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: Patrick A. Thomas

Patrick A. Thomas's wiki page.

Patrick A. Thomas has authored 43 sequences. Here are the ten most recent ones:

A356518 Maximal numerators in approximations to the Aurifeuillian factors of p^p +- 1.

Original entry on oeis.org

2, 28, 1706, 25082, 816634, 157704814
Offset: 1

Author

Patrick A. Thomas, Aug 10 2022

Keywords

Comments

If R=(p^p+-1)/(p+-1) then the left Aurifeuillian factor of R is (1/e)*sqrt(R/(1+z)), where z = Sum_{n>=1} r(n)/p^n and r(n) is a rational number in Q; a(n) is the numerator of r(n). r(7) appears to be about 0.572082, but there is insufficient precision to identify a(7).

Examples

			The r(n) are 2/3, 28/45, 1706/2835, 25082/42525, 816634/1403325, ...
		

Crossrefs

Cf. A356519 (denominators), A352711, A352732, A352400, A352401.

A356519 Denominators in approximations to the Aurifeuillian factors of p^p +- 1.

Original entry on oeis.org

3, 45, 2835, 42525, 1403325, 273648375
Offset: 1

Author

Patrick A. Thomas, Aug 10 2022

Keywords

Comments

a(1) = 3 = 4! / 2^3
a(2) = 3^2 * 5 = 6! / 2^4
a(3) = 3^4 * 5 * 7 = 8! / 2^7 * 3^2
a(4) = 3^5 * 5^2 * 7 = 10! / 2^8 * 3
a(5) = 3^6 * 5^2 * 7 * 11 = 12! / 2^10 * 3

Crossrefs

Cf. A356518 (numerators), A352711, A352732, A352400, A352401.

A352401 The right Aurifeuillian factor of p^p + 1 for primes p congruent to 3 (mod 4).

Original entry on oeis.org

7, 911, 407353, 870542161121, 2498077661567473, 63472256064447557254913, 54382651771205224279713471565249817, 767102704711961850109296220485687497279, 6066304600323886604542912453739225327712511596287519
Offset: 1

Author

Patrick A. Thomas, Jun 08 2022

Keywords

Comments

For prime factorizations of p^p + 1 see A125136.

Examples

			870542161121 is the larger Aurifeuillian factor of 19^19 + 1, and 19 is the 4th term of A002145, so a(4) = 870542161121.
		

A352400 a(n) is the left Aurifeuillian factor of p^p + 1 for A002145(n), where A002145 lists the primes congruent to 3 (mod 4).

Original entry on oeis.org

1, 113, 58367, 113631466919, 348275601426959, 8403855868042458448127, 7248206084007410402911299180581471, 105318477338066161993242388018074119617, 830220061043693789623432394289631761145130727636121
Offset: 1

Author

Patrick A. Thomas, Jun 08 2022

Keywords

Comments

For prime factorizations of p^p + 1 see A125136.

Examples

			105318477338066161993242388018074119617 is the smaller Aurifeuillian factor of 47^47 + 1, and 47 is the 8th term of A002145, so it is a(8).
		

Formula

If R is (p^p+1)/(p+1), where p == 3 (mod 4) and p > 7, then an approximation of the left Aurifeuillian factor of R is (1/e) * sqrt(R/(1+z)), where z =
2/(3p) + 28/(45p^2) + 1706/(2835p^3) if p=1,79,109,121,151 or 169 (mod 210),
2/(3p) + 28/(45p^2) + 86/(2835p^3) if p=19,31,61,139,181 or 199 (mod 210),
2/(3p) - 8/(45p^2) + 194/(2835p^3) if p=37,43,67,127,163 or 193 (mod 210),
2/(3p) - 8/(45p^2) - 1426/(2835p^3) if p=13,73,97,103,157 or 187 (mod 210),
-2/(3p) - 8/(45p^2) + 1426/(2835p^3) if p=23,53,107,113,137 or 197 (mod 210),
-2/(3p) - 8/(45p^2) - 194/(2835p^3) if p=17,47,83,143,167 or 173 (mod 210),
-2/(3p) + 28/(45p^2) - 86/(2835p^3) if p=11,29,71,149,179 or 191 (mod 210),
-2/(3p) + 28/(45p^2) - 1706/(2835p^3) if p=41,59,89,101,131 or 209 (mod 210).

A352732 The right Aurifeuillian factor of p^p - 1, for primes p congruent to 1 (mod 4).

Original entry on oeis.org

71, 13993643, 19152352117, 813955076015309926319, 46959719470144429555105032871, 491873569944394295636860313807677, 1848593595048531176470116001230356265643249547, 1000403244183535565720394723140528028235711874491322863, 33027769942300819203735411144251223948236849608414254057770836237073
Offset: 1

Author

Patrick A. Thomas, Mar 30 2022

Keywords

Comments

For prime factorizations of p^p - 1 see A125135.

Examples

			813955076015309926319 is the larger Aurifeuillian factor of 29^29-1, and 29 is the 4th term of A002144, so a(4) = 813955076015309926319.
		

Crossrefs

A352711 The left Aurifeuillian factor of p^p - 1 for primes p congruent to 1 (mod 4).

Original entry on oeis.org

11, 1803647, 2699538733, 112663560435723374699, 6243610407478181159725577611, 67643278270835231300426724641533, 253382315888712050791030544452181354268272663, 133904013361225746608283522164245432446284642589451147, 4429523820749528526448423858097183945539957285504166342434080091097
Offset: 1

Author

Patrick A. Thomas, Mar 30 2022

Keywords

Comments

For prime factorizations of p^p - 1 see A125135.
Named after the French mathematician Léon-François-Antoine Aurifeuille (1822-1882). - Bernard Schott, Nov 04 2022

Examples

			112663560435723374699 is the smaller Aurifeuillian factor of 29^29-1, and 29 is the 4th term of A002144, so a(4) = 112663560435723374699.
		

Formula

If R is (p^p-1)/(p-1), where p == 1 (mod 4) and p > 5, then an approximation of the left Aurifeuillian factor of R is (1/e) * sqrt(R/(1+z)), where z =
2/(3p) + 28/(45p^2) + 1706/(2835p^3) if p=1,79,109,121,151 or 169 (mod 210),
2/(3p) + 28/(45p^2) + 86/(2835p^3) if p=19,31,61,139,181 or 199 (mod 210),
2/(3p) - 8/(45p^2) + 194/(2835p^3) if p=37,43,67,127,163 or 193 (mod 210),
2/(3p) - 8/(45p^2) - 1426/(2835p^3) if p=13,73,97,103,157 or 187 (mod 210),
-2/(3p) - 8/(45p^2) + 1426/(2835p^3) if p=23,53,107,113,137 or 197 (mod 210),
-2/(3p) - 8/(45p^2) - 194/(2835p^3) if p=17,47,83,143,167 or 173 (mod 210),
-2/(3p) + 28/(45p^2) - 86/(2835p^3) if p=11,29,71,149,179 or 191 (mod 210),
-2/(3p) + 28/(45p^2) - 1706/(2835p^3) if p=41,59,89,101,131 or 209 (mod 210).

A333828 The 20-adic integer x = ...70D9AE7F1DI8 satisfying x^5 = x.

Original entry on oeis.org

8, 18, 13, 1, 15, 7, 14, 10, 9, 13, 0, 7, 6, 6, 13, 5, 14, 16, 0, 4, 11, 8, 8, 10, 8, 8, 3, 12, 6, 7, 19, 8, 10, 12, 11, 1, 1, 15, 2, 12, 8, 8, 10, 19, 4, 10, 19, 7, 17, 8, 12, 14, 9, 19, 11, 18, 16, 14, 19, 9, 4, 2, 16, 11, 0, 12, 11, 1, 6, 11, 12, 3, 3, 16, 11
Offset: 0

Author

Patrick A. Thomas, Apr 07 2020

Keywords

Comments

Letters A through J represent the base-20 digits 10 through 19, respectively.
Conjecture: There exists a nontrivial n-adic integer x satisfying x^5 = x, and x^2, x^3, and x^4 are not x, if and only if n has a prime factor of the form 4k+1. Further, there is one nontrivial pair (x and -x) for each different prime factor of the form 4k+1.

Examples

			8^25 in base 20 ends in the digits 13, 18, 8 (or ...DI8 in extended hexadecimal notation).
		

Crossrefs

Formula

The last n+1 digits of 8^(5^n) in base 20, for all n.

A331550 15-adic integer x = ...65762C0520697E8CA1A31469 satisfying x^3 = x.

Original entry on oeis.org

9, 6, 4, 1, 3, 10, 1, 10, 12, 8, 14, 7, 9, 6, 0, 2, 5, 0, 12, 2, 6, 7, 5, 6, 9, 8, 7, 0, 4, 2, 10, 1, 2, 9, 8, 0, 11, 13, 6, 11, 6, 6, 12, 5, 2, 9, 0, 1, 5, 1, 10, 9, 11, 8, 8, 14, 0, 12, 6, 0, 1, 1, 12, 14, 2, 13, 5, 13, 14, 9, 10, 12, 14, 9, 6, 6, 0, 12, 12, 7
Offset: 0

Author

Patrick A. Thomas, Jan 20 2020

Keywords

Comments

The base-15 version of A091664. A, B, C, D, and E are the standard notations for the hexadecimal digits 10, 11, 12, 13, and 14, respectively. x+1 is a base-15 automorph.

Examples

			x   = ...65762C0520697E8CA1A31469.
x^2 = ...8978C2E9CE8570624D4BDA86 = A331549.
x^3 = ...65762C0520697E8CA1A31469 = x.
		

Crossrefs

Programs

  • PARI
    \\ See A331548 with initial b=9 instead of b=3.
    
  • PARI
    Vecrev(digits(lift((9+O(15^99))^5^99),15)) \\ M. F. Hasler, Jan 26 2020

Formula

x = 15-adic lim_{n->infinity} 9^(5^n).

A331549 15-adic integer x = ...8978C2E9CE8570624D4BDA86 satisfying x^2 = x.

Original entry on oeis.org

6, 8, 10, 13, 11, 4, 13, 4, 2, 6, 0, 7, 5, 8, 14, 12, 9, 14, 2, 12, 8, 7, 9, 8, 5, 6, 7, 14, 10, 12, 4, 13, 12, 5, 6, 14, 3, 1, 8, 3, 8, 8, 2, 9, 12, 5, 14, 13, 9, 13, 4, 5, 3, 6, 6, 0, 14, 2, 8, 14, 13, 13, 2, 0, 12, 1, 9, 1, 0, 5, 4, 2, 0, 5, 8, 8, 14, 2, 2, 7
Offset: 0

Author

Patrick A. Thomas, Jan 20 2020

Keywords

Comments

The base-15 version of A018248. A, B, C, D, and E are the standard representations of the hexadecimal digits 10, 11, 12, 13, and 14, respectively.

Examples

			x^2 = ...8978C2E9CE8570624D4BDA86 = x.
		

Crossrefs

Cf. A018248.

Programs

  • PARI
    \\ See A331548, with initial b=6 instead of b=3.
    
  • PARI
    A331549_vec(n)=Vecrev(digits(lift(chinese(Mod(0,3^n),Mod(1,5^n))),15)) \\ or simpler but slower: Vecrev(digits(lift(Mod(3^4,15^n)^5^(n-1)),15)) \\ M. F. Hasler, Jan 26 2020

Formula

x = 15-adic lim_{n->infinity} 3^(4*(5^n)).

A331548 15-adic integer x = ...2AA66B44A40E43797853AD13 satisfying x^5 = x; also x^3 = -x; (x^2)^3 = x^2 = A331550; (x^4)^2 = x^4 = A331549.

Original entry on oeis.org

3, 1, 13, 10, 3, 5, 8, 7, 9, 7, 3, 4, 14, 0, 4, 10, 4, 4, 11, 6, 6, 10, 10, 2, 8, 1, 9, 9, 0, 4, 8, 3, 10, 11, 5, 9, 11, 0, 8, 0, 10, 9, 2, 6, 0, 8, 11, 5, 8, 5, 7, 1, 6, 10, 5, 12, 14, 0, 0, 6, 10, 6, 12, 8, 2, 12, 4, 6, 1, 6, 14, 6, 7, 8, 13, 5, 5, 3, 4, 3, 0
Offset: 0

Author

Patrick A. Thomas, Jan 20 2020

Keywords

Comments

The base-15 version of A120817. A, B, C, D, and E are the standard notations for the hexadecimal digits 10, 11, 12, 13, and 14, respectively.
Conjecture: If k is the number of prime factors congruent to 1 (mod 4) of an integer n, then there are exactly k n-adic integers x satisfying x^5 = x, while not satisfying x^h = x for h = 2, 3, or 4. This does not count -x, which also satisfies, in each case. - Patrick A. Thomas, Mar 31 2020

Examples

			x equals the limit of the (n+1) trailing digits of 3^(5^n):
3^(5^0) = (3), 3^(5^1) = 1(13), 3^(5^2) = 1708EB01(D13), ...
x   = ...2AA66B44A40E43797853AD13.
x^2 = ...65762C0520697E8CA1A31469 = A331550.
x^3 = ...C44883AA4AE0AB75769B41DC = -x.
x^4 = ...8978C2E9CE8570624D4BDA86 = A331549.
x^5 = ...2AA66B44A40E43797853AD13 = x.
		

Crossrefs

Programs

  • PARI
    \\ after Paul D. Hanna's program in A120817
    {a(n)=local(b=3, v=[]); for(k=1, n+1, b=b^5%15^k; v=concat(v, (15*b\15^k))); v[n+1]}
    for(k=0,80,print1(a(k),", ")) \\ Hugo Pfoertner, Jan 26 2020
    
  • PARI
    (A331548_vec(n)=Vecrev(digits(lift(Mod(3,15^n)^5^(n-1)),15)))(99) \\ M. F. Hasler, Jan 26 2020

Formula

x = 15-adic lim_{n->infinity} 3^(5^n).