A208643 -id:A208643 - cp's OEIS frontend

A210144 a(n) = least integer m>1 such that the product of the first k primes for k=1,...,n are pairwise distinct modulo m.

2, 3, 5, 11, 11, 23, 29, 37, 37, 41, 47, 47, 47, 47, 47, 73, 131, 131, 131, 131, 131, 151, 151, 151, 151, 199, 223, 223, 271, 271, 271, 281, 281, 281, 281, 281, 281, 281, 281, 281, 353, 353, 457, 457, 457, 457, 457, 457, 457, 457, 457, 641, 641, 641, 641, 641, 643, 643, 643, 643

Offset: 1

Zhi-Wei Sun, Mar 17 2012

nonn

Conjecture: all the terms are primes and a(n) 1.

			a(3)=5 because 2, 2*3=6, 2*3*5=30 are distinct modulo m=5 but not distinct modulo m=2,3,4.

Jason Yuen, Table of n, a(n) for n = 1..10000 (first 1172 terms from Zhi-Wei Sun)
R. Mestrovic, Euclid's theorem on the infinitude of primes: a historical survey of its proofs (300 BC--2012) and another new proof, arXiv preprint arXiv:1202.3670 [math.HO], 2012-2023. - From _N. J. A. Sloane_, Jun 13 2012
Zhi-Wei Sun, A function taking only prime values, a message to Number Theory List, Feb. 21, 2012.
Zhi-Wei Sun, On functions taking only prime values, J. Number Theory 133(2013), no.8, 2794-2812.

Cf. A000040, A207982, A208494, A208643, A214196.

R[n_,m_]:=Union[Table[Mod[Product[Prime[j],{j,1,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,1,1000}]

A210186 a(n) = least integer m>1 such that m divides none of P_i + P_j with 0

Original entry on oeis.org

2, 3, 5, 7, 11, 19, 23, 23, 23, 47, 59, 61, 71, 71, 71, 101, 101, 101, 101, 101, 101, 113, 113, 113, 113, 113, 113, 113, 113, 113, 223, 223, 223, 223, 223, 223, 223, 223, 223, 223, 223, 223, 223, 223, 223, 223, 223, 223, 487, 487, 661, 661, 661, 661, 661, 661, 661, 661, 661, 719, 719, 719, 719, 719, 719, 811, 811, 811, 811, 811, 811, 811, 811, 811, 811

Offset: 1

Zhi-Wei Sun, Mar 18 2012

nonn

Conjecture: all the terms are primes and a(n) < n^2 for all n > 1.

			We have a(3)=5 since 2+2*3, 2+2*3*5, 2*3+2*3*5 are pairwise distinct modulo m=5 but not pairwise distinct modulo m=2,3,4.

Zhi-Wei Sun, Table of n, a(n) for n = 1..258
Romeo Meštrović, Euclid's theorem on the infinitude of primes: a historical survey of its proofs (300 BC--2012) and another new proof, arXiv preprint arXiv:1202.3670 [math.HO], 2012-2018. - From _N. J. A. Sloane_, Jun 13 2012
Zhi-Wei Sun, A function taking only prime values, message to Number Theory List, Feb. 21, 2012.
Zhi-Wei Sun, On functions taking only prime values, J. Number Theory, Vol. 133, No. 8 (2013), pp. 2794-2812.

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

P[n_]:=Product[Prime[k],{k,1,n}]
R[n_,m_]:=Product[If[Mod[P[k]+P[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,2,Max[2,n^2]}]; Print[n];Label[aa];Continue,{n,1,300}]

A210393 a(n) = least integer m>1 such that S_k! for 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, 3, 7, 13, 19, 29, 43, 61, 79, 101, 131, 167, 199, 239, 293, 331, 389, 443, 503, 571, 641, 719, 797, 877, 971, 1063, 1163, 1277, 1373, 1481, 1601, 1721, 1861, 1997, 2131, 2281, 2437, 2591, 2753, 2927, 3089, 3271, 3457, 3659, 3847, 4049, 4231, 4441, 4663, 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.

			a(3)=7 since 2!,(2+3)!,(2+3+5)! are pairwise distinct modulo m=7 but not pairwise distinct modulo m=2,3,4,5,6.

Zhi-Wei Sun, Table of n, a(n) for n = 1..720
Zhi-Wei Sun, On functions taking only prime values, J. Number Theory 133(2013), no.8, 2794-2812.

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

s[n_]:=s[n]=Sum[Prime[k],{k,1,n}]
f[n_]:=f[n]=s[n]!
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,Max[2,s[n-1]],s[n]}];
   Print[n];Label[aa];Continue,{n,1,720}]

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

Original entry on oeis.org

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}]

A349459 Least positive integer m such that the n numbers k^2*(k^2-1) (k=1..n) are pairwise distinct modulo m.

Original entry on oeis.org

1, 5, 7, 11, 13, 23, 23, 23, 23, 41, 41, 41, 41, 41, 41, 101, 101, 107, 107, 107, 107, 107, 107, 107, 107, 107, 107, 107, 223, 223, 223, 223, 223, 223, 223, 223, 223, 223, 229, 239, 239, 239, 383, 383, 383, 383, 383, 383, 383, 383, 401, 401, 557, 557, 557, 557, 557, 557, 557, 557, 557, 557, 557, 733, 733, 733, 733, 733, 733, 733

Offset: 1

Zhi-Wei Sun, Nov 18 2021

nonn

Conjecture: For any integer n > 1, the term a(n) is the least prime p > 2*n with p dividing a^2 + b^2 - 1 for no 1 <= a < b <= n.
This has been verified for all n = 2..15000.
Note that a^2*(a^2-1)-b^2*(b^2-1) = (a-b)*(a+b)*(a^2+b^2-1). For any positive integers m and n > 1, if k^2*(k^2-1) (k=1..n) are pairwise distinct modulo m, then it is easy to see that m > 2*n.

			a(2) = 5 since the two numbers 1^2*(1^2-1)=0 and 2^2*(2^2-1) = 12 are distinct modulo 5, but they are congruent modulo each of 1,2,3,4.

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), no.8, 2794-2812.

Cf. A000290, A208643.

f[k_]:=f[k]=k^2*(k^2-1);
U[m_,n_]:=U[m,n]=Length[Union[Table[Mod[f[k],m],{k,1,n}]]]
tab={};s=1;Do[m=s;Label[bb];If[U[m,n]==n,s=m;tab=Append[tab,s];Goto[aa]];m=m+1;Goto[bb];Label[aa],{n,1,70}];Print[tab]

A349530 Least positive integer m such that the n numbers k*(k^4+1) (k=1..n) are pairwise distinct modulo m^2.

Original entry on oeis.org

1, 3, 3, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 25, 25, 25, 25, 25

Offset: 1

Zhi-Wei Sun, Nov 21 2021

nonn

Conjecture 1: Suppose that 5^a < sqrt(n) <= 5^(a+1). Then a(n) = 3*5^a if sqrt(n) <= sqrt(3)*5^a, and a(n) = 5^(a+1) otherwise.
Conjecture 2: Let f(n) be the least positive integer m such that the n numbers 18k*(k^4+1) (k=1..n) are pairwise distinct modulo m^2. Then f(n) is the least power of 5 not smaller than sqrt(n), except for 25 < n <= 45 (and in this case f(n) = 19).
Conjecture 3: Let n be a positive integer not among 26, 27, 28, 626, 627, 628, 629, 630, and define D(n) as the least positive integer m such that the n numbers k*(k^4+1) (k=1..n) are pairwise distinct modulo m. If 5^a < n <= 3*5^a, then D(n) = 3*5^a. If 3*5^a < n <= 5^(a+1), then D(n) = 5^(a+1).
We have verified the above conjectures for n up to 10^5.

			a(2) = 3 since the two numbers 1*(1^4+1) = 2 and 2*(2^4+1) = 34 are distinct modulo 3^2, but they are congruent modulo each of 1^2 and 2^2.

Zhi-Wei Sun, On functions taking only prime values, J. Number Theory 133(2013), no.8, 2794-2812.
Zhi-Wei Sun, New Conjectures in Number Theory and Combinatorics (in Chinese), Harbin Institute of Technology Press, 2021.
Quan-Hui Yang and Lilu Zhao, On a conjecture of Sun involving powers of three, arXiv:2111.02746 [math.NT], 2021.

Cf. A000351, A002523, A208643, A349459, A349537.

f[k_]:=f[k]=k*(k^4+1);
U[m_,n_]:=U[m,n]=Length[Union[Table[Mod[f[k],m^2],{k,1,n}]]]
tab={};s=1;Do[m=s;Label[bb];If[U[m,n]==n,s=m;tab=Append[tab,s];Goto[aa]];m=m+1;Goto[bb];Label[aa],{n,1,80}];Print[tab]

A349537 Least positive integer m such that the n numbers 33k^2(k^3+1) (k = 1..n) are pairwise distinct modulo m.

Original entry on oeis.org

1, 4, 7, 7, 13, 13, 13, 13, 13, 31, 41, 41, 61, 61, 61, 61, 61, 61, 61, 73, 101, 137, 137, 137, 137, 137, 137, 137, 137, 233, 233, 233, 233, 233, 233, 349, 349, 349, 349, 349, 349, 349, 349, 349, 349, 349, 349, 349, 349, 349, 349, 349, 349, 349, 547, 547, 547, 547, 547, 547, 547, 547, 547, 547, 859, 859, 859, 859, 859, 859

Offset: 1

Zhi-Wei Sun, Nov 21 2021

nonn

Conjecture: a(n) is prime for each n > 2.

We have verified this for n up to 10^4.

			a(2) = 4 since 33*1^2*(1^3+1) = 66 and 33*2^2*(2^3+1) = 1188 are incongruent modulo 4, but they are congruent modulo each of 1, 2 and 3.

Zhi-Wei Sun, On functions taking only prime values, J. Number Theory 133(2013), no.8, 2794-2812.
Zhi-Wei Sun, New Conjectures in Number Theory and Combinatorics (in Chinese), Harbin Institute of Technology Press, 2021.
Quan-Hui Yang and Lilu Zhao, On a conjecture of Sun involving powers of three, arXiv:2111.02746 [math.NT], 2021.

Cf. A000040, A000290, A001093, A208643, A349530, A349459.

f[k_]:=f[k]=33*k^2*(k^3+1);
U[m_,n_]:=U[m,n]=Length[Union[Table[Mod[f[k],m],{k,1,n}]]]
tab={};s=1;Do[m=s;Label[bb];If[U[m,n]==n,s=m;tab=Append[tab,s];Goto[aa]];m=m+1;Goto[bb];Label[aa],{n,1,70}];Print[tab]

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 *)

cp's OEIS Frontend

A210144 a(n) = least integer m>1 such that the product of the first k primes for k=1,...,n are pairwise distinct modulo m.

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A210186 a(n) = least integer m>1 such that m divides none of P_i + P_j with 0

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A210393 a(n) = least integer m>1 such that S_k! for k=1,...,n are pairwise distinct modulo m where S_k is the sum of the first k primes.

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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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A349459 Least positive integer m such that the n numbers k^2*(k^2-1) (k=1..n) are pairwise distinct modulo m.

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A349530 Least positive integer m such that the n numbers k*(k^4+1) (k=1..n) are pairwise distinct modulo m^2.

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A349537 Least positive integer m such that the n numbers 33k^2(k^3+1) (k = 1..n) are pairwise distinct modulo m.