A034939 -id:A034939 - cp's OEIS frontend

A199206 a(n) = (1+(A034939(n))^2)/5^n.

1, 1, 2, 26, 53, 365, 73, 13369, 31226, 40193, 1346393, 10982633, 59784881, 12459674, 197301737, 4044371993, 16335212753, 165277755905, 33055551181, 627510050434, 2556360882389, 25830314642530, 5166062928506, 433477841994649, 1561599571802482

Offset: 0

Zak Seidov, Nov 04 2011

nonn

			a(0)=(1+(A034939(0))^2)/5^0=(1+(0)^2)/1=1,
a(1)=(1+(A034939(1))^2)/5^1=(1+(2)^2)/5=1,
a(2)=(1+(A034939(2))^2)/5^2=(1+(7)^2)/25=2,
a(3)=(1+(A034939(3))^2)/5^3=(1+(57)^2)/5^3=3250/125=26.

Zak Seidov, Table of n, a(n) for n = 0..100

Cf. A034939.

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

Original entry on oeis.org

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

A257833 Table T(k, n) of smallest bases b > 1 such that p = prime(n) satisfies b^(p-1) == 1 (mod p^k), read by antidiagonals.

Original entry on oeis.org

5, 8, 9, 7, 26, 17, 18, 57, 80, 33, 3, 18, 182, 242, 65, 19, 124, 1047, 1068, 728, 129, 38, 239, 1963, 1353, 1068, 2186, 257, 28, 158, 239, 27216, 34967, 32318, 6560, 513, 28, 333, 4260, 109193, 284995, 82681, 110443, 19682, 1025, 14, 42, 2819, 15541, 861642, 758546, 2387947, 280182, 59048, 2049

Offset: 2

Felix Fröhlich, May 10 2015

			T(3, 5) = 124, since prime(5) = 11 and the smallest b such that b^10 == 1 (mod 11^3) is 124.
Table starts
  k\n|    1     2       3        4       5       6         7
  ---+----------------------------------------------------------
   2 |    5     8       7       18       3      19        38 ...
   3 |    9    26      57       18     124     239       158 ...
   4 |   17    80     182     1047    1963     239      4260 ...
   5 |   33   242    1068     1353   27216  109193     15541 ...
   6 |   65   728    1068    34967  284995  861642    390112 ...
   7 |  129  2186   32318    82681  758546 6826318  21444846 ...
   8 |  257  6560  110443  2387947 9236508 6826318 112184244 ...
   9 |  513 19682  280182 14906455 ....
  10 | 1025 59048 3626068 ....
  ...

Chai Wah Wu, Table of n, a(n) for n = 2..10000

Column 1 of table is A000051.
Column 2 of table is A024023 (with offset 2).
Column 3 of table is A034939 (with offset 2).
Rows k=2-10 give: A039678, A249275, A353937, A353938, A353939, A353940, A353941, A353942, A353943.

for(k=2, 10, forprime(p=2, 25, b=2; while(Mod(b, p^k)^(p-1)!=1, b++); print1(b, ", ")); print(""))

T(k,n) = my(p=prime(n), v=List([2])); if(n==1, return(2^k+1)); for(i=1, k, w=List([]); for(j=1, #v, forstep(b=v[j], p^i-1, p^(i-1), if(Mod(b, p^i)^p==b, listput(w, b)))); v=Vec(w)); vecmin(v); \\ Jinyuan Wang, May 17 2022

from itertools import count, islice
from sympy import prime
from sympy.ntheory.residue_ntheory import nthroot_mod
def A257833_T(n,k): return 2**k+1 if n == 1 else int(nthroot_mod(1,(p:= prime(n))-1,p**k,True)[1])
def A257833_gen(): # generator of terms
    yield from (A257833_T(n,i-n+2) for i in count(1) for n in range(i,0,-1))
A257833_list = list(islice(A257833_gen(),50)) # Chai Wah Wu, May 17 2022

A218709 a(n) is smallest number such that a(n)^2 + 1 is divisible by 13^n.

Original entry on oeis.org

0, 5, 70, 239, 239, 143044, 1999509, 6826318, 6826318, 822557039, 52199939826, 603633907222, 11356596271444, 11356596271444, 1828607235824962, 13920898306972194, 13920898306972194, 2675587335039691558, 49226908181248336040, 513050126578538629605

Offset: 0

Michel Lagneau, Nov 04 2012

nonn

			a(4) = 239 because 239^2+1 = 2*13^4.

Cf. A002522, A049532, A034939.

b=5;n13=13;jo=Join[{0,b},Table[n13=13*n13;b=PowerMod[b,13,n13];b=Min[b,n13-b],{99}]]

A218710 a(n) is smallest number such that a(n)^2 + 1 is divisible by 17^n.

Original entry on oeis.org

0, 4, 38, 1985, 27493, 390112, 390112, 96940388, 3379649772, 24306922095, 450044583893, 5597937117454, 28673959190179, 3524407382568745, 13428985415474682, 13428985415474682, 5711417117604156904, 91610966633729580058, 6709533061724423693474

Offset: 0

Michel Lagneau, Nov 04 2012

nonn

			a(4) = 27493 because 27493^2+1 =  2 * 5 ^ 2 * 17 ^ 4 * 181.

Cf. A002522, A049532, A034939, A218709.

b=4;n17=17;jo=Join[{0,b},Table[n17=17*n17;b=PowerMod[b,17,n17];b=Min[b,n17-b],{99}]]

A218712 a(n) is the smallest number such that a(n)^2 + 1 is divisible by 29^n.

Original entry on oeis.org

0, 12, 41, 10133, 34522, 7745569, 253879357, 7986582530, 61012922706, 4563230639355, 67972499239990, 1330094199140593, 47471944863682723, 5000878909740249297, 5000878909740249297, 590115586441858677665, 77072583141941801290876, 423420364218752896284166

Offset: 0

Michel Lagneau, Nov 04 2012

nonn

			a(4) = 34522 because 34522^2+1 =  5 * 29 ^ 4 * 337.

Robert Israel, Table of n, a(n) for n = 0..682

Cf. A002522, A049532, A034939, A218709, A218710.

R:= 0,12: U:= [12,17]:
for n from 2 to 30 do
  qs:= map(u -> (u^2+1)/29^(n-1), U);
  ys:= [seq(-qs[i]/(2*U[i]) mod 29,i=1..2)];
  U:= U + ys*29^(n-1) mod 29^n;
  R:= R,min(U);
od:
R; # Robert Israel, Jan 13 2025

b=12;n29=29;jo=Join[{0,b},Table[n29=29*n29;b=PowerMod[b,29,n29];b=Min[b,n29-b],{99}]]

A218713 a(n) is smallest number such that a(n)^2 + 1 is divisible by 37^n.

Original entry on oeis.org

0, 6, 117, 9466, 800982, 6423465, 756360062, 24900904028, 1019349744435, 15069267560119, 794839706330581, 71333925879937154, 2419512779032508628, 116073623326088126430, 359642847542169431827, 144552623583462302226851, 3523356323886506075746572

Offset: 0

Michel Lagneau, Nov 04 2012

nonn

			a(3) = 9466 because 9466^2+1 =  29 * 37 ^ 3 * 61.

Cf. A002522, A049532, A034939, A218709, A218710, A218712.

b=6;n37=37;jo=Join[{0,b},Table[n37=37*n37;b=PowerMod[b,37,n37];b=Min[b,n37-b],{99}]]

A218714 a(n) is smallest number such that a(n)^2 + 1 is divisible by 41^n.

Original entry on oeis.org

0, 9, 378, 11389, 1251967, 46464143, 2363588163, 92615568742, 287369842623, 112076323050317, 2403749863808044, 56094387104417648, 1156752450536914530, 43970228150195457632, 10132163897314954464899, 503212117431217218892992, 19164391897329672149556204

Offset: 0

Michel Lagneau, Nov 04 2012

nonn

			a(3) = 11389 because 11389^2+1 =  2 * 41 ^ 3 * 941.

Cf. A002522, A049532, A034939, A218709, A218710, A218712, A218713.

b=9;n41=41;jo=Join[{0,b},Table[n41=41*n41;b=PowerMod[b,41,n41];b=Min[b,n41-b],{99}]]

A034935 Successive approximations to 5-adic integer sqrt(-1).

Original entry on oeis.org

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

Offset: 0

N. J. A. Sloane

This is the root congruent to 2 mod 5.

J. H. Conway, The Sensual Quadratic Form, p. 118.
K. Mahler, Introduction to p-Adic Numbers and Their Functions, Cambridge, 1973, p. 35.

Seiichi Manyama, Table of n, a(n) for n = 0..1121

Cf. A034939, A048898, A048899.

Table[ PowerMod[2, 5^n, 5^n], {n, 0, 24}] // Union (* Jean-François Alcover, Dec 03 2012, from formula given by Joe K. Crump *)

PARI
```
sqrt(-1+O(5^40))
```

{a(n) = local(k, x, y); for(i = 0, n, until( x != (y = truncate( sqrt( -1 + O(5^(k++))))), x = y));x} /* Michael Somos, Mar 03 2008 */

Successive values of 2^(5^x) mod 5^x. - Joe K. Crump (joecr(AT)carolina.rr.com), Jan 20 2001

A218562 Numbers k such that k^2 + 1 is divisible by a cube.

Original entry on oeis.org

57, 68, 182, 193, 239, 307, 318, 432, 443, 557, 568, 682, 693, 807, 818, 932, 943, 1057, 1068, 1182, 1193, 1307, 1318, 1432, 1443, 1557, 1568, 1682, 1693, 1807, 1818, 1932, 1943, 1958, 1985, 2057, 2068, 2182, 2193, 2307, 2318, 2432, 2436, 2443, 2557, 2568

Offset: 1

Michel Lagneau, Nov 02 2012

nonn

			239 is in the sequence because 239^2 + 1 =  2 * 13 ^ 4.
1985 is in the sequence because 1985^2 + 1 = 2 * 17 ^ 3 * 401.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A002522, A049532, A034939.

Cf. A000578 (cubes).

Select[Range[2,2600],Max[Transpose[FactorInteger[#^2+1]][[2]]]>2&]

cp's OEIS Frontend

A199206 a(n) = (1+(A034939(n))^2)/5^n.

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

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A257833 Table T(k, n) of smallest bases b > 1 such that p = prime(n) satisfies b^(p-1) == 1 (mod p^k), read by antidiagonals.

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A218709 a(n) is smallest number such that a(n)^2 + 1 is divisible by 13^n.

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Mathematica

A218710 a(n) is smallest number such that a(n)^2 + 1 is divisible by 17^n.

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Mathematica

A218712 a(n) is the smallest number such that a(n)^2 + 1 is divisible by 29^n.

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Maple

Mathematica

A218713 a(n) is smallest number such that a(n)^2 + 1 is divisible by 37^n.

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Mathematica

A218714 a(n) is smallest number such that a(n)^2 + 1 is divisible by 41^n.

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Mathematica

A034935 Successive approximations to 5-adic integer sqrt(-1).

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