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

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

Author

Antti Karttunen, Oct 08 2015

Keywords

• nonn

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.

Crossrefs

Cf. A049820, A155043, A262508, A262509, A262909, A263083.

Programs

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)))))))

Formula

Other identities. For all n >= 1:

a(n) <= A262909(n).