A269591 -id:A269591 - cp's OEIS frontend

A114525 Triangle of coefficients of the Lucas (w-)polynomials.

2, 0, 1, 2, 0, 1, 0, 3, 0, 1, 2, 0, 4, 0, 1, 0, 5, 0, 5, 0, 1, 2, 0, 9, 0, 6, 0, 1, 0, 7, 0, 14, 0, 7, 0, 1, 2, 0, 16, 0, 20, 0, 8, 0, 1, 0, 9, 0, 30, 0, 27, 0, 9, 0, 1, 2, 0, 25, 0, 50, 0, 35, 0, 10, 0, 1, 0, 11, 0, 55, 0, 77, 0, 44, 0, 11, 0, 1, 2, 0, 36, 0, 105, 0, 112, 0, 54, 0, 12, 0, 1

Offset: 0

Eric W. Weisstein, Dec 06 2005

Unsigned version of A108045.

The row reversed triangle is A162514. - Paolo Bonzini, Jun 23 2016

			2, x, 2 + x^2, 3*x + x^3, 2 + 4*x^2 + x^4, 5*x + 5*x^3 + x^5, ... give triangle
  n\k   0  1  2  3  4  5  6  7  8  9 10 ...
  0:    2
  1:    0  1
  2:    2  0  1
  3:    0  3  0  1
  4:    2  0  4  0  1
  5:    0  5  0  5  0  1
  6:    2  0  9  0  6  0  1
  7:    0  7  0 14  0  7  0  1
  8:    2  0 16  0 20  0  8  0  1
  9:    0  9  0 30  0 27  0  9  0  1
  10:   2  0 25  0 50  0 35  0 10  0  1
  n\k   0  1  2  3  4  5  6  7  8  9 10 ...
  .... reformatted by _Wolfdieter Lang_, Feb 10 2023

G. C. Greubel, Rows n = 0..100 of triangle, flattened
Alexander Burstein and Louis W. Shapiro, Pseudo-involutions in the Riordan group and Chebyshev polynomials, arXiv:2502.13673 [math.CO], 2025.
B. Johnson, Fibonacci numbers and matrices, 2009.
Jong Hyun Kim, Hadamard products and tilings, JIS 12 (2009) 09.7.4.
M. Pétréolle, Characterization of Cyclically Fully commutative elements in finite and affine Coxeter Groups, arXiv preprint arXiv:1403.1130 [math.GR], 2014.
Eric Weisstein's World of Mathematics, Fibonacci Polynomial
Eric Weisstein's World of Mathematics, Lucas Polynomial

Cf. A108045 (signed version).
Cf. A034807, A162514.
Cf. Sequences L(n,x): A000032(x = 1), A002203 (x = 2), A006497 (x = 3), A014448 (x = 4), A087130 (x = 5), A085447 (x = 6), A086902 (x = 7), A086594 (x = 8), A087798 (x = 9), A086927 (x = 10), A001946 (x = 11), A086928 (x = 12), A088316 (x = 13), A090300 (x = 14), A090301 (x = 15), A090305 (x = 16), A090306 (x = 17), A090307 (x = 18), A090308 (x = 19), A090309 (x = 20), A090310 (x = 21), A090313 (x = 22), A090314 (x = 23), A090316 (x = 24), A087281 (x = 29), A087287 (x = 76), A089772 (x = 199).

Lucas := proc(n,x)
    option remember;
    if  n=0 then
        2;
    elif n =1 then
        x ;
    else
        x*procname(n-1,x)+procname(n-2,x) ;
    end if;
end proc:
A114525 := proc(n,k)
    coeftayl(Lucas(n,x),x=0,k) ;
end proc:
seq(seq(A114525(n,k),k=0..n),n=0..12) ; # R. J. Mathar, Aug 16 2019

row[n_] := CoefficientList[LucasL[n, x], x];
Table[row[n], {n, 0, 12}] // Flatten (* Jean-François Alcover, Aug 11 2018 *)

From Peter Bala, Mar 18 2015: (Start)
The Lucas polynomials L(n,x) satisfy the recurrence L(n+1,x) = x*L(n,x) + L(n-1,x) with L(0,x) = 2 and L(1,x) = x.
O.g.f.: Sum_{n >= 0} L(n,x)*t^n = (2 - x*t)/(1 - t^2 - x*t) = 2 + x*t + (x^2 + 2)*t^2 + (3*x + x^3)*t^3 + ....
L(n,x) = trace( [ x, 1; 1, 0 ]^n ).
exp( Sum_{n >= 1} L(n,x)*t^n/n ) = Sum_{n >= 0} F(n+1,x)*t^n, where F(n,x) denotes the n-th Fibonacci polynomial. (see Appendix A3 in Johnson).
exp( Sum_{n >= 1} L(n,x)*L(2*n,x)*t^n/n ) = 1/( F(1,x)*F(2*x)*F(3,x) ) * Sum_{n >= 0} F(n+1,x)*F(n+2,x)*F(n+3,x)*t^n.
exp( Sum_{n >= 1} L(3*n,x)/L(n,x)*t^n/n ) = Sum_{n >= 0} L(2*n + 1,x)*t^n.
L(n,1) = Lucas(n) = A000032(n); L(n,4) = Lucas(3*n) = A014448(n); L(n,11) = Lucas(5*n) = A001946(n); L(n,29) = Lucas(7*n) = A087281(n); L(n,76) = Lucas(9*n) = A087287(n); L(n,199) = Lucas(11*n) = A089772(n). The general result is L(n,Lucas(2*k + 1)) = Lucas((2*k + 1)*n). (End)
From Jeremy Dover, Jun 10 2016: (Start)
Read as a triangle T(n,k), n >= 0, n >= k >= 0, T(n,k) = (Binomial((n+k)/2,k) + Binomial((n+k-2)/2,k))*(1+(-1)^(n-k))/2.
T(n,k) = A046854(n-1,k-1) + A046854(n-1,k) + A046854(n-2,k) for even n+k with n+k > 0, assuming A046854(n,k) = 0 for n < 0, k < 0, k > n.
T(n,k) is the number of binary strings of length n with exactly k pairs of consecutive 0's and no pair of consecutive 1's, where the first and last bits are considered consecutive. (End)
From Peter Bala, Sep 03 2022: (Start)
L(n,x) = 2*(i)^n*T(n,-i*x/2), where i = sqrt(-1) and T(n,x) is the n-th Chebyshev polynomial of the first kind.
d/dx(L(n,x)) = n*F(n,x), where F(n,x) denotes the n-th Fibonacci polynomial.
Let P_n(x,y) = (L(n,x) - L(n,y))/(x - y). Then {P_n(x,y): n >= 1} is a fourth-order linear divisibility sequence of polynomials in the ring Z[x,y]: if m divides n in Z then P_m(x,y) divides P_n(x,y) in Z[x,y].
P_n(1,1) = A045925(n); P_n(1,4) = A273622; P_n(2,2) = A093967(n).
L(2*n,x)^2 - L(2*n-1,x)*L(2*n+1,x) = x^2 + 4 for n >= 1.
Sum_{n >= 1} L(2*n,x)/( L(2*n-1,x) * L(2*n+1,x) ) = 1/x^2 and
Sum_{n >= 1} (-1)^(n+1)/( L(2*n,x) + x^2/L(2*n,x) ) = 1/(x^2 + 4), both valid for all nonzero real x. (End)
From Peter Bala, Nov 18 2022: (Start)
L(n,x) = Sum_{k = 0..floor(n/2)} (n/(n-k))*binomial(n-k,k)*x^(n-2*k) for n >= 1.
For odd m, L(n, L(m,x)) = L(n*m, x).
For integral x, the sequence {u(n)} := {L(n,x)} satisfies the Gauss congruences: u(m*p^r) == u(m*p^(r-1)) (mod p^r) for all positive integers m and r and all primes p.
Let p be an odd prime and let 0 <= k <= p - 1. Let alpha_k = the p-adic limit_{n -> oo} L(p^n,k). Then alpha_k is a well-defined p-adic-integer and the polynomial L(p,x) - x of degree p factorizes as L(p,x) - x = Product_{k = 0..p-1} (x - alpha_k). For example, L(5,x) - x = x^5 + 5*x^3 + 4*x = x*(x - A269591)*(x - A210850)*(x - A210851)*(x - A269592) in the ring of 5-adic integers. (End)
The formula for L(n,x) given in the first line of the preceding section, with L(0, x) = 2, is rewritten L(n, x) = Sum_{k = 0..floor(n/2)} A034807(n, k)*x^(n - 2*k). See the formula by Alexander Elkins in A034807. - Wolfdieter Lang, Feb 10 2023

A268922 One of the two successive approximations up to 5^n for the 5-adic integer sqrt(-4). These are the 1 mod 5 numbers, except for n = 0.

Original entry on oeis.org

0, 1, 11, 11, 261, 2136, 2136, 64636, 220886, 1392761, 7252136, 46314636, 241627136, 974049011, 2194752136, 8298267761, 99851002136, 710202564636, 710202564636, 12154294361511, 31227780689636

Offset: 0

Wolfdieter Lang, Mar 02 2016

The other approximation for the 5-adic integer sqrt(-4) with numbers 4 (mod 5) is given in A269590.
For the two approximations of the 5-adic integer sqrt(-1) see A048899 and A048898. For comments and some proofs see A210848.
For the digits of this 5-adic integer sqrt(-4), that is the scaled first differences, see A269591. This 5-adic number has the digits read from the right to the left ...32234111132111101130213043113443324032021 = u.
  The companion 5-adic number -u has digits ....12210333312333343314231401331001120412424. See A269592.
The recurrence given below (for n >= 1) has been derived from the following facts (i) x^2 + 4 == 0 (mod 5) has the two distinct solutions x(1) = 1 and x(2) = 5 - x(1) = 4. This guarantees the existence of a unique solution x = x1(n) of  x^2 + 4 == 0 (mod 5^n) , for n >= 2 , which satisfies also x1(n) == 1 (mod 5). See e.g., Theorem 50, p. 87 of the Nagell reference. The same is true for the solution x = x2(n) with x2(n) == 4 (mod 5^n). (ii) As a consequence of Hensel's lemma (see e.g., the Wikipedia reference under Hensel lifting) one knows that  x1(n) (which is treated here) satisfies the congruence x1(n) == x1(n-1) (mod 5^(n-1)) with x1(1) = x1 = 1. (A similar statement holds for x2(n) with input x2(1) = x(2) = 4. This is used in A269590). These two facts allow one to derive a recurrence for x1(n) (and for x2(n)).

			n=2:  11^2 + 4 = 125  == 0 (mod 5^2), and 125 is the only solution from {0, 1, ..., 24} which is congruent to 1 modulo 5.
n=3:  the only solution of x1^2 + 4 == 0 (mod 5^3) with x1 from {0, ..., 124} and x1 ==  1 (mod 4) is also 11. The number 114 satisfies also the first congruence but not the second one: 114 == 2 (mod 4).

Trygve Nagell, Introduction to Number Theory, Chelsea Publishing Company, New York, 1964, p. 87.

Seiichi Manyama, Table of n, a(n) for n = 0..1430
Wolfdieter Lang, Note on a Recurrence for Approximation Sequences of p-adic Square Roots
Wikipedia, Hensel's lemma.

Cf. A000032, A048899, A048898, A144837, A210848, A269590, A269591, A269592.

with(padic):D1:=op(3,op([evalp(RootOf(x^2+4),5,20)][1])): 0,seq(sum('D1[k]*5^(k-1)','k'=1..n), n=1..20);
# alternative program - see A144837
a := proc (n) option remember; if n = 1 then 1 else irem( a(n-1)^5 + 5*a(n-1)^3 + 5*a(n-1), 5^n) end if; end: seq(a(n), n = 1..20); # Peter Bala, Nov 14 2022

a(n) = truncate(sqrt(-4+O(5^(n)))); \\ Michel Marcus, Mar 04 2016

Recurrence for n >= 1: a(n) = modp( a(n-1) + 2*(a(n-1)^2 + 4), 5^n), n >= 2, with a(1) = 1. Here modp(a, m) is used to pick the representative of the residue class a modulo m from the smallest nonnegative complete residue system {0, 1, ... , m-1}.
a(n) = 5^n - A269590(n), n >= 1.
a(n) == Lucas(5^n) (mod 5^n). - Peter Bala, Nov 10 2022

A269592 Digits of one of the two 5-adic integers sqrt(-4). Here the ones related to A269590.

Original entry on oeis.org

4, 2, 4, 2, 1, 4, 0, 2, 1, 1, 0, 0, 1, 3, 3, 1, 0, 4, 1, 3, 2, 4, 1, 3, 3, 4, 3, 3, 3, 3, 2, 1, 3, 3, 3, 3, 0, 1, 2, 2, 1, 2, 0, 0, 4, 1, 3, 0, 4, 1, 1, 3, 4, 3, 1, 1, 2, 1, 1, 1, 0, 0, 1, 3, 1, 3

Offset: 0

Wolfdieter Lang, Mar 02 2016

This is the scaled first difference sequence of A269590.
The digits of the other 5-adic integer sqrt(-4), are given in A269591. See also A268922 for the two 5-adic numbers -u and u.
a(n) is the unique solution of the linear congruence 2*A269590(n)*a(n) + A269594(n) == 0 (mod 5), n>=1. Therefore only the values 0, 1, 2, 3 and 4 appear. See the Nagell reference given in A268922, eq. (6) on p. 86, adapted to this case. a(0) = 4 follows from the formula given below.

			a(4) = -212*(2*364)^3 (mod 5) = -2*(2*(-1))^3 (mod 5) = 1.

Seiichi Manyama, Table of n, a(n) for n = 0..10000
BCMATH Congruence Programs, Finding a p-adic square root of a quadratic residue (mod p), p an odd prime..

Cf. A269590, A269591 (companion), A269594.

a(n) = truncate(-sqrt(-4+O(5^(n+1))))\5^n; \\ Michel Marcus, Mar 04 2016

a(n) = (b(n+1) - b(n))/5^n, n>=0, with b(n) = A269590(n), n >= 0.
a(n) = -A269594(n)*(2*A269590(n))^3 (mod 5), n >= 1. Solution of the linear congruence see, e.g., Nagell, Theorem 38 pp. 77-78.
A269590(n+1) = sum(a(k)*5^k, k=0..n), n>=0.
a(n) = 4 - A269591(n) if n > 0 and a(0) = 5 - A269591(0) = 5-4 = 1. - Michel Marcus, Mar 31 2016

A269590 One of the two successive approximations up to 5^n for the 5-adic integer sqrt(-4). These are the 4 mod 5 numbers (except for n=0).

Original entry on oeis.org

0, 4, 14, 114, 364, 989, 13489, 13489, 169739, 560364, 2513489, 2513489, 2513489, 246654114, 3908763489, 22219310364, 52736888489, 52736888489, 3104494700989, 6919191966614

Offset: 0

Wolfdieter Lang, Mar 02 2016

The other approximation for the 5-adic integer sqrt(-4) with numbers 1 (mod 5) is given in A268922. See this also for more details and references.

For the digits of this 5-adic integer sqrt(-4), that is the scaled first differences, see A269592. See also A268922 for the 5-adic numbers u and -u written from the right to the left.

Seiichi Manyama, Table of n, a(n) for n = 0..1432

Cf. A048899, A048898, A269591, A268922, A269592.

with(padic): D2:=op(3,op([evalp(RootOf(x^2+4),5,20)][2])):
0,seq(sum('D2[k]*5^(k-1)','k'=1..n),n=1..20);
# alternative program
a := proc(n) option remember; if n = 1 then 4 else irem( a(n-1)^5 + 5*a(n-1)^3 + 5*a(n-1), 5^n) end if; end: seq(a(n), n = 1..20); # Peter Bala, Nov 14 2022

a(n) = if (n==0, 0, 5^n - truncate(sqrt(-4+O(5^(n))))); \\ Michel Marcus, Mar 07 2016

Recurrence for n >= 1: a(n) = modp( a(n-1) + 3*(a(n-1)^2 + 4), 5^n), n >= 2, with a(1) = 4. Here modp(a, m) is used to pick the representative of the residue class a modulo m from the smallest nonnegative complete residue system {0, 1, ... , m-1}.
a(n) = 5^n - A268922(n), n >= 1.
a(n) == Lucas(3*(5^n)) (mod 5^n). - Peter Bala, Nov 14 2022

A309443 Coefficients in 5-adic expansion of 4^(1/3).

Original entry on oeis.org

4, 1, 2, 4, 4, 3, 3, 4, 0, 4, 2, 1, 1, 1, 4, 2, 2, 3, 3, 2, 3, 4, 2, 3, 2, 0, 3, 4, 2, 1, 4, 3, 3, 3, 4, 4, 0, 3, 2, 0, 0, 2, 4, 2, 3, 4, 4, 1, 4, 4, 1, 3, 1, 2, 2, 0, 3, 0, 1, 1, 3, 2, 0, 0, 0, 1, 2, 1, 4, 2, 1, 0, 4, 0, 2, 1, 4, 0, 0, 3, 1, 0, 4, 1, 2, 4, 2, 0, 1, 4, 4

Offset: 0

Seiichi Manyama, Aug 03 2019

Robert Israel, Table of n, a(n) for n = 0..10000

Cf. A309444.
Digits of p-adic integers:
A269591, A269592 (5-adic, sqrt(-4));
A210850, A210851 (5-adic, sqrt(-1));
A290566 (5-adic, 2^(1/3));
A290563 (5-adic, 3^(1/3)).

op([1,3], padic:-rootp(x^3-4,5,101)); # Robert Israel, Aug 04 2019

Vecrev(digits(truncate((4+O(5^100))^(1/3)), 5))

require 'OpenSSL'
def f_a(ary, a)
  (0..ary.size - 1).inject(0){|s, i| s + ary[i] * a ** i}
end
def df(ary)
  (1..ary.size - 1).map{|i| i * ary[i]}
end
def A(c_ary, k, m, n)
  x = OpenSSL::BN.new((-f_a(df(c_ary), k)).to_s).mod_inverse(m).to_i % m
  f_ary = c_ary.map{|i| x * i}
  f_ary[1] += 1
  d_ary = []
  ary = [0]
  a, mod = k, m
  (n + 1).times{|i|
    b = a % mod
    d_ary << (b - ary[-1]) / m ** i
    ary << b
    a = f_a(f_ary, b)
    mod *= m
  }
  d_ary
end
def A309443(n)
  A([-4, 0, 0, 1], 4, 5, n)
end
p A309443(100)

A318962 Digits of one of the two 2-adic integers sqrt(-7) that ends in 01.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1

Offset: 0

Jianing Song, Sep 06 2018

Over the 2-adic integers there are 2 solutions to x^2 = -7, one ends in 01 and the other ends in 11. This sequence gives the former one. See A318960 for detailed information.

			...10110001110011100100110001100000010110101.

Jianing Song, Table of n, a(n) for n = 0..1000

Cf. A318960.
Digits of p-adic integers:
this sequence, A318963 (2-adic, sqrt(-7));
A271223, A271224 (3-adic, sqrt(-2));
A269591, A269592 (5-adic, sqrt(-4));
A210850, A210851 (5-adic, sqrt(-1));
A290566 (5-adic, 2^(1/3));
A290563 (5-adic, 3^(1/3));
A290794, A290795 (7-adic, sqrt(-6));
A290798, A290799 (7-adic, sqrt(-5));
A290796, A290797 (7-adic, sqrt(-3));
A212152, A212155 (7-adic, (1+sqrt(-3))/2);
A051277, A290558 (7-adic, sqrt(2));
A286838, A286839 (13-adic, sqrt(-1));
A309989, A309990 (17-adic, sqrt(-1)).
Also there are numerous sequences related to digits of 10-adic integers.

a(n) = truncate(-sqrt(-7+O(2^(n+2))))\2^n

a(0) = 1, a(1) = 0; for n >= 2, a(n) = 0 if A318960(n)^2 + 7 is divisible by 2^(n+2), otherwise 1.
a(n) = 1 - A318963(n) for n >= 1.
For n >= 2, a(n) = (A318960(n+1) - A318960(n))/2^n.

Corrected by Jianing Song, Aug 28 2019

A318963 Digits of one of the two 2-adic integers sqrt(-7) that ends in 11.

Original entry on oeis.org

1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 0, 1, 0, 0, 0, 1, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0

Offset: 0

Jianing Song, Sep 06 2018

Over the 2-adic integers there are 2 solutions to x^2 = -7, one ends in 01 and the other ends in 11. This sequence gives the latter one. See A318961 for detailed information.

			...01001110001100011011001110011111101001011.

Jianing Song, Table of n, a(n) for n = 0..1000

Cf. A318961.
Digits of p-adic integers:
A318962, this sequence (2-adic, sqrt(-7));
A271223, A271224 (3-adic, sqrt(-2));
A269591, A269592 (5-adic, sqrt(-4));
A210850, A210851 (5-adic, sqrt(-1));
A290566 (5-adic, 2^(1/3));
A290563 (5-adic, 3^(1/3));
A290794, A290795 (7-adic, sqrt(-6));
A290798, A290799 (7-adic, sqrt(-5));
A290796, A290797 (7-adic, sqrt(-3));
A212152, A212155 (7-adic, (1+sqrt(-3))/2);
A051277, A290558 (7-adic, sqrt(2));
A286838, A286839 (13-adic, sqrt(-1));
A309989, A309990 (17-adic, sqrt(-1)).
Also there are numerous sequences related to digits of 10-adic integers.

a(n) = if(n==1, 1, truncate(sqrt(-7+O(2^(n+2))))\2^n)

a(0) = a(1) = 1; for n >= 2, a(n) = 0 if A318961(n)^2 + 7 is divisible by 2^(n+2), otherwise 1.
a(n) = 1 - A318962(n) for n >= 1.
For n >= 2, a(n) = (A318961(n+1) - A318961(n))/2^n.

Corrected by Jianing Song, Aug 28 2019

A324025 Digits of one of the two 5-adic integers sqrt(6) that is related to A324023.

Original entry on oeis.org

1, 3, 0, 4, 2, 1, 2, 3, 1, 3, 3, 0, 3, 3, 2, 3, 2, 2, 2, 4, 3, 3, 1, 4, 0, 1, 2, 0, 0, 0, 3, 3, 1, 4, 1, 0, 1, 2, 4, 1, 4, 1, 1, 0, 2, 4, 4, 3, 0, 2, 3, 4, 1, 1, 4, 3, 4, 2, 4, 2, 1, 1, 2, 4, 4, 3, 2, 3, 1, 1, 0, 1, 4, 2, 3, 4, 4, 4, 4, 0, 3, 3, 1, 2, 3, 2, 3, 1

Offset: 0

Jianing Song, Sep 07 2019

This square root of 6 in the 5-adic field ends with digit 1. The other, A324026, ends with digit 4.

			The solution to x^2 == 6 (mod 5^4) such that x == 1 (mod 5) is x == 516 (mod 5^4), and 516 is written as 4031 in quinary, so the first four terms are 1, 3, 0 and 4.

Wikipedia, p-adic number

Cf. A324023, A324024.
Digits of 5-adic square roots:
A324029, A324030 (sqrt(-6));
A269591, A269592 (sqrt(-4));
A210850, A210851 (sqrt(-1));
this sequence, A324026 (sqrt(6)).

PARI
```
a(n) = truncate(sqrt(6+O(5^(n+1))))\5^n
```

a(n) = (A324023(n+1) - A324023(n))/5^n.
For n > 0, a(n) = 4 - A324026(n).
Equals A210850*A324030 = A210851*A324029, where each A-number represents a 5-adic number.

A324026 Digits of one of the two 5-adic integers sqrt(6) that is related to A324024.

Original entry on oeis.org

4, 1, 4, 0, 2, 3, 2, 1, 3, 1, 1, 4, 1, 1, 2, 1, 2, 2, 2, 0, 1, 1, 3, 0, 4, 3, 2, 4, 4, 4, 1, 1, 3, 0, 3, 4, 3, 2, 0, 3, 0, 3, 3, 4, 2, 0, 0, 1, 4, 2, 1, 0, 3, 3, 0, 1, 0, 2, 0, 2, 3, 3, 2, 0, 0, 1, 2, 1, 3, 3, 4, 3, 0, 2, 1, 0, 0, 0, 0, 4, 1, 1, 3, 2, 1, 2, 1, 3

Offset: 0

Jianing Song, Sep 07 2019

This square root of 6 in the 5-adic field ends with digit 4. The other, A324025, ends with digit 1.

			The solution to x^2 == 6 (mod 5^4) such that x == 4 (mod 5) is x == 109 (mod 5^4), and 109 is written as 414 in quinary, so the first four terms are 4, 1, 4 and 0.

Wikipedia, p-adic number

Cf. A324023, A324024.
Digits of 5-adic square roots:
A324029, A324030 (sqrt(-6));
A269591, A269592 (sqrt(-4));
A210850, A210851 (sqrt(-1));
A324025, this sequence (sqrt(6)).

a(n) = truncate(-sqrt(6+O(5^(n+1))))\5^n

a(n) = (A324024(n+1) - A324024(n))/5^n.
For n > 0, a(n) = 4 - A324025(n).
Equals A210850*A324029 = A210851*A324030, where each A-number represents a 5-adic number.

A309989 Digits of one of the two 17-adic integers sqrt(-1).

Original entry on oeis.org

4, 2, 10, 5, 12, 16, 12, 8, 13, 3, 14, 0, 6, 1, 0, 15, 1, 8, 14, 5, 7, 16, 14, 1, 5, 13, 9, 6, 5, 12, 16, 15, 9, 16, 14, 12, 16, 1, 3, 6, 4, 10, 15, 5, 16, 12, 2, 1, 5, 4, 0, 15, 2, 11, 14, 9, 5, 1, 11, 16, 15, 7, 5, 6, 14, 3, 12, 0, 0, 11, 12, 13, 9, 5, 4, 16, 13

Offset: 0

Jianing Song, Aug 26 2019

This square root of -1 in the 17-adic field ends with digit 4. The other, A309990, ends with digit 13 (D when written as a 17-adic number).

			The solution to x^2 == -1 (mod 17^4) such that x == 4 (mod 17) is x == 27493 (mod 17^4), and 27493 is written as 5A24 in heptadecimal, so the first four terms are 4, 2, 10 and 5.

Seiichi Manyama, Table of n, a(n) for n = 0..10000
Wikipedia, p-adic number

Cf. A286877, A286878.
Digits of p-adic square roots:
A318962, A318963 (2-adic, sqrt(-7));
A271223, A271224 (3-adic, sqrt(-2));
A269591, A269592 (5-adic, sqrt(-4));
A210850, A210851 (5-adic, sqrt(-1));
A290794, A290795 (7-adic, sqrt(-6));
A290798, A290799 (7-adic, sqrt(-5));
A290796, A290797 (7-adic, sqrt(-3));
A051277, A290558 (7-adic, sqrt(2));
A321074, A321075 (11-adic, sqrt(3));
A321078, A321079 (11-adic, sqrt(5));
A322091, A322092 (13-adic, sqrt(-3));
A286838, A286839 (13-adic, sqrt(-1));
A322087, A322088 (13-adic, sqrt(3));
this sequence, A309990 (17-adic, sqrt(-1)).

a(n) = truncate(sqrt(-1+O(17^(n+1))))\17^n

a(n) = (A286877(n+1) - A286877(n))/17^n.

For n > 0, a(n) = 16 - A309990(n).

cp's OEIS Frontend

A114525 Triangle of coefficients of the Lucas (w-)polynomials.

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A268922 One of the two successive approximations up to 5^n for the 5-adic integer sqrt(-4). These are the 1 mod 5 numbers, except for n = 0.

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A269592 Digits of one of the two 5-adic integers sqrt(-4). Here the ones related to A269590.

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A269590 One of the two successive approximations up to 5^n for the 5-adic integer sqrt(-4). These are the 4 mod 5 numbers (except for n=0).

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A309443 Coefficients in 5-adic expansion of 4^(1/3).

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A318962 Digits of one of the two 2-adic integers sqrt(-7) that ends in 01.

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A318963 Digits of one of the two 2-adic integers sqrt(-7) that ends in 11.

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