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 1-4 of 4 results.

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.

A322086 One of the two successive approximations up to 13^n for 13-adic integer sqrt(3). Here the 9 (mod 13) case (except for n = 0).

Original entry on oeis.org

0, 9, 61, 1075, 9863, 9863, 3722793, 56817692, 245063243, 2692255406, 23901254152, 1540344664491, 12293307028713, 198677988008561, 804428201193067, 24428686515388801, 75614579529479558, 741031188712659399, 26692278946856673198, 813880127610558425101, 11047322160238681199840
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 9 modulo 13.
A322085 is the approximation (congruent to 4 mod 13) of another square root of 3 over the 13-adic field.

Examples

			9^2 = 81 = 6*13 + 3.
61^2 = 3721 = 22*13^2 + 3.
1075^2 = 1155625 = 526*13^3 + 3.

Links

Crossrefs

Cf. A286840, A286841, A322085, A322088, A322089, A322090.

Programs

  • Maple
    S:= map(t -> op([1,3],t),[padic:-evalp(RootOf(x^2-3,x),13,30)]):
S9:= op(select(t -> t[1]=9, S)):
seq(add(S9[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 - A322085(n).
a(n) = Sum_{i=0..n-1} A322088(i)*13^i.
a(n) = A286840(n)*A322090(n) mod 13^n = A286841(n)*A322089(n) mod 13^n.

A322092 Digits of one of the two 13-adic integers sqrt(-3).

Original entry on oeis.org

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

Views

Author

Jianing Song, Nov 26 2018

Keywords

Comments

This square root of -3 in the 13-adic field ends with digit 7. The other, A322091, ends with digit 6.

Examples

			...96AA70B9168BB38376AB76C879074A34388526097.

Links

Crossrefs

Cf. A114525, A286838, A286839, A322087, A322088, A322090, A322091.

Programs

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

Formula

a(n) = (A322090(n+1) - A322090(n))/13^n.
For n > 0, a(n) = 12 - A322091(n).
Equals A286838*A322087 = A286839*A322088.
This 13-adic integer is the 13-adic limit as n -> oo of the integer sequence {L(13^n,7)}, where L(n,x) denotes the n-th Lucas polynomial, the n-th row polynomial of A114525. - Peter Bala, Dec 05 2022

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
Showing 1-4 of 4 results.

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