cp's OEIS Frontend

A316625 Terms in A259663, in ascending order.

%I A316625 #23 Aug 23 2018 16:29:37
%S A316625 1,3,5,7,11,13,15,19,21,23,31,35,47,53,55,63,79,85,87,95,99,127,143,
%T A316625 151,191,213,223,227,255,271,319,341,351,383,407,483,511,575,663,739,
%U A316625 767,783,853,863,895,1023,1175,1251,1279,1365,1407,1535,1599,1807,1887,2047
%N A316625 Terms in A259663, in ascending order.
%C A316625 See A259663 for discussion of these terms in relation to Collatz sequences.
%C A316625 There are k terms in the interval [2^k, 2^(k+1)], k >= 1; terms in each interval are of the form 2^k + a(n) for some n.
%C A316625 The sequence is a permutation (without repeating terms) of the following numbers:
%C A316625 2^i-1 and 7*2^i-1 when i is odd, i >= 1;
%C A316625 3^2^i-1 and 5^2^i-1 when i is even, i >= 2;
%C A316625 For fixed k >= 4: least residues of 3^j*(2^(2^(k-3) + i*2^(k-2) - j)) - 1 mod 2^(2^(k-3) + i*2^(k-2) + k-j), i >= 0, 0 <= j < 2^(k-3) + i*2^(k-2) .  (See example).
%e A316625 k=5, i=1 -- terms are least residues of 3^j*2^(12-j)-1 mod 2^(17-j), 0 <= j < 12:
%e A316625 j=0: 4096-1 mod 131072 = 4095;
%e A316625 j=1: 3*2048-1 mod 65536 = 6143;
%e A316625 j=2: 9*1024-1 mod 32768 = 9215;
%e A316625 j=3: 27*512-1 mod 16384 = 13823;
%e A316625 j=4: 81*256-1 mod 8192 = 20735 mod 8192 == 4351;
%e A316625 j=5: 243*128-1 mod 4096 = 31103 mod 4096 == 2431;
%e A316625 j=6: 729*64-1 mod 2048 = 46655 mod 2048 == 1599;
%e A316625 j=7: 2187*32-1 mod 1024 = 69983 mod 1024 == 351;
%e A316625 j=8: 6561*16-1 mod 512 = 104975 mod 512 == 15;
%e A316625 j=9: 19683*8-1 mod 256 = 157463 mod 256 == 23;
%e A316625 j=10: 59049*4-1 mod 128 = 236195 mod 128 == 35;
%e A316625 j=11: 177147*2-1 mod 64 = 354293 mod 64 == 53.
%e A316625 Note: k=5, i=0 is equivalent to starting with j=0: 15 mod 512.
%Y A316625 Cf. A259663.
%K A316625 nonn
%O A316625 1,2
%A A316625 _Bob Selcoe_, Jul 08 2018
%E A316625 More terms from _Michel Marcus_, Jul 10 2018