A006972 -id:A006972 - cp's OEIS frontend

A306685 Composite squarefree numbers k such that k^2-1 is divisible by p-1 and p+1, where p are all the prime factors of k.

35, 1189, 3059, 6479, 8569, 30889, 39689, 51271, 84419, 133399, 272251, 321265, 430199, 544159, 564719, 569449, 585311, 608399, 1033241, 1212751, 1930499, 3131029, 7056721, 7110179, 7639919, 8740601, 11255201, 15857855, 17966519, 18996769, 22427999, 32871761, 34966009

Offset: 1

Paolo P. Lava, Mar 05 2019

nonn

			Prime factors of 35 are 5, 7 and 35^2-1 = 1224, 1124/4 = 306, 1124/6 = 204, 1124/8 = 153.
Prime factors of 1189 are 29, 41 and 1189^2-1 = 1413720, 1413720/28 = 50490, 1413720/30 = 47124, 1413720/40 = 35343, 1413720/42 = 33660.

Amiram Eldar, Table of n, a(n) for n = 1..100

Cf. A002997, A006972, A208728, A225711, A304291, A306723.

with(numtheory): P:=proc(q) local a,k,ok,n; for n from 2 to q do
if not isprime(n) and issqrfree(n) then a:=factorset(n); ok:=1;
for k from 1 to nops(a) do if frac((n^2-1)/(a[k]+1))>0 or frac((n^2-1)/(a[k]-1))>0 then ok:=0; break; fi; od; if ok=1 then print(n); fi; fi; od; end: P(10^9);

csfQ[n_]:=CompositeQ[n]&&SquareFreeQ[n]&&Union[Mod[n^2-1,Flatten[{#+1, #-1}&/@ FactorInteger[n][[All,1]]]]]=={0}; Select[Range[35*10^6],csfQ] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Apr 07 2020 *)

isok(n) = {if (issquarefree(n) && !isprime(n) && (n>1), my(f = factor(n)[,1], x = n^2-1); for (k=1, #f, if ((x % (f[k]-1)) || (x % (f[k]+1)), return (0));); return (1);); return (0);} \\ Michel Marcus, Mar 12 2019

More terms from Giovanni Resta, Mar 06 2019

A329948 Carmichael numbers m that have at least 3 prime factors p such that p+1 | m+1.

Original entry on oeis.org

9857524690572481, 33439671284716801, 96653613831890401, 270136961300544031, 528096456788419441, 650643395658753601, 710238404427321601, 1822922951416158241, 4011563714063821201, 4525693104167627041, 4631812281009523441, 7049793086137296001, 8605736094003523201, 10449416165574628801, 11175581620177915681, 12746447178170148001, 12769123623410580481, 17705945296667070001

Offset: 1

Daniel Suteu, Nov 25 2019

nonn

It is not known whether any Carmichael number (A002997) is also Lucas-Carmichael number (A006972). If such a number exists, then it would be a term of this sequence.

			m = 9857524690572481 is a term because it is a Carmichael number and it has at least 3 prime factors p, {13, 61, 433}, such that p+1 | m+1.

Amiram Eldar, Table of n, a(n) for n = 1..179 (terms below 10^22, calculated using data from Claude Goutier)
Claude Goutier, Compressed text file carm10e22.gz containing all the Carmichael numbers up to 10^22.
Wikipedia, Carmichael number.
Wikipedia, Lucas-Carmichael number.

Cf. A002997, A006972.

use bigint; use ntheory ':all'; sub isok { my $m = $[0]; is_carmichael($m) && (grep { ($m+1) % ($+1) == 0 } factor($m)) >= 3 };

A202160 a(n) = smallest k having at least five prime divisors d such that (d + n) | (k + n).

Original entry on oeis.org

588455, 179998, 460317, 6265805, 1236235, 287274, 949025, 1436932, 794871, 2013650, 3797365, 1169688, 3739827, 1587586, 6872565, 7706270, 1529983, 7351242, 2528045, 5247970, 487179, 10920965, 1316497, 121894476, 1404455, 5814874, 12223653, 2260412, 8022531

Offset: 1

Michel Lagneau, Dec 13 2011

nonn

The sequence of numbers k composite and squarefree, prime p | k ==> p+n | k+n is given by A029591 (least quasi-Carmichael number of order -n).
If k is squarefree, for n = 1, we obtain Lucas-Carmichael numbers: A006972.
In this sequence, the majority of terms are not squarefree.

			a(3) = 460317 because the prime divisors of 460317 are 3, 11, 13, 29, 37  =>
(3 + 3) | (460317 + 3) = 460320 = 6*76720;
(11 + 3) | 460320 = 14*32880;
(13 + 3) | 460320 = 16*28770;
(29+3)  |  460320 = 32*14385;
(37+3) | 460320 = 40*11508.

Amiram Eldar, Table of n, a(n) for n = 1..125

Cf. A006972, A029591, A202157, A202158, A202159.

with(numtheory):for n from 1 to 23 do:i:=0:for k from 1 to 10^8 while(i=0) do:x:=factorset(k):n1:=nops(x):y:=k+n: j:=0:for m from 1 to n1 do:if  n1>=2 and irem(y,x[m]+n)=0 then j:=j+1:else fi:od:if j>4 then i:=1: printf ( "%d %d \n",n,k):else fi:od:od:

numd[n_, k_] := Module[{p=FactorInteger[k][[;;,1]], c=0}, Do[If[Divisible[n+k, n+p[[i]]], c++], {i,1,Length[p]}]; c]; a[n_]:=Module[{k=1}, While[numd[n, k] <= 4, k++]; k]; Array[a, 30] (* Amiram Eldar, Sep 09 2019 *)

A217371 Array read by rows in which row n lists the prime factors of the n-th Lucas-Carmichael number.

Original entry on oeis.org

3, 7, 19, 5, 11, 17, 5, 13, 31, 5, 11, 53, 7, 23, 31, 7, 19, 43, 5, 17, 83, 5, 7, 11, 23, 7, 23, 79, 5, 7, 11, 47, 5, 41, 101, 11, 23, 83, 11, 31, 67, 5, 11, 13, 41, 5, 7, 17, 53, 11, 59, 71, 7, 11, 23, 29, 19, 29, 109, 11, 23, 251, 11, 41, 149, 5, 7, 11, 191, 23, 47, 71

Offset: 1

Gerasimov Sergey, Oct 01 2012

			Array begins:
3, 7, 19,
5, 11, 17,
5, 13, 31,
5, 11, 53,
7, 23, 31,
7, 19, 43,
5, 17, 83,
5, 7, 11, 23.

Jason Kimberley, Rows i = 1..1000 of irregular triangle, flattened
John Smith, Lucas-Carmichael number, PlanetMath.org
Wikipedia, Lucas-Carmichael number
Index entries for sequences related to Carmichael numbers.

Cf. A006972, A027746.

The n-th row is the A006972(n)-th row of A027746. - Jason Kimberley, Nov 04 2012

A238162 Least common multiple of the prime factors of n, each increased by 1.

Original entry on oeis.org

3, 4, 3, 6, 12, 8, 3, 4, 6, 12, 12, 14, 24, 12, 3, 18, 12, 20, 6, 8, 12, 24, 12, 6, 42, 4, 24, 30, 12, 32, 3, 12, 18, 24, 12, 38, 60, 28, 6, 42, 24, 44, 12, 12, 24, 48, 12, 8, 6, 36, 42, 54, 12, 12, 24, 20, 30, 60, 12, 62, 96, 8, 3, 42, 12, 68, 18, 24, 24, 72, 12, 74, 114, 12, 60, 24, 84, 80, 6, 4, 42, 84, 24, 18, 132, 60, 12, 90, 12, 56, 24, 32, 48, 60, 12, 98, 24, 12, 6

Offset: 2

Joseph L. Pe, Feb 18 2014

nonn

If n is prime, then a(n) = n + 1. - Wesley Ivan Hurt, Apr 05 2014

If n is a composite squarefree number and a(n) divides n+1, then n is a Lucas-Carmichael number (A006972). - Daniel Suteu, Oct 02 2022

			The prime factors of 6 are 2 and 3, which become 3 and 4 when respectively increased by 1, and lcm(3, 4) = 12. Therefore, a(6) = 12.

Daniel Suteu, Table of n, a(n) for n = 2..10000

Cf. A006972.

a(n) = my(f=factor(n)); lcm(vector(#f~, k, f[k, 1]+1)); \\ Daniel Suteu, Oct 02 2022

A287119 Squarefree composite numbers n such that p^2 - 1 divides n^2 - 1 for every prime p dividing n.

Original entry on oeis.org

8569, 39689, 321265, 430199, 564719, 585311, 608399, 7056721, 11255201, 17966519, 18996769, 74775791, 75669551, 136209151, 321239359, 446660929, 547674049, 866223359, 1068433631, 1227804929, 1291695119, 2315403649, 2585930689, 7229159729, 7809974369, 8117634239

Offset: 1

Thomas Ordowski, May 20 2017

nonn

Such numbers are odd and have at least three prime factors.

Problem: are there infinitely many such numbers?

Giovanni Resta, Table of n, a(n) for n = 1..50

Cf. A002997, A006972, A175530, A175531.

Subsequence of A120944.

isok(n) = {if (issquarefree(n) && !isprime(n), my(f = factor(n)); for (k=1, #f~, if ((n^2-1) % (f[k,1]^2-1), return (0));); return (1););} \\ Michel Marcus, May 20 2017

More terms from Michel Marcus, May 20 2017

a(14)-a(26) from Giovanni Resta, May 20 2017

A292021 Lucas-Carmichael numbers that are congruent to 1 (mod 4).

Original entry on oeis.org

20705, 80189, 120581, 1162349, 7274249, 8734109, 10260809, 14658349, 49412285, 90393029, 105818129, 110066669, 125532329, 256074029, 362868329, 366648281, 395032609, 434886605, 503733257, 558705449, 563601257, 574342145, 640057109, 939989609, 962529749

Offset: 1

Amiram Eldar, Sep 07 2017

nonn

Most Lucas-Carmichael numbers are congruent to 3 (mod 4). Of the 9967 numbers less than 10^12 only 198 are congruent to 1 (mod 4).

Analogous to A185321 - Carmichael numbers that are congruent to 3 (mod 4).

Amiram Eldar, Table of n, a(n) for n = 1..198 (terms below 10^12)

Intersection of A006972 and A016813.

Cf. A110885, A185321.

a=Select[Range[2, 10^6],!PrimeQ[#] && Union[Transpose[FactorInteger[#]][[2]]] == {1} && Union[Mod[# + 1, Transpose[FactorInteger[#]][[1]] + 1]]=={0} &] ;Select[a,Mod[#,4]==1 &] (* after Richard Pinch and Jeffrey Shallit at A006972 *)

A292539 Primes p1 such that p2 = 2p1 + 1 and p3 = p1p2 - 2 are also primes, so p1p2*p3 is a Lucas-Carmichael number of the form k^2 - 1.

Original entry on oeis.org

3, 5, 11, 29, 53, 83, 173, 239, 281, 359, 431, 719, 761, 809, 911, 1031, 1103, 1223, 1289, 1451, 1481, 1511, 1559, 1931, 2069, 2339, 2351, 2393, 2693, 2699, 2819, 2969, 3359, 3491, 3539, 3851, 4019, 4211, 4409, 5039, 6113, 6269, 6329, 6491, 6521, 6551, 6581

Offset: 1

Amiram Eldar, Sep 18 2017

nonn

All the primes, except the first, are of the form p1 = 6k - 1, p2 = 12k - 1, p3 = 72k^2 - 18k - 1, with k = 1, 2, 5, 9, 14, 29, 40, 47, 60, 72, 120, 127, 135, 152, 172, 184, ...
The generated Lucas-Carmichael numbers are 399, 2915, 63503, 2924099, 32148899, 192099599, 3603600899, 13105670399, 25027872803, ...
Subsequence of A005384 (Sophie Germain primes).

			p1 = 3 is in the sequence since with p2 = 2*3 + 1 = 7 and p3 = 3*7 - 2 = 19 they are all primes. 3*7*19 = 399 is a Lucas-Carmichael number.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A006972, A216925, A290947, A292538.

aQ[n_] := AllTrue[{n, 2n+1, 2 n^2+n-2}, PrimeQ]; Select[Range[10^3], aQ]
Select[Prime[Range[1000]],AllTrue[{2#+1,#(2#+1)-2},PrimeQ]&] (* Harvey P. Dale, Aug 16 2024 *)

is(n) = if(!ispseudoprime(n), return(0), my(p=2*n+1); if(!ispseudoprime(p), return(0), if(ispseudoprime(n*p-2), return(1)))); 0 \\ Felix Fröhlich, Sep 18 2017

A300959 Number of prime factors of the n-th Lucas-Carmichael number.

Original entry on oeis.org

3, 3, 3, 3, 3, 3, 3, 4, 3, 4, 3, 3, 3, 4, 4, 3, 4, 3, 3, 3, 4, 3, 3, 4, 3, 3, 3, 4, 3, 4, 3, 3, 4, 4, 4, 3, 3, 4, 3, 4, 4, 3, 4, 3, 5, 4, 3, 3, 3, 3, 4, 3, 3, 4, 3, 4, 4, 3, 3, 3, 5, 4, 4, 3, 3, 4, 3, 4, 3, 3, 4, 4, 4, 3, 3, 4, 4, 3, 4, 4, 4, 4, 3, 5, 3, 4, 3

Offset: 1

Tim Johannes Ohrtmann, Mar 17 2018

nonn

The number of prime factors is always >= 3.

Tim Johannes Ohrtmann, Table of n, a(n) for n = 1..10000

Cf. A006972 (Lucas-Carmichael numbers).
Cf. A216925, A216926, A216927, A217002, A217003, A217091 (Lucas-Carmichael numbers with 3 to 8 prime factors).
Cf. A216928 (Least Lucas-Carmichael number with n prime factors).

islc(n)=my(f=factor(n)); for(i=1, #f[, 1], if((n+1)%(f[i, 1]+1) || f[i, 2]>1, return(0))); #f[, 1]>1; \\ from A006972
lista(nn) = for (n=1, nn, if (islc(n), print1(omega(n), ", "))); \\ Michel Marcus, Mar 17 2018

a(n) = A001221(A006972(n)).

A306723 Composite squarefree numbers k such that k^2+1 is divisible by p-1, where p are all the prime factors of k.

Original entry on oeis.org

33, 36003, 426747, 220067817

Offset: 1

Paolo P. Lava, Mar 06 2019

Tested up to 3*10^10. - Giovanni Resta, Mar 06 2019

			Prime factors of 33 are 3, 11 and 33^2+1 = 1090, 1090/2 = 545, 1090/10 = 109.
Prime factors of 220067817 are 3, 59, 131, 9491 and 220067817^2+1 = 48429844079145490, 48429844079145490/2 = 24214922039572745, 48429844079145490/58 = 834997311709405, 48429844079145490/130 = 372537262147273, 48429844079145490/9490 = 5103250166401.

Cf. A002997, A006972, A208728, A225711, A304291, A306685.

with(numtheory): P:=proc(q) local a,k,ok,n;
for n from 1 to q do if not isprime(n) and issqrfree(n) then a:=factorset(n); ok:=1; for k from 1 to nops(a) do if frac((n^2+1)/(a[k]+1))>0 then ok:=0; break; fi; od; if ok=1 then print(n); fi; fi; od; end: P(10^20);

isok(n) = {if (issquarefree(n) && !isprime(n) && (n>1), my(f = factor(n)[,1], x = n^2+1); for (k=1, #f, if ((x % (f[k]-1)), return (0));); return (1);); return (0);} \\ Michel Marcus, Mar 12 2019

a(4) from Giovanni Resta, Mar 06 2019

cp's OEIS Frontend

A306685 Composite squarefree numbers k such that k^2-1 is divisible by p-1 and p+1, where p are all the prime factors of k.

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A329948 Carmichael numbers m that have at least 3 prime factors p such that p+1 | m+1.

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Perl

A202160 a(n) = smallest k having at least five prime divisors d such that (d + n) | (k + n).

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Mathematica

A217371 Array read by rows in which row n lists the prime factors of the n-th Lucas-Carmichael number.

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Formula

A238162 Least common multiple of the prime factors of n, each increased by 1.

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PARI

A287119 Squarefree composite numbers n such that p^2 - 1 divides n^2 - 1 for every prime p dividing n.

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A292021 Lucas-Carmichael numbers that are congruent to 1 (mod 4).

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Mathematica

A292539 Primes p1 such that p2 = 2p1 + 1 and p3 = p1*p2 - 2 are also primes, so p1*p2*p3 is a Lucas-Carmichael number of the form k^2 - 1.

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A292539 Primes p1 such that p2 = 2p1 + 1 and p3 = p1p2 - 2 are also primes, so p1p2*p3 is a Lucas-Carmichael number of the form k^2 - 1.