cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-4 of 4 results.

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).

Original entry on oeis.org

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

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Author

David W. Wilson

Keywords

Comments

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

Examples

			376^3 = 53157376 which ends with 376.

References

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

Links

Crossrefs

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

Programs

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

A216093 a(n) = 10^n - (5^(2^n) mod 10^n).

Original entry on oeis.org

5, 75, 375, 9375, 9375, 109375, 7109375, 87109375, 787109375, 1787109375, 81787109375, 81787109375, 81787109375, 40081787109375, 740081787109375, 3740081787109375, 43740081787109375, 743740081787109375
Offset: 1

Views

Author

V. Raman, Sep 01 2012

Keywords

Comments

a(n)^3 mod 10^n = a(n).
a(n) is the unique positive integer less than 10^n such that a(n) is divisible by 5^n and a(n) + 1 is divisible by 2^n. - Eric M. Schmidt, Sep 01 2012
a(n+1) + a(n)^2 == 0 (mod 10^(n+1)). - Robert Israel, Apr 24 2017

Links

Crossrefs

Cf. A007185, A016090, A216092, A091663, A018248.

Programs

Formula

2^(4*5^(n-1)) mod 10^n - 1.

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

Views

Author

Jianing Song, Sep 06 2018

Keywords

Comments

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.

Examples

			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.
...

Links

Crossrefs

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).

Programs

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

Formula

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.

Extensions

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

Views

Author

Jianing Song, Sep 06 2018

Keywords

Comments

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.

Examples

			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.
...

Links

Crossrefs

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).

Programs

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

Formula

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.

Extensions

Offset corrected by Jianing Song, Aug 28 2019
Showing 1-4 of 4 results.

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