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 1-10 of 20 results. Next

A136374 a(n) = (A128677(n) - 1)/(2*A000040(n)).

Original entry on oeis.org

3, 4, 2, 1, 3, 1224, 551, 1, 697, 66, 60, 31, 12, 7, 24641820, 3343240, 122610, 134, 4, 101, 80, 1, 979, 518, 1414
Offset: 2

Views

Author

Alexander Adamchuk, Dec 27 2007

Keywords

Comments

All terms are integer since for n > 1, p = A000040(n) divides (A128677(n)-1)/2.

Examples

			a(2) = (A128677(2) - 1)/(2*A000040(2)) = (19 - 1)/(2*3) = 3.
		

Crossrefs

Formula

a(n) = (A128677(n) - 1)/(2*A000040(n)).

Extensions

a(16)-a(26) from Max Alekseyev, May 14 2010

A128678 Numbers k such that k^3 divides 4^(k^2) + 1.

Original entry on oeis.org

1, 5, 205, 505, 20705, 40505, 168305, 408545, 1342505, 1660705, 2084645, 4091005, 10201505, 16322105, 16750345, 16998805, 37217545, 49985405, 55042705, 117165905, 135593005, 167731205, 186678805, 210549145, 218301905, 418261705
Offset: 1

Views

Author

Alexander Adamchuk, Mar 31 2007

Keywords

Comments

5 divides a(n) for n > 1.

Crossrefs

Programs

  • Mathematica
    Select[Range[5*10^6], Mod[ 4^(#^2)+1, #^3]==0 &] (* G. C. Greubel, Jan 17 2018 *)

Extensions

More terms from Ryan Propper, Jan 13 2008

A128683 Numbers k such that k^3 divides 10^(k^2) + 1.

Original entry on oeis.org

1, 11, 253, 11891, 35167, 45023, 47927, 96569, 276859, 640343, 1035529, 1102321, 1652849, 2221087, 6367757, 16339367, 19588019, 26499979, 30096121, 48669863, 51809087, 74901409, 88023001, 89007677, 104391089, 105080767, 143938531
Offset: 1

Views

Author

Alexander Adamchuk, Mar 31 2007

Keywords

Comments

11 divides a(n) for n > 1.

Crossrefs

Programs

  • Mathematica
    Select[Range[5*10^6], Mod[ 10^(#^2)+1, #^3]==0 &] (* G. C. Greubel, Jan 18 2018 *)

Extensions

More terms from J. Mulder (jasper.mulder(AT)planet.nl), Jan 26 2010
Terms a(16) onward from Max Alekseyev and Alexander Adamchuk, May 12 2010

A128679 Numbers k such that k^3 divides 5^(k^2) + 1.

Original entry on oeis.org

1, 3, 21, 609, 903, 2667, 9429, 15501, 18543, 26187, 77343, 108507, 114681, 159663, 212541, 273441, 318507, 405447, 537747, 797349, 1197483, 1357503, 1398747, 1731387, 2354961, 3146703, 3325749, 4451181, 4630227, 4665801, 5192943
Offset: 1

Views

Author

Alexander Adamchuk, Mar 31 2007

Keywords

Comments

3 divides a(n) for n > 1. 7 divides a(n) for n > 2.

Crossrefs

Cf. A127263 (numbers k such that k^3 divide 2^(k^2)+1).
Cf. A128677 (least k > p such that (k*p)^3 divides (p-1)^(k*p)^2+1, where p = prime(n) > 2).

Programs

  • Mathematica
    Select[Range[5*10^6], Mod[ 5^(#^2)+1, #^3]==0 &] (* G. C. Greubel, Jan 18 2018 *)

Extensions

More terms from Ryan Propper, Dec 31 2007

A128680 Numbers k such that k^3 divides 6^(k^2) + 1.

Original entry on oeis.org

1, 7, 203, 1379, 11977, 39991, 577129, 2359469, 4780447, 6521291, 6640739, 9904979, 21511301, 28434413, 34050611, 113694413, 189117439, 222281549, 250163599, 282046373, 391803601, 941748059, 1269166759, 1308225583
Offset: 1

Views

Author

Alexander Adamchuk, Mar 31 2007

Keywords

Comments

7 divides a(n) for n > 1.

Crossrefs

Cf. A127263 (numbers k such that k^3 divide 2^(k^2)+1).
Cf. A128677 (least k > p such that (k*p)^3 divides (p-1)^(k*p)^2+1, where p = prime(n) > 2).

Programs

  • Mathematica
    Select[Range[5*10^6], Mod[ 6^(#^2)+1, #^3]==0 &] (* G. C. Greubel, Jan 18 2018 *)

Extensions

More terms from Ryan Propper, Jan 13 2008

A128681 Numbers k such that k^3 divides 8^(k^2) + 1.

Original entry on oeis.org

1, 3, 9, 57, 171, 1467, 13131, 27873, 32547, 97641, 249489, 261633, 784899, 1218897, 1255347, 2140353, 4971027, 8246019, 9961491, 12914001, 14913081, 15915483, 23159043, 23851593, 24738057, 29884473, 40666707, 83512353, 127938537
Offset: 1

Views

Author

Alexander Adamchuk, Mar 31 2007

Keywords

Comments

3 divides a(n) for n > 1.

Crossrefs

Cf. A127263 (numbers k such that k^3 divide 2^(k^2)+1).
Cf. A128677 (least k > p such that (k*p)^3 divides (p-1)^(k*p)^2+1, where p = prime(n) > 2).

Programs

  • Mathematica
    Select[Range[5*10^6], Mod[ 8^(#^2)+1, #^3]==0 &] (* G. C. Greubel, Jan 18 2018 *)

Extensions

More terms from Ryan Propper, Jan 13 2008

A128682 Numbers k such that k^3 divides 9^(k^2) + 1.

Original entry on oeis.org

1, 5, 505, 5905, 596405, 1971005, 199071505, 515151005, 581457505, 618181105, 2327756905, 11855474405, 19555113485, 31361954905, 101901244565, 235103447405, 305354088005, 395331063745, 600905819065, 608393336905, 686701313405, 730071885005
Offset: 1

Views

Author

Alexander Adamchuk, Mar 31 2007

Keywords

Comments

5 divides a(n) for n > 1.

Crossrefs

Programs

  • Mathematica
    Select[Range[5*10^6], Mod[ 9^(#^2)+1, #^3]==0 &] (* G. C. Greubel, Jan 18 2018 *)

Extensions

a(5)-a(6) from J. Mulder (jasper.mulder(AT)planet.nl), Jan 26 2010
Terms a(7) onward from Max Alekseyev, May 14 2010

A128684 Numbers k such that k^3 divides 11^(k^2) + 1.

Original entry on oeis.org

1, 3, 111, 24753, 1770231, 23401797, 39402669, 65498547, 298100379, 440377293, 1788478509, 5218600731, 8786795187, 14606175981, 24342867543, 45340639419, 98204136339, 5032808252679, 5293823142513, 6241891310439, 7013907181779
Offset: 1

Views

Author

Alexander Adamchuk, Mar 31 2007

Keywords

Comments

3 divides a(n) for n > 1.

Crossrefs

Programs

  • Mathematica
    Select[Range[5*10^6], Mod[ 11^(#^2)+1, #^3]==0 &] (* G. C. Greubel, Jan 18 2018 *)

Extensions

a(5) from J. Mulder (jasper.mulder(AT)planet.nl), Jan 26 2010
Terms from a(6) onward from Max Alekseyev, May 14 2010

A128685 Numbers k such that k^3 divides 12^(k^2) + 1.

Original entry on oeis.org

1, 13, 1027, 52741, 468481, 2890927, 4166539, 37009999, 228383233, 999884119, 1345997497, 1900627417, 2126334691, 4499474941, 5037873529, 11728490839, 104180336299, 105897267463, 150149565943, 926550776281, 4056529870783
Offset: 1

Views

Author

Alexander Adamchuk, Mar 31 2007

Keywords

Comments

13 divides a(n) for n > 1.

Crossrefs

Programs

  • Mathematica
    Select[Range[5*10^6], Mod[ 12^(#^2)+1, #^3]==0 &] (* G. C. Greubel, Jan 18 2018 *)
    Select[Range[5*10^6],PowerMod[12,#^2,#^3]==#^3-1&] (* The program generates the first 7 terms of the sequence. *) (* Harvey P. Dale, Aug 10 2025 *)

Extensions

16 more terms from Ryan Propper, Dec 03 2007
Terms a(17) onward from Max Alekseyev, May 14 2010

A127263 Numbers k such that k^3 divides 2^(k^2)+1.

Original entry on oeis.org

1, 3, 57, 32547, 9961491, 297381939, 1338104811, 3942759027, 5688011361, 8920514307, 9146532873, 40253706489, 243640690617, 764039295291, 1127102902923, 1556475424971, 2251315404417, 3005607686883, 5222670270483
Offset: 1

Views

Author

Max Alekseyev, Mar 27 2007, Mar 29 2007, Apr 18 2007

Keywords

Comments

If k belongs to this sequence, then so does (2^(k^2)+1)/k^2.
From Alexander Adamchuk, May 14 2010: (Start)
3 divides a(n) for n>1.
19 divides a(n) for n>2. (End)

Crossrefs

Programs

  • Mathematica
    Select[Range[100000], Divisible[2^(#^2) + 1, #^3] &] (* Robert Price, Mar 23 2020 *)

Extensions

a(7) from Ryan Propper, Jan 01 2008
a(8)-a(19) from Max Alekseyev, May 14 2010
Showing 1-10 of 20 results. Next