A007936
Some permutation of digits is a square.
Original entry on oeis.org
1, 4, 9, 10, 16, 18, 25, 36, 40, 46, 49, 52, 61, 63, 64, 81, 90, 94, 100, 106, 108, 112, 121, 136, 144, 148, 160, 163, 169, 180, 184, 196, 205, 211, 225, 234, 243, 250, 252, 256, 259, 265, 279, 289, 295, 297, 298, 306, 316, 324, 342, 360, 361, 400, 406, 409, 414
Offset: 1
- M. Le, On Smarandache Pseudo-Powers of Third Kind, Smarandache Notions Journal, Vol. 10, No. 1-2-3, 1999, 150-151.
Cf.
A235993 (leading zeros not allowed).
-
Select[Range[500],Length[Select[FromDigits/@Permutations[ IntegerDigits[#]], IntegerQ[Sqrt[#]]&]]>0&] (* Harvey P. Dale, Mar 30 2011 *)
A235994
Numbers having at least one anagram which is a cube.
Original entry on oeis.org
1, 8, 27, 46, 64, 72, 125, 126, 152, 162, 215, 216, 251, 261, 279, 297, 334, 343, 433, 512, 521, 612, 621, 729, 792, 927, 972, 1000, 1133, 1269, 1278, 1279, 1287, 1296, 1297, 1313, 1331, 1349, 1394, 1439, 1493, 1629, 1692, 1728, 1729, 1782, 1792, 1827, 1872
Offset: 1
126 is in the sequence because 216 = 6^3.
-
Select[Range[2000],AnyTrue[Surd[FromDigits/@Select[ Permutations[ IntegerDigits[#]],#[[1]]>0&],3],IntegerQ]&] (* The program uses the AnyTrue function from Mathematica version 10 *) (* Harvey P. Dale, Jul 15 2016 *)
-
from itertools import count, takewhile
def hash(n): return "".join(sorted(str(n)))
def aupto_digits(d):
cubes = takewhile(lambda x:x<10**d, (i**3 for i in count(1)))
C = set(map(hash, cubes))
return [k for k in range(1, 10**d) if hash(k) in C]
print(aupto_digits(4)) # Michael S. Branicky, Feb 18 2024
A337252
Digits of 2^n can be rearranged with no leading zeros to form t^2, for t not a power of 2.
Original entry on oeis.org
8, 10, 12, 14, 20, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 60, 62, 64, 66, 68, 70, 72, 74, 76, 78, 80, 82, 84, 86, 88, 90, 92, 94, 96, 98, 100, 102, 104, 106, 108, 110, 112, 114, 116, 118, 120, 122, 124, 126, 128, 130, 132, 134, 136, 138, 140, 142, 144, 146, 148, 150
Offset: 1
Here are the squares corresponding to the first few powers of 2:
2^8, 25^2
2^10, 49^2
2^12, 98^2
2^14, 178^2
2^20, 1028^2
2^26, 8291^2
2^28, 19112^2
2^30, 33472^2
2^32, 51473^2
2^34, 105583^2
2^36, 129914^2
2^38, 640132^2
2^40, 1081319^2
2^42, 1007243^2
2^44, 3187271^2
2^46, 4058042^2
2^48, 10285408^2
2^50, 32039417^2
2^52, 44795066^2
2^54, 100241288^2
From _Robert Israel_, Aug 21 2020: (Start)
2^56, 142847044^2
2^58, 318068365^2 (End)
From _Chai Wah Wu_, Aug 21 2020: (Start)
2^60, 1000562716^2
2^62, 1000709692^2
2^64, 3164169028^2
2^66, 4498215974^2
2^68, 10061077457^2
2^70, 31624545442^2
2^72, 34960642066^2
2^74, 100786105136^2
2^76, 105467328383^2
2^78, 316579648042^2
2^80, 1000556206526^2
2^82, 1001129296612^2
2^84, 3179799285956^2
2^86, 3333501503458^2
2^88, 10000006273742^2
2^90, 31624717039768^2
2^92, 31640399136637^2
2^94, 100001179435324^2
2^96, 100609261981363^2
2^98, 316227945405958^2
2^100, 1000000068136465^2
2^102, 1000000012839623^2
2^104, 3162279442052185^2
2^106, 3162295238497457^2
2^108, 10006109951303125^2
2^110, 31622778376826465^2
2^112, 31626290060004883^2
2^114, 100005555418898327^2
2^116, 100061093137010524^2
2^118, 316229698532373214^2
2^120, 1000000611139735223^2
2^122, 1005540208662183694^2
2^124, 3179814811220058566^2
2^126, 9994442844707576056^2
2^128, 31605185913938432804^2
2^130, 31799720491491676612^2
2^132, 99999944438762188450^2
2^134, 316052017518707374894^2
2^136, 100055595656929586657^2
2^138, 316227783779026656472^2
2^140, 3162277642424057210351^2
2^142, 1000056109592630240914^2
2^144, 3162279417006463372135^2
2^146, 3162279434557126331437^2
2^148, 10005559566228010636663^2
2^150, 99999999444438629490484^2 (End)
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filter:= proc(n) local L,X,S,t,s,x,b;
b:= 2^(n/2);
L:= sort(convert(2^n,base,10));
S:= map(t -> rhs(op(t)), [msolve(X^2=2^n,9)]);
for t from floor(10^((nops(L)-1)/2)/9) to floor(10^(nops(L)/2)/9) do
for s in S do
x:= 9*t+s;
if x = b then next fi;
if sort(convert(x^2,base,10))=L then return true fi;
od od;
false
end proc:
select(filter, [seq(i,i=2..58,2)]); # Robert Israel, Aug 21 2020
-
from math import isqrt
def ok(n, verbose=True):
s = str(2**n)
L, target, hi = len(s), sorted(s), int("".join(sorted(s, reverse=True)))
if '0' not in s: lo = int("".join(target))
else:
lownzd, targetcopy = min(set(s) - {'0'}), target[:]
targetcopy.remove(lownzd)
rest = "".join(targetcopy)
lo = int(lownzd + rest)
for r in range(isqrt(lo), isqrt(hi)+1):
rr = r*r
if sorted(str(rr)) == target:
brr = bin(rr)[2:]
if brr != '1' + '0'*(len(brr)-1):
if verbose: print(f"2^{n}, {r}^2")
return r
return 0
print(list(filter(ok, range(2, 73, 2)))) # Michael S. Branicky, Aug 10 2021
Showing 1-3 of 3 results.
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