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 .55618 .)
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 _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun