A366973 -id:A366973 - cp's OEIS frontend

A366982 a(n) is the smallest odd k > 1 such that n^((k+1)/2) == n (mod k).

3, 3, 7, 3, 3, 5, 3, 3, 7, 3, 3, 5, 3, 3, 5, 3, 3, 9, 3, 3, 5, 3, 3, 7, 3, 3, 5, 3, 3, 5, 3, 3, 7, 3, 3, 5, 3, 3, 11, 3, 3, 5, 3, 3, 5, 3, 3, 11, 3, 3, 5, 3, 3, 7, 3, 3, 5, 3, 3, 5, 3, 3, 9, 3, 3, 5, 3, 3, 13, 3, 3, 5, 3, 3, 5, 3, 3, 7, 3, 3, 5, 3, 3, 17, 3, 3

Offset: 0

Thomas Ordowski, Oct 30 2023

nonn

If this sequence is bounded, then it is periodic with period P = LCM(A), where A is the set of all (pairwise distinct) terms.
Note that n^((1729+1)/2) == n (mod 1729) for every n >= 0, where 1729 is the smallest absolute Euler pseudoprime (A033181).
Thus a(n) <= 1729. So, as said, this sequence is periodic.
What is its period?
If the largest term of this sequence is indeed 1729, it should be expected that its period P may be longer than the period of Euler primary pretenders (A309316), namely P > 41#*571#/4 (248 digits).

Antti Karttunen, Table of n, a(n) for n = 0..65537

Cf. A033181, A309316, A366930, A366973.

a[n_] := Module[{k = 3}, While[PowerMod[n, (k + 1)/2, k] != Mod[n, k], k += 2]; k]; Array[a, 100, 0] (* Amiram Eldar, Oct 30 2023 *)

a(n) = my(k=3); while (Mod(n, k)^((k+1)/2) != n, k+=2); k; \\ Michel Marcus, Oct 31 2023

More terms from Amiram Eldar, Oct 30 2023

A366930 a(n) is the smallest odd composite k such that n^((k+1)/2) == n (mod k).

Original entry on oeis.org

9, 9, 341, 121, 341, 65, 15, 21, 9, 9, 9, 33, 33, 21, 21, 15, 15, 9, 9, 9, 21, 15, 21, 33, 25, 15, 9, 9, 9, 21, 15, 15, 25, 33, 21, 9, 9, 9, 57, 39, 15, 21, 21, 21, 9, 9, 9, 65, 21, 21, 21, 15, 39, 9, 9, 9, 21, 21, 57, 145, 15, 15, 9, 9, 9, 33, 15, 33, 25, 21

Offset: 0

Thomas Ordowski, Nov 01 2023

nonn

If this sequence is bounded, then it is periodic with period P = LCM(A), where A is the set of all (pairwise distinct) terms.
Note that n^((1729+1)/2) == n (mod 1729) for every n >= 0, where 1729 is the smallest absolute Euler pseudoprime (A033181).
Thus a(n) <= 1729. So, as said, this sequence is periodic.
What is its period?
The period P of this sequence may be longer than the period of Euler primary pretenders (A309316), namely P > 41#*571#/4 (248 digits).

Cf. A033181, A309316, A366973, A366982.

a[n_] := Module[{k = 9}, While[PrimeQ[k] || PowerMod[n, (k + 1)/2, k] != Mod[n, k], k += 2]; k]; Array[a, 100, 0] (* Amiram Eldar, Nov 01 2023 *)

a(n) = my(k=3); while (isprime(k) || Mod(n, k)^((k+1)/2) != n, k+=2); k; \\ Michel Marcus, Nov 01 2023

a(n) >= A309316(n).

A373666 Smallest positive integer whose square can be written as the sum of n positive perfect squares whose square roots differ by no more than 1.

Original entry on oeis.org

1, 5, 3, 2, 5, 3, 4, 10, 3, 4, 7, 6, 4, 9, 6, 4, 25, 6, 5, 10, 6, 5, 16, 6, 5, 12, 6, 7, 11, 6, 7, 20, 6, 7, 15, 6, 7, 31, 9, 7, 13, 9, 7, 14, 9, 7, 36, 9, 7, 15, 9, 8, 22, 9, 8, 17, 9, 8, 16, 9, 8, 49, 9, 8, 20, 9, 10, 50, 9, 10, 17, 9, 10, 19, 9, 10, 28, 9

Offset: 1

Charles L. Hohn, Jun 12 2024

nonn

Shortest possible integer length of the diagonal of an n-dimensional hyperrectangle where each edge has a positive integer length, and edge lengths differ by no more than 1.

			a(1) = 1 because 1^2 = 1^2.
a(2) = 5 because 5^2 = 3^2 + 4^2.
a(3) = 3 because 3^2 = 1^2 + 2*(2^2).
a(4) = 2 because 2^2 = 4*(1^2).
a(5) = 5 because 5^2 = 4*(2^2) + 3^2.
a(6) = 3 because 3^2 = 5*(1^2) + 2^2.
a(7) = 4 because 4^2 = 4*(1^2) + 3*(2^2).

Charles L. Hohn, Table of n, a(n) for n = 1..10000

Cf. A351061, A366973.

a(n) = my(d=ceil(sqrt(n))); while(true, my(b=sqrtint(floor(d^2/n))); if ((d^2-b^2*n)%(b*2+1)==0, return(d), d++)) \\ Charles L. Hohn, Jul 02 2024

a366973(n) = {for(i=2, oo, my(p=prime(i)); for(j=0, (p-1)/2, if(n%p==j^2%p, return(p))))}
bstep(np, p) = {my(t=np+if(np%2, p)); while(!issquare(t), t+=p*2); sqrtint(t)/2}
a(n) = my(p=a366973(n), b=sqrtint(n*((p-1)/2)^2-1)+1, bp=b%p, s=bstep(n%p, p)); b-bp+if(bp<=s, s, bp<=p-s, p-s, p+s) \\ Charles L. Hohn, Sep 27 2024

a(n) = min(d) such that d^2 - n*b^2 == 0 (mod 2*b + 1) and d >= ceiling(sqrt(n)) where b = floor(sqrt(d^2/n)).

A367034 a(n) is the smallest odd number k > 1 for which the Jacobi symbol (n / k) >= 0.

Original entry on oeis.org

3, 3, 7, 3, 3, 5, 3, 3, 7, 3, 3, 5, 3, 3, 5, 3, 3, 9, 3, 3, 5, 3, 3, 7, 3, 3, 5, 3, 3, 5, 3, 3, 7, 3, 3, 5, 3, 3, 9, 3, 3, 5, 3, 3, 5, 3, 3, 9, 3, 3, 5, 3, 3, 7, 3, 3, 5, 3, 3, 5, 3, 3, 9, 3, 3, 5, 3, 3, 9, 3, 3, 5, 3, 3, 5, 3, 3, 7, 3, 3, 5, 3, 3, 9, 3, 3, 5, 3, 3, 5, 3, 3, 7, 3, 3, 5, 3, 3, 7, 3, 3, 5, 3, 3, 5

Offset: 0

Amiram Eldar and Thomas Ordowski, Nov 02 2023

This sequence is periodic with period P = 3*5*7 = 105.

All terms are in {3, 5, 7, 9}.

Antti Karttunen, Table of n, a(n) for n = 0..11025

Cf. A366973.

a[n_] := Module[{k = 3}, While[JacobiSymbol[n, k] < 0, k += 2]; k]; Array[a, 105, 0]

a(n) = my(k=3); while(kronecker(n,k)<0, k+=2); k; \\ Michel Marcus, Nov 02 2023

a(n + 105) = a(n).

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A366982 a(n) is the smallest odd k > 1 such that n^((k+1)/2) == n (mod k).

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A366930 a(n) is the smallest odd composite k such that n^((k+1)/2) == n (mod k).

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A373666 Smallest positive integer whose square can be written as the sum of n positive perfect squares whose square roots differ by no more than 1.

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A367034 a(n) is the smallest odd number k > 1 for which the Jacobi symbol (n / k) >= 0.

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