A045878 -id:A045878 - cp's OEIS frontend

A045877 Rotating digits of a(n)^2 right once still yields a square.

1, 2, 3, 16, 21, 31, 129, 221, 247, 258, 1062, 1593, 1964, 2221, 13516, 17287, 18516, 19821, 22221, 28064, 29631, 103764, 182362, 222221, 273543, 1246713, 1509437, 1635219, 1856538, 2222221, 2253804, 2749249, 2784807, 11619096, 11949507

Offset: 1

Erich Friedman

Squares resulting in leading zeros excluded.

(2*10^k-11)/9 are terms, i.e. A165402 is a subsequence. - Chai Wah Wu, Apr 23 2022

			13516^2 = 18268225{6} -> {6}18268225 = 24865^2.

Chai Wah Wu, Table of n, a(n) for n = 1..1838
Eric Weisstein's World of Mathematics, Square Number

Cf. A000290, A045878, A035126, A035128, A165402.

from itertools import count, islice
from sympy.solvers.diophantine.diophantine import diop_DN
def A045877_gen(): # generator of terms
    for l in count(0):
        l1, l2 = 10**(l+1), 10**l
        yield from sorted(set(abs(x) for z in (diop_DN(10,m*(1-l1)) for m in range(10)) for x, y in z if l1 >= x**2 >= l2))
A045877_list = list(islice(A045877_gen(),30)) # Chai Wah Wu, Apr 23 2022

More terms from Patrick De Geest, Nov 15 1998

A035129 Rotating digits of a(n)^3 left once still yields a cube.

Original entry on oeis.org

1, 2, 8, 457

Offset: 1

Patrick De Geest, Nov 15 1998

Cubes resulting in leading zeros excluded.

			457^3 = 95443993 -> 54439939 = 379^3.

Eric Weisstein's World of Mathematics, Cubic Number

Cf. A045878, A035128.

A035127 Squares which when digits are rotated left once remain square.

Original entry on oeis.org

1, 4, 9, 144, 196, 625, 11664, 14884, 46656, 96100, 1493284, 4112784, 6385729, 9253764, 139287204, 149377284, 187799616, 618268225, 634284225, 678758809, 929884036, 14938217284, 43325589904, 61076696769, 97482577284

Offset: 1

Patrick De Geest, Nov 15 1998

Those resulting in leading zeros are excluded.

			2527^2 = 6385729 -> 3857296 = 1964^2.

Eric Weisstein's World of Mathematics, Square Number

Subsequence of A000290.

Cf. A045878, A035131.

okQ[n_]:=Module[{idn=IntegerDigits[n]},idn[[2]]!=0&&IntegerQ[Sqrt[ FromDigits[RotateLeft[idn]]]]]; Join[{1,4,9},Select[Range[4,320000]^2, okQ]] (* Harvey P. Dale, Apr 30 2011 *)

from itertools import count, islice
from sympy.solvers.diophantine.diophantine import diop_DN
def A035127_gen(): # generator of terms
    for l in count(0):
        l1, l2 = 10**(l+1), 10**l
        yield from sorted(set(y**2 for z in (diop_DN(10,m*(1-l1)) for m in range(10)) for x, y in z if l1 >= x**2 >= l2))
A035127_list = list(islice(A035127_gen(),20)) # Chai Wah Wu, Apr 23 2022

A245584 Let f(m) put the leftmost digit of the positive integer m at its end; a(n) is the sequence of all positive integers m with f^2(m)=f(m^2).

Original entry on oeis.org

1, 2, 3, 12, 122, 1222, 12222, 122222, 1222222, 12222222, 122222222

Offset: 1

Reiner Moewald, Jul 26 2014

			122^2=14884 and 221^2=48841.

Cf. A045878, A090843.

f[m_Integer] := Module[{w}, w := IntegerDigits[m]; FromDigits[Rest[AppendTo[w, First[w]]]]]; a245584[n_Integer] :=
Select[Range[n], If[f[#]^2 == f[#^2] && ! Mod[#, 10] == 0, True, False] &]; a245584[10^5] (* Michael De Vlieger, Aug 17 2014 *)

import math
max = 10000
print('los')
for n in range(1, max):
   nst = str(n*n)
   nnewst = nst[1:] + nst[0]
   d = int(nnewst)
   e = int(math.sqrt(d))
   est = str(e)
   enewst = est[len(est)-1] + est[:len(est)-1]
   if (e * e == d) and (nnewst[0] != "0") and (str(n) == enewst):
      print(n, '  ',  e)
print('End.')

One can easily prove that all integers of the form 12...2 are elements of the sequence.

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A045877 Rotating digits of a(n)^2 right once still yields a square.

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A035129 Rotating digits of a(n)^3 left once still yields a cube.

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A035127 Squares which when digits are rotated left once remain square.

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A245584 Let f(m) put the leftmost digit of the positive integer m at its end; a(n) is the sequence of all positive integers m with f^2(m)=f(m^2).

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