[math-fun] Prime[1.] [Was Chernick numbers]
DanA>Yes, it's fairly immediate. (Though the quoted formula is more explicit.) --Dan P.S. Can someone please tell me why my naïve Mma summation NSum[1/(Prime[k]^2 - 1), {k,1,Infinity}] gets the output: ----- Prime::intpp: Positive integer argument expected in Prime[16.]. Prime::intpp: Positive integer argument expected in Prime[17.]. Prime::intpp: Positive integer argument expected in Prime[18.]. General::stop: Further output of Prime::intpp will be suppressed during this calculation. Out[20]= 0.550863 ----- I mean, who asked it to put a float into Prime[] ? (Not me.) (The actual value is about .55[168] .) The problem is Mathematica perpetuating the *evil* myth that floats are fuzzy. They should be regarded as *exact* dyadic rationals liable to fuzzy operations. Thus in Macsyma, prime(1.) was 2. . In Mma 10, In[2]:= NSum[1/(Prime[k]^2 - 1), {k, 1, Infinity}, NSumTerms -> 9999] During evaluation of In[2]:= Prime::intpp: Positive integer argument expected in Prime[10000.]. >> During evaluation of In[2]:= Prime::intpp: Positive integer argument expected in Prime[10001.]. >> During evaluation of In[2]:= Prime::intpp: Positive integer argument expected in Prime[10002.]. >> During evaluation of In[2]:= General::stop: Further output of Prime::intpp will be suppressed during this calculation. >> During evaluation of In[2]:= SequenceLimit::seqlim: The general form of the sequence could not be determined, and the result may be incorrect. Out[2]= 0.551692946 --rwg On Nov 28, 2014, at 7:53 PM, Bill Gosper <billgosper@gmail.com> wrote: "J. Chernick[6] < http://en.wikipedia.org/wiki/Carmichael_number#cite_note-Chernick1939-6> proved a theorem in 1939 which can be used to construct a subset < http://en.wikipedia.org/wiki/Subset> of Carmichael numbers. The number [image: (6k + 1)(12k + 1)(18k + 1)] is a Carmichael number if its three factors are all prime." (I used to think these were the only Carmichael numbers.) Isn't it fairly immediate that this can be simplified to p*(2p-1)*(3p-2), p>3, all three factors prime, since 2p-1 can't be prime when mod(p,6)=5? --rwg
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Bill Gosper