A249275 -id:A249275 - cp's OEIS frontend

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.

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

A353937 Smallest b > 1 such that b^(p-1) == 1 (mod p^4) for p = prime(n).

Original entry on oeis.org

17, 80, 182, 1047, 1963, 239, 4260, 2819, 19214, 2463, 15714, 51344, 20677, 3038, 224444, 189323, 11550, 397575, 201305, 15384, 840838, 1372873, 1576656, 278454, 1721322, 48072, 281007, 119551, 252595, 1001934, 3489507, 2489004, 598987, 3082551, 6136759, 3928984

Offset: 1

Felix Fröhlich, May 12 2022

nonn

Robert Israel, Table of n, a(n) for n = 1..10000

Row k = 4 of A257833.

Cf. similar sequences for p^k: A039678 (k=2), A249275 (k=3), A353938 (k=5), A353939 (k=6), A353940 (k=7), A353941 (k=8), A353942 (k=9), A353943 (k=10).

f:= proc(j) local p,b,i;
  p:= ithprime(j);
  b:= numtheory:-primroot(p^4) &^ (p^3) mod p^4;
  min(seq(b &^i mod p^4, i=1..p-2))
end proc:
f(1):= 17:
map(f, [$1..40]); # Robert Israel, Dec 19 2024

a[n_] := Module[{p = Prime[n], b = 2}, While[PowerMod[b, p - 1, p^4] != 1, b++]; b]; Array[a, 20] (* Amiram Eldar, May 12 2022 *)

a(n) = my(p=prime(n)); for(b=2, oo, if(Mod(b, p^4)^(p-1)==1, return(b)))

from sympy import prime
from sympy.ntheory.residue_ntheory import nthroot_mod
def A353937(n): return 2**4+1 if n == 1 else int(nthroot_mod(1,(p:= prime(n))-1,p**4,True)[1]) # Chai Wah Wu, May 17 2022

A353938 Smallest b > 1 such that b^(p-1) == 1 (mod p^5) for p = prime(n).

Original entry on oeis.org

33, 242, 1068, 1353, 27216, 109193, 15541, 133140, 495081, 1115402, 2754849, 1353359, 649828, 3228564, 2359835, 4694824, 7044514, 28538377, 1111415, 77588426, 16178110, 2553319, 9571390, 158485540, 18664438, 146773512, 45639527, 448251412, 48834112, 141076650

Offset: 1

Felix Fröhlich, May 12 2022

nonn

Row k = 5 of A257833.

Cf. similar sequences for p^k: A039678 (k=2), A249275 (k=3), A353937 (k=4), A353939 (k=6), A353940 (k=7), A353941 (k=8), A353942 (k=9), A353943 (k=10).

a[n_] := Module[{p = Prime[n], b = 2}, While[PowerMod[b, p - 1, p^5] != 1, b++]; b]; Array[a, 12] (* Amiram Eldar, May 12 2022 *)

a(n) = my(p=prime(n)); for(b=2, oo, if(Mod(b, p^5)^(p-1)==1, return(b)))

from sympy import prime
from sympy.ntheory.residue_ntheory import nthroot_mod
def A353938(n): return 2**5+1 if n == 1 else int(nthroot_mod(1,(p:= prime(n))-1,p**5,True)[1]) # Chai Wah Wu, May 17 2022

A353939 Smallest b > 1 such that b^(p-1) == 1 (mod p^6) for p = prime(n).

Original entry on oeis.org

65, 728, 1068, 34967, 284995, 861642, 390112, 333257, 2818778, 42137700, 8078311, 33518159, 92331463, 21583010, 138173066, 8202731, 390421192, 1006953931, 77622331, 270657300, 5915704483, 522911165, 2507851273, 1329885769, 2789067613, 3987072867, 7938255646

Offset: 1

Felix Fröhlich, May 12 2022

nonn

Row k = 6 of A257833.

Cf. similar sequences for p^k: A039678 (k=2), A249275 (k=3), A353937 (k=4), A353938 (k=5), A353940 (k=7), A353941 (k=8), A353942 (k=9), A353943 (k=10).

a[n_] := Module[{p = Prime[n], b = 2}, While[PowerMod[b, p - 1, p^6] != 1, b++]; b]; Array[a, 9] (* Amiram Eldar, May 12 2022 *)

a(n) = my(p=prime(n)); for(b=2, oo, if(Mod(b, p^6)^(p-1)==1, return(b)))

from sympy import prime
from sympy.ntheory.residue_ntheory import nthroot_mod
def A353939(n): return 2**6+1 if n == 1 else int(nthroot_mod(1,(p:= prime(n))-1,p**6,True)[1]) # Chai Wah Wu, May 17 2022

a(25)-a(27) from Jinyuan Wang, May 17 2022

A353940 Smallest b > 1 such that b^(p-1) == 1 (mod p^7) for p = prime(n).

Original entry on oeis.org

129, 2186, 32318, 82681, 758546, 6826318, 21444846, 44702922, 178042767, 393747520, 1548729003, 4741156070, 2203471139, 3242334565, 16609835418, 114175761515, 30338830655, 20115543070, 114457309347, 370162324382, 57877856575, 12692933349, 280646695286, 127762186531

Offset: 1

Felix Fröhlich, May 12 2022

nonn

Row k = 7 of A257833.

Cf. similar sequences for p^k: A039678 (k=2), A249275 (k=3), A353937 (k=4), A353938 (k=5), A353939 (k=6), A353941 (k=8), A353942 (k=9), A353943 (k=10).

a(n) = my(p=prime(n)); for(b=2, oo, if(Mod(b, p^7)^(p-1)==1, return(b)))

from sympy import prime
from sympy.ntheory.residue_ntheory import nthroot_mod
def A353940(n): return 2**7+1 if n == 1 else int(nthroot_mod(1,(p:= prime(n))-1,p**7,True)[1]) # Chai Wah Wu, May 17 2022

a(9)-a(11) from Amiram Eldar, May 12 2022

More terms from Jinyuan Wang, May 17 2022

A353941 Smallest b > 1 such that b^(p-1) == 1 (mod p^8) for p = prime(n).

Original entry on oeis.org

257, 6560, 110443, 2387947, 9236508, 6826318, 112184244, 674273372, 571782680, 8827420195, 46142113101, 85760131222, 287369842623, 120773832179, 83209719751, 1684374218587, 6358345589299, 6305601215112, 5800992744105, 33960226045484, 56924554232879, 11856046381401

Offset: 1

Felix Fröhlich, May 12 2022

nonn

Row k = 8 of A257833.

Cf. similar sequences for p^k: A039678 (k=2), A249275 (k=3), A353937 (k=4), A353938 (k=5), A353939 (k=6), A353940 (k=7), A353942 (k=9), A353943 (k=10).

a(n) = my(p=prime(n)); for(b=2, oo, if(Mod(b, p^8)^(p-1)==1, return(b)))

from sympy import prime
from sympy.ntheory.residue_ntheory import nthroot_mod
def A353941(n): return 2**8+1 if n == 1 else int(nthroot_mod(1,(p:= prime(n))-1,p**8,True)[1]) # Chai Wah Wu, May 17 2022

a(7)-a(8) from Amiram Eldar, May 12 2022

More terms from Jinyuan Wang, May 17 2022

A353942 Smallest b > 1 such that b^(p-1) == 1 (mod p^9) for p = prime(n).

Original entry on oeis.org

513, 19682, 280182, 14906455, 676386984, 822557039, 8185328614, 1835323405, 147534349327, 430099398783, 746688111476, 3054750102760, 9430469115218, 42562034654367, 92084372092298, 28307243117603, 17362132628379, 430275700643181, 478910674129864, 69114209866295

Offset: 1

Felix Fröhlich, May 12 2022

nonn

Row k = 9 of A257833.

Cf. similar sequences for p^k: A039678 (k=2), A249275 (k=3), A353937 (k=4), A353938 (k=5), A353939 (k=6), A353940 (k=7), A353941 (k=8), A353943 (k=10).

a(n) = my(p=prime(n)); for(b=2, oo, if(Mod(b, p^9)^(p-1)==1, return(b)))

from sympy import prime
from sympy.ntheory.residue_ntheory import nthroot_mod
def A353942(n): return 2**9+1 if n == 1 else int(nthroot_mod(1,(p:= prime(n))-1,p**9,True)[1]) # Chai Wah Wu, May 17 2022

a(5)-a(6) from Amiram Eldar, May 12 2022

More terms from Jinyuan Wang, May 17 2022

A353943 Smallest b > 1 such that b^(p-1) == 1 (mod p^10) for p = prime(n).

Original entry on oeis.org

1025, 59048, 3626068, 135967276, 1509748675, 14149342837, 109522148350, 649340249056, 191730243526, 45941644105613, 6359301533362, 24287026146320, 265934493600922, 927939012431924, 1377672497815095, 4440230734662684, 10400007512898615, 12198961352308417

Offset: 1

Felix Fröhlich, May 12 2022

nonn

Row k = 10 of A257833.

Cf. similar sequences for p^k: A039678 (k=2), A249275 (k=3), A353937 (k=4), A353938 (k=5), A353939 (k=6), A353940 (k=7), A353941 (k=8), A353942 (k=9).

a(n) = my(p=prime(n)); for(b=2, oo, if(Mod(b, p^10)^(p-1)==1, return(b)))

from sympy.ntheory.residue_ntheory import nthroot_mod
from sympy import prime
def A353943(n): return 2**10+1 if n == 1 else int(nthroot_mod(1,(p:= prime(n))-1,p**10,True)[1]) # Chai Wah Wu, May 17 2022

a(5)-a(6) from Amiram Eldar, May 12 2022

More terms from Jinyuan Wang, May 17 2022

A254444 Largest k such that p = prime(n) satisfies b^(p-1) == 1 (mod p^k) for some base b with 1 < b < p.

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 2, 1, 2, 3, 2, 2, 2, 1, 1, 2, 1, 2, 1, 1, 1, 2, 2, 1, 2, 2, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 2, 1, 1, 2, 2, 2, 2, 1, 2, 1, 2, 1, 2, 2, 2, 2, 2, 2, 2, 1, 1, 2, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1

Offset: 2

Felix Fröhlich, May 04 2015

nonn

a(n) > 1 iff p is in A134307.
Meyer proved in 1902 that for any prime p exactly p - 1 bases b with b < p^k exist such that b^(p-1) == 1 (mod p^k) (cf. Keller, Richstein, 2005, page 930).
a(30) = 3 is the first term with a value > 2, corresponding to prime(30) = 113 (see the comment from 2011 in A134307). This is the first case where A249275(n) < A000040(n).
Do the values of this sequence have an upper bound or, more formally, does this sequence have a supremum?

			With p = 113: For all bases b with 1 < b < 113, p (trivially) satisfies b^112 == 1 (mod 113^k) for k = 1 and for no k > 1, with the single exception of b = 68, where p satisfies the congruence for k = 3 (and hence for k = 1 and k = 2). Since 3 is the largest value of k for all 1 < b < 113, a(30) = 3.

Felix Fröhlich, Table of n, a(n) for n = 2..10000
W. Keller and J. Richstein, Solutions of the congruence a^p-1 == 1 (mod p^r), Math. Comp., 74 (2005), 927-936.

Cf. A134307.

forprime(p=3, 400, k=1; maxk=0; for(b=2, p-1, while(Mod(b, p^k)^(p-1)==1, k++); if(k-1 > maxk, maxk=k-1)); print1(maxk, ", "))

A275338 Smallest prime p where a base b with 1 < b < p exists such that b^(p-1) == 1 (mod p^n).

Original entry on oeis.org

3, 11, 113

Offset: 1

Felix Fröhlich, Jul 28 2016

Smallest prime p such that A254444(i) >= n, where i is the index of p in A000040.
For n > 1, a(n) is a term of A134307.
For n > 1, if A000040(i) is a term of the sequence, then A249275(i) < A000040(i).
For n > 1, smallest prime p such that T(n, i) < p, where i is the index of p in A000040 and T = A257833.
a(4) > 5*10^8 if it exists (see Fischer link).

			For n = 3: p = 113 satisfies 68^(p-1) == 1 (mod p^3) and there is no smaller prime p such that p satisfies b^(p-1) == 1 (mod p^3) for some b with 1 < b < p, so a(3) = 113.

R. Fischer, Thema: Fermatquotient B^(P-1) == 1 (mod P^3).

Cf. A134307, A254444, A257833.

a(n) = forprime(p=1, , for(b=2, p-1, if(Mod(b, p^n)^(p-1)==1, return(p))))

cp's OEIS Frontend

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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A353937 Smallest b > 1 such that b^(p-1) == 1 (mod p^4) for p = prime(n).

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A353938 Smallest b > 1 such that b^(p-1) == 1 (mod p^5) for p = prime(n).

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A353939 Smallest b > 1 such that b^(p-1) == 1 (mod p^6) for p = prime(n).

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A353940 Smallest b > 1 such that b^(p-1) == 1 (mod p^7) for p = prime(n).

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A353941 Smallest b > 1 such that b^(p-1) == 1 (mod p^8) for p = prime(n).

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PARI

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A353942 Smallest b > 1 such that b^(p-1) == 1 (mod p^9) for p = prime(n).

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PARI

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A353943 Smallest b > 1 such that b^(p-1) == 1 (mod p^10) for p = prime(n).

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PARI

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A254444 Largest k such that p = prime(n) satisfies b^(p-1) == 1 (mod p^k) for some base b with 1 < b < p.

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