A210393 -id:A210393 - cp's OEIS frontend

A210394 a(n) = least integer m>1 such that m divides none of S_i!+S_j! with 0

2, 3, 7, 11, 19, 31, 43, 67, 79, 101, 131, 163, 199, 241, 283, 331, 383, 443, 503, 571, 641, 719, 797, 877, 967, 1061, 1163, 1277, 1373, 1481, 1597, 1721, 1871, 1999, 2129, 2281, 2437, 2593, 2749, 2927, 3089, 3271, 3457, 3643, 3833, 4057, 4229, 4441, 4673, 4889

Offset: 1

Zhi-Wei Sun, Mar 20 2012

nonn

When n>1, we have S_n!+S_{n-1}!=0 (mod m) for all m=1,...,S_{n-1} and hence a(n)>S_{n-1}. Zhi-Wei Sun conjectured that a(n) is always a prime not exceeding S_n.

			We have a(3)=7, since  m=7 divides none of 2!+(2+3)!,2!+(2+3+5)!,(2+3)!+(2+3+5)! but this fails for m=2,3,4,5,6.

Zhi-Wei Sun, Table of n, a(n) for n = 1..225
Zhi-Wei Sun, On functions taking only prime values, J. Number Theory 133(2013), no.8, 2794-2812.

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

s[n_]:=s[n]=Sum[Prime[k],{k,1,n}]
f[n_]:=s[n]!
R[n_,m_]:=Product[If[Mod[f[k]+f[j],m]==0,0,1],{k,2,n},{j,1,k-1}]
Do[Do[If[R[n,m]==1,Print[n," ",m];Goto[aa]],{m,Max[2,s[n-1]],s[n]}];
   Print[n];Label[aa];Continue,{n,1,225}]

A181901 a(n) = least positive integer m such that 2(s_k)^2 for k=1,...,n are pairwise distinct modulo m where s_k = Sum_{j=1..k} (-1)^(k-j)*p_j and p_j is the j-th prime.

Original entry on oeis.org

1, 4, 7, 9, 13, 17, 19, 23, 25, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233

Offset: 1

Zhi-Wei Sun, Mar 31 2012

nonn

On Mar 28 2012, Zhi-Wei Sun conjectured that a(n) is the (n+1)-th prime p_{n+1} with the only exceptions being a(1)=1, a(2)=4, a(4)=9 and a(9)=25. He has shown that 2(s_k)^2 (k=1,...,n) are indeed pairwise distinct modulo p_{n+1} and hence a(n) does not exceed p_{n+1}.
Note that the sequence 0,s_1,s_2,s_3,... is A008347 introduced by N. J. A. Sloane and J. H. Conway.
Compare a(n) with the sequence A210640.
The conjecture was verified for n up to 2*10^5 by the author in 2018, and for n up to 3*10^5 by Chang Zhang (a student at Nanjing Univ.) in June 2020. - Zhi-Wei Sun, Jun 22 2020

			We have a(4)=9 since 2(s_1)^2=8, 2(s_2)^2=2, 2(s_3)^2=32, 2(s_4)^2=18 are pairwise distinct modulo 9 but not pairwise distinct modulo any of 1,...,8.

Zhi-Wei Sun, Table of n, a(n) for n = 1..600
Zhi-Wei Sun, An amazing recurrence for primes, a message to Number Theory List, March 31, 2012.
Zhi-Wei Sun, On functions taking only prime values, J. Number Theory 133(2013), no.8, 2794-2812.

Cf. A000040, A008347, A210640, A210393, A210394, A210186, A210144, A208494, A208643, A207982.

s[n_]:=Sum[(-1)^k*Prime[k],{k,1,n}]
f[n_]:=2*s[n]^2
R[n_,m_]:=Union[Table[Mod[f[k],m],{k,1,n}]]
Do[Do[If[Length[R[n,m]]==n,Print[n," ",m];Goto[aa]],{m,1,Prime[n+1]}];
   Print[n];Label[aa];Continue,{n,1,600}]

A210640 a(n) = least integer m > 1 such that 2S_k^2 (k=1,...,n) are pairwise distinct modulo m, where S_k is the sum of the first k primes.

Original entry on oeis.org

2, 4, 9, 13, 17, 28, 37, 37, 37, 37, 37, 61, 61, 61, 151, 151, 151, 151, 151, 151, 151, 227, 227, 227, 227, 227, 307, 307, 307, 337, 433, 433, 433, 433, 433, 433, 433, 433, 433, 433, 433, 509, 509, 509, 509, 509, 643, 727, 727, 761, 761, 761, 971, 971, 971

Offset: 1

Zhi-Wei Sun, Mar 26 2012

nonn

If a(n) is an odd prime p_k then a(n)|2(S_k^2-S_{k-1}^2) and hence k>n. Zhi-Wei Sun conjectured that a(n) is a prime smaller than n^2 unless n divides 6.

			We have a(3)=9, because 2*2^2=8, 2*(2+3)^2=50, 2(2+3+5)^2=200 are pairwise distinct modulo m=9 but not pairwise distinct modulo any of 2, 3, 4, 5, 6, 7, 8.

Zhi-Wei Sun, Table of n, a(n) for n = 1..5258
Zhi-Wei Sun, On functions taking only prime values, J. Number Theory 133(2013), no.8, 2794-2812.
Zhi-Wei Sun, On sums of consecutive primes, a message to Number Theory List, March 23, 2012.

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

s[n_] := s[n] = Sum[Prime[k], {k,1,n}]; f[n_] := f[n] = 2*s[n]^2; R[n_,m_] := Union[Table[Mod[f[k],m], {k,1,n}]]; Do[Do[If[Length[R[n,m]] == n, Print[n," ",m]; Goto[aa]], {m,2,Max[2,n^2]}]; Print[n]; Label[aa]; Continue, {n,1,5000}]

A182003 a(n) = least prime p_k such that n = p_k - p_{k-1} + ... + (-1)^{k-j}p_j for some 1 <= j <= k, with p_k the k-th prime.

Original entry on oeis.org

3, 2, 3, 5, 5, 11, 7, 11, 11, 17, 11, 17, 13, 23, 17, 23, 17, 31, 19, 41, 23, 41, 23, 47, 29, 47, 37, 59, 29, 59, 31, 59, 43, 67, 37, 67, 37, 67, 43, 73, 41, 83, 43, 83, 47, 101, 47, 97, 53, 97, 59, 97, 53, 103, 61, 109, 67, 127, 59, 131

Offset: 1

Zhi-Wei Sun, Apr 05 2012

a(n) is obviously at least n. In March 2012, Zhi-Wei Sun conjectured that a(n) always exists and does not exceed 2.2*n for n > 1; he even thought that 2.2*n might be replaced by 2*n + 2.5*sqrt(n) for n > 1.
IFF n=p, a(n)=1.
For odd n's, lim_{n->inf.} a(n)/n = 1; for even n's, lim_{n->inf.} a(n)/n = 2.

			We have a(4) = 5 since 4 = 5 - 3 + 2 and 3 - 2 < 4.

Robert G. Wilson v, Table of n, a(n) for n = 1..10000 (first 5000 terms from Zhi-Wei Sun)
Zhi-Wei Sun, On functions taking only prime values, preprint, arxiv:1202.6589 [math.NT], 2012-2013.
Zhi-Wei Sun, A representation problem involving consecutive primes, a message to Number Theory List, Mar 31 2012.
Zhi-Wei Sun, An amazing recurrence for primes, a message to Number Theory List, Mar 31 2012.

Cf. A000040, A181901, A210640, A210393, A210394, A210186, A210144, A208494, A208643, A207982.

s[0_]:=0; s[n_]:=s[n]=Prime[n]-s[n-1]; Do[Do[If[s[j]-(-1)^(j-i)*s[i]==n, Print[n," ",Prime[j]]; Goto[aa]], {j, 1, PrimePi[3n]}, {i, 0, j-1}]; Print[n]; Label[aa]; Continue, {n, 1, 5000}]
f[n_] := Block[{j = PrimePi[n]}, While[ !MemberQ[ Accumulate@ Table[(-1)^(j - i) Prime[i], {i, j, 1, -1}], n], j++]; Prime[j]]; Array[f, 60] (* Robert G. Wilson v, Apr 06 2012 *)

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

Original entry on oeis.org

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}]

cp's OEIS Frontend

A210394 a(n) = least integer m>1 such that m divides none of S_i!+S_j! with 0

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A181901 a(n) = least positive integer m such that 2(s_k)^2 for k=1,...,n are pairwise distinct modulo m where s_k = Sum_{j=1..k} (-1)^(k-j)*p_j and p_j is the j-th prime.

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A210640 a(n) = least integer m > 1 such that 2S_k^2 (k=1,...,n) are pairwise distinct modulo m, where S_k is the sum of the first k primes.

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A182003 a(n) = least prime p_k such that n = p_k - p_{k-1} + ... + (-1)^{k-j}p_j for some 1 <= j <= k, with p_k the k-th prime.

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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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