A210851 -id:A210851 - cp's OEIS frontend

A048898 One of the two successive approximations up to 5^n for the 5-adic integer sqrt(-1).

0, 2, 7, 57, 182, 2057, 14557, 45807, 280182, 280182, 6139557, 25670807, 123327057, 123327057, 5006139557, 11109655182, 102662389557, 407838170807, 3459595983307, 3459595983307, 79753541295807, 365855836217682, 2273204469030182, 2273204469030182, 49956920289342682

Offset: 0

Michael Somos, Jul 26 1999

This is the root congruent to 2 mod 5.
Or, residues modulo 5^n of a 5-adic solution of x^2+1=0.
The radix-5 expansion of a(n) is obtained from the n rightmost digits in the expansion of the following pentadic integer:
  ...422331102414131141421404340423140223032431212 = u
The residues modulo 5^n of the other 5-adic solution of x^2+1=0 are given by A048899 which corresponds to the pentadic integer -u:
  ...022113342030313303023040104021304221412013233 = -u
The digits of u and -u are given in A210850 and A210851, respectively. - Wolfdieter Lang, May 02 2012
For approximations for p-adic square roots see also the W. Lang link under A268922. - Wolfdieter Lang, Apr 03 2016
From Jianing Song, Sep 06 2022: (Start)
For n > 0, a(n)-1 is one of the four solutions to x^4 == -4 (mod 5^n), the one that is congruent to 1 modulo 5.
For n > 0, a(n)+1 is one of the four solutions to x^4 == -4 (mod 5^n), the one that is congruent to 3 modulo 5. (End)

			a(0)=0 because 0 satisfies any equation in integers modulo 1.
a(1)=2 because 2 is one solution of x^2+1=0 modulo 5. (The other solution is 3, which gives rise to A048899.)
a(2)=7 because the equation (5y+a(1))^2+1=0 modulo 25 means that y is 1 modulo 5.

J. H. Conway, The Sensual Quadratic Form, p. 118, Mathematical Association of America, 1997, The Carus Mathematical Monographs, Number 26.
K. Mahler, Introduction to p-Adic Numbers and Their Functions, Cambridge, 1973, p. 35.

Seiichi Manyama, Table of n, a(n) for n = 0..1431 (terms 0..200 from Vincenzo Librandi)
Peter Bala, Using Lucas polynomials to find the p -adic square roots of -1, -2 and -3, Dec 2022.
G. P. Michon, On the witnesses of a composite integer, Numericana.
G. P. Michon, Introduction to p-adic integers, Numericana.

The two successive approximations up to p^n for p-adic integer sqrt(-1): this sequence and A048899 (p=5), A286840 and A286841 (p=13), A286877 and A286878 (p=17).
Cf. A000351 (powers of 5), A034939(n) = Min(a(n), A048899(n)).
Different from A034935.

[n le 2 select 2*(n-1) else Self(n-1)^5 mod 5^(n-1): n in [1..30]]; // Vincenzo Librandi, Feb 29 2016

a[0] = 0; a[1] = 2; a[n_] := a[n] = Mod[a[n-1]^5, 5^n]; Table[a[n], {n, 0, 21}] (* Jean-François Alcover, Nov 24 2011, after PARI *)
Join[{0}, RecurrenceTable[{a[1] == 2, a[n] == Mod[a[n-1]^5, 5^n]}, a, {n, 25}]] (* Vincenzo Librandi, Feb 29 2016 *)

PARI
```
{a(n) = if( n<2, 2, a(n-1)^5) % 5^n}
```

a(n) = lift(sqrt(-1 + O(5^n))); \\ Kevin Ryde, Dec 22 2020

If n>0, a(n) = 5^n - A048899(n).
From Wolfdieter Lang, Apr 28 2012: (Start)
Recurrence: a(n) = a(n-1)^5 (mod 5^n), a(1) = 2, n>=2. See the J.-F. Alcover Mathematica program and the PARI program below.
a(n) == 2^(5^(n-1)) (mod 5^n), n>=1.
a(n)*a(n-1) + 1 == 0 (mod 5^(n-1)), n>=1.
(a(n)^2 + 1)/5^n = A210848(n), n>=0.
(End)
Another recurrence: a(n) = modp(a(n-1) + a(n-1)^2 + 1, 5^n), n >= 2, a(1) = 2. Here modp(a, m) is the representative from {0, 1, ..., |m|-1} of the residue class a modulo m. Note that a(n) is in the residue class of a(n-1) modulo 5^(n-1) (see Hensel lifting). - Wolfdieter Lang, Feb 28 2016
a(n) == L(5^n,2) (mod 5^n), where L(n,x) denotes the n-th Lucas polynomial of A114525. - Peter Bala, Nov 20 2022

Additional comments from Gerard P. Michon, Jul 15 2009
Edited by N. J. A. Sloane, Jul 25 2009
Name clarified by Wolfdieter Lang, Feb 19 2016

A048899 One of the two successive approximations up to 5^n for 5-adic integer sqrt(-1).

Original entry on oeis.org

0, 3, 18, 68, 443, 1068, 1068, 32318, 110443, 1672943, 3626068, 23157318, 120813568, 1097376068, 1097376068, 19407922943, 49925501068, 355101282318, 355101282318, 15613890344818, 15613890344818, 110981321985443, 110981321985443, 9647724486047943, 9647724486047943

Offset: 0

Michael Somos, Jul 26 1999

This is the root congruent to 3 (mod 5) for n>0.
The other case with the 2 (mod 5) numbers (except for n=0) is given in A048898. - Wolfdieter Lang, Feb 19 2016
From Jianing Song, Sep 06 2022: (Start)
For n > 0, a(n)-1 is one of the four solutions to x^4 == -4 (mod 5^n), the one that is congruent to 2 modulo 5.
For n > 0, a(n)+1 is one of the four solutions to x^4 == -4 (mod 5^n), the one that is congruent to 4 modulo 5. (End)

			a(2) = 18 because the two roots of x^2 + 1 == 0 (mod 5^2) are 7 and 18 and 18 == 3 (mod 5). For 7 see A048898(2).

J. H. Conway, The Sensual Quadratic Form, p. 118, Mathematical Association of America, 1997, The Carus Mathematical Monographs, Number 26.
K. Mahler, Introduction to p-Adic Numbers and Their Functions, Cambridge, 1973, p. 35.

Seiichi Manyama, Table of n, a(n) for n = 0..1431
Peter Bala, Using Lucas polynomials to find the p-adic square roots of -1, -2 and -3, Dec 2022.

Cf. A048898, A066601, A114525, A210851.

[n le 2 select 3*(n-1) else Self(n-1)^5 mod 5^(n-1): n in [1..30]]; // Vincenzo Librandi, Feb 29 2016

Join[{0}, RecurrenceTable[{a[1] == 3, a[n] == Mod[a[n-1]^5, 5^n]}, a, {n, 25}]] (* Vincenzo Librandi, Feb 29 2016 *)

PARI
```
{a(n) = if( n<2, 3, a(n - 1)^5) % 5^n}
```

a(n) = lift(-sqrt(-1 + O(5^n))); \\ Kevin Ryde, Dec 22 2020

a(n) = 5^n - A048898(n), n>=1.
a(n) = A066601(5^n), n>=0.
0 <= a(n) < 5^n. 5^n divides a(n)^2 + 1.
From Wolfdieter Lang, Apr 28 2012: (Start)
Recurrence: a(n) = a(n-1)^5 (mod 5^n), a(1) = 3, n>=2. See the Pari program below, and the J.- F. Alcover Mathematica program for A048898.
a(n) = 3^(5^(n-1)) (mod 5^n), n>=1. Compare with the above given formula involving A066601.
a(n)*a(n-1) + 1 == 0 (mod 5^(n-1)), n>=1.
(a(n)^2 + 1)/5^n = A210849(n), n>=0.
(End)
Another recurrence: a(n) = modp(a(n-1) + 4*(a(n-1)^2 + 1), 5^n), n >= 2, a(1) = 3. Here modp(a, m) is the representative from {0, 1, ... ,|m|-1} of the residue class a modulo m. Note that a(n) is in the residue class of a(n-1) modulo 5^(n-1) (see Hensel lifting). - Wolfdieter Lang, Feb 28 2016
a(n) == L(5^n,3) (mod 5^n), where L(n,x) denotes the n-th Lucas polynomial of A114525. - Peter Bala, Nov 20 2022

Example corrected by Wolfdieter Lang, Apr 28 2012

Name clarified by Wolfdieter Lang, Feb 19 2016

A210850 Digits of one of the two 5-adic integers sqrt(-1).

Original entry on oeis.org

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

Offset: 0

Wolfdieter Lang, Apr 30 2012

See A048898 for the successive approximations to this 5-adic integer, called there u.
The digits of -u, the other 5-adic integer sqrt(-1), are given in A210851.
a(n) is the unique solution of the linear congruence 2*A048898(n)*a(n) + A210848(n) == 0 (mod 5), n>=1. Therefore only the values 0, 1, 2, 3 and 4 appear. See the Nagell reference given in A210848, eq. (6) on p. 86 adapted to this case. a(0)=2 follows from the formula given below.
If n>0, a(n) == A210848(n) (mod 5), since A048898(n) == 2 (mod 5). - Álvar Ibeas, Feb 21 2017
If a(n)=0 then A048899(n+1) and A048899(n) coincide.
a(n) + A210851(n) = 4 for n >= 1. - Robert Israel, Mar 04 2016
From Jianing Song, Sep 06 2022: (Start)
With a(0) = 1, this is the digits of one of the four 4th root of -4 in the ring of 5-adic integers, the one that is congruent to 1 modulo 5.
With a(0) = 3, this is the digits of one of the four 4th root of -4 in the ring of 5-adic integers, the one that is congruent to 3 modulo 5. (End)
This square root of -1 in the 5-adic integers is equal to the 5-adic limit of the sequence {L(5^n,2)}, where L(n,x) denotes the n-th Lucas polynomial, the n-th row polynomial of A114525. - Peter Bala, Dec 02 2022

			a(4) = 3 because 2*182*3 + 53 = 1145 == 0 (mod 5).
A048898(5) = 2057 = 2*5^0 + 1*5^1 + 2*5^2 + 1*5^3 + 3*5^4.
a(8) = 0, therefore A048898(9) = A048898(8) = Sum_{k=0..7} a(k)*5^k = 280182.

Robert Israel, Table of n, a(n) for n = 0..10000
Peter Bala, Using Lucas polynomials to find the p-adic square roots of -1, -2 and -3, Dec 2022.

Cf. A048898, A048899, A114525, A210848, A210849, A210851.

R:= select(t -> padic:-ratvaluep(t,1)=2,[padic:-rootp(x^2+1,5,10001)]):
op([1,1,3],R); # Robert Israel, Mar 04 2016

Table[Floor[First@Select[PowerModList[-1,1/2,5^(k+1)],Mod[#,5]==2&]/5^k],{k,0,99}] (* Giorgos Kalogeropoulos, Feb 28 2023 *)

a(n) = truncate(sqrt(-1+O(5^(n+1))))\5^n; \\ Michel Marcus, Mar 05 2016

a(n) = (b(n+1) - b(n))/5^n, n >= 0, with b(n):=A048898(n) computed from its recurrence. A Maple program for b(n) is given there.

A048898(n+1) = Sum_{k=0..n} a(k)*5^k, n >= 0.

Keyword "base" added by Jianing Song, Feb 17 2021

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

Original entry on oeis.org

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

A269591 Digits of one of the two 5-adic integers sqrt(-4).

Original entry on oeis.org

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

Offset: 0

Wolfdieter Lang, Mar 02 2016

This is the scaled first difference sequence of A268922.
The digits of the other 5-adic integer sqrt(-4), are given in A269592. See also A268922 for the two 5-adic numbers sqrt(-4), called u and -u.
a(n) is the unique solution of the linear congruence 2*A268922(n)*a(n) + A269593(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) = 1 follows from the formula given below.

			a(4) = 3 because 2*261*3 + 109 = 1675 == 0 (mod 5).
a(4) = - 109*(2*261)^3 (mod 5) = -(-1)*(2*1)^3 (mod 5) = 8 (mod 5) = 3.
A268922(5) = 2136 = 1*5^0 + 2*5^1 + 0*5^2 + 2*5^3 + 3*5^4.

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. A210850, A210851, A268922, A269592 (companion), A269593.

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) = A268922(n), n >= 0.
a(n) = -A269593(n)*(2*A268922(n))^3 (mod 5), n >= 1. Solution of the linear congruence see, e.g., Nagell, Theorem 38 pp. 77-78.
A268922(n+1) = sum(a(k)*5^k, k=0..n), n >= 0.

A286838 Digits of one of the two 13-adic integers sqrt(-1) (digits 0, 1, ... , 9, 10, 11, 12 are used instead of 0, 1, ... , 9, A, B, C).

Original entry on oeis.org

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

Offset: 0

Seiichi Manyama, Aug 01 2017

Seiichi Manyama, Table of n, a(n) for n = 0..10000

Digits of one of the two p-adic integers sqrt(-1): A210850 and A210851 (p=5), this sequence and A286839 (p=13).

def A(k, m, n)
  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 = b ** m
    mod *= m
  }
  d_ary
end
def A286838(n)
  A(5, 13, n)
end
p A286838(100)

Keyword "base" added by Jianing Song, Feb 17 2021

A286839 Digits of one of the two 13-adic integers sqrt(-1) (digits 0, 1, ... , 9, 10, 11, 12 are used instead of 0, 1, ... , 9, A, B, C).

Original entry on oeis.org

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

Offset: 0

Seiichi Manyama, Aug 01 2017

Seiichi Manyama, Table of n, a(n) for n = 0..10000

Digits of one of the two p-adic integers sqrt(-1): A210850 and A210851 (p=5), A286838 and this sequence (p=13).

def A(k, m, n)
  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 = b ** m
    mod *= m
  }
  d_ary
end
def A286839(n)
  A(8, 13, n)
end
p A286839(100)

a(n) = 12 - A286838(n) for n > 0.

Keyword "base" added by Jianing Song, Feb 17 2021

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)

A325490 Digits of one of the four 5-adic integers 6^(1/4) that is congruent to 2 mod 5.

Original entry on oeis.org

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

Offset: 0

Jianing Song, Sep 07 2019

One of the two square roots of A324026, where an A-number represents a 5-adic number. The other square root is A325491.

For k not divisible by 5, k is a fourth power in 5-adic field if and only if k == 1 (mod 5). If k is a fourth power in 5-adic field, then k has exactly 4 fourth-power roots.

			The unique number k in [1, 5^3] and congruent to 2 modulo 5 such that k^4 - 6 is divisible by 5^3 is k = 22 = (42)_5, so the first three terms are 2, 4 and 0.

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

Cf. A210850, A210851, A324026, A325484, A325485, A325486, A325487.
Digits of p-adic fourth-power roots:
A325489, this sequence, A325491, A325492 (5-adic, 6^(1/4));
A324085, A324086, A324087, A324153 (13-adic, 3^(1/4)).

S:= select(t -> op([1,3,1],t)=2, [padic:-rootp(_Z^4-6,5,100)]):
op([1,1,3],S); # Robert Israel, Mar 23 2023

a(n) = lift(sqrtn(6+O(5^(n+1)), 4) * sqrt(-1+O(5^(n+1))))\5^n

Equals A325489*A210850 = A325492*A210851.
a(n) = (A325485(n+1) - A325485(n))/13^n.
For n > 0, a(n) = 4 - A325491(n).

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

cp's OEIS Frontend

A048898 One of the two successive approximations up to 5^n for the 5-adic integer sqrt(-1).

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A048899 One of the two successive approximations up to 5^n for 5-adic integer sqrt(-1).

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A210850 Digits of one of the two 5-adic integers sqrt(-1).

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

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A269591 Digits of one of the two 5-adic integers sqrt(-4).

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A286838 Digits of one of the two 13-adic integers sqrt(-1) (digits 0, 1, ... , 9, 10, 11, 12 are used instead of 0, 1, ... , 9, A, B, C).

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A286839 Digits of one of the two 13-adic integers sqrt(-1) (digits 0, 1, ... , 9, 10, 11, 12 are used instead of 0, 1, ... , 9, A, B, C).