A174590 a(n) = (k-1)/lambda(k), the index of the n-th Carmichael number k.

7, 23, 48, 22, 47, 5, 45, 21, 44, 163, 342, 162, 43, 31, 1777, 314, 337, 161, 1753, 70, 2868, 1745, 421, 2487, 1363, 159, 39, 645, 950, 67, 198, 1358, 949, 158, 2303, 134, 305, 1692, 1733, 5731, 2794, 7107, 1732, 345, 1689, 2654, 1671, 1829, 947, 1353, 1557

Offset: 1

Author

Michel Lagneau, Mar 23 2010, Mar 31 2010

Keywords

• nonn

Comments

The index of a Carmichael number k is i(k) = (k-1)/lambda(k).

Or, i(k) = (k-1)/lcm(p_1-1,p_2-1,...,p_j-1), where k = p_1*p_2*...*p_j. - Thomas Ordowski, Oct 15 2015

For composite k, lambda(k) divides k-1 iff k is a Carmichael number. - Thomas Ordowski, Oct 23 2015

Examples

			a(1)= 7 because A002997(1) = 561, and (561 - 1)/lambda(561) = 560/80 = 7.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000
• J. M. Chick, Carmichael number variable relations: three-prime Carmichael numbers up to 10^24, arXiv:0711.2915 [math.NT] 2007-2008, Table 1, p. 34.
• Andrew Granville and Carl Pomerance, Two contradictory conjectures concerning Carmichael numbers, Math. Comp. 71 (2002), no. 238, 883-908.
• R. G. E. Pinch, Carmichael numbers up to 10^18, April 2006.
• Richard Pinch, Carmichael Numbers up to 10^20, Abstract, ANTS 7.
• Richard Pinch, Carmichael Numbers up to 10^20, Poster, ANTS 7.

Crossrefs

Cf. A002322 (the Carmichael lambda function), A002997, A011773, A306414.

Programs

Maple

with(numtheory) : for n from 2 to 2000000 do: if type(n,prime)=false and issqrfree(n)=true then  x:=factorset(n):n1:=nops(x):ii:=0:for j from 1 to n1 do:if irem(n-1, x[j]-1)=0  then ii:=ii+1:else fi:od:if ii=n1 then z:=(n-1)/lambda(n):printf(`%d, `,z):else fi:fi:od:

Mathematica

carNums = Select[Range[561, 3 10^6, 2], CompositeQ[#] && Mod[#, CarmichaelLambda[#]] == 1&];
a[n_] := (carNums[[n]] - 1)/CarmichaelLambda[carNums[[n]]];
Array[a, 60] (* Jean-François Alcover, Sep 05 2018 *)

PARI

t(n)=my(f=factor(n)); for(i=1, #f[, 1], if(f[i, 2]>1||(n-1)%(f[i, 1]-1), return(0))); 1;
for(n=1, 1e7, if(n%2 && !isprime(n) && t(n) && n>1, print1((n-1)/(lcm(znstar(n)[2])), ", "))) \\ Altug Alkan, Oct 15 2015

Formula

a(n) = (A002997(n) - 1) / lambda(A002997(n)).

a(n) = (A002997(n) - 1) / A306414(n). - Jianing Song, Dec 12 2021

Extensions

Edited by Michel Lagneau, Jul 31 2012

Further edits from N. J. A. Sloane, Oct 31 2015