A002942 -id:A002942 - cp's OEIS frontend

A350810 a(n) = ceiling((n-R(n^2))^2/(n+R(n^2))), where R(n^2) is the digit reversal of n^2.

0, 1, 3, 50, 39, 48, 75, 27, 3, 8, 92, 407, 923, 651, 479, 606, 933, 372, 114, 11, 92, 422, 859, 607, 456, 602, 850, 410, 81, 12, 96, 4106, 9703, 6410, 5117, 6814, 9521, 4329, 1139, 5, 1742, 4547, 9353, 6261, 5069, 5976, 8882, 3891, 904, 1, 919, 3919, 8925, 6032, 5041, 6147, 9254

Offset: 1

Claude H. R. Dequatre, Jan 17 2022

This sequence gives both at large and small scales well-structured graphs; specific and periodic patterns are visible in separated layers.

			For n = 1, R(n^2) = 1, thus a(1) = ceiling((1-1)^2/(1+1)) = 0.
For n = 10, R(n^2) = 1, thus a(10) = ceiling((10-1)^2/(10+1)) = 8.
For n = 21, R(n^2) = 144, thus a(21) = ceiling((21-144)^2/(21+144)) = 92.

Rémy Sigrist, Table of n, a(n) for n = 1..10000

Cf. A000290, A002942, A004086.

Table[Ceiling[(n-FromDigits[Reverse[IntegerDigits[n^2]]])^2/(n+FromDigits[Reverse[IntegerDigits[n^2]]])],{n,57}] (* Stefano Spezia, Jan 18 2022 *)

a(n) = my(x = fromdigits(Vecrev(digits(n^2))));r = ceil((n-x)^2/(n+x));
for(n = 1,2000,print1(a(n)", "))

def R(n): return int(str(n)[::-1])
def a(n):
    Rn2 = R(n**2)
    q, r = divmod((n-Rn2)**2, n+Rn2)
    return q if r == 0 else q + 1
print([a(n) for n in range(1, 67)]) # Michael S. Branicky, Jan 17 2022

A113800 Numbers k such that k^2 plus the reverse of k^2 gives a perfect power.

Original entry on oeis.org

2, 231, 1010, 102010, 451429, 1000100, 9426681, 1000001000, 8803095102, 10002000100, 56017891104, 1000000010000, 4811618419542

Offset: 1

Giovanni Resta, Jan 22 2006

Sequence is infinite since it contains all the numbers of the form (1+100^k)*10^k and (1+100^k)^2*10^k, for k >= 1. - Giovanni Resta, Sep 24 2013

			451429^2 = 203788142041 and 203788142041 + 140241887302 = 7007^3,
9426681^2 = 88862314675761 and 88862314675761 + 16757641326888 = 10277157^2.

Cf. A000290, A002942, A061226.

ppQ[n_] := n != 1 && GCD @@ (Transpose[FactorInteger[n]][[2]]) > 1; rev[n_] := FromDigits@Reverse@IntegerDigits@n ; lst = {}; Do[If[ppQ[n^2 + rev[n^2]], AppendTo[lst, n]], {n, 10^6}]; lst

a(8)-a(13) from Giovanni Resta, Sep 24 2013

A356417 Numbers whose reversal is a square.

Original entry on oeis.org

0, 1, 4, 9, 10, 18, 40, 46, 52, 61, 63, 90, 94, 100, 121, 144, 148, 163, 169, 180, 400, 423, 441, 460, 484, 487, 520, 522, 526, 610, 630, 652, 675, 676, 691, 900, 925, 927, 940, 961, 982, 1000, 1042, 1062, 1089, 1210, 1251, 1273, 1297, 1405, 1426, 1440, 1480

Offset: 1

Daniel Blam, Aug 06 2022

Every power of 10 is a term in the sequence.
If k is in the sequence then so is 10*k. - David A. Corneth, Aug 06 2022
If all multiples of 10 were omitted, A074896 would result. - Jon E. Schoenfield, Aug 06 2022

			0 is a term, 0 -> 0 (0^2).
10 is a term, 10 -> 01 (leading zero is dropped) (1^2).
691 is a term, 691 -> 196 (14^2).
1800 is a term, 1800 -> 0081 (leading zeros are dropped) (9^2).

Cf. A002942, A004086, A074896.

Select[Range[0, 1500], IntegerQ[Sqrt[IntegerReverse[#]]] &] (* Amiram Eldar, Aug 06 2022 *)

isok(k) = issquare(fromdigits(Vecrev(digits(k)))); \\ Michel Marcus, Aug 06 2022

from sympy import integer_nthroot
def ok(n):
    return integer_nthroot(int(str(n)[::-1]), 2)[1]
print([k for k in range(1500) if ok(k)])

from math import isqrt
from itertools import count, islice
def A356417_gen(): # generator of terms
    yield 0
    for l in count(1):
        nlist = []
        for m in range(1,isqrt(10**l)+1):
            if m%10:
                s = str(m**2)
                nlist.append(int(s[::-1])*10**(l-len(s)))
        yield from sorted(nlist)
A356417_list = list(islice(A356417_gen(),100)) # Chai Wah Wu, Aug 07 2022

A383027 a(n) is the number whose reversed n-digit square is the largest among all reversals of n-digit squares.

Original entry on oeis.org

3, 7, 17, 83, 167, 583, 1833, 8167, 10583, 41833, 191833, 489417, 1989417, 9958167, 25041833, 74958167, 125041833, 831989417, 1625041833, 9874958167, 15831989417, 65831989417, 309874958167, 834168010583, 1809874958167, 6809874958167, 29165831989417, 51809874958167, 179165831989417

Offset: 1

Zhining Yang, Apr 13 2025

			a(3)=17 because 982 (reverse of 17^2=289) is the largest reverse of all 3-digit squares.

Cf. A002942.

a[n_]:=Module[{s=Ceiling[Sqrt[10^(n-1)]]},max=IntegerReverse[s^2];
For[k=s,k<=Floor[Sqrt[(10^n-1)]],k++,t=IntegerReverse[k^2];
If[t>max,max=t;s=k]];s];Table[a[n],{n,10}]

cp's OEIS Frontend

A350810 a(n) = ceiling((n-R(n^2))^2/(n+R(n^2))), where R(n^2) is the digit reversal of n^2.

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A113800 Numbers k such that k^2 plus the reverse of k^2 gives a perfect power.

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A356417 Numbers whose reversal is a square.

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Python

A383027 a(n) is the number whose reversed n-digit square is the largest among all reversals of n-digit squares.

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Mathematica