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Least of 3 consecutive prime powers in arithmetic progression.
1

%I #19 Jan 07 2018 21:35:47

%S 1,2,3,7,9,23,25,27,61,79,151,199,239,257,331,361,367,557,587,601,619,

%T 647,941,971,1097,1117,1181,1217,1499,1669,1741,1747,1901,2281,2411,

%U 2671,2897,2957,3301,3307,3631,3721,3727,4007,4093,4397,4451,4591,4651,4679,4987

%N Least of 3 consecutive prime powers in arithmetic progression.

%C This sequence is motivated by the article by L. Panaitopol. Actually he defines q(n) = A000961(n-1), and Q(n) = q(n+1)-2*q(n)+q(n-1). Then he asks if the sequence of indices n such that Q(n)=0 is infinite.

%H Laurentiu Panaitopol, <a href="https://projecteuclid.org/euclid.rmjm/1181070157">Some of the properties of the sequence of powers of prime numbers</a>, Rocky Mountain Journal of Mathematics, Volume 31, Number 4, Winter 2001.

%e In A000961, 7 is followed by 8 and 9, a 3-term arithmetic progression with a common difference 1.

%e 9 is followed by 11 and 13, a 3-term arithmetic progression with a common difference 2.

%o (PARI) ispp(n) = isprimepower(n) || (n==1);

%o lista(nn) = {v = select(x->ispp(x), vector(nn,i,i)); for (k=2, #v-1, if (v[k+1] - 2*v[k] + v[k-1] == 0, print1(v[k-1], ", ")););}

%Y Cf. A000961 (prime powers), A057820 (common differences of consecutive prime powers).

%K nonn

%O 1,2

%A _Michel Marcus_, Nov 16 2014