A257649 Squares that are the concatenation of two integers (without leading zeros) the sum of which is also a square.

36, 81, 169, 196, 324, 361, 576, 729, 841, 1156, 1521, 1681, 1764, 2809, 3249, 3481, 4356, 5625, 6084, 6241, 6724, 7396, 7569, 7744, 7921, 8281, 9216, 12321, 12544, 12769, 12996, 13689, 15129, 16384, 17424, 18769, 19881, 24964, 25600, 31684, 32041, 34596, 36864, 38416, 39601

Offset: 1

Author

Reiner Moewald, Jul 25 2015

Keywords

• nonn
• base

Comments

Squares that can be split up in more than one way, e.g., 729 (72 + 9 and 7 + 29), appear only once.

The number of such squares is infinite, since 39...960...01 (the numbers of the digits 9 and 0 is equal) can be split up into 3 and 9...960..01 with 3 + 9...960...01 = (100...0-2)^2 and 39...960...01 = (2*100...0 - 1)^2.

From Robert G. Wilson v, Aug 06 2015: (Start)

Number of terms < 10^k: 0, 2, 9, 27, 66, 149, 370, 910, 2164, 5325, 12916, 29448, ..., .

Terms which are members of A257649 in more than one way: 729, 7569, 15129, 56169, 86436, 123201, ..., .

Terms which are members of A257649 in more than two way: 881377344, 3784833441, 39999600001, 54444755556, 71111288889, 89999400001, 159999200001, 321111488889, 751111688889, ..., .

Least term which is a member of A257649 in k ways: 36, 729, 881377344, 399999960000001, ..., . (End)

Examples

			36 = 6^2 and 3 + 6 = 9 = 3^2.
		

Links

• Robert G. Wilson v, Table of n, a(n) for n = 1..10000 (first 909 terms from Reiner Moewald)

Crossrefs

Subsequence of A052041.

Programs

Mathematica

f[n_] := Block[{a, b, c, k = 1, idn = IntegerDigits@ n, lng, lst = {}}, lng = Length@ idn; While[k < lng, a = FromDigits[ Take[idn, {1, k}]]; b = FromDigits[ Take[idn, {k + 1, lng}]]; c = a*10^(lng - k) + b; If[b > 0 && Floor[1 + Log10@ b] == lng - k && IntegerQ@ Sqrt[a + b], AppendTo[lst, c]]; k++]; Length@ lst]; k = 1; lst = {}; While[k < 201, If[ f[k^2] > 0, AppendTo[lst, k^2]]; k++]; lst (* Robert G. Wilson v, Aug 06 2015 *)
ctiQ[n_]:=AnyTrue[Total/@Select[Table[FromDigits/@TakeDrop[IntegerDigits[n],d],{d,IntegerLength[ n]-1}],IntegerLength[#[[1]]]+IntegerLength[#[[2]]] ==IntegerLength[ n]&],IntegerQ[ Sqrt[#]]&]; Select[Range[200]^2,ctiQ] (* Harvey P. Dale, Jun 04 2023 *)

Python

import math
print("Start")
list =[]
for i in range(1,1000):
   a = i*i
   b = str(a)
   l = len(b)
   for j in range(1, l):
      a_1 = b[:j]
      a_2 = b[j:]
      c = int(a_1)+int(a_2)
      sqrt_c = int(math.sqrt(int(c)))
      if (sqrt_c * sqrt_c == c) and (int(a_2[:1]) > 0):
         if not a in list:
            list.append(a)
print(list)
print("End")