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A262908 a(n) = largest k such that A049820(k + A262509(n)) <= A262509(n).

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%I A262908 #19 Apr 15 2023 15:02:04
%S A262908 53,49,69,55,53,31,47,39,25,35,31,39,37,51,33,43,33,69,65,57,43,41,57,
%T A262908 49,33,33,43,41,37,33,37,39,35,27,41,27,43,75,177,171,173,155,45,133,
%U A262908 107,121,111,139,78,119,123,47,65,79,77,97,81,151,149,145,111,197,375,71,59,81,259,257
%N A262908 a(n) = largest k such that A049820(k + A262509(n)) <= A262509(n).
%C A262908 For all nonzero terms a(n), A263083(n) = a(n) + A262509(n) and A155043(A263083(n)) < A155043(A262509(n)) because at each A262509(n) the "distance to zero", A155043 obtains a unique value A262508(n), thus no A049820-iteration trajectory starting from any k larger than A262509(n) and using a greater or equal number of steps to reach zero may bypass A262509(n) [i.e., without going through A262509(n)], because then A262508(n) would not be unique anymore. See also comments in A262909.
%F A262908 Other identities. For all n >= 1:
%F A262908 a(n) <= A262909(n).
%o A262908 (Scheme)
%o A262908 (define (A262908 n) (let ((t (A262509 n))) (let loop ((k (A002183 (+ 2 (A261100 t))))) (cond ((<= (A049820 (+ t k)) t) k) (else (loop (- k 1)))))))
%Y A262908 Cf. A049820, A155043, A262508, A262509, A262909, A263083.
%K A262908 nonn
%O A262908 1,1
%A A262908 _Antti Karttunen_, Oct 08 2015