A045326 -id:A045326 - cp's OEIS frontend

A081727 Length of periods of Euler numbers modulo n.

1, 1, 2, 1, 2, 2, 6, 4, 6, 2, 10, 2, 6, 6, 2, 8, 8, 6, 18, 2, 6, 10, 22, 4, 10, 6, 18, 6, 14, 2, 30, 16, 10, 8, 6, 6, 18, 18, 6, 4, 20, 6, 42, 10, 6, 22, 46, 8, 42, 10, 8, 6, 26, 18, 10, 12, 18, 14, 58, 2, 30, 30, 6, 32, 6, 10, 66, 8, 22, 6, 70, 12, 36, 18, 10

Offset: 1

Benoit Cloitre, Apr 06 2003

nonn

Terms after a(126) need more Euler numbers to check the period length. There are 85 unknown terms starting from a(127) till a(500) when 242 Euler numbers are used. - Hakan Icoz, Sep 06 2020

			A000364 modulo 5 gives : 1,1,0,1,0,1,0,1,0,1,0,... with period (1,0) of length 2, hence a(5)=2.

Hakan Icoz, Table of n, a(n) for n = 1..125

Cf. A000364, A045326.

a(n) = n-1 if n=2, 3, 7, 11, 19, 23, 31...is a prime == 2 or 3 (mod 4) (A045326).

More terms from Hakan Icoz, Sep 06 2020

A331047 T(n,k) = -(-1)^k*ceiling(k/2)^2 mod p, where p is the n-th prime congruent to 2 or 3 mod 4; triangle T(n,k), n>=1, 0<=k<=p-1, read by rows.

Original entry on oeis.org

0, 1, 0, 1, 2, 0, 1, 6, 4, 3, 2, 5, 0, 1, 10, 4, 7, 9, 2, 5, 6, 3, 8, 0, 1, 18, 4, 15, 9, 10, 16, 3, 6, 13, 17, 2, 11, 8, 7, 12, 5, 14, 0, 1, 22, 4, 19, 9, 14, 16, 7, 2, 21, 13, 10, 3, 20, 18, 5, 12, 11, 8, 15, 6, 17, 0, 1, 30, 4, 27, 9, 22, 16, 15, 25, 6, 5

Offset: 1

Alois P. Heinz, Jan 08 2020

Row n is a permutation of {0, 1, ..., A045326(n)-1}.

			Triangle T(n,k) begins:
  0, 1;
  0, 1,  2;
  0, 1,  6, 4,  3, 2,  5;
  0, 1, 10, 4,  7, 9,  2,  5, 6, 3,  8;
  0, 1, 18, 4, 15, 9, 10, 16, 3, 6, 13, 17, 2, 11, 8, 7, 12, 5, 14;
  ...

Alois P. Heinz, Rows n = 1..75, flattened

Columns k=0-2 give: A000004, A000012, A281664 (for n>1).
Last elements of rows give A190105(n-1) for n>1.
Row lengths give A045326.
Row sums give A000217(A281664(n)).
Cf. A000040, A230077, A331103.

b:= proc(n) option remember; local p;
      p:= 1+`if`(n=1, 1, b(n-1));
      while irem(p, 4)<2 do p:= nextprime(p) od; p
    end:
T:= n-> (p-> seq(modp(-(-1)^k*ceil(k/2)^2, p), k=0..p-1))(b(n)):
seq(T(n), n=1..8);

b[n_] := b[n] = Module[{p}, p = 1+If[n == 1, 1, b[n-1]]; While[Mod[p, 4]<2, p = NextPrime[p]]; p];
T[n_] := With[{p = b[n]}, Table[Mod[-(-1)^k*Ceiling[k/2]^2, p], {k, 0, p-1}]];
Table[T[n], {n, 1, 8}] // Flatten (* Jean-François Alcover, Oct 29 2021, after Alois P. Heinz *)

A331103 T(n,k) = -(-1)^k*k^2 mod p, where p is the n-th prime congruent to 2 or 3 mod 4; triangle T(n,k), n>=1, 0<=k<=p-1, read by rows.

Original entry on oeis.org

0, 1, 0, 1, 2, 0, 1, 3, 2, 5, 4, 6, 0, 1, 7, 9, 6, 3, 8, 5, 2, 4, 10, 0, 1, 15, 9, 3, 6, 2, 11, 12, 5, 14, 7, 8, 17, 13, 16, 10, 4, 18, 0, 1, 19, 9, 7, 2, 10, 3, 5, 12, 15, 6, 17, 8, 11, 18, 20, 13, 21, 16, 14, 4, 22, 0, 1, 27, 9, 15, 25, 26, 18, 29, 19, 24

Offset: 1

Alois P. Heinz, Jan 09 2020

Row n is a permutation of {0, 1, ..., A045326(n)-1}.

			Triangle T(n,k) begins:
  0, 1;
  0, 1,  2;
  0, 1,  3, 2, 5, 4, 6;
  0, 1,  7, 9, 6, 3, 8,  5,  2, 4, 10;
  0, 1, 15, 9, 3, 6, 2, 11, 12, 5, 14, 7, 8, 17, 13, 16, 10, 4, 18;
  ...

Alois P. Heinz, Rows n = 1..75, flattened

Columns k=0-1 give: A000004, A000012.
Last elements of rows give A281664.
Row lengths give A045326.
Row sums give A000217(A281664(n)).
Cf. A331047.

b:= proc(n) option remember; local p;
      p:= 1+`if`(n=1, 1, b(n-1));
      while irem(p, 4)<2 do p:= nextprime(p) od; p
    end:
T:= n-> (p-> seq(modp(-(-1)^k*k^2, p), k=0..p-1))(b(n)):
seq(T(n), n=1..8);

b[n_] := b[n] = Module[{p}, p = 1+If[n == 1, 1, b[n-1]]; While[Mod[p, 4]<2, p = NextPrime[p]]; p];
T[n_] := With[{p = b[n]}, Table[Mod[-(-1)^k*k^2, p], {k, 0, p - 1}]];
Table[T[n], {n, 1, 8}] // Flatten (* Jean-François Alcover, Oct 29 2021, after Alois P. Heinz *)

A081728 Length of periods of Euler numbers modulo prime(n).

Original entry on oeis.org

1, 2, 2, 6, 10, 6, 8, 18, 22, 14, 30, 18, 20, 42, 46, 26, 58, 30, 66, 70, 36, 78, 82, 44, 48, 50, 102, 106, 54, 56, 126, 130, 68, 138, 74, 150, 78, 162, 166, 86, 178, 90, 190, 96, 98, 198, 210, 222, 226, 114, 116, 238, 120, 250, 128, 262, 134, 270, 138, 140, 282, 146

Offset: 1

Benoit Cloitre, Apr 06 2003

nonn

As proved by Kummer, if the actual signed Euler numbers (A122045) are used, then the period is prime(n)-1 for n>1. - T. D. Noe, Mar 16 2007

			A000364 modulo 5=prime(3) gives : 1,1,0,1,0,1,0,1,0,1,0,... with period (1,0) of length 2, hence a(3)=2.

Cf. A000364, A045326, A080148.

f[n_] := Block[{p = Prime[n], t, d = Divisors[p - 1], dk, k = 1},t = Mod[Table[Abs@EulerE[2i], {i, 2, p}], p];While[dk = d[[k]];Nand @@ Equal @@@ Partition[Partition[t, dk], 2, 1], k++ ];dk];Array[f, 63] (* Ray Chandler, Mar 15 2007 *)

a(n)=prime(n)-1 if prime(n) == 2 or 3 (mod 4)

More terms from John W. Layman, Jul 29 2005

Extended by Ray Chandler, Mar 15 2007

A277859 Least k > 1 such that 1^(k-1) + 2^(k-1) + 3^(k-1) + … + (k-1)^(k-1) - n == 0 (mod k).

Original entry on oeis.org

2, 3, 2, 4, 2, 7, 2, 3, 2, 11, 2, 4, 2, 3, 2, 4, 2, 19, 2, 3, 2, 23, 2, 4, 2, 3, 2, 4, 2, 31, 2, 3, 2, 5, 2, 4, 2, 3, 2, 4, 2, 9, 2, 3, 2, 47, 2, 4, 2, 3, 2, 4, 2, 5, 2, 3, 2, 59, 2, 4, 2, 3, 2, 4, 2, 45, 2, 3, 2, 15, 2, 4, 2, 3, 2, 4, 2, 9, 2, 3, 2, 83, 2, 4, 2

Offset: 1

Paolo P. Lava, Nov 02 2016

a(2*n-1) = 2.

a(n) = n + 1 for some prime n + 1 congruent to {2, 3} mod 4.

			a(8) = 3 because:
1^(2-1) - 8 = -7 but -7 mod 2 = 1;
1^(3-1) + 2^(3-1) - 8 = -3 and  -3 mod 3 = 0;

Paolo P. Lava, Table of n, a(n) for n = 1..1000

Cf. A045326, A191677.

P:=proc(q) local j,k,n; for n from 1 to q do for k from 2 to q do
if (add(j^(k-1),j=1..k-1)-n) mod k=0 then print(k); break; fi; od; od; end: P(10^3);

cp's OEIS Frontend

A081727 Length of periods of Euler numbers modulo n.

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A331047 T(n,k) = -(-1)^k*ceiling(k/2)^2 mod p, where p is the n-th prime congruent to 2 or 3 mod 4; triangle T(n,k), n>=1, 0<=k<=p-1, read by rows.

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A331103 T(n,k) = -(-1)^k*k^2 mod p, where p is the n-th prime congruent to 2 or 3 mod 4; triangle T(n,k), n>=1, 0<=k<=p-1, read by rows.

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A081728 Length of periods of Euler numbers modulo prime(n).

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Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

Formula

Extensions

A277859 Least k > 1 such that 1^(k-1) + 2^(k-1) + 3^(k-1) + … + (k-1)^(k-1) - n == 0 (mod k).

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