A102859 -id:A102859 - cp's OEIS frontend

A104379 a(n) = square root of digital reversal of A102859(n)^2.

0, 1, 2, 3, 1, 11, 21, 31, 2, 12, 22, 26, 3, 13, 99, 33, 1, 101, 201, 301, 11, 111, 211, 311, 21, 121, 221, 31, 2, 102, 202, 12, 112, 212, 22, 122, 26, 264, 3, 103, 307, 13, 113, 99, 836, 33, 1, 1001, 2001, 3001, 101, 1101, 2101, 3101, 201, 1201, 2201, 301, 1301, 11, 1011

Offset: 1

Zak Seidov, Mar 03 2005

The sequence is inverse to A102859.

Index entry for sequences related to reversing digits of squares

Cf. A102859.

r[n_] := FromDigits@ Reverse@ IntegerDigits@ n; Sqrt@ r[#^2] & /@ Select[Range[0, 1101], IntegerQ@ Sqrt@ r[#^2] &] (* Michael De Vlieger, Sep 21 2015 *)

0 inserted, and name replaced with formula, by Jon E. Schoenfield, Sep 20 2015

A061457 Squares whose reversal is also a square.

Original entry on oeis.org

0, 1, 4, 9, 100, 121, 144, 169, 400, 441, 484, 676, 900, 961, 1089, 9801, 10000, 10201, 10404, 10609, 12100, 12321, 12544, 12769, 14400, 14641, 14884, 16900, 40000, 40401, 40804, 44100, 44521, 44944, 48400, 48841, 67600, 69696, 90000, 90601

Offset: 1

Amarnath Murthy, May 03 2001

The corresponding square roots are in A102859.

			169 and 961 are both squares.
1089 = 33^2 and 9801 = 99^2 so 1089 and 9801 belong to the sequence.

Jon E. Schoenfield, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Harry J. Smith)
Index entry for sequences related to reversing digits of squares

Cf. A102859 (square roots), A033294 (exclude final digit 0).

[n^2: n in [0..306] | IsSquare(Seqint(Reverse(Intseq(n^2))))]; // Bruno Berselli, Apr 30 2011

Select[Range[0,400]^2,IntegerQ[Sqrt[IntegerReverse[#]]]&] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Jul 29 2019 *)

{ for(m=0, 1000, my(r=fromdigits(Vecrev(digits(m^2)))); if(issquare(r), print1(m^2, ", ") )) } \\ Harry J. Smith, Jul 23 2009

from itertools import count, islice
from sympy import integer_nthroot
def A061457_gen(): # generator of terms
    return filter(lambda n:integer_nthroot(int(str(n)[::-1]),2)[1], (n**2 for n in count(0)))
A061457_list = list(islice(A061457_gen(),30)) # Chai Wah Wu, Nov 18 2022

a(n) = A102859(n)^2.

More terms from Larry Reeves (larryr(AT)acm.org), May 17 2001

A284986 Integers k such that k^2 is not a palindrome and the square root of the reverse of k^2 yields an integer with the same number of digits as k that is not the reverse of k.

Original entry on oeis.org

33, 99, 3168, 6501, 20508, 21468, 110922, 219111, 303577, 304877, 1100922, 1109211, 1110922, 1119111, 2191011, 2191111, 3080367, 3140793, 10110922, 11009111, 11009122, 11009221, 11019111, 11091011, 11091022, 11091111, 11091121, 11091122, 11091202, 11091211

Offset: 1

Geoffrey Marnell, Apr 06 2017

Computer-based iteration shows that there are very few integers such that the square root of the reverse of the square of the integer yields an integer with the same number of digits. Between 1 and 100,000 there are just 168. Curiously, 153 of these 168 yield an integer that is the mirror (or reverse) of the starting number. Two examples: 21^2 = 441 and sqrt(144) = 12; 22^2 = 484 and sqrt(484) = 22. Intuition suggests that this mirroring is more likely if the square is a palindrome (as in the second example). But this is not the case. (The square is not a palindrome in 132 of the 153 cases.) The opposite appears to be the true for those integers that do not yield an equi-digit mirror: more cases occur where the square is a palindrome (9) than when it is not (6). The sequence is of those integers that generate neither a palindrome on squaring nor a mirror when the square root is taken of the reverse of the square (up to 10^6).

G. Marnell, Mathematical Doodlings: Curiosities, Conjectures and Challenges, Burdock Books, 2017, pages 96-102.

Giovanni Resta, Table of n, a(n) for n = 1..9794 (terms < 3*10^13)
Index entry for sequences related to reversing digits of squares

Subsequence of A102859.

rev[n_] := FromDigits@ Reverse@ IntegerDigits@n; Select[Range[10^5], (r = rev[#^2]) != #^2 && IntegerQ[q = Sqrt[r]] && q != rev[#] && Equal @@ IntegerLength[{q, #}] &] (* Giovanni Resta, Apr 15 2017 *)

a(11)-a(30) from Giovanni Resta, Apr 15 2017

A359347 Roots of reversible pandigital square numbers.

Original entry on oeis.org

1111103, 3011111, 11110922, 11111003, 11111030, 11111120, 11111210, 11112110, 11211110, 12111110, 21111110, 21911111, 30011111, 30111110, 101110922, 101111112, 101111121, 101111211, 102111111, 110109212, 110911211, 110921111, 111109220, 111110030, 111110103

Offset: 1

Martin Renner, Dec 27 2022

Reversible pandigital square numbers are perfect squares containing each digit from 0 to 9 at least once and still remain square numbers (not necessarily of the same length) when reversing the digits.

			1111103^2 = 1234549876609 <~> 9066789454321 = 3011111^2.
11110922^2 = 123452587690084  <~> 480096785254321 = 21911111^2.

Martin Renner, Table of n, a(n) for n = 1..458

Cf. A102859, A359346.

from math import isqrt
from itertools import count, islice
def c(n): return len(set(s:=str(n)))==10 and isqrt(r:=int(s[::-1]))**2==r
def agen(): yield from (k for k in count(10**6) if c(k*k))
print(list(islice(agen(), 25))) # Michael S. Branicky, Dec 27 2022

a(n) = sqrt(A359346(n)).

A129914 Irregular square reversible numbers. Numbers which when squared and written backwards give a square again, but don't satisfy reverse(n^2) = reverse(n)^2.

Original entry on oeis.org

26, 33, 99, 260, 264, 307, 330, 836, 990, 2285, 2600, 2636, 2640, 3070, 3168, 3300, 6501, 8360, 9900, 20508, 21468, 22850, 22865, 24846, 26000, 26360, 26400, 30693, 30700, 31680, 33000, 65010

Offset: 1

Robert J. Lemke Oliver (lemkeoliver(AT)gmail.com), Jun 05 2007

			33^2 = 1089 reversed is 9801 = 99^2.

Jon E. Schoenfield, Table of n, a(n) for n = 1..321 (all terms < 10^9).
Index entry for sequences related to reversing digits of squares

Cf. A102859, A002942.

A234472 Numbers that when raised to the fourth power and written backwards give squares.

Original entry on oeis.org

0, 1, 10, 11, 100, 101, 110, 1000, 1001, 1010, 1100, 10000, 10001, 10010, 10100, 11000, 100000, 100001, 100010, 100100, 101000, 110000, 1000000, 1000001, 1000010, 1000100, 1001000, 1010000, 1100000, 10000000, 10000001, 10000010, 10000100, 10001000, 10010000

Offset: 1

Colin Barker, Dec 26 2013

It seems that the numbers contain only the digits 0 and 1, and that the reversed fourth power and the square root of the reversed fourth power are both palindromes.

If the above comment is correct, and also if (as it appears) no more than two ones are among the digits of any term, this Mathematica program quickly generates the terms of the sequence: Flatten[Table[Select[ FromDigits/@Permutations[PadRight[PadRight[{},k,1],8,0]],IntegerQ[ Sqrt[ IntegerReverse[#^4]]]&],{k,0,2}]]//Sort - Harvey P. Dale, May 05 2020

			101 is in the sequence because 101^4 = 104060401 and 104060401 = 10201^2.
110 is in the sequence because 110^4 = 146410000 and 14641 = 121^2.

Cf. A102859, A043681.

[n: n in [0..10^7] | IsSquare(Seqint(Reverse(Intseq(n^4))))]; // Bruno Berselli, Dec 27 2013

Select[Range[0,10^7],IntegerQ[Sqrt[IntegerReverse[#^4]]]&] (* Harvey P. Dale, May 05 2020 *)

revint(n) = m=n%10; n\=10; while(n>0, m=m*10+n%10; n\=10); m
s=[]; for(i=0, 1000000, if(issquare(revint(i^4)), s=concat(s, i))); s

from itertools import count, islice
from sympy import integer_nthroot
def A234472_gen(startvalue=0): # generator of terms >= startvalue
    return filter(lambda n:integer_nthroot(int(str(n**4)[::-1]),2)[1], count(max(startvalue,0)))
A234472_list = list(islice(A234472_gen(),10)) # Chai Wah Wu, Nov 18 2022

A289140 Positive numbers k such that rev(k)^2 + rev(k^2) is a square, where rev(n) = A004086(n) is the digital reverse of n.

Original entry on oeis.org

998586, 3632658, 9985860, 36326580, 74471091, 99664458, 99858600, 363265800, 634826115, 743193501, 744710910, 756335085, 759317343, 996644580, 998586000, 3632658000, 6348261150, 7177621788, 7431935010, 7447109100, 7563350850, 7593173430, 9966445800

Offset: 1

Giovanni Resta, Jun 26 2017

Every term must be a multiple of 3.

			998586 is a term since rev(998586^2) + 685899^2 = 1079100^2.

Cf. A004086, A061230, A061231, A068536, A102859, A113798, A202386, A256515.

rev[n_] := FromDigits@ Reverse@ IntegerDigits@ n; Parallelize@ Select[3 Range[4 10^6], IntegerQ@ Sqrt[rev[#^2] + rev[#]^2] &]

isok(n) = issquare(fromdigits(Vecrev(digits(n)))^2 + fromdigits(Vecrev(digits(n^2)))); \\ Michel Marcus, Jun 29 2017

cp's OEIS Frontend

A104379 a(n) = square root of digital reversal of A102859(n)^2.

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A061457 Squares whose reversal is also a square.

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A284986 Integers k such that k^2 is not a palindrome and the square root of the reverse of k^2 yields an integer with the same number of digits as k that is not the reverse of k.

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A359347 Roots of reversible pandigital square numbers.

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A129914 Irregular square reversible numbers. Numbers which when squared and written backwards give a square again, but don't satisfy reverse(n^2) = reverse(n)^2.

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A234472 Numbers that when raised to the fourth power and written backwards give squares.

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Magma

Mathematica

PARI

Python

A289140 Positive numbers k such that rev(k)^2 + rev(k^2) is a square, where rev(n) = A004086(n) is the digital reverse of n.

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