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.

Previous Showing 11-15 of 15 results.

A296938 Rational primes that decompose in the field Q(sqrt(17)).

Original entry on oeis.org

2, 13, 19, 43, 47, 53, 59, 67, 83, 89, 101, 103, 127, 137, 149, 151, 157, 179, 191, 223, 229, 239, 251, 257, 263, 271, 281, 293, 307, 331, 349, 353, 359, 373, 383, 389, 409, 421, 433, 443, 457, 461, 463, 467, 491, 509, 523, 557, 563, 569, 577, 587, 593, 599
Offset: 1

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Author

N. J. A. Sloane, Dec 26 2017

Keywords

Comments

From Jianing Song, Apr 21 2022: (Start)
Primes p such that kronecker(17, p) = kronecker(p, 17) = 1, where kronecker() is the kronecker symbol. That is to say, primes p that are quadratic residues modulo 17.
Primes p such that p^8 == 1 (mod 17).
Primes p == 1, 2, 4, 8, 9, 13, 15, 16 (mod 17). (End)

Crossrefs

Cf. A011584 (kronecker symbol modulo 17).
Rational primes that decompose in the quadratic field with discriminant D: A139513 (D=-20), A191019 (D=-19), A191018 (D=-15), A296920 (D=-11), A033200 (D=-8), A045386 (D=-7), A002144 (D=-4), A002476 (D=-3), A045468 (D=5), A001132 (D=8), A097933 (D=12), A296937 (D=13), this sequence (D=17).
Cf. A038890 (inert rational primes in the field Q(sqrt(17))).

Programs

A344234 Irregular triangle read by rows: row n gives the pairs of proper solutions (X, Y), with gcd(X, Y) = 1 and X >= 0, of the Diophantine equation 2*X^2 + 2*X*Y + 3*Y^2 = A344232(n), for n >= 1.

Original entry on oeis.org

1, 0, 0, 1, 1, -1, 1, 1, 2, -1, 1, -2, 2, 1, 3, -1, 1, 2, 3, -2, 1, -3, 2, -3, 3, 1, 4, -1, 1, 3, 4, -3, 1, -4, 3, 2, 3, -4, 5, -2, 4, 1, 5, -1, 5, 2, 7, -2, 4, 3, 7, -3, 1, 5, 6, 1, 6, -5, 7, -1, 3, 4, 7, -4, 1, -6, 5, -6
Offset: 1

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Author

Wolfdieter Lang, May 17 2021

Keywords

Comments

The length of row n is r(n) = 2*A343240(b(n)), if A344232(n) = A343238(b(n)), for n >= 1. This sequence begins 2*(1, 2, 2, 1, 2, 2, 2, 2, 2, 4, 2, 2, 2, 4, 2, 2, 2, 4, 2, 2, ...).
See A344231 for references and links on parallel forms and half-reduced right neighbor forms (R-transformations), and also for the remark on the equivalent reduced form [2, -2, 3].
The number of proper solutions (X, Y), with X > 0, is 1 for n = 1 and 4. X = 0 only for n = 2, but the solution (0, -1) = (-0, -1) is not listed here.
For other n each distinct odd prime from {1, 3, 7, 9} (mod 20), i.e., from A139513, that divides A344232(n) contributes a factor of 2 to the listed number of solutions. See A343238 and A343240 for the multiplicities.
Only solutions with nonnegative X are listed. There is also the corresponding solution (-X, -Y). Hence the total number of signed solution is twice the number considered here.

Examples

			The irregular triangle T(n, m) begins (A(n) = A344232(n)):
n   A(n) \ m  1  2   3  4   5  6   7  8 ...
1,   2:       1  0
2,   3:       0  1   1 -1
3,   7:       1  1   2 -1
4,  10:       1 -2
5,  15:       2  1   3 -1
6,  18:       1  2   3 -2
7,  23:       1 -3   2 -3
8,  27:       3  1   4 -1
9,  35:       1  3   4 -3
10, 42:       1 -4   3  2   3 -4  5 -2
11, 43:       4  1   5 -1
12, 47:       2  3   5 -3
13, 58:       1  4   5 -4
14, 63:       2 -5   3 -5   5  1   6 -1
15, 67:       1 -5   4 -5
16, 82:       5  2   7 -2
17, 83:       4  3   7 -3
18, 87:       1  5   6  1   6 -5   7 -1
19, 90:       3  4   7 -4
20, 98:       1 -6   5 -6
...
n = 2: The prime 3 is a member of A139513, hence 2^1 = 2 solutions are listed. There are also the corresponding (-X, -Y) solutions.
n = 4: 10 = A344232(4) = A343238(8) = 2*5, A343240(8) = 1, hence there is 1 pair of proper solution with X >= 0. This is because neither 2 nor 5  are primes from A139513. There is also the solution (-1, 2).
n = 6: Prime 3 is a member of A139513, not prime 2. This there are 2 solutions listed. The solution (3, 0) does not appear; it is not proper.
n = 10: 42 = A344232(10) = A343238(19) = 2*3*7,  A343240(19) = 2^2 = 4, hence there are 4 pairs of proper solution with X >= 0. 3 and 7 are primes from A139513.
		

Crossrefs

Formula

T(n, m) gives for m = 2^j - 1, the nonnegative X(n, j) solution, and for m = 2*j the Y(n, j) solution of 2*T(n, 2*j-1)^2 + 2*T(n, 2*j-1)*T(n, 2*j) + 3*T(n, 2*j)^2 = A344232(n), for j = 1, 2, ..., r(n), for n >= 1. For n = 2 the solution (0, -1) is not listed here.

A344233 Irregular triangle read by rows: row n gives the pairs of proper solutions (X, Y), with gcd(X, Y) = 1 and X >= 0, of the Diophantine equation X^2 + 5*Y^2 = A344231(n), for n >= 1.

Original entry on oeis.org

1, 0, 0, 1, 1, 1, 1, -1, 2, 1, 2, -1, 3, 1, 3, -1, 1, 2, 1, -2, 4, 1, 4, -1, 3, 2, 3, -2, 5, 1, 5, -1, 6, 1, 6, -1, 5, 2, 5, -2, 1, 3, 1, -3, 2, 3, 2, -3, 7, 1, 7, -1, 4, 3, 3, -3, 8, 1, 8, -1, 7, 2, 7, -2, 5, 3, 5, -3, 1, 4, 1, -4, 9, 1, 9, -1, 3, 4, 3, -4, 7, 3, 7, -3
Offset: 1

Views

Author

Wolfdieter Lang, May 17 2021

Keywords

Comments

The length of row n is r(n) = 2*A343240(b(n)), if A344231(n) = A343238(b(n)), for n >= 1. This sequence begins 2*(1, 1, 2, 2, 2, 4, 2, 2, 2, 2, 2, 2, 2, 2, 4, 2, 2, 2, 2, 2, ...).
The number of proper solutions (X, Y), with X > 0, is 1 for n = 1. X = 0 only for n = 2, and for n >= 2 only one half of the solutions are listed, namely those with X >= 0. There are also the solutions with (-X, -Y). Thus the total number of solutions for n >= 2 is actually r(n) given above.
For n >= 2 each distinct odd primes from {1, 3, 7, 9} (mod 20), i.e., from
A139513, that divides A344231(n) contributes a factor 2 to the number of solutions listed here. See A343238 and the corresponding A343240.

Examples

			The irregular triangle T(n, m) begins (A(n) = A344231(n)):
n   A(n) \ m  1  2   3  4   5  6   7  8 ...
1,   1:       1  0
2,   5:       0  1
3,   6:       1  1   1 -1
4,   9:       2  1   2 -1
5,  14:       3  1   3 -1
6,  21:       1  2   1 -2   4  1   4 -1
7,  29:       3  2   3 -2
8,  30:       5  1   5 -1
9,  41:       6  1   6 -1
10, 45:       5  2   5 -2
11, 46:       1  3   1 -3
12, 49:       2  3   2 -3
13, 54:       7  1   7 -1
14, 61:       4  3   3 -3
15, 69:       8  1   8 -1   7  2   7 -2
16, 70:       5  3   5 -3
17, 81:       1  4   1 -4
18, 86:       9  1   9 -1
19, 89:       3  4   3 -4
20, 94:       7  3   7 -3
...
n = 2: Prime 5 is not a member of A139513, therefore only 1 solution appears here (see the remark above on the solution (0, -1)).
n = 4: Prime 3 a member A139513. Thus there are 2^1 = 2 solutions are listed. The solution (3, 0) does not appear; it is not proper.
n = 6: 21 = A344231(6) = A343238(12) = 3*7. hence A343240(12) = 2^2 = 4 and there are 4 pairs of proper solutions with X >= 0.
		

Crossrefs

Formula

T(n, m) gives for m = 2^j-1, the nonnegative X(n, j) solution, and for m = 2*j the Y(n, j) solution of T(n, 2*j-1)^2 + 5*T(n, 2*j)^2 = A344231(n), for j = 1 ..r(n), for n >= 1. For n = 2 (X(2) = 0) the solution (0, -1) is not listed.

A363415 a(n) = the real part of Product_{k = 0..n} 1 + k*sqrt(-5).

Original entry on oeis.org

1, 1, -9, -54, 426, 6426, -50274, -1465884, 10992996, 552727476, -3792193524, -312571718424, 1853425616616, 248005863100296, -1173524207653224, -263102748395914224, 865735128320476176, 359884863190774985616, -584551982838131141904, -616984573598760535235424, -155177934223071790979424
Offset: 0

Views

Author

Peter Bala, Jun 01 2023

Keywords

Comments

Compare with A105750(n) = the real part of Product_{k = 0..n} 1 + k*sqrt(-1).
Moll (2012) studied the prime divisors of the terms of A105750 and divided the primes into three classes. Numerical calculation suggests that a similar division also holds in this case.
Type 1: primes that do not divide any element of the sequence {a(n)}.
We conjecture that the set of type 1 primes begins {5, 11, 13, 31, 53, 79, 97, 113, 137, 157, 179, 193, 197, ...}.
Type 2: primes p such that the p-adic valuation v_p(a(n)) has asymptotically linear behavior. An example is given below.
We conjecture that the set of type 2 primes begins {2, 3, 7, 23, 29, 41, 43, 47, 61, 67, 83, 89, 101, ...} and consists of primes p == 1, 3, 7 or 9 (mod 20), equivalently, rational primes that split in the field extension Q(sqrt(-5)) of Q, together with the prime p = 2. See A139513.
It can be shown that the 2-adic valuation v_2(a(n)) = floor((n+1)/4).
Moll's conjecture 5.5 extends to this sequence: for type two primes p > 2, v_p(a(n)) ~ n/(p - 1) as n -> oo.
Type 3: primes p such that the sequence of p-adic valuations {v_p(a(n)) : n >= 0} exhibits an oscillatory behavior (this phrase is not precisely defined). An example is given below.
We conjecture that the set of type 3 primes begins {17, 19, 37, 59, 71, 73, 131, 151, 173, 191, 199, ...}.
Taken together, the type 1 and type 3 primes appear to consist of primes p == 11, 13, 17 or 19 (mod 20), equivalently, primes that remain inert in the field extension Q(sqrt(-5)) of Q, together with the prime p = 5, which ramifies in Q(sqrt(-5)). See A003626.

Examples

			Type 2 prime p = 3: the sequence of 3-adic valuations [v_3(a(n)) : n = 0..80] = [0, 0, 2, 3, 1, 3, 3, 3, 4, 4, 4, 5, 5, 5, 7, 7, 7, 8, 8, 8, 9, 9, 9, 12, 12, 12, 13, 13, 13, 14, 14, 14, 16, 16, 16, 17, 17, 17, 18, 18, 18, 20, 20, 20, 21, 21, 21, 22, 22, 22, 25, 25, 25, 26, 26, 26, 27, 27, 27, 29, 29, 29, 30, 30, 30, 31, 31, 31, 33, 33, 33, 34, 34, 34, 35, 35, 35, 39, 39, 40, 40].
Note that v_3(a(80)) = 40 = 80/(3 - 1), in agreement with the asymptotic behavior for type 2 primes conjectured above.
Type 3 prime p = 17: the sequence of 17-adic valuations [v_17(a(n)) : n = 0..100] = [0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0], showing the oscillatory behavior for type 3 primes conjectured above.
		

Crossrefs

Programs

  • Maple
    a := proc(n) option remember; if n = 0 then 1 elif n = 1 then 1 else (
    (2*n - 1)*a(n-1) - n*(5*n^2 - 10*n + 6)*a(n-2) )/(n - 1) end if; end:
    seq(a(n), n = 0..20);

Formula

a(n) = Sum_{k = 0..floor((n+1)/2)} (-5)^k*Stirling1(n+1,n+1-2*k).
a(n+1)/a(n) = 1 - (5*n + 5)*tan(Sum_{k = 1..n} arctan(sqrt(5)*k))/sqrt(5).
P-recursive: (n - 1)*a(n) = (2*n - 1)*a(n-1) - n*(5*n^2 - 10*n + 6)*a(n-2) with
a(0) = a(1) = 1.

A139538 Odd numbers of the form x^2+5y^2 not divisible by 5, with both x and y >== 1.

Original entry on oeis.org

9, 21, 29, 41, 49, 61, 69, 81, 89, 101, 109, 129, 141, 149, 161, 181, 189, 201, 229, 241, 249, 261, 269, 281, 301, 309, 321, 329, 349, 369, 381, 389, 401, 409, 421, 441, 449, 461, 469, 489, 501, 509, 521, 529, 541, 549, 569, 581, 601, 609, 621, 641, 661, 669
Offset: 1

Views

Author

Artur Jasinski, Apr 25 2008, Apr 28 2008

Keywords

Comments

All numbers in this sequence are divisible by primes A139513

References

  • Dirichlet & Dedekins Lectures on Number Theory (English Translation 1999) p. 119.

Programs

  • Mathematica
    a = {}; Do[Do[k = x^2 + 5 y^2; If[OddQ[k], If[Mod[k, 5] != 0, AppendTo[a, k]]], {x, 1, 100}], {y, 1, 100}]; Union[a] (*Artur Jasinski*)
    With[{upto=1000},Select[Union[#[[1]]^2+5#[[2]]^2&/@Tuples[Range[Floor[ Sqrt[ upto-5]]],2]],OddQ[#]&&Mod[#,5]!=0&&#<=upto&]] (* Harvey P. Dale, Jul 25 2019 *)

Extensions

Definition clarified by Harvey P. Dale, Jul 25 2019
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