A208643 -id:A208643 - cp's OEIS frontend

A210642 a(n) = least integer m > 1 such that k! == n! (mod m) for no 0 < k < n.

2, 2, 3, 4, 5, 9, 7, 13, 17, 17, 11, 13, 13, 19, 23, 17, 17, 29, 19, 23, 31, 31, 23, 41, 31, 29, 31, 37, 29, 31, 31, 37, 41, 41, 59, 37, 37, 59, 43, 41, 41, 59, 43, 67, 53, 53, 47, 53, 67, 59, 61, 53, 53, 79, 59, 59, 67, 73, 59, 67

Offset: 1

Zhi-Wei Sun, Mar 26 2012

nonn

Conjecture: a(n) is a prime not exceeding 2n with the only exceptions a(4)=4 and a(6)=9.
Note that a(n) is at least n and there is at least a prime in the interval [n,2n] by the Bertrand Postulate first confirmed by Chebyshev.
Compare this sequence with A208494.

			We have a(4)=4, because 4 divides none of 4!-1!=23, 4!-2!=22, 4!-3!=18, and both 2 and 3 divide 4!-3!=18.

Zhi-Wei Sun, Table of n, a(n) for n = 1..2500
Oliver Gerard, Re: A new conjecture on primes, a message to Number Theory List, March 23, 2012.
Zhi-Wei Sun, A new conjecture on primes, a message to Number Theory List, March 20, 2012.
Zhi-Wei Sun, On functions taking only prime values, J. Number Theory 133(2013), no.8, 2794-2812.

Cf. A000040, A208494, A210640, A210393, A210394, A210186, A210144, A208643, A207982.

R[n_,m_]:=If[n==1,1,Product[If[Mod[n!-k!, m]==0, 0, 1], {k, 1, n-1}]] Do[Do[If[R[n,m]==1, Print[n, " ", m]; Goto[aa]], {m,Max[2,n],2n}]; Print[n]; Label[aa]; Continue,{n,1,2500}]

A222114 Least integer m>1 such that 6p_k(p_k-1) (k=1,...,n) are pairwise incongruent modulo m, where p_k denotes the k-th prime.

Original entry on oeis.org

2, 5, 5, 13, 19, 29, 31, 37, 37, 37, 61, 61, 61, 89, 97, 97, 97, 109, 131, 139, 149, 157, 157, 157, 173, 181, 193, 193, 193, 193, 241, 241, 241, 271, 271, 271, 271, 317, 331, 331, 331, 349, 349, 367, 367, 367, 397, 397, 397, 397, 397, 397, 457, 457, 457, 457, 457, 457, 523, 523

Offset: 1

Zhi-Wei Sun, May 13 2013

nonn

Conjecture: For each n=3,4,..., a(n) is the first prime p>=p_n dividing none of those p_i+p_j-1 (1<=i

			a(2)=5 since 6*p_1*(p_1-1)=12 and 6*p_2*(p_2-1)=36 are incongruent modulo 5 but 12 is congruent to 36 modulo any of 2, 3, 4.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, On functions taking only prime values, J. Number Theory, 133 (2013), 2794-2812.

Cf. A000040, A208643, A181901.

R[n_,m_]:=Union[Table[Mod[6Prime[k](Prime[k]-1),m],{k,1,n}]]
s=2
Do[Do[If[Length[R[n,m]]==n,s=m;Print[n," ",m];Goto[aa]],{m,s,n^2}];
Print[n];Label[aa];Continue,{n,1,100}]

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A210642 a(n) = least integer m > 1 such that k! == n! (mod m) for no 0 < k < n.

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A222114 Least integer m>1 such that 6p_k(p_k-1) (k=1,...,n) are pairwise incongruent modulo m, where p_k denotes the k-th prime.

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cp's OEIS Frontend

A210642 a(n) = least integer m > 1 such that k! == n! (mod m) for no 0 < k < n.

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A222114 Least integer m>1 such that 6*p_k*(p_k-1) (k=1,...,n) are pairwise incongruent modulo m, where p_k denotes the k-th prime.

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A222114 Least integer m>1 such that 6p_k(p_k-1) (k=1,...,n) are pairwise incongruent modulo m, where p_k denotes the k-th prime.