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 1-8 of 8 results.

A128357 Quotients A128356(n)/prime(n).

Original entry on oeis.org

10, 7, 311, 127, 23, 157, 343927, 7805561, 47, 9629, 311, 25679, 821, 1470086279, 12409, 71233, 1232333, 2443783, 2939291, 71711, 352883, 181113265579, 167, 105199, 3881, 1314520253, 619, 20759, 117503, 1162660843, 1880415721, 263
Offset: 1

Views

Author

Alexander Adamchuk, Mar 02 2007, Mar 09 2007

Keywords

Comments

A128356 = {20, 21, 1555, 889, 253, 2041, 5846759, ...} = Least number k>1 (that is not the power of prime p) such that k divides (p+1)^k-1, where p = prime(n). Most listed terms are primes, except a(7) = 20231*17 and a(8) = 410819*19. a(15) = 12409. a(16) = 71233.
Note that all prime listed terms of {a(n)} coincide with terms of A128456 = {2, 7, 311, 127, 23, 157, 7563707819165039903, 75368484119, 47, 9629, 311, 25679, 821, ...} = least prime factor of ((p+1)^p - 1)/p^2, where p = prime(n).

Crossrefs

Cf. A128356 (least number k > 1 (that is not a power of prime p) such that k divides (p+1)^k-1, where p = prime(n)).
Cf. A014960, A128360, A128358, A014960, A014956, A014951, A014949, A014946, A014945, A067945.
Cf. A128456 (least prime factor of ((p+1)^p - 1)/p^2, where p = prime(n)).

Extensions

Terms a(14) onwards from Max Alekseyev, Feb 08 2010

A128360 Numbers k such that k divides 20^k - 1.

Original entry on oeis.org

1, 19, 361, 6859, 130321, 2476099, 47045881, 148305659, 893871739, 2817807521, 4234136149, 10350100679, 16983563041, 53538342899, 80448586831, 196651912901, 322687697779, 815211156289, 1017228515081, 1432001198261, 1528523149789
Offset: 1

Views

Author

Alexander Adamchuk, Mar 02 2007

Keywords

Comments

19 divides a(n) for n > 1. All powers of 19 are terms. a(n) = 19^(n-1) for all to n < 8, while a(8) = A128356(8) = 148305659 = 410819*19^2.
Prime divisors of a(n) in the order of appearance are {19, 410819, 617311, 1508981, ...}. - Alexander Adamchuk, May 16 2010

Crossrefs

Cf. A001029, A128358, A014960, A014956, A014951, A014949, A014946, A014945, A067945, A128357, A128356, A128400, A177920.

Programs

Extensions

a(9)-a(11) from Stefan Steinerberger, May 09 2007
a(12)-a(15) from Alexander Adamchuk, May 16 2010
Edited and a(16)-a(21) added by Max Alekseyev, Oct 02 2010

A128456 Quotients A128452(p+1)/p for prime p = A000040(n).

Original entry on oeis.org

2, 7, 311, 127, 23, 157, 7563707819165039903, 75368484119, 47, 9629, 311, 25679, 821, 758771382833029, 12409, 71233, 18438666190697, 2443783, 2939291, 71711, 352883, 181113265579, 167, 105199, 3881, 1314520253, 619, 20759, 117503, 1162660843
Offset: 1

Views

Author

Alexander Adamchuk, Mar 05 2007, Mar 09 2007

Keywords

Comments

a(n) coincides with A128357(n) from n = 2 up to n = 6.

Crossrefs

Cf. A128452, A128357, A128356, A127103, A127104, A127105, A127106, A127107, A127102, A127101, A127100, A127092, A128393, A128394, A128395, A128396, A128397, A128398, A128399, A128400, A128401, A128402, A128403, A128404.

Formula

a(n) = A128452(A000040(n)+1)/A000040(n).
a(n) = A020639(((p+1)^p - 1)/p^2), i.e., the smallest prime factor of ((p+1)^p - 1)/p^2, where p = A000040(n).

Extensions

Terms a(14) onward from Max Alekseyev, May 05 2010

A128452 Least number k > n such that k^2 divides n^k - 1.

Original entry on oeis.org

4, 21, 6, 1555, 8, 889, 10, 111, 12, 253, 14, 2041, 16, 21, 18, 128583032925805678351, 20, 1432001198261, 22, 39, 24, 1081, 26, 55, 28, 171, 30, 279241, 32, 9641, 34, 1191, 36, 55, 38, 950123, 40, 1641, 42, 33661, 44, 32627169461820247, 46, 63, 48, 583223, 50
Offset: 3

Views

Author

Alexander Adamchuk, Mar 05 2007, Mar 09 2007

Keywords

Comments

For prime p, p divides a(p+1). Quotients a(p+1)/p for prime p = A000040(n) are listed in A128456(n) which coincides with A128357(n) for n from 2 to 6.
a(n) divides n^(n-1) - 1.

Crossrefs

Cf. A128456, A128357, A128356, A127103, A127104, A127105, A127106, A127107, A127102, A127101, A127100, A127092, A128393, A128394, A128395, A128396, A128397, A128398, A128399, A128400, A128401, A128402, A128403, A128404.

Formula

a(2n-1) = 2n.

Extensions

More terms from Alexander Adamchuk, Mar 09 2007
Terms a(22) onward from Max Alekseyev, May 05 2010

A137666 Largest prime factor of A137664(n) = (p + 1)^p - 1 for p = prime(n).

Original entry on oeis.org

2, 7, 311, 337, 266981089, 29914249171, 7563707819165039903, 192696104561, 58769065453824529, 847499019384726257346113954958447091, 18158209813151, 138233050898929517126243814850350442620694127
Offset: 1

Views

Author

Alexander Adamchuk, Feb 04 2008

Keywords

Comments

a(n) is also the largest prime factor of A137665(n) = A137664(n)/prime(n)^2. p^2 divides A137664(n) = (p + 1)^p - 1, p = prime(n). Least prime factors of A137664(n) are listed in A128456.
a(n) = A128456(n) = A137665(n) = ((p + 1)^p - 1)/p^2 for n = {1,2,3,7,595,...} corresponding to p = prime(n) = {2,3,5,17,4357,...} = A127837.

Links

Crossrefs

Cf. A128452, A128456, A128356, A128357, A137664, A137665, A128466, A127837.

Programs

  • Mathematica
    FactorInteger[#][[-1,1]]&/@((#+1)^#-1&/@Prime[Range[12]]) (* Harvey P. Dale, Apr 07 2018 *)

A134360 Primes dividing terms of A128358.

Original entry on oeis.org

17, 20231, 83233, 3828719, 8519143, 39951841, 68916943, 76672279, 144326023, 721630111, 35915273857, 39392517233, 42845049953, 83709099119
Offset: 1

Views

Author

Alexander Adamchuk, Jan 31 2008

Keywords

Comments

A prime p is in this sequence if the multiplicative order of 18 modulo p equals the product of smaller primes from this sequence. - Max Alekseyev, Sep 24 2009

Examples

			A128358 begins with {1, 17, 17^2, 17^3, 17^4, 17^5, 20231*17^2, 83233*17^2, ...}. Thus 17, 20231, 83233 are in this sequence.

Crossrefs

Cf. A128356, A128357, A128358.

Extensions

Edited and extended by Max Alekseyev, Jul 28 2009
a(11)-a(14) from Max Alekseyev, May 26 2010

A137664 a(n) = (p+1)^p - 1 where p = prime(n).

Original entry on oeis.org

8, 63, 7775, 2097151, 743008370687, 793714773254143, 2185911559738696531967, 5242879999999999999999999, 55572324035428505185378394701823, 6863037736488299999999999999999999999999999
Offset: 1

Views

Author

Alexander Adamchuk, Feb 04 2008

Keywords

Comments

p^2 divides a(n) = (p+1)^p - 1, p = prime(n).
Quotients a(n)/prime(n)^2 are listed in A137665(n) = {2, 7, 311, 42799, 6140565047, 4696537119847, 7563707819165039903, ...}.
Least prime factors of A137665(n) = a(n)/prime(n)^2 are listed in A128456(n) = {2, 7, 311, 127, 23, 157, 7563707819165039903, ...}.
Largest prime factors A137665(n) = a(n)/prime(n)^2 are listed in A137666(n) = {2, 7, 311, 337, 266981089, 29914249171, 7563707819165039903, ...}.

Crossrefs

Cf. A128452, A128456, A128356, A128357, A137665, A137666.

Programs

  • Mathematica
    Table[ (Prime[n] + 1)^Prime[n] - 1, {n,1,15} ]
(#+1)^#-1&/@Prime[Range[10]] (* Harvey P. Dale, Jan 10 2025 *)

Formula

a(n) = (prime(n) + 1)^prime(n) - 1.

A137665 Quotients ((p+1)^p - 1)/p^2 for p = prime(n).

Original entry on oeis.org

2, 7, 311, 42799, 6140565047, 4696537119847, 7563707819165039903, 14523213296398891966759, 105051652240885643072548950287, 8160568057655529131985731272294887039239, 47525417447024678661670292427038339608998847, 20681861558186805237407813095538883147812221153173966103
Offset: 1

Views

Author

Alexander Adamchuk, Feb 04 2008

Keywords

Comments

p^2 divides a(n) = (p+1)^p - 1, p = prime(n). (p+1)^p - 1 = A137664(n) = {8, 63, 7775, 2097151, 743008370687, 793714773254143, 2185911559738696531967, ...}.
Least prime factors of a(n) are listed in A128456(n) = {2, 7, 311, 127, 23, 157, 7563707819165039903, ...}.
Largest prime factors a(n) are listed in A137666.
a(n) is prime for n = {1, 2, 3, 7, 595, ...} corresponding to p = prime(n) = {2, 3, 5, 17, 4357, ...} = A127837.
Primes in this sequence are A128466.

Crossrefs

Cf. A128452, A128456, A128356, A128466, A127837, A128357, A060073, A137664, A137666.

Programs

  • Mathematica
    Table[ ((Prime[n] + 1)^Prime[n] - 1)/Prime[n]^2, {n,1,15} ]
  • PARI
    a(n) = my(p=prime(n)); polcyclo(p,p+1)/p \\ Hugo Pfoertner, Jul 21 2024

Formula

a(n) = ((prime(n) + 1)^prime(n) - 1)/prime(n)^2;
a(n) = A137664(n)/prime(n)^2.
Showing 1-8 of 8 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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