0,3

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.

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

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

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]]]]]

Sequence in context: A122683 A072862 A062295 this_sequence A075056 A022779 A008105

Adjacent sequences: A068799 A068800 A068801 this_sequence A068803 A068804 A068805

base,fini,full,nonn

Amarnath Murthy (amarnath_murthy(AT)yahoo.com), Mar 06 2002

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