A099795 -id:A099795 - cp's OEIS frontend

A099796 a(n) = (A094998(n)-1) / A099795(n).

1, 1, 2, 5, 10, 3, 10, 4, 3, 28, 17, 18, 30, 20, 41, 42, 14, 19, 30, 37, 63, 50, 7, 12, 83, 30, 91, 19, 69, 91, 97, 56, 22, 80, 39, 137, 44, 9, 154, 19, 37, 141, 141, 168, 126, 183, 200, 205, 136, 55, 95, 204, 126, 213, 230, 68, 63, 158, 202, 162, 102, 182, 104, 38, 165, 119

Offset: 1

Ray Chandler, Oct 29 2004

nonn

Cf. A094998, A099794, A099795.

Cf. A059955. [From R. J. Mathar, Sep 02 2008]

A254939 a(n) = (A099795(n)^-1 mod p)*A099795(n), where p = prime(n).

Original entry on oeis.org

1, 4, 36, 120, 2520, 277200, 5045040, 183783600, 4655851200, 80313433200, 32607253879200, 2743667504978400, 58772246027695200, 5038384364010597600, 56517528952814529600, 34089489546705963770400, 7391221142626702144764000

Offset: 1

Bruno Berselli, Feb 12 2015 - proposed by Umberto Cerruti (Department of Mathematics "Giuseppe Peano", University of Turin, Italy)

nonn

The sequence lists the smallest nonnegative solutions z to the system of congruences z == 1 (mod p), z == 0 (mod v(p-1)), where p is a prime and v(p-1) = lcm(1,...,p-1).

			5045040 is the seventh term of the sequence because the modular inverse of A099795(7) mod A000040(7) is 7 and 7*A099795(7) = 7*720720 = 5045040.

Umberto Cerruti, Il Teorema Cinese dei Resti (in Italian), 2015. The sequence is on page 21.
Eric Weisstein's World of Mathematics, Modular Inverse.

Cf. A000040, A056604, A099795, A254924, A255010.

[Modinv(Lcm([1..p-1]),p)*Lcm([1..p-1]): p in PrimesUpTo(60)];

with(numtheory): P:=proc(q)  local a, n;  a:=[];
for n from 1 to q do a:=[op(a),n]; if isprime(n+1) then
print(lcm(op(a))*(lcm(op(a))^(-1) mod (n+1))); fi;
od; end: P(10^3); # Paolo P. Lava, Feb 16 2015

r[k_] := LCM @@ Range[k]; u[k_] := PowerMod[r[k - 1], -1, k] r[k - 1]; Table[u[Prime[n]], {n, 1, 20}]

a099795(n) = lcm(vector(prime(n)-1, k, k));
a(n) = {my(m = a099795(n)); m*lift(1/Mod(m, prime(n)));} \\ Michel Marcus, Feb 13 2015

a(n) = A255010(n)*A099795(n).

A255010 a(n) = A099795(n)^-1 mod prime(n).

Original entry on oeis.org

1, 2, 3, 2, 1, 10, 7, 15, 20, 1, 14, 19, 11, 23, 6, 11, 45, 42, 37, 34, 10, 29, 76, 77, 14, 71, 12, 88, 40, 22, 30, 75, 115, 59, 110, 14, 113, 154, 13, 154, 142, 40, 50, 25, 71, 16, 11, 18, 91, 174, 138, 35, 115, 38, 27, 195, 206, 113, 75, 119, 181, 111, 203

Offset: 1

Bruno Berselli, Feb 13 2015 - proposed by Umberto Cerruti (Department of Mathematics "Giuseppe Peano", University of Turin, Italy)

nonn

By the definition, a(n)*A099795(n) == 1 (mod prime(n)).

a(n) is 1 with the primes 2, 11, 29, 787, 15773 (see A178629).

Robert Israel, Table of n, a(n) for n = 1..10000
Umberto Cerruti, Il Teorema Cinese dei Resti (in Italian), 2015. The sequence is on page 21.
Eric Weisstein's World of Mathematics, Modular Inverse

Cf. A000040, A099795, A178629, A254924, A254939.

[Modinv(Lcm([1..p-1]),p): p in PrimesUpTo(400)];

with(numtheory): P:=proc(q)  local a, n;  a:=[];
for n from 1 to q do a:=[op(a),n]; if isprime(n+1) then print(lcm(op(a))^(-1) mod (n+1)); fi;
od; end: P(10^3); # Paolo P. Lava, Feb 16 2015

r[k_] := LCM @@ Range[k]; t[k_] := PowerMod[r[k - 1], -1, k]; Table[t[Prime[n]], {n, 1, 70}]

a(n) = lift(1/Mod(lcm(vector(prime(n)-1, k, k)), prime(n))); \\ Michel Marcus, Feb 13 2015

[inverse_mod(lcm([1..p-1]),p) for p in primes(400)]

a(n) = A254939(n)/A099795(n).

A094998 a(n) is smallest positive integer multiple of n-th prime, say kprime(n), such that kprime(n) == 1 (mod j) for each integer j with 1 < j < prime(n).

Original entry on oeis.org

2, 3, 25, 301, 25201, 83161, 7207201, 49008961, 698377681, 2248776129601, 39594522567601, 2599263952084801, 160287943711896001, 4381203794791824001, 386203114510899285601, 130159869178331861668801

Offset: 1

Mark Troll (mtroll(AT)u.washington.edu), Oct 22 2004

nonn

			For n = 1 there are no such j, so the condition is vacuously satisfied and we can take k=1, getting a(1)=2. - _N. J. A. Sloane_, Feb 10 2015

Dimitri Papadopoulos, Table of n, a(n) for n = 1..167

Cf. A099794, A099795, A099796.

/* By definition (slow): */
S:=[]; for n in [1..9] do k:=1; while not forall{j: j in [2..NthPrime(n)-1] | IsOne(k*NthPrime(n) mod j)} do k:=k+1; end while; Append(~S, k*NthPrime(n)); end for; S; /* or */
[p eq 2 select p else Modinv(p, Lcm([1..p-1]))*p: p in PrimesUpTo(60)]; // Bruno Berselli, Feb 08 2015

a[1] = 2; a[2] = 3; a[n_] := Module[{p, m, r, r0, r1}, p = Prime[n]; m = LCM @@ Range[2, p - 1]; r = Reduce[k > 0 && p*k + m*j == 1, {k, j}, Integers]; r0 = r /. C[] -> 0; r1 = r /. C[] -> 1; If[r0 === False, r1[[1, 2]], Min[r0[[1, 2]], r1[[1, 2]]]]*p]; Table[a[n], {n, 1, 20}] (* Jean-François Alcover, Feb 09 2015 *)

a(n) = {p=prime(n); k=1; for(n=2, p-1, k=lcm(k,n)); for(j=1, p, if((j*k+1)/p==ceil((j*k+1)/p), t=j*k+1; break())); return(t);} \\ Dimitri Papadopoulos, Dec 28 2018

a(n) = A099795(n) * A099796(n) + 1 = A099794(n) * prime(n).

log(a(n)) = prime(n) (approximately, empirical observation). - Dimitri Papadopoulos, Dec 27 2018

Edited and extended by Ray Chandler, Oct 29 2004

Added "positive" to definition. - N. J. A. Sloane, Feb 10 2015

A099794 a(n) = smallest integer k such that k*prime(n) == 1 mod j for each integer j with 1

Original entry on oeis.org

1, 1, 5, 43, 2291, 6397, 423953, 2579419, 30364247, 77544004469, 1277242663471, 70250377083373, 3909462041753561, 101888460343995907, 8217087542785091183, 2455846588270412484317, 38974424104246263663539

Offset: 1

Ray Chandler, Oct 29 2004

nonn

Cf. A094998, A099795, A099796.

/* By definition (slow): */
S:=[]; for n in [1..9] do k:=1; while not forall{j: j in [2..NthPrime(n)-1] | IsOne(k*NthPrime(n) mod j)} do k:=k+1; end while; Append(~S, k); end for; S; /* or */
[p eq 2 select 1 else Modinv(p, Lcm([1..p-1])): p in PrimesUpTo(60)];// Bruno Berselli, Feb 08 2015

a[1] = a[2] = 1; a[n_] := Module[{p, m, r, r0, r1}, p = Prime[n]; m = LCM @@ Range[2, p-1]; r = Reduce[k>0 && p*k + m*j == 1, {k, j}, Integers]; r0 = r /. C[] -> 0; r1 = r /. C[] -> 1 ; If[r0 === False, r1[[1, 2]], Min[r0[[1, 2]], r1[[1, 2]]]]]; Table[a[n], {n, 1, 20}] (* Jean-François Alcover, Feb 09 2015 *)

a(n) = A094998(n) / prime(n).

A354882 a(n) is the smallest number k that is divisible by all numbers d with d < p = prime(n), and such that all of k+1, k-1, k+p, k-p are prime.

Original entry on oeis.org

12, 60, 93240, 2383920, 298378080, 5133688560, 73329656400, 2168462696400, 1215784751781600, 150901712773812000, 133573286426580000, 657837749787992373600, 10597036678652724300000, 2761248653283183065402400, 2053281233421697855815439200

Offset: 3

Florian Baur, Jun 10 2022

nonn

Suggested by Charles Kusniec on the mersenneforum.org message board (see "Links" section): a number c(n) = k(n) + r(n), where k(n) = m * A099795(n), r(n) = {-1, 1, p, -p} and p = prime(n), is not divisible by any d with d < p and the density of primes among c(n) is expected to be higher than for random numbers of that magnitude.
The probability that an arbitrary c(n) is prime is higher than that of an arbitrary number of the same magnitude. Let t(n) denote that probability, then t(n) = 1/(((1/2)*(2/3)*(4/5)*(6/7)*...*(prime(n-1)-1)/prime(n-1)) * log(c(n))). According to Mertens's third theorem the value asymptotically approaches 1/((e^gamma/log(prime(n-1))) * log(c(n))). log(c(n)) can be approximated as prime(n-1). This yields t(n) ~ log(prime(n-1))/(e^gamma*prime(n-1)). The probability that c(n) is prime for a random value of m is t(n)^4. Thus, the sequence is expected to grow with (prime(n-1)/log(prime(n-1)))^4*A099795(n). For n < 201 the arithmetic mean of m(n)*t(n)^4 is 1.1. - Florian Baur, Jul 12 2023
For all n < 171, a(n) > a(n-1) with the exception of a(11) < a(10). This occurs whenever m(n-1) > prime(n) * m(n).
The principle can be extended to r(n,i) = {-1,1,-p,p,-q_i,q_i} where q_i = prime(n+i). Such a sequence b for i = 1 would have b(3) = 12, as 5, 7, 11, 13, 17, 19 are all prime. This is the only number k for which all three of k+-1, k+-5, and k+-7 are prime. To satisfy the first requirement for k > 6, we need k == {0, 2, 8} (mod 10). Under this condition, one of k-5 or k+-7 will be divisible by 5. Since 12 - 7 = 5 is the only prime that is divisible by 5, k = 12 is the only k satisfying the condition.
If q, with a(n) + 1 < q < a(n) + prime(n)^2, is prime, the difference r(n) = q - a(n) is also prime. Proof: Per definition a(n) is divisible by all d < prime(n). It follows that, if r is divisible by any d, then so is q = a(n) + r, whence q is not prime. Thus, if q is prime, then r is either also prime or only has prime factors f >= prime(n), i.e., r >= prime(n)^2. See "Fortunate Numbers" (A005235).
There is no a(1) and a(2). Since prime(1) = 2, both k+1 and k+2 need to be prime. This is only true for k = 1, but 1 - 1 = 0 is not prime. For a(2) we have prime(2) = 3 and one of k+1, k-1, k+3 is divisible by 3.

			a(3): The 3rd prime is 5. The smallest number divisible by all d < 5 is 12. Since 12 - 1 = 11, 12 + 1 = 13, 12 + 5 = 17, 12 - 5 = 7 are all prime, a(3) = 12.
a(5): The 5th prime is 11. The smallest number divisible by all d < 11 is 2520. However, 2520 - 1 = 2519 is not prime. The smallest number satisfying all conditions is 93240, since 93240 - 1, 93240 + 1, 93240 + 11, 93240 - 11 are all prime and 93240 is divisible by all d < 11. Thus, a(5) = 93240.

Florian Baur, Table of n, a(n) for n = 3..200
Charles Kusniec, An idea for a new class of some numbers, mersenneforum.org

A354882(n) = { my(s = 1, p = prime(n), c = lcm([1..p-1])); while(!(isprime(s*c+1) & isprime(s*c-1) & isprime(s*c+p) & isprime(s*c-p)), s++); return(s*c)} \\ Florian Baur, Jul 17 2023

a(n) = m(n) * A099795(n). Specifically, m(3) = m(4) = 1. For all other n < 201, 25 < m(n) < 333054037 and m(n) cannot have prime(n) as a factor. - Florian Baur, Jul 12 2023

A377356 a(n) = Product{i = 1..(n-1)} prime(i)^e_i, where prime(i)^e_i is the smallest power of prime(i) which exceeds prime(n).

Original entry on oeis.org

1, 4, 72, 1800, 529200, 64033200, 21643221600, 6254891042400, 2258015666306400, 17917354312141284000, 15068494976510819844000, 28961647344853795740168000, 39648495215104846368289992000, 66649120456591246745095476552000, 123234223724237215231681536144648000, 1905570801447880059127491593404692024000

Offset: 1

David James Sycamore, Oct 26 2024

nonn

a(n) is the product of powers of primes p, for all p < prime(n), where each prime power is the smallest which exceeds prime(n), (compare with A099795). Every term may be expressed as a product of primorial powers, (A002110(n-1)^2 being the greatest primorial power divisor of a(n)).
From Michael De Vlieger, Oct 26 2024: (Start)
This sequence adds 1 to all exponents of prime power factors of A099795(n) for n > 1.
Proper subset of A001694, all terms are powerful. (End)

			For n = 5, a(5) = 529200, since prime(5) = 11, thus we have 2^4*3^3*5^2*7^2 = 16*27*25*49 = 529200. We may express this instead as 210*2520 = A002110(4)*A099795(5) = 210^2*6^1*2^1 = 529200.
From _Michael De Vlieger_, Oct 26 2024: (Start)
Table of first 12 terms showing exponents of prime power factors of a(n), where "." represents 0.
                                    Exponents of primes
                                            1 1 1 1 2 2 3
   n                          a(n)  2 3 5 7 1 3 7 9 3 9 1
  -------------------------------------------------------
   1                            1   . . . . . . . . . . .
   2                            4   2 . . . . . . . . . .
   3                           72   3 2 . . . . . . . . .
   4                         1800   3 2 2 . . . . . . . .
   5                       529200   4 3 2 2 . . . . . . .
   6                     64033200   4 3 2 2 2 . . . . . .
   7                  21643221600   5 3 2 2 2 2 . . . . .
   8                6254891042400   5 3 2 2 2 2 2 . . . .
   9             2258015666306400   5 3 2 2 2 2 2 2 . . .
  10         17917354312141284000   5 4 3 2 2 2 2 2 2 . .
  11      15068494976510819844000   5 4 3 2 2 2 2 2 2 2 .
  12   28961647344853795740168000   6 4 3 2 2 2 2 2 2 2 2 (End)

Michael De Vlieger, Table of n, a(n) for n = 1..193

Cf. A000040, A001694, A002110, A007947, A099795.

Array[Product[Prime[i]^(1 + Floor[Log[Prime[i], Prime[#]]]), {i, # - 1}] &, 12] (* Michael De Vlieger, Oct 26 2024 *)

a(n) = A002110(n-1)*A099795(n); A007947(a(n)) = rad(a(n)) = A002110(n-1).

More terms from Michael De Vlieger, Oct 26 2024

A109923 a(n) = lcm(1,2,3,...,prime(n))/(1 + 2 + ... + prime(n)).

Original entry on oeis.org

1, 4, 15, 420, 3960, 80080, 1225224, 19399380, 5354228880, 145568097675, 7600186994400, 254425307479200, 9957281351799600, 392482839950100900, 114779426083185063200, 5474978624167927514640, 312603618218620377448800

Offset: 2

Amarnath Murthy, Jul 16 2005

			a(4)=15 because the 4th prime is 7 and lcm(1,2,3,4,5,6,7)/(1+2+3+4+5+6+7) = 420/28 = 15.

Cf. A006254, A034953, A056604, A099795, A109922.

a:=n->lcm(seq(i,i=1..ithprime(n)))/sum(j,j=1..ithprime(n)): seq(a(n),n=2..20); # Emeric Deutsch, Jul 16 2005

a(n) = lcm(vector(prime(n), k, k))/sum(k=1, prime(n), k); \\ Michel Marcus, Mar 07 2018

a(n) = A099795(n)/A006254(n-1). - Andrey Zabolotskiy, Mar 07 2018

a(n) = A056604(n)/A034953(n). - Michel Marcus, Mar 07 2018

More terms from Emeric Deutsch, Jul 16 2005

cp's OEIS Frontend

A099796 a(n) = (A094998(n)-1) / A099795(n).

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A254939 a(n) = (A099795(n)^-1 mod p)*A099795(n), where p = prime(n).

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A255010 a(n) = A099795(n)^-1 mod prime(n).

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A094998 a(n) is smallest positive integer multiple of n-th prime, say k*prime(n), such that k*prime(n) == 1 (mod j) for each integer j with 1 < j < prime(n).

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A099794 a(n) = smallest integer k such that k*prime(n) == 1 mod j for each integer j with 1

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A354882 a(n) is the smallest number k that is divisible by all numbers d with d < p = prime(n), and such that all of k+1, k-1, k+p, k-p are prime.

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A377356 a(n) = Product{i = 1..(n-1)} prime(i)^e_i, where prime(i)^e_i is the smallest power of prime(i) which exceeds prime(n).

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A109923 a(n) = lcm(1,2,3,...,prime(n))/(1 + 2 + ... + prime(n)).

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A094998 a(n) is smallest positive integer multiple of n-th prime, say kprime(n), such that kprime(n) == 1 (mod j) for each integer j with 1 < j < prime(n).