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.

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A177916 Numbers k such that k^3 divides 16^(k^2) - 1.

Original entry on oeis.org

1, 3, 5, 15, 21, 39, 55, 57, 105, 111, 155, 165, 195, 205, 219, 273, 285, 327, 399, 465, 505, 555, 609, 615, 741, 777, 903, 915, 1095, 1155, 1255, 1265, 1365, 1443, 1515, 1533, 1635, 1705, 1995, 2067, 2109, 2145, 2255, 2265, 2289, 2373, 2667, 2715, 2847
Offset: 1

Views

Author

Alexander Adamchuk, May 14 2010

Keywords

Comments

From Robert Israel, Apr 24 2015: (Start)
The only primes in the sequence are 3 and 5.
If m and n are in the sequence and are coprime, then m*n is in the sequence.
Are all members of the sequence divisible by 3 or 5? (End)

Crossrefs

Programs

  • Magma
    [n: n in [1..2000] | Denominator((16^(n^2) - 1)/n^3) eq 1]; // Vincenzo Librandi, Apr 24 2015
  • Maple
    A177916:=n->`if`((16&^(n^2)-1) mod n^3 = 0, n, NULL): seq(A177916(n), n=1..1000); # Wesley Ivan Hurt, Apr 23 2015
  • Mathematica
    Select[Range[3000], IntegerQ[(16^(#^2) - 1) / #^3 ] &] (* Vincenzo Librandi, Apr 24 2015 *)
    Join[{1},Select[Range[3000],PowerMod[16,#^2,#^3]==1&]] (* Harvey P. Dale, Dec 05 2024 *)

A177920 Numbers k such that k^3 divides 20^(k^2) - 1.

Original entry on oeis.org

1, 19, 7805561, 1432001198261, 3661225986659, 58130944174609, 187470481770989
Offset: 1

Views

Author

Alexander Adamchuk, May 14 2010

Keywords

Comments

19 divides a(n) for n > 1.

Crossrefs

Programs

  • Mathematica
    Join[{1}, Select[Range[10000000], PowerMod[20, #^2, #^3] == 1 &]] (* Robert Price, Apr 04 2020 *)

Extensions

a(4)-a(7) from Max Alekseyev, Oct 02 2010

A128395 Numbers k such that k^2 divides 15^k-1.

Original entry on oeis.org

1, 2, 4, 7, 8, 14, 16, 28, 56, 112, 136, 272, 452, 812, 904, 952, 1624, 1808, 1904, 3164, 3248, 6328, 11912, 12656, 15368, 18632, 23824, 27608, 30736, 37264, 47908, 55216, 60248, 83384, 91756, 95816, 102604, 107576, 113936, 120496, 130424, 166768
Offset: 1

Views

Author

Alexander Adamchuk, Mar 01 2007

Keywords

Crossrefs

Programs

  • Mathematica
    a={1};For[n=1,n<200000,n++,If[PowerMod[15,n,n^2]==1, AppendTo[a, n]]]; a (* Stefan Steinerberger, Jun 10 2007 *)
    Join[{1},Select[Range[167000],PowerMod[15,#,#^2]==1&]] (* Harvey P. Dale, Sep 14 2020 *)
  • PARI
    is(k) = Mod(15, k^2)^k == 1; \\ Amiram Eldar, May 21 2024

Extensions

More terms from Stefan Steinerberger, Jun 10 2007

A128402 Numbers k such that k^2 divides 22^k-1.

Original entry on oeis.org

1, 3, 7, 21, 39, 273, 507, 3081, 3549, 21567, 40053, 78117, 280371, 343239, 546819, 1015521, 2056899, 2402673, 5998317, 6171243, 7108647, 8740173, 12338859, 14398293, 18988203, 27115881, 41988219, 43198701, 47727771, 55431363
Offset: 1

Views

Author

Alexander Adamchuk, Mar 01 2007

Keywords

Crossrefs

Programs

  • Maple
    select(t -> 22 &^ t - 1 mod t^2 = 0, [seq(2*k+1,k=0..10^6)]); # Robert Israel, Jan 23 2015
  • Mathematica
    a={}; Do[r=(22^n-1)/n^2; If[r==IntegerPart[r], AppendTo[a, n]], {n, 1, 10^3}]; a (* Vladimir Joseph Stephan Orlovsky, Aug 07 2008 *)
  • PARI
    { forstep(m=11,10^8,2, if( Mod(22,m^2)^m==1, print(m) ) ) } \\ Max Alekseyev, Oct 18 2008

Extensions

a(14)-a(30) from Max Alekseyev, Oct 18 2008

A128404 Numbers k such that k^2 divides 24^k-1.

Original entry on oeis.org

1, 23, 1081, 2870377, 7009273, 15954479, 134907719, 329435831, 537539141, 15001987199, 874750261127, 1991103024721, 4272172921319, 4862143429729, 7933540182019, 12816504745411, 41113262272969, 67084347257659
Offset: 1

Views

Author

Alexander Adamchuk, Mar 01 2007

Keywords

Comments

23 divides all terms except the first.

Crossrefs

Programs

Extensions

a(5)-a(6) from Farideh Firoozbakht, Mar 05 2007
a(7)-a(10) from Ryan Propper, Feb 23 2008
Terms a(11) onward from Max Alekseyev, May 06 2010

A177905 Numbers k such that k^3 divides 5^(k^2) - 1.

Original entry on oeis.org

1, 2, 4, 6, 12, 26, 42, 52, 68, 78, 84, 114, 156, 186, 204, 222, 228, 372, 444, 546, 798, 876, 884, 1092, 1218, 1252, 1302, 1378, 1428, 1482, 1554, 1596, 1806, 2418, 2436, 2604, 2652, 2756, 2886, 2964, 3108, 3534, 3606, 3612, 3756, 3876, 4134, 4218, 4836
Offset: 1

Views

Author

Alexander Adamchuk, May 14 2010

Keywords

Crossrefs

Programs

A177907 Numbers k such that k^3 divides 7^(k^2) - 1.

Original entry on oeis.org

1, 2, 3, 4, 6, 8, 10, 12, 20, 24, 30, 40, 50, 57, 60, 68, 78, 100, 110, 111, 114, 120, 136, 150, 156, 200, 204, 220, 222, 228, 258, 300, 312, 330, 340, 390, 408, 440, 444, 456, 516, 550, 570, 600, 660, 680, 780, 820, 876, 888, 930, 1010, 1020, 1032, 1086, 1100
Offset: 1

Views

Author

Alexander Adamchuk, May 14 2010

Keywords

Crossrefs

Programs

  • Mathematica
    Select[Range[1100],Divisible[7^(#^2)-1,#^3]&] (* Harvey P. Dale, Nov 05 2011 *)

A177909 Numbers k such that k^3 divides 9^(k^2) - 1.

Original entry on oeis.org

1, 2, 4, 8, 10, 20, 40, 68, 82, 110, 136, 164, 220, 328, 340, 410, 440, 610, 680, 772, 820, 1010, 1210, 1220, 1510, 1544, 1640, 2020, 2420, 2440, 2530, 2788, 3020, 3740, 3860, 4040, 4510, 4840, 5060, 5576, 6040, 6710, 6806, 7004, 7370, 7480, 7720, 8020, 9020
Offset: 1

Views

Author

Alexander Adamchuk, May 14 2010

Keywords

Examples

			9^(2^2) - 1 = 6560, which is divisible by 2^3, so 2 is in the sequence.
9^(4^2) - 1 = 1853020188851840, which is divisible by 4^3, so 4 is in the sequence.
9^(6^2) - 1 = 22528399544939174411840147874772640, which is not divisible by 6, and certainly not by 6^3, so 6 is not in the sequence.
		

Crossrefs

Programs

A177911 Numbers k such that k^3 divides 11^(k^2) - 1.

Original entry on oeis.org

1, 2, 4, 5, 6, 10, 12, 20, 30, 42, 60, 68, 78, 84, 114, 122, 156, 204, 210, 222, 228, 244, 340, 366, 390, 420, 444, 546, 570, 610, 732, 780, 798, 820, 876, 930, 1010, 1020, 1092, 1110, 1140, 1164, 1218, 1220, 1428, 1482, 1554, 1596, 1806, 1830, 1860, 2020, 2220
Offset: 1

Views

Author

Alexander Adamchuk, May 14 2010

Keywords

Crossrefs

Programs

  • Mathematica
    Join[{1},Select[Range[2300],PowerMod[11,#^2,#^3]==1&]] (* Harvey P. Dale, Aug 24 2024 *)

A177912 Numbers k such that k^3 divides 12^(k^2) - 1.

Original entry on oeis.org

1, 11, 253, 11891, 77209, 768361, 1775807, 17666737, 36112967, 61488361, 83462929, 154261943, 173185331, 591303757, 830336639, 971656873, 2936791979, 4139054953, 5393125859, 8139710557, 8142849781, 45667873031, 53653880269
Offset: 1

Views

Author

Alexander Adamchuk, May 14 2010

Keywords

Comments

11 divides a(n) for n > 1.

Crossrefs

Programs

  • Mathematica
    Join[{1}, Select[Range[3000000], PowerMod[12, #^2, #^3] == 1 &]] (* Robert Price, Mar 31 2020 *)

Extensions

More terms from Max Alekseyev, Oct 02 2010
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