A130448 Squares whose decimal representation contains no proper subsequence which is a positive square.

1, 4, 9, 25, 36, 576, 676, 7056, 80656, 665856, 2027776, 2802276, 22282727076, 77770707876, 78807087076, 7888885568656, 8782782707776, 72822772707876, 555006880085056, 782280288087076, 827702888070276, 888288787822276, 2282820800707876, 7880082008070276, 80077778877070276, 88778000807227876, 782828878078078276, 872727072820287876

Offset: 1

Author

Martin Raab, Aug 07 2007

Keywords

• base
• nonn
• fini

Comments

Is this sequence finite and if so what is the last term?

Yes, the sequence must be finite. This follows from a well-known result: there are no infinite antichains for the subsequence ordering. - Jeffrey Shallit, Mar 05 2014

If it exists, a(29) is greater than 10^18. - Mauro Fiorentini, Jan 24 2017

If it exists, a(51) is greater than 2*10^30. - Giovanni Resta, Jan 08 2018

Examples

			576 is in the list because none of its proper subsequences 5, 7, 6, 57, 76 or 56 are squares.
		

Links

• Martin Raab, Table of n, a(n) for n = 1..55 (terms 1..50 from Giovanni Resta, terms 51..55 from Martin Raab)
• Michael S. Branicky, Python program
• Wikipedia, Subsequence

Crossrefs

Cf. A131599, A238334 (substrings).

Programs

Mathematica

fQ[n_] := Module[{d = IntegerDigits[n], ds, sq}, ds = FromDigits /@ Union[Most[Rest[Subsets[d]]]]; sq = Select[ds, # > 0 && IntegerQ[Sqrt[#]] &, 1]; sq == {}]; Select[Range[0, 100000]^2, fQ] (* T. D. Noe, Mar 05 2014 *)

PARI

isok(n) = {my(d = digits(n)); for (k = 1, #d, for (j= 1, #d - k + 1, if (j != #d, sd = vector(j, i, d[k+i-1]); nsd = fromdigits(sd); if (nsd && issquare(nsd), return(0));););); return (1);} \\ Michel Marcus, Apr 21 2018

Python

# see linked program for faster version
from math import isqrt
from itertools import chain, combinations as C, count, islice
def issquare(n): return isqrt(n)**2 == n
def psets(s): # nonempty proper subsets of s
    return chain.from_iterable(C(s, r) for r in range(1, len(s)))
def cond(s):
    ss = ("".join(t) for t in psets(s) if t[0] != "0")
    return not any(issquare(int(u)) for u in ss)
def agen(): yield from (k**2 for k in count(1) if cond(str(k**2)))
print(list(islice(agen(), 15))) # Michael S. Branicky, Feb 23 2023

Extensions

Edited by N. J. A. Sloane, Sep 14 2007

a(23)-a(28) from Mauro Fiorentini, Jan 24 2017