^s * 8.94427190999916. So for s>=10, the decimal expansion of (n+1)^2 - n^2 has 2 consecutive 9's. (In fact 4 for large s, but 2 is enough.) Since n^2 has no 0's this implies that n^2 and (n+1)^2 have the same digit in the position of the first of the 2 9's. The same idea works for other values of e, but the consecutive 9's occur later.

Examples

			29997529 is a term since 29997529 and 30008484 are two consecutive squares with no common digits.
		

Programs

Maple

s := X->convert(convert(X,base,10),set); seq(`if`((s(n^2) intersect s((n+1)^2))={},n^2,printf("")),n=1..350000);

Mathematica

For[lastn=-1; r=0, r<500, r++, For[d=1, d<10, d++, n=Ceiling[Sqrt[d*10^r]]-1; If[n>lastn, lastn=n; If[Intersection[IntegerDigits[n^2], IntegerDigits[(n+1)^2]]=={}, Print[n^2]]]]]
First /@ Select[Partition[Range[0, 6000]^2, 2, 1], Intersection @@ IntegerDigits /@ # == {} &] (* Jayanta Basu, Aug 06 2013 *)

Extensions

Edited by Dean Hickerson and Francois Jooste (phukraut(AT)hotmail.com), Mar 19 2002