A048653 Numbers k such that the decimal digits of k^2 can be partitioned into two or more nonzero squares.

7, 12, 13, 19, 21, 35, 37, 38, 41, 44, 57, 65, 70, 107, 108, 112, 119, 120, 121, 125, 129, 130, 190, 191, 204, 205, 209, 210, 212, 223, 253, 285, 305, 306, 315, 342, 343, 345, 350, 369, 370, 379, 380, 408, 410, 413, 440, 441, 475, 487, 501, 538, 570, 642, 650

Offset: 1

Author

Felice Russo

Keywords

• base
• nice
• nonn

Examples

			12 is present because 12^2=144 can be partitioned into three squares 1, 4 and 4.
108^2 = 11664 = 1_16_64, 120^2 = 14400 = 1_4_400, so 108 and 120 are in the sequence.
		

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A048646, A048375, A010052, A000290; subsequence of A128783.

Programs

Haskell

a048653 n = a048653_list !! (n-1)
a048653_list = filter (f . show . (^ 2)) [1..] where
   f zs = g (init $ tail $ inits zs) (tail $ init $ tails zs)
   g (xs:xss) (ys:yss)
     | h xs      = h ys || f ys || g xss yss
     | otherwise = g xss yss
     where h ds = head ds /= '0' && a010052 (read ds) == 1
   g   = False
-- Reinhard Zumkeller, Oct 11 2011

Mathematica

(* This non-optimized program is not suitable to compute a large number of terms. *) split[digits_, pos_] := Module[{pos2}, pos2 = Transpose[{Join[ {1}, Most[pos+1]], pos}]; FromDigits[Take[digits, {#[[1]], #[[2]]}]]& /@ pos2]; sel[n_] := Module[{digits, ip, ip2, accu, nn}, digits = IntegerDigits[n^2]; ip = IntegerPartitions[Length[digits]]; ip2 = Flatten[ Permutations /@ ip, 1]; accu = Accumulate /@ ip2; nn = split[ digits, #]& /@ accu; SelectFirst[nn, Length[#]>1 && Flatten[ IntegerDigits[#] ] == digits && AllTrue[#, #>0 && IntegerQ[Sqrt[#]]&]&] ]; k = 1; Reap[Do[If[(s = sel[n]) != {}, Print["a(", k++, ") = ", n, " ", n^2, " ", s]; Sow[n]], {n, 1, 10^4}]][[2, 1]] (* Jean-François Alcover, Sep 28 2016 *)

Python

from math import isqrt
def issquare(n): return isqrt(n)**2 == n
def ok(n, c):
    if n%10 in {2, 3, 7, 8}: return False
    if issquare(n) and c > 1: return True
    d = str(n)
    for i in range(1, len(d)):
        if d[i] != '0' and issquare(int(d[:i])) and ok(int(d[i:]), c+1):
            return True
    return False
def aupto(lim): return [r for r in range(lim+1) if ok(r*r, 1)]
print(aupto(650)) # Michael S. Branicky, Jul 10 2021

Extensions

Corrected and extended by Naohiro Nomoto, Sep 01 2001

Definition clarified by Harvey P. Dale, May 09 2021