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A294497
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Squares k (not ending in 0) such that the integer that is built up by concatenating the floors of the square roots of the two-digit numbers into which the original number is separated (from right to left) is the square root of the original number.
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1
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1, 4, 9, 16, 25, 36, 49, 64, 81, 225, 625, 1225, 2025, 2601, 2704, 2809, 2916, 3025, 3136, 3249, 3364, 3481, 4225, 5625, 7225, 9025, 22801, 23104, 23409, 50625, 63001, 63504, 75625, 123201, 180625, 203401, 225625, 390625, 432964, 455625, 573049, 680625, 732736, 765625, 2175625, 6260004, 6270016
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OFFSET
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1,2
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COMMENTS
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If k has an odd number of digits, all digits after the first digit are paired; see first example below.
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LINKS
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EXAMPLE
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75625 is a term because partitioning its digits as (7)(56)(25), taking the square root of each part and truncating, and then concatenating the results, gives floor(sqrt(7))|floor(sqrt(56))|floor(sqrt(25)) = 275 = sqrt(75625);
180625 is a term because floor(sqrt(18))|floor(sqrt(06))|floor(sqrt(25)) = 425 = sqrt(180625).
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MATHEMATICA
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#^2 & /@ Select[Range[10^4], And[Mod[#, 10] != 0, FromDigits@ Map[Floor@ Sqrt@ FromDigits@ # &, Partition[PadLeft[#, 2 Ceiling[Length@ #/2]], 2, 2]] &@ IntegerDigits[#^2] == #] &] (* Michael De Vlieger, Nov 23 2017 *)
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PROG
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(PARI) is(n) = if(issquare(n) == 0||n % 10 == 0, return(0)); my(sq = i = 0, cn = n); while(cn > 0, sq += 10^i * sqrtint(cn % 100); cn \= 100; i++); sq ^ 2 == n \\ David A. Corneth, Jan 18 2018
(Python)
import math
for k in range(1, 1000000000):
p = 0
z = 0
n = k*k
while n >= 100:
z = z + int(math.floor(math.sqrt(n % 100)) * math.pow(10, p))
n = int((n - (n % 100)) / 100)
p = p + 1
z = z + int(math.floor(math.sqrt(n)) * math.pow(10, p))
if z == k and k % 10 > 0:
print(k * k, k)
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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