A253597 Least Lucas-Carmichael number divisible by the n-th prime.

399, 935, 399, 935, 2015, 935, 399, 4991, 51359, 2015, 1584599, 20705, 5719, 18095, 2915, 46079, 162687, 22847, 46079, 16719263, 12719, 7055, 80189, 104663, 20705, 482143, 196559, 60059, 90287, 162687, 3441239, 13971671

Offset: 2

Author

Tim Johannes Ohrtmann, Jan 05 2015

Keywords

• nonn

Comments

Has any odd prime number at least one Lucas-Carmichael multiple?

Examples

			a(2) = 399 because this is the least Lucas-Carmichael number which is divisible by 3 (the second prime number).
		

Links

• Tim Johannes Ohrtmann and Charles R Greathouse IV, Table of n, a(n) for n = 2..1000 (terms up to 202 from Ohrtmann)

Crossrefs

Cf. A006972, A065091, A253598.

Programs

Mathematica

LucasCarmichaelQ[n_] := Block[{fi = FactorInteger@ n}, !PrimeQ@ n && Times @@ (Last@# & /@ fi) == 1 && Plus @@ Mod[n + 1, 1 + First@# & /@ fi] == 0]; f[n_] := Block[{k = p = Prime@ n}, While[ !LucasCarmichaelQ@ k, k += p]; k]; Array[f, 35, 2] (* Robert G. Wilson v, Feb 11 2015 *)

PARI

is_A006972(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
a(n) = pn = prime(n); ln = 1; until (is_A006972(ln) && (ln % pn == 0), ln++); ln;

PARI

is_A006972(n)=my(f=factor(n)); for(i=1, #f~, if((n+1)%(f[i, 1]+1) || f[i, 2]>1, return(0))); #f~>1
a(n)=my(p=prime(n), c=p^2+p, t=p); while(!is_A006972(t+=c),); t \\ Charles R Greathouse IV, Feb 03 2015

Formula

a(n) >> n^2 log^2 n. - Charles R Greathouse IV, Feb 03 2015