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A244561 Odd integers m such that for every integer k > 0, m*2^k+1 has a divisor in the set {3, 5, 7, 13, 17, 241}. 7
271129, 271577, 482719, 575041, 603713, 903983, 965431, 1518781, 1624097, 1639459, 2131043, 2131099, 2541601, 2931767, 2931991, 3083723, 3098059, 3555593, 3608251, 4067003, 4573999, 6134663, 6135559, 6557843, 6676921, 6678713, 6742487, 6799831, 7400371, 7523267, 7523281, 7761437, 7765021, 7892569, 8007257, 8629967, 8840599, 8871323, 9208337, 9454129, 9454157, 9854491, 9854603, 9930469, 9937637, 10192733, 10422109, 10675607 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,1
COMMENTS
For n > 48, a(n) = a(n-48) + 11184810; the first 48 values are in the data.
The set {3, 5, 7, 13, 17, 241} is the set of prime divisors of 2^24 - 1. Hence for every p in the set the multiplicative order of 2 modulo p divides 24. Note that twice the product of {3, 5, 7, 13, 17, 241} is 11184810. - Jeppe Stig Nielsen, Mar 10 2019
Subset of provable Sierpiński numbers A076336. - Jeppe Stig Nielsen, Mar 10 2019
LINKS
FORMULA
For n > 48, a(n) = a(n-48) + 11184810.
PROG
(PARI) D=[3, 5, 7, 13, 17, 241]; P=2*lcm(D); M=lcm(apply(d->znorder(Mod(2, d)), D)); forstep(k=1, +oo, 2, if(k%P==1, print(); print()); for(n=0, M-1, for(i=1, #D, k*Mod(2, D[i])^n+1==0 && next(2)); next(2)); print1(k, ", ")) \\ Jeppe Stig Nielsen, Mar 10 2019
CROSSREFS
Sequence in context: A075467 A257647 A206001 * A137715 A258893 A237181
KEYWORD
nonn
AUTHOR
Pierre CAMI, Jun 30 2014
STATUS
approved

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Last modified May 21 17:21 EDT 2024. Contains 372738 sequences. (Running on oeis4.)