A224474 -id:A224474 - cp's OEIS frontend

A033819 Trimorphic numbers: n^3 ends with n. Also m-morphic numbers for all m > 5 such that m-1 is not divisible by 10 and m == 3 (mod 4).

0, 1, 4, 5, 6, 9, 24, 25, 49, 51, 75, 76, 99, 125, 249, 251, 375, 376, 499, 501, 624, 625, 749, 751, 875, 999, 1249, 3751, 4375, 4999, 5001, 5625, 6249, 8751, 9375, 9376, 9999, 18751, 31249, 40625, 49999, 50001, 59375, 68751, 81249, 90624, 90625

Offset: 1

David W. Wilson

n is in this sequence iff it occurs in one of A002283, A007185, A016090, A198971, A199685, A216092, A216093, A224473, A224474, A224475, A224476, A224477, and A224478. - Eric M. Schmidt, Apr 08 2013

Let q(n) = floor(a(n)^3 / 10^A055642(a(n))), where A055642(n) is the number of digits in the decimal expansion of n. As well, let na and nb denote the indices of the preceding and next terms that begin with a 9. Then (1/q(n)) * (a(n)^4 - a(n)^3 - a(n)^2 + a(n)) - 2*a(n)^2 + a(n) + q(n) + 1 = a(na+nb-n)^2 - a(na+nb-n) - q(na+nb-n). - Christopher Hohl, Apr 08 2019

			376^3 = 53157376 which ends with 376.

S. Premchaud, A class of numbers, Math. Student, 48 (1980), 293-300.

Eric M. Schmidt, Table of n, a(n) for n = 1..1000
Robert Dawson, On Some Sequences Related to Sums of Powers, J. Int. Seq., Vol. 21 (2018), Article 18.7.6.
Shyam Sunder Gupta, Elegance of Squares, Cubes, and Higher Powers, Exploring the Beauty of Fascinating Numbers, Springer (2025) Ch. 2, 29-81.
Eric Weisstein's World of Mathematics, Trimorphic Number
Index entries for sequences related to automorphic numbers

Cf. A074194, A215558 (cubes of the terms).

[n: n in [0..10^5] | Intseq(n^3)[1..#Intseq(n)] eq Intseq(n)]; // Bruno Berselli, Apr 04 2013

Do[x=Floor[N[Log[10, n], 25]]+1; If[Mod[n^3, 10^x] == n, Print[n]], {n, 1, 10000}]
Select[Range[100000],PowerMod[#,3,10^IntegerLength[#]]==#&](* Harvey P. Dale, Nov 04 2011 *)
Select[Range[0, 10^5], 10^IntegerExponent[#^3-#, 10]>#&] (* Jean-François Alcover, Apr 04 2013 *)

A318960 One of the two successive approximations up to 2^n for 2-adic integer sqrt(-7). This is the 1 (mod 4) case.

Original entry on oeis.org

1, 5, 5, 21, 53, 53, 181, 181, 181, 181, 181, 181, 181, 16565, 49333, 49333, 49333, 49333, 573621, 1622197, 1622197, 1622197, 10010805, 10010805, 10010805, 77119669, 211337397, 479772853, 479772853, 479772853, 2627256501, 6922223797, 15512158389, 15512158389

Offset: 2

Jianing Song, Sep 06 2018

nonn

a(n) is the unique number k in [1, 2^n] and congruent to 1 (mod 4) such that k^2 + 7 is divisible by 2^(n+1).

The 2-adic integers are very different from p-adic ones where p is an odd prime. For example, provided that there is at least one solution, the number of solutions to x^n = a over p-adic integers is gcd(n, p-1) for odd primes p and gcd(n, 2) for p = 2. For odd primes p, x^2 = a is solvable iff a is a quadratic residue modulo p, while for p = 2 it's solvable iff a == 1 (mod 8). If gcd(n, p-1) > 1 and gcd(a, p) = 1, then the solutions to x^n = a differ starting at the rightmost digit for odd primes p, while for p = 2 they differ starting at the next-to-rightmost digit. As a result, the formulas and the program here are different from those in other entries related to p-adic integers.

			The unique number k in [1, 4] and congruent to 1 modulo 4 such that k^2 + 7 is divisible by 8 is 1, so a(2) = 1.
a(2)^2 + 7 = 8 which is not divisible by 16, so a(3) = a(2) + 2^2 = 5.
a(3)^2 + 7 = 32 which is divisible by 32, so a(4) = a(3) = 5.
a(4)^2 + 7 = 32 which is divisible by 64, so a(5) = a(4) + 2^4 = 21.
a(5)^2 + 7 = 448 which is divisible by 128, so a(6) = a(5) + 2^5 = 53.
...

Jianing Song, Table of n, a(n) for n = 2..999 (offset corrected by Jianing Song)
G. P. Michon, Introduction to p-adic integers, Numericana.

Cf. A318962.
Expansions of p-adic integers:
this sequence, A318961 (2-adic, sqrt(-7));
A268924, A271222 (3-adic, sqrt(-2));
A268922, A269590 (5-adic, sqrt(-4));
A048898, A048899 (5-adic, sqrt(-1));
A290567 (5-adic, 2^(1/3));
A290568 (5-adic, 3^(1/3));
A290800, A290802 (7-adic, sqrt(-6));
A290806, A290809 (7-adic, sqrt(-5));
A290803, A290804 (7-adic, sqrt(-3));
A210852, A212153 (7-adic, (1+sqrt(-3))/2);
A290557, A290559 (7-adic, sqrt(2));
A286840, A286841 (13-adic, sqrt(-1));
A286877, A286878 (17-adic, sqrt(-1)).
Also expansions of 10-adic integers:
A007185, A010690 (nontrivial roots to x^2-x);
A216092, A216093, A224473, A224474 (nontrivial roots to x^3-x).

PARI
```
a(n) = truncate(-sqrt(-7+O(2^(n+1))))
```

a(2) = 1; for n >= 3, a(n) = a(n-1) if a(n-1)^2 + 7 is divisible by 2^(n+1), otherwise a(n-1) + 2^(n-1).
a(n) = 2^n - A318961(n).
a(n) = Sum_{i=0..n-1} A318962(i)*2^i.

Offset corrected by Jianing Song, Aug 28 2019

A318961 One of the two successive approximations up to 2^n for 2-adic integer sqrt(-7). This is the 3 (mod 4) case.

Original entry on oeis.org

3, 3, 11, 11, 11, 75, 75, 331, 843, 1867, 3915, 8011, 16203, 16203, 16203, 81739, 212811, 474955, 474955, 474955, 2572107, 6766411, 6766411, 23543627, 57098059, 57098059, 57098059, 57098059, 593968971, 1667710795, 1667710795, 1667710795, 1667710795, 18847579979

Offset: 2

Jianing Song, Sep 06 2018

nonn

a(n) is the unique number k in [1, 2^n] and congruent to 3 (mod 4) such that k^2 + 7 is divisible by 2^(n+1).

The 2-adic integers are very different from p-adic ones where p is an odd prime. For example, provided that there is at least one solution, the number of solutions to x^n = a over p-adic integers is gcd(n, p-1) for odd primes p and gcd(n, 2) for p = 2. For odd primes p, x^2 = a is solvable iff a is a quadratic residue modulo p, while for p = 2 it's solvable iff a == 1 (mod 8). If gcd(n, p-1) > 1 and gcd(a, p) = 1, then the solutions to x^n = a differ starting at the rightmost digit for odd primes p, while for p = 2 they differ starting at the next-to-rightmost digit. As a result, the formulas and the program here are different from those in other entries related to p-adic integers.

			The unique number k in [1, 4] and congruent to 3 modulo 4 such that k^2 + 7 is divisible by 8 is 3, so a(2) = 3.
a(2)^2 + 7 = 16 which is divisible by 16, so a(3) = a(2) = 3.
a(3)^2 + 7 = 16 which is not divisible by 32, so a(4) = a(3) + 2^3 = 11.
a(4)^2 + 7 = 128 which is divisible by 64, so a(5) = a(4) = 11.
a(5)^2 + 7 = 128 which is divisible by 128, so a(6) = a(5) = 11.
...

Jianing Song, Table of n, a(n) for n = 2..999 (offset corrected by Jianing Song)
G. P. Michon, Introduction to p-adic integers, Numericana.

Cf. A318963.
Expansions of p-adic integers:
A318960, this sequence (2-adic, sqrt(-7));
A268924, A271222 (3-adic, sqrt(-2));
A268922, A269590 (5-adic, sqrt(-4));
A048898, A048899 (5-adic, sqrt(-1));
A290567 (5-adic, 2^(1/3));
A290568 (5-adic, 3^(1/3));
A290800, A290802 (7-adic, sqrt(-6));
A290806, A290809 (7-adic, sqrt(-5));
A290803, A290804 (7-adic, sqrt(-3));
A210852, A212153 (7-adic, (1+sqrt(-3))/2);
A290557, A290559 (7-adic, sqrt(2));
A286840, A286841 (13-adic, sqrt(-1));
A286877, A286878 (17-adic, sqrt(-1)).
Also expansions of 10-adic integers:
A007185, A010690 (nontrivial roots to x^2-x);
A216092, A216093, A224473, A224474 (nontrivial roots to x^3-x).

a(n) = if(n==2, 3, truncate(sqrt(-7+O(2^(n+1)))))

a(2) = 3; for n >= 3, a(n) = a(n-1) if a(n-1)^2 + 7 is divisible by 2^(n+1), otherwise a(n-1) + 2^(n-1).
a(n) = 2^n - A318960(n).
a(n) = Sum_{i=0..n-1} A318963(i)*2^i.

Offset corrected by Jianing Song, Aug 28 2019

A224476 (2*16^(5^n) + (10^n)/2 - 1) mod 10^n: a sequence of trimorphic numbers ending (for n > 1) in 1.

Original entry on oeis.org

6, 1, 251, 3751, 68751, 718751, 9218751, 24218751, 74218751, 8574218751, 13574218751, 663574218751, 5163574218751, 30163574218751, 980163574218751, 2480163574218751, 37480163574218751, 987480163574218751, 487480163574218751, 65487480163574218751

Offset: 1

Eric M. Schmidt, Apr 07 2013

a(n) is the unique positive integer less than 10^n such that a(n) + 2^(n-1) + 1 is divisible by 2^n and a(n) - 1 is divisible by 5^n.

Eric M. Schmidt, Table of n, a(n) for n = 1..1000
Eric Weisstein's World of Mathematics, Trimorphic Number
Index entries for sequences related to automorphic numbers

Cf. A033819. Converges to the 10-adic number A063006. The other trimorphic numbers ending in 1 are included in A199685 and A224474.

def A224476(n) : return crt(2^(n-1)-1, 1, 2^n, 5^n)

a(n) = (A224474(n) + 10^n/2) mod 10^n.

A181539 Smallest number m > 1 such that m^2 == 1 (mod 10^n).

Original entry on oeis.org

9, 49, 249, 1249, 18751, 218751, 781249, 24218751, 74218751, 1425781249, 13574218751, 163574218751, 163574218751, 19836425781249, 19836425781249, 2480163574218751, 12519836425781249, 12519836425781249, 487480163574218751, 15487480163574218751, 215487480163574218751, 215487480163574218751

Offset: 1

Kevin Batista (kevin762401(AT)yahoo.com), Oct 29 2010

nonn

a(n) > 10^floor(n/2).
All terms have last digit 1 or 9.
Squares of terms are listed in A085877.
Decimal representation of each term is formed by the reverse concatenation of initial terms of either A063006 or A091661.
Except for 3, there are no solutions for n>1 and m^2 == -1 (mod 10^n). See comment in A063006 under extensions. - Robert G. Wilson v, Jan 26 2013
If a(n)<(10^n)/2 then (10^n-a(n))^2 is also congruent (modulo 10^n), it is just not the least. - Robert G. Wilson v, Jan 26 2013

			1249^2 = 1560001 == 1 (mod 10^4), and there is no smaller m > 1 such that m^2 == 1 (mod 10^4). Hence a(4) = 1249.

Max Alekseyev, Table of n, a(n) for n = 1..1000

Cf. A085610, A181607.
Cf. A063006, A091661 (the two nontrivial 10-adic square roots of 1).
Cf. A224473, A224474 (approximation of the two nontrivial 10-adic square roots of 1 up to powers of 10).

install(Zn_quad_roots, GGG);
a181539(n) = vecsort(Zn_quad_roots(10^n,0,-1)[2])[2]; \\ Max Alekseyev, Oct 13 2024

Let b(n) = A224474(n) (or equivalently b(n) = A224473(n)), then for n >= 3, there are eight solutions in [0,10^n) to x^2 == 1 (mod 10^n), namely x = 1, 5*10^(n-1) - 1, 5*10^(n-1) + 1, 10^n - 1, b(n), 10^n - b(n), |b(n) - 5*10^(n-1)|, and 10^n - |b(n) - 5*10^(n-1)|, so a(n) = min{b(n), |b(n) - 5*10^(n-1)|, 10^n - b(n)} < 25*10^(n-2). - Jianing Song, Sep 23 2024

a(2) through a(4), a(7) through a(11) corrected, comment added, example replaced by Klaus Brockhaus, Nov 01 2010
Edited by N. J. A. Sloane, Oct 29 2010, Nov 09 2010
Definition to avoid the constant sequence a(n)=1 constrained by R. J. Mathar, Nov 18 2010
a(1) corrected, terms a(13) onward added by Max Alekseyev, Dec 10 2012

A181607 Numbers n with k digits such that n^2 == 1 (mod 10^k).

Original entry on oeis.org

1, 9, 49, 51, 99, 249, 251, 499, 501, 749, 751, 999, 1249, 3751, 4999, 5001, 6249, 8751, 9999, 18751, 31249, 49999, 50001, 68751, 81249, 99999, 218751, 281249, 499999, 500001, 718751, 781249, 999999, 4218751, 4999999, 5000001, 5781249, 9218751

Offset: 1

Robert G. Wilson v, Nov 01 2010

Least term of n digits: 1, 49, 249, 1249, 18751, 218751, 4218751, ..., .
If n of k digits is present then 10^k-n is present.
The union of A002283, A198971, A199685, A224473, A224474, A224475, and A224476 (except that this sequence omits 0, 4, and 6). - Eric M. Schmidt, Jan 26 2016

Eric M. Schmidt, Table of n, a(n) for n = 1..1000

Cf. A007185, A016090, A033819.

Table[ Select[ Range[10^(k - 1), 10^k - 1], Mod[ #^2, 10^k] == 1 &], {k, 7}] // Flatten

cp's OEIS Frontend

A033819 Trimorphic numbers: n^3 ends with n. Also m-morphic numbers for all m > 5 such that m-1 is not divisible by 10 and m == 3 (mod 4).

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A318960 One of the two successive approximations up to 2^n for 2-adic integer sqrt(-7). This is the 1 (mod 4) case.

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A318961 One of the two successive approximations up to 2^n for 2-adic integer sqrt(-7). This is the 3 (mod 4) case.

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A224476 (2*16^(5^n) + (10^n)/2 - 1) mod 10^n: a sequence of trimorphic numbers ending (for n > 1) in 1.

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A181539 Smallest number m > 1 such that m^2 == 1 (mod 10^n).

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Extensions

A181607 Numbers n with k digits such that n^2 == 1 (mod 10^k).

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Mathematica