OFFSET
1,1
COMMENTS
The terms > 1 of A058411 can be considered as primitive elements of this sequence, obtained by multiplying those by powers of 10 (cf. formula). These terms of A058411 have at least 2 nonzero digits, and therefore their square has at least one digit 2. - M. F. Hasler, Nov 15 2017
LINKS
M. F. Hasler, Table of n, a(n) for n = 1..1380 (first 50 terms from Colin Barker)
FORMULA
MATHEMATICA
Select[Range[4*10^5], And[#[[2]] > 0, Union@ Take[RotateLeft[#, 2], 7] == {0}] &@ DigitCount[#^2] &] (* Michael De Vlieger, Nov 16 2017 *)
PROG
(PARI) L=List(); for(n=1, 10000, if(vecmax(digits(n^2))==2, listput(L, n))); Vec(L)
(PARI) A277959(LIM=1e15, L=List(), N=1)={while(LIM>N=next_A058411(N), my(t=N); until(LIM<t*=10, listput(L, t))); Set(L)} \\ M. F. Hasler, Nov 15 2017
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Colin Barker, Nov 06 2016
EXTENSIONS
Edited by M. F. Hasler, Nov 16 2017
STATUS
approved