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%I #40 Aug 11 2015 17:25:08
%S 1423,2143,2341,4231,12253,21523,22153,22531,23251,25321,32251,35221,
%T 36343,36433,43633,52321,64333,114451,144511,224461,244261,246241,
%U 365557,415141,424261,426421,446221,446461,451411,462421,466441,541141,555637,556537,556573
%N Primes with nonzero digits such that sum of cubes of digits equal to square of sums.
%C Largest term of this sequence is the 20-digit prime 99151111111111111111.
%C The Pagni article mentioned below has no bearing on this problem because it deals with the well-known identity sum_{i=1..n} i^3 = (sum_{i=1..n} i)^2. However, the article is interesting. - _T. D. Noe_, Jul 26 2013
%C This sequence has exactly 14068465 provable primes. This result required about one hour of Mathematica on fairly fast computer having 16 GB of memory. - _T. D. Noe_, Jul 30 2013
%H T. D. Noe, <a href="/A225567/b225567.txt">Table of n, a(n) for n = 1..1201</a> (terms < 10^7)
%H David Pagni, <a href="http://www.jstor.org/stable/3620410">82.27 An interesting number fact</a>, The Mathematical Gazette 82:494 (1998), pp. 271-273.
%H C. Rivera, <a href="http://www.primepuzzles.net/puzzles/puzz_158.htm">PP&P Puzzle 158: Sum of Cubes equal to Square of Sum</a>
%e a(5) = 12253 since 1^3 + 2^3 + 2^3 + 5^3 + 3^3 = (1 + 2 + 2 + 5 + 3)^2.
%t (* let tz[[i]] be numbers computed in A227073 *) Select[tz, PrimeQ] (* _T. D. Noe_, Jul 30 2013 *)
%t pQ[n_]:=Module[{idn=IntegerDigits[n]},FreeQ[idn,0]&&Total[idn^3] == Total[ idn]^2]; Select[Prime[Range[50000]],pQ] (* _Harvey P. Dale_, Sep 17 2013 *)
%o (PARI)forprime(n=1, 10^7, v=digits(n); if(sum(i=1, length(v), v[i]^3)==sum(i=1, length(v), v[i])^2 & setsearch(Set(v),0)!=1, print1(n", ")))
%Y Cf. A055012 (sum of cubes of digits), A118881 (square of sum of the digits).
%Y Cf. A227072, A227073.
%K nonn,base,fini,easy
%O 1,1
%A _Balarka Sen_, Jul 26 2013
%E Corrected by _T. D. Noe_, Jul 26 2013
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