A238712 -id:A238712 - cp's OEIS frontend

A246422 Numbers in which cubes may end (in base 10).

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 16, 17, 19, 21, 23, 24, 25, 27, 28, 29, 31, 32, 33, 36, 37, 39, 41, 43, 44, 47, 48, 49, 51, 52, 53, 56, 57, 59, 61, 63, 64, 67, 68, 69, 71, 72, 73, 75, 76, 77, 79, 81, 83, 84, 87, 88, 89, 91, 92, 93, 96, 97, 99, 101, 103, 104, 107, 109, 111, 112, 113, 117

Offset: 1

Derek Orr, Aug 25 2014

			33 is a member because 77^3 = 456533 is a cube ending in 33.

Ivan Neretin, Table of n, a(n) for n = 1..5065

Cf. A238712, A246449.

Union@Flatten@Table[Mod[i^3, 10^n], {n, 3}, {i, 10^n}] (* Ivan Neretin, Aug 30 2015*)

v=[];for(k=1,10^3,for(m=1,3, v=concat(v,k^3%10^m)));v=vecsort(v,,8)

a=[]; for(m=1, 3, a=setunion(a, Set(vector(10^m, n, n^3)%10^m))); a

from itertools import count, islice
from sympy import nthroot_mod
def A246422_gen(startvalue=0): # generator of terms >= startvalue
    return filter(lambda n:len(nthroot_mod(n,3,10**(len(str(n))))),count(max(startvalue,0)))
A246422_list = list(islice(A246422_gen(),20)) # Chai Wah Wu, Feb 16 2023

Corrected by Ivan Neretin, Mar 03 2016

A316347 a(n) = n^2 mod(10^m), where m is the number of digits in n (written in base 10).

Original entry on oeis.org

0, 1, 4, 9, 6, 5, 6, 9, 4, 1, 0, 21, 44, 69, 96, 25, 56, 89, 24, 61, 0, 41, 84, 29, 76, 25, 76, 29, 84, 41, 0, 61, 24, 89, 56, 25, 96, 69, 44, 21, 0, 81, 64, 49, 36, 25, 16, 9, 4, 1, 0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 0, 21, 44, 69, 96, 25, 56, 89, 24, 61, 0, 41, 84, 29

Offset: 0

Christopher D Chamness, Jun 29 2018

The set of the terms is the same as that of A238712.

			n = 13 has 2 digits in base 10, thus a(13) = 169 mod 100 = 69.

Georg Fischer, Table of n, a(n) for n = 0..10000, Jan 16 2019 (terms a(0..719585) initially submitted by Christopher D Chamness).

Cf. A238712.

a(n) = n^2 % 10 ^ #digits(n) \\ David A. Corneth, Jun 30 2018

my $mod = 10;
foreach my $i(0..10000) {
     print "$i " . (($i * $i) % $mod) . "\n";
     if (length($i + 1) > length($i)) { $mod *= 10; }
} # Georg Fischer, Jan 16 2019

i=0
while True:
     m=i
     j=i**2
     l=0
     while True:
          m=m//10
          l+=1
          if m==0:
               break
     mod_num = 10**l
     print(j%mod_num)
     i+=1

A246448 Numbers n such that a square will never end in the digits of n.

Original entry on oeis.org

2, 3, 7, 8, 10, 11, 12, 13, 14, 15, 17, 18, 19, 20, 22, 23, 26, 27, 28, 30, 31, 32, 33, 34, 35, 37, 38, 39, 40, 42, 43, 45, 46, 47, 48, 50, 51, 52, 53, 54, 55, 57, 58, 59, 60, 62, 63, 65, 66, 67, 68, 70, 71, 72, 73, 74, 75, 77, 78, 79, 80, 82, 83, 85, 86, 87, 88, 90, 91, 92, 93, 94, 95

Offset: 1

Derek Orr, Aug 26 2014

Complement of A238712.

			No square will ever end in a 2, thus 2 is a member of this sequence.

Cf. A238712.

b(n)=v=[];for(k=10^(n-1),10^n,v=concat(v,k^2%10^n));v=vecsort(v,,8);v
for(n=1,500,d=digits(n);if(!vecsearch(b(#d),n),print1(n,", ")))

A380428 Numbers k for which nonnegative integers x and y exist such that k is the concatenation of x and y as well as k = (x + y)^2.

Original entry on oeis.org

81, 100, 2025, 3025, 88209, 494209, 4941729, 7441984, 24502500, 25502500, 52881984, 60481729, 300814336, 493817284, 6049417284, 6832014336, 20408122449, 21948126201, 33058148761, 35010152100, 43470165025, 101558217124, 108878221089, 123448227904, 127194229449, 152344237969

Offset: 1

Felix Huber, Jan 25 2025

Subsequence of A000290.
From David A. Corneth, Apr 26 2025: (Start)
If y has q digits then a term m is of the form (x + y) = 10^q * x + y. Choosing some y we can solve for x (the equation is a quadratic with respect to x) and see if it produces a term.
y comes from A238712.
The sequence is infinite; it contains (25*100^i +- 5*10^i)^2 = concat(25*100^i +- 5*10^i, 25*100^i) for all i >= 0.
Neither x nor y can have a leading 0. (End)

			2025 is in the sequence because (20 + 25)^2 = 2025.
100 is in the sequence because (10 + 0)^2 = 100.
88209 is in the sequence because (88 + 209)^2 = 88209.
From _David A. Corneth_, Apr 26 2025: (Start)
9801 is not in the sequence even though (98 + 01)^2 = 9801 but 01 has a leading 0 which is disallowed.
If a term m ends in y = 209 where y has three digits we have 10^3*x + y = (x + y)^2. Solving for x gives x = 88 or x = 494 corresponding to terms 88209 and 494209. (End)

David A. Corneth, Table of n, a(n) for n = 1..96 (terms <= 10^20)

Cf. A000290, A238712.

Cf. A115527, A115528, A115529, A115530, A115531, A115532, A115533, A115534, A115535, A115536, A115537, A115538, A115539, A115540, A115541, A115542, A115543, A115544, A115545, A115546, A115547, A115548, A115549, A115550, A115551, A115552, A115553, A115554, A115555, A115556.

A380428:=proc(n)
    option remember;
    local a,i,k,x,y;
    if n=1 then
        81
    elif n=2 then
        100
    else
        for a from isqrt(procname(n-1))+1 do
            k:=length(a^2);
            for i to k-1 do
                x:=floor(a^2/10^i);
                y:=a^2-x*10^i;
                if x+y=a and length(x)+length(y)=k then
                    return a^2
                fi
            od
        od
    fi;
end proc;
seq(A380428(n),n=1..26);

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A246422 Numbers in which cubes may end (in base 10).

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A316347 a(n) = n^2 mod(10^m), where m is the number of digits in n (written in base 10).

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A246448 Numbers n such that a square will never end in the digits of n.

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A380428 Numbers k for which nonnegative integers x and y exist such that k is the concatenation of x and y as well as k = (x + y)^2.

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Maple