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%I #19 Apr 11 2024 03:55:39
%S 3,213,13653,873813,55924053,3579139413,229064922453,14660155037013,
%T 938249922368853,60047995031606613,3843071682022823253,
%U 245956587649460688213,15741221609565484045653,1007438183012190978921813,64476043712780222650996053,4126466797617934249663747413,264093875047547791978479834453
%N Multiples of 3 which on one operation of the Collatz function T (N -> 3N+1/2^r) yield the number 5.
%H Michael De Vlieger, <a href="/A072196/b072196.txt">Table of n, a(n) for n = 1..554</a>
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (65,-64).
%F a(n) = (10*64^(n-1)-1)/3. - _Henry Bottomley_, Dec 02 2002 [Formula adapted to a change of offset by _Georg Fischer_, Apr 10 2024]
%e (3*3+1)/2=5, (3*213+1)/2^7=5, etc. Thus multiples of 3 act as generators on the numbers in the Collatz domain.
%p seq((10*64^(n-1)-1)/3, n=1..13); # _Georg Fischer_, Apr 10 2024
%t Array[(10*64^(# - 1) - 1)/3 &, 13] (* _Michael De Vlieger_, Apr 10 2024 *)
%Y Cf. A139391, A072197.
%K nonn,easy
%O 1,1
%A N. Rathankar (rathankar(AT)yahoo.com), Jul 03 2002
%E More terms from _Henry Bottomley_, Dec 02 2002
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