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A285173 Numbers n such that A002496(n+1) < A002496(n)^(1+1/n). 0

%I #39 May 06 2017 17:28:39

%S 3,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,

%T 29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,

%U 52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70

%N Numbers n such that A002496(n+1) < A002496(n)^(1+1/n).

%C Conjecture: The sequence is the complement of 1, 2, 4, 351 in A000027 (cf. Sun, 2013, Conjecture 2.6. (i)).

%D Zhi-Wei Sun, Conjectures involving arithmetical sequences, in: Number Theory: Arithmetic in Shangri-La (eds., S. Kanemitsu, H. Li and J. Liu), Proceedings of the 6th China-Japan Seminar (Shanghai, August 15-17, 2011), World Scientific Publishing, Singapore, 2013, pp. 244-258.

%H Zhi-Wei Sun, <a href="https://arxiv.org/abs/1208.2683">Conjectures involving arithmetical sequences</a>, arXiv:1208.2683 [math.CO], 2012.

%o (PARI) a002496(n) = my(k=1, i=0); while(i < n, if(ispseudoprime(k^2+1), i++); if(i==n, return(k^2+1)); k++)

%o is(n) = a002496(n+1) < a002496(n)^(1+1/n)

%Y Cf. A000027, A002496.

%K nonn

%O 1,1

%A _Felix Fröhlich_, May 06 2017

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