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.

User: Thomas M. Bridge

Thomas M. Bridge's wiki page.

Thomas M. Bridge has authored 9 sequences.

A231960 Powers of 3 together with multiples of 6.

Original entry on oeis.org

1, 3, 6, 9, 12, 18, 24, 27, 30, 36, 42, 48, 54, 60, 66, 72, 78, 81, 84, 90, 96, 102, 108, 114, 120, 126, 132, 138, 144, 150, 156, 162, 168, 174, 180, 186, 192, 198, 204, 210, 216, 222, 228, 234, 240, 243, 246, 252, 258
Offset: 1

Views

Author

Thomas M. Bridge, Nov 15 2013

Keywords

Comments

Union of A000244 and A008588 (without 0).
Also, 1 and 3*(even numbers (A005843) UNION powers of 3 (A008588)).
Also, numbers m such that m divides A057083(m-1), see the Smyth reference.

Links

Crossrefs

Cf. A029744.

Programs

  • Sage
    def is_in_A231960(n):
    return 6.divides(n) or n==3^valuation(n,3)

Extensions

Edited by Ralf Stephan, Feb 28 2014

A231959 Numbers n dividing the Lucas sequence u(n) defined by u(i) = 3*u(i-1) - u(i-2) with initial conditions u(0)=0, u(1)=1.

Original entry on oeis.org

1, 5, 6, 12, 18, 24, 25, 30, 36, 48, 54, 55, 60, 72, 84, 90, 96, 108, 120, 125, 144, 150, 162, 168, 180, 192, 216, 240, 252, 270, 275, 276, 288, 300, 306, 324, 330, 336, 342, 360, 384, 420, 432, 450, 480, 486, 504, 540, 552, 576, 588, 600, 605, 612, 625
Offset: 1

Views

Author

Thomas M. Bridge, Nov 15 2013

Keywords

Comments

All terms except 1 are divisible by either 5 or 6. The sequence contains every nonnegative integer power of 5. There are infinitely many multiples of 6 in the sequence and infinitely many consecutive integers in the sequence (for example, 5,6 or 24,25, or 54,55).

Links

Crossrefs

Cf. A000351 (powers of 5 (subsequence)).
Cf. A001906 (Lucas sequence).

Programs

  • Mathematica
    nn = 1000; s = LinearRecurrence[{3, -1}, {1, 3}, nn]; t = {}; Do[If[Mod[s[[n]], n] == 0, AppendTo[t, n]], {n, nn}]; t (* T. D. Noe, Nov 22 2013 *)

A231958 Numbers n dividing the Lucas sequence u(n) defined by u(i) = 2*u(i-1) - 5*u(i-2) with initial conditions u(0)=0, u(1)=1.

Original entry on oeis.org

1, 2, 4, 8, 12, 16, 24, 32, 36, 48, 56, 64, 72, 96, 108, 112, 128, 132, 144, 156, 168, 192, 216, 224, 256, 264, 272, 288, 312, 324, 336, 384, 392, 396, 432, 448, 468, 496, 504, 512, 528, 544, 552, 576, 624, 648, 672, 768, 784, 792, 816, 864, 896, 936, 972
Offset: 1

Views

Author

Thomas M. Bridge, Nov 15 2013

Keywords

Comments

All terms except 1 and 2 are divisible by 4. The sequence contains every nonnegative integer power of 2. There are infinitely many multiples of 12 in the sequence.

Links

Crossrefs

Cf. A000079 (powers of 2 (subsequence)).
Cf. A045873 (Lucas sequence).

Programs

  • Mathematica
    nn = 2000; s = LinearRecurrence[{2, -5}, {1, 2}, nn]; t = {}; Do[If[Mod[s[[n]], n] == 0, AppendTo[t, n]], {n, nn}]; t (* T. D. Noe, Nov 20 2013 *)

A231388 Numbers n dividing the Lucas sequence u(n), defined by u(i) = 2*u(i-1) - 3*u(i-2) with initial conditions u(0)=0, u(1)=1.

Original entry on oeis.org

1, 2, 4, 8, 16, 32, 56, 64, 112, 128, 224, 256, 272, 392, 448, 512, 544, 728, 784, 896, 992, 1024, 1088, 1456, 1568, 1792, 1904, 1984, 2048, 2176, 2408, 2744, 2912, 3136, 3584, 3808, 3968, 4096, 4352, 4624, 4816, 5096, 5488, 5824, 6176, 6272, 6944, 7168, 7616
Offset: 1

Views

Author

Thomas M. Bridge, Nov 08 2013

Keywords

Comments

Except for 1 and 2, all other terms are divisible by 4. This sequence contains every nonnegative power of 2.

Examples

			For n=0,...,5 we have u(n)=0,1,2,1,-4,-11. Clearly n=1,2,4 divide their respective u(n).

Links

Crossrefs

Cf. A000079 (powers of 2 (subsequence)).
Cf. A088137 (Lucas sequence).

Programs

  • Mathematica
    nn = 10000; s = LinearRecurrence[{2, -3}, {1, 2}, nn]; t = {}; Do[If[Mod[s[[n]], n] == 0, AppendTo[t, n]], {n, nn}]; t (* T. D. Noe, Nov 08 2013 *)

A231404 Integers n dividing the Lucas sequence u(n), where u(i) = 2*u(i-1) - 4*u(i-2) with initial conditions u(0)=0, u(1)=1.

Original entry on oeis.org

1, 2, 3, 4, 6, 8, 9, 12, 15, 16, 18, 21, 24, 27, 30, 32, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 64, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, 99, 102, 105, 108, 111, 114, 117, 120, 123, 126, 128, 129, 132, 135, 138, 141, 144, 147, 150, 153, 156, 159, 162, 165
Offset: 1

Views

Author

Thomas M. Bridge, Nov 08 2013

Keywords

Comments

The sequence consists of all nonnegative powers of 2, together with all positive multiples of 3. There are infinitely many pairs of consecutive integers in this sequence.

Examples

			For n=0,...,4 we have u(n)= 0,1,2,0,-8. Clearly n=1,2,3,4 are in the sequence.

Links

Crossrefs

Cf. A088138 (Lucas sequence).
Equal to union of A008585 (multiples of 3) and A000079 (powers of 2).

Programs

  • Mathematica
    nn = 500; s = LinearRecurrence[{2, -4}, {1, 2}, nn]; t = {}; Do[If[Mod[s[[n]], n] == 0, AppendTo[t, n]], {n, nn}]; t (* T. D. Noe, Nov 08 2013 *)

A231290 Numbers k dividing u(k), where the Lucas sequence is defined u(i) = u(i-1) - 5*u(i-2) with initial conditions u(0)=0, u(1)=1.

Original entry on oeis.org

1, 12, 19, 24, 36, 48, 72, 84, 96, 108, 144, 168, 192, 216, 228, 252, 288, 324, 336, 361, 384, 432, 456, 504, 576, 588, 648, 672, 684, 744, 756, 768, 816, 864, 912, 972, 1008, 1092, 1152, 1176, 1296, 1344, 1368, 1488, 1512, 1536, 1596, 1632, 1728, 1764
Offset: 0

Views

Author

Thomas M. Bridge, Nov 06 2013

Keywords

Comments

Contains every nonnegative power of 19. All terms that are not a power of 19 are multiples of 12.

Links

Crossrefs

Cf. A001029 (powers of 19 (subsequence)).

Programs

  • Mathematica
    nn = 3000; s = LinearRecurrence[{1, -5}, {1, 1}, nn]; t = {}; Do[If[Mod[s[[n]], n] == 0, AppendTo[t, n]], {n, nn}]; t (* T. D. Noe, Nov 08 2013 *)

A228440 Numbers n dividing u(n), where the Lucas sequence is defined u(i) = u(i-1) - 3*u(i-2) with initial conditions u(0)=0, u(1)=1.

Original entry on oeis.org

1, 11, 121, 253, 1331, 2783, 5819, 11891, 14641, 29161, 30613, 64009, 130801, 133837, 161051, 273493, 320771, 336743, 558877, 640343, 670703, 704099, 895873, 1438811, 1472207, 1771561, 3008423, 3078251, 3528481, 3544453, 3704173, 6147647, 6290339, 7027801
Offset: 1

Views

Author

Thomas M. Bridge, Nov 02 2013

Keywords

Comments

Since the absolute value of the discriminant of the characteristic polynomial is prime (=11), the sequence contains every nonnegative integer power of 11 (A001020 is subsequence). Other terms are formed on multiplication of 11^k by sporadic primes.

Examples

			u(1)=1 and u(11)=253. Clearly n divides u(n) for these terms.

Links

Crossrefs

Cf. A214733 (Lucas sequence u(n) ignoring sign).
Cf. A001020 (powers of 11).

Programs

  • Mathematica
    nn = 10000; s = LinearRecurrence[{1, -3}, {1, 1}, nn]; t = {}; Do[
If[Mod[s[[n]], n] == 0, AppendTo[t, n]], {n, nn}]; t (* T. D. Noe, Nov 06 2013 *)

Extensions

a(27)-a(34) from Lars Blomberg, Feb 15 2016

A231114 Numbers k dividing u(k), where the Lucas sequence is defined u(i) = u(i-1) - 4*u(i-2) with initial conditions u(0)=0, u(1)=1.

Original entry on oeis.org

1, 3, 5, 9, 15, 25, 27, 45, 75, 81, 125, 135, 171, 225, 243, 375, 405, 435, 465, 513, 625, 675, 729, 855, 1125, 1215, 1305, 1395, 1539, 1875, 2025, 2175, 2187, 2325, 2565, 3125, 3249, 3375, 3645, 3725, 3915, 4005, 4185, 4275, 4617, 5625, 6075, 6327, 6525, 6561
Offset: 1

Views

Author

Thomas M. Bridge, Nov 06 2013

Keywords

Comments

Every term (except leading term) is divisible by at least one of 3 or 5.
Furthermore, this sequence contains 3^i*5^j for all i, j >= 0, that is, A003593 is a subsequence.

Examples

			The sequence u(i) begins 0, 1, 1, -3, -7, 5, 33. Only for k = 1, 3, 5 does k divides u(k).

Links

Crossrefs

Cf. A003593 (subsequence), A106853 (Lucas sequence).

Programs

  • Mathematica
    nn = 10000; s = LinearRecurrence[{1, -4}, {1, 1}, nn]; t = {}; Do[If[Mod[s[[n]], n] == 0, AppendTo[t, n]], {n, nn}]; t (* T. D. Noe, Nov 06 2013 *)

A228439 Numbers k dividing u(k), where the Lucas sequence is defined u(i) = u(i-1) - 2*u(i-2) with initial conditions u(0)=0, u(1)=1.

Original entry on oeis.org

1, 7, 49, 343, 2401, 4753, 16807, 33271, 76783, 117649, 232897, 461041, 537481, 823543, 1630279, 3227287, 3762367, 5764801, 7447951, 11411953, 11527201, 19358969, 22591009, 26336569, 40353607, 44720977, 52135657, 79883671, 80690407
Offset: 1

Views

Author

Thomas M. Bridge, Nov 02 2013

Keywords

Comments

Since the absolute value of the discriminant of the characteristic polynomial is prime (=7), the sequence contains every nonnegative integer power of 7. Other terms are formed on multiplication of 7^k by sporadic primes.

Examples

			For k = 0, 1 , ..., 10, there is u(k) = 0,1,1,-1,-3,-1,5,7,-3,-17,-11. Clearly only k = 1 and k = 7 satisfy k divides u(k).

Links

Crossrefs

Cf. A107920 (Lucas Sequence u(n)=u(n-1)-2u(n-2)).

Programs

  • Mathematica
    nn = 10000; s = LinearRecurrence[{1, -2}, {1, 1}, nn]; t = {}; Do[If[Mod[s[[n]], n] == 0, AppendTo[t, n]], {n, nn}]; t (* T. D. Noe, Nov 08 2013 *)

Extensions

a(19)-a(29) from Amiram Eldar, May 28 2024

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