A263083 -id:A263083 - cp's OEIS frontend

A263081 a(n) = largest k for which A155043(k) < A262508(n); a(n) = A262509(n) + A262909(n).

124340, 124340, 124340, 124340, 124340, 124340, 124340, 124340, 124340, 124340, 124340, 124340, 124340, 124340, 124340, 124340, 124340, 124340, 124340, 124340, 124340, 124340, 124340, 124340, 124340, 124340, 124340, 124340, 124340, 124340, 124340, 124340, 124340, 124340, 124340, 124340, 124340, 124340, 24684000, 24684000, 24684000, 24684000, 24684000, 24684000, 24684000

Offset: 1

Antti Karttunen, Oct 09 2015

nonn

a(n) = largest k for which A155043(k) < A155043(A262509(n)).
If a(n) > A262509(n) then it must be a leaf (see comments in A262909 for why). Particularly, we have A045765(40722) = 124340, A045765(8191770) = 24684000.
Terms of sequence (together with the corresponding values in A262508) give particularly clean values for the boundaries that are used for example in the C++-program which computes A262896.

Antti Karttunen, Table of n, a(n) for n = 1..68

Cf. A045765, A155043, A262508, A262509, A262896, A262909, A263077, A263083.

(define (A263081 n) (+ (A262509 n) (A262909 n)))

a(n) = A263077(A262509(n)).

a(n) = A262509(n) + A262909(n).

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

Original entry on oeis.org

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

Antti Karttunen, Oct 08 2015

nonn

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.

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

(define (A262908 n) (let ((t (A262509 n))) (let loop ((k (A002183 (+ 2 (A261100 t))))) (cond ((<= (A049820 (+ t k)) t) k) (else (loop (- k 1)))))))

Other identities. For all n >= 1:

a(n) <= A262909(n).

cp's OEIS Frontend

A263081 a(n) = largest k for which A155043(k) < A262508(n); a(n) = A262509(n) + A262909(n).

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