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%I #11 Oct 09 2023 15:50:18
%S 0,1024,2049,4099,8199,16399,32799,65599,131199,262399,524799,1049599,
%T 2097152,2099199,4194305,4196352,4198399,8388611,8392705,8394752,
%U 8396799,16777223,16785411,16789505,16791552,16793599,33554447,33570823,33579011,33583105,33585152,33587199,67108895,67141647,67158023,67166211,67170305
%N Numbers k such that A163511(k) is an eleventh power.
%C Equivalently, numbers k for which A332214(k), and also A332817(k) are eleventh powers.
%C The sequence is defined inductively as:
%C (a) it contains 0 and 1024,
%C and
%C (b) for any nonzero term a(n), (2*a(n)) + 1 and 2048*a(n) are also included as terms.
%C When iterating n -> 2n+1 mod 2047, starting from 64 we get 1024, 2, 5, 11, 23, 47, 95, 191, 383, 767, 1535, and then cycle starts again from 1024 (see A153893), while on the other hand, x^11 mod 2047 obtains values: 0, 1, 230, 322, 344, 368, 390, 482, 622, 712, 713, 942, 967, 1013, 1034, 1080, 1105, 1334, 1335, 1425, 1565, 1657, 1679, 1703, 1725, 1817, 2046. These sets have no terms in common, therefore there are no eleventh powers in this sequence after the initial 0.
%o (PARI)
%o A163511(n) = if(!n, 1, my(p=2, t=1); while(n>1, if(!(n%2), (t*=p), p=nextprime(1+p)); n >>= 1); (t*p));
%o isA366391v(n) = ispower(A163511(n),11);
%o (PARI) isA366391(n) = if(n<=1024, !(n%1024), if(n%2, isA366391((n-1)/2), if(n%2048, 0, isA366391(n>>11))));
%Y Positions of multiples of 11 in A365805.
%Y Sequence A243071(n^11), n >= 1, sorted into ascending order.
%Y Cf. A008455, A153893, A163511, A332214, A332817.
%Y Cf. also A365801, A365802, A365808, A366287.
%K nonn
%O 1,2
%A _Antti Karttunen_, Oct 09 2023
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