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 1-6 of 6 results.

A324082 One of the four successive approximations up to 13^n for 13-adic integer 3^(1/4).This is the 3 (mod 13) case (except for n = 0).

Original entry on oeis.org

0, 3, 68, 575, 13757, 156562, 4612078, 52880168, 178377202, 9967145854, 137221138330, 1240089073122, 22746013801566, 279024950148857, 2399150696294628, 2399150696294628, 104770936724476142, 3431853982640375347, 98586429095835092610, 1335595905567366417029
Offset: 0

Views

Author

Jianing Song, Sep 01 2019

Keywords

Comments

For n > 0, a(n) is the unique number k in [1, 13^n] and congruent to 3 mod 13 such that k^4 - 3 is divisible by 13^n.
For k not divisible by 13, k is a fourth power in 13-adic field if and only if k == 1, 3, 9 (mod 13). If k is a fourth power in 13-adic field, then k has exactly 4 fourth-power roots.

Examples

			The unique number k in [1, 13^2] and congruent to 3 modulo 13 such that k^4 - 3 is divisible by 13^2 is k = 68, so a(2) = 68.
The unique number k in [1, 13^3] and congruent to 3 modulo 13 such that k^4 - 3 is divisible by 13^3 is k = 575, so a(3) = 575.

Links

Crossrefs

Cf. A286840, A286841, A322085, A324077, A324083, A324084, A324085, A324086, A324087, A324153.

Programs

  • PARI
    a(n) = lift(sqrtn(3+O(13^n), 4))

Formula

a(n) = A324077(n)*A286841(n) mod 13^n = A324084(n)*A286840(n) mod 13^n.
For n > 0, a(n) = 13^n - A324083(n).
a(n)^2 == A322086(n) (mod 13^n).

A324083 One of the four successive approximations up to 13^n for 13-adic integer 3^(1/4).This is the 10 (mod 13) case (except for n = 0).

Original entry on oeis.org

0, 10, 101, 1622, 14804, 214731, 214731, 9868349, 637353519, 637353519, 637353519, 552071320915, 552071320915, 23850156443396, 1538225689404661, 48786742317796129, 560645672458703699, 5218561936740962586, 13868977856122300519, 126324384808079693648
Offset: 0

Views

Author

Jianing Song, Sep 01 2019

Keywords

Comments

For n > 0, a(n) is the unique number k in [1, 13^n] and congruent to 10 mod 13 such that k^4 - 3 is divisible by 13^n.
For k not divisible by 13, k is a fourth power in 13-adic field if and only if k == 1, 3, 9 (mod 13). If k is a fourth power in 13-adic field, then k has exactly 4 fourth-power roots.

Examples

			The unique number k in [1, 13^2] and congruent to 10 modulo 13 such that k^4 - 3 is divisible by 13^2 is k = 101, so a(2) = 101.
The unique number k in [1, 13^3] and congruent to 10 modulo 13 such that k^4 - 3 is divisible by 13^3 is k = 1622, so a(3) = 1622.

Links

Crossrefs

Cf. A286840, A286841, A322085, A324077, A324082, A324084, A324085, A324086, A324087, A324153.

Programs

  • PARI
    a(n) = lift(-sqrtn(3+O(13^n), 4))

Formula

a(n) = A324077(n)*A286840(n) mod 13^n = A324084(n)*A286841(n) mod 13^n.
For n > 0, a(n) = 13^n - A324082(n).
a(n)^2 == A322086(n) (mod 13^n).

A322088 Digits of one of the two 13-adic integers sqrt(3).

Original entry on oeis.org

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

Views

Author

Jianing Song, Nov 26 2018

Keywords

Comments

This square root of 3 in the 13-adic field ends with digit 9. The other, A322087, ends with digit 4.

Examples

			...10B47929C097954C47A7C773116286B233BA04649.

Links

Crossrefs

Cf. A286838, A286839, A322086, A322087, A322091, A322092.

Programs

  • PARI
    a(n) = truncate(-sqrt(3+O(13^(n+1))))\13^n

Formula

a(n) = (A322086(n+1) - A322086(n))/13^n.
For n > 0, a(n) = 12 - A322087(n).
Equals A286838*A322092 = A286839*A322091.
This 13-adic integer is the 13-adic limit as n -> oo of the integer sequence {2*T(13^n,9/2)}, where T(n,x) denotes the n-th Chebyshev polynomial. - Peter Bala, Dec 04 2022

A322085 One of the two successive approximations up to 13^n for 13-adic integer sqrt(3). Here the 4 (mod 13) case (except for n = 0).

Original entry on oeis.org

0, 4, 108, 1122, 18698, 361430, 1104016, 5930825, 570667478, 7912243967, 113957237697, 251815729546, 11004778093768, 104197118583692, 3132948184506222, 26757206498701956, 589802029653700283, 7909384730668678534, 85763128005100719931, 648040162764887685576
Offset: 0

Views

Author

Jianing Song, Nov 26 2018

Keywords

Comments

For n > 0, a(n) is the unique solution to x^2 == 3 (mod 13^n) in the range [0, 13^n - 1] and congruent to 4 modulo 13.
A322086 is the approximation (congruent to 9 mod 13) of another square root of 3 over the 13-adic field.

Examples

			4^2 = 16 = 1*13 + 3.
108^2 = 11664 = 69*13^2 + 3.
1122^2 = 1258884 = 573*13^3 + 3.

Links

Crossrefs

Cf. A286840, A286841, A322086, A322087, A322089, A322090.

Programs

  • Maple
    S:= map(t -> op([1,3],t),[padic:-evalp(RootOf(x^2-3,x),13,30)]):
S4:= op(select(t -> t[1]=4, S)):
seq(add(S4[i]*13^(i-1),i=1..n-1),n=1..31); # Robert Israel, Jun 13 2019
  • PARI
    a(n) = truncate(sqrt(3+O(13^n)))

Formula

For n > 0, a(n) = 13^n - A322086(n).
a(n) = Sum_{i=0..n-1} A322087(i)*13^i.
a(n) = A286840(n)*A322089(n) mod 13^n = A286841(n)*A322090(n) mod 13^n.

A322089 One of the two successive approximations up to 13^n for 13-adic integer sqrt(-3). Here the 6 (mod 13) case (except for n = 0).

Original entry on oeis.org

0, 6, 45, 2073, 15255, 300865, 2899916, 22207152, 273201220, 7614777709, 92450772693, 1333177199334, 4917497987408, 191302178967256, 1705677711928521, 48954194340319989, 202511873382592260, 3529594919298491465, 38131258596823843197, 38131258596823843197, 8809653000849500507259
Offset: 0

Views

Author

Jianing Song, Nov 26 2018

Keywords

Comments

For n > 0, a(n) is the unique solution to x^2 == -3 (mod 13^n) in the range [0, 13^n - 1] and congruent to 6 modulo 13.
A322090 is the approximation (congruent to 7 mod 13) of another square root of -3 over the 13-adic field.

Examples

			6^2 = 36 = 3*13 - 3.
45^2 = 2025 = 12*13^2 - 3.
2073^2 = 4297329 = 1956*13^3 - 3.

Links

Crossrefs

Cf. A114525, A286840, A286841, A322085, A322086, A322090, A322091.

Programs

  • PARI
    a(n) = truncate(sqrt(-3+O(13^n)))

Formula

For n > 0, a(n) = 13^n - A322090(n).
a(n) = Sum_{i=0..n-1} A322091(i)*13^i.
a(n) = A286840(n)*A322086(n) mod 13^n = A286841(n)*A322085(n) mod 13^n.
a(n) == L(13^n,6) (mod 13^n) == (3 + sqrt(10))^(13^n) + (3 - sqrt(10))^(13^n) (mod 13^n), where L(n,x) denotes the n-th Lucas polynomial, the n-th row polynomial of A114525. - Peter Bala, Dec 05 2022

A322090 One of the two successive approximations up to 13^n for 13-adic integer sqrt(3). Here the 7 (mod 13) case (except for n = 0).

Original entry on oeis.org

0, 7, 124, 124, 13306, 70428, 1926893, 40541365, 542529501, 2989721664, 45407719156, 458983194703, 18380587135073, 111572927624997, 2231698673770768, 2231698673770768, 462904735800587581, 5120821000082846468, 74324148355133549932, 1423789031778622267480, 10195310774031298931542
Offset: 0

Views

Author

Jianing Song, Nov 26 2018

Keywords

Comments

For n > 0, a(n) is the unique solution to x^2 == -3 (mod 13^n) in the range [0, 13^n - 1] and congruent to 7 modulo 13.
A322089 is the approximation (congruent to 6 mod 13) of another square root of -3 over the 13-adic field.

Examples

			7^2 = 49 = 4*13 - 3.
124^2 = 15376 = 91*13^2 - 3 = 7*13^3 - 3.
13306^2 = 177049636 = 6199*13^4 - 3.

Links

Crossrefs

Cf. A114525, A286840, A286841, A322085, A322086, A322089, A322092.

Programs

  • PARI
    a(n) = truncate(-sqrt(-3+O(13^n)))

Formula

For n > 0, a(n) = 13^n - A322089(n).
a(n) = Sum_{i=0..n-1} A322092(i)*13^i.
a(n) = A286840(n)*A322085(n) mod 13^n = A286841(n)*A322086(n) mod 13^n.
a(n) == L(13^n,7) (mod 13^n) == ((7 + sqrt(53))/2)^(13^n) + ((7 - sqrt(53))/2)^(13^n) (mod 13^n), where L(n,x) denotes the n-th Lucas polynomial, the n-th row polynomial of A114525. - Peter Bala, Dec 05 2022
Showing 1-6 of 6 results.

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

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