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A262908
a(n) = largest k such that A049820(k + A262509(n)) <= A262509(n).
3
53, 49, 69, 55, 53, 31, 47, 39, 25, 35, 31, 39, 37, 51, 33, 43, 33, 69, 65, 57, 43, 41, 57, 49, 33, 33, 43, 41, 37, 33, 37, 39, 35, 27, 41, 27, 43, 75, 177, 171, 173, 155, 45, 133, 107, 121, 111, 139, 78, 119, 123, 47, 65, 79, 77, 97, 81, 151, 149, 145, 111, 197, 375, 71, 59, 81, 259, 257
OFFSET
1,1
COMMENTS
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.
FORMULA
Other identities. For all n >= 1:
a(n) <= A262909(n).
PROG
(Scheme)
(define (A262908 n) (let ((t (A262509 n))) (let loop ((k (A002183 (+ 2 (A261100 t))))) (cond ((<= (A049820 (+ t k)) t) k) (else (loop (- k 1)))))))
CROSSREFS
KEYWORD
nonn
AUTHOR
Antti Karttunen, Oct 08 2015
STATUS
approved