A280695 -id:A280695 - cp's OEIS frontend

A280694 Largest Lucas number (A000032) dividing n.

1, 2, 3, 4, 1, 3, 7, 4, 3, 2, 11, 4, 1, 7, 3, 4, 1, 18, 1, 4, 7, 11, 1, 4, 1, 2, 3, 7, 29, 3, 1, 4, 11, 2, 7, 18, 1, 2, 3, 4, 1, 7, 1, 11, 3, 2, 47, 4, 7, 2, 3, 4, 1, 18, 11, 7, 3, 29, 1, 4, 1, 2, 7, 4, 1, 11, 1, 4, 3, 7, 1, 18, 1, 2, 3, 76, 11, 3, 1, 4, 3, 2, 1, 7, 1, 2, 29, 11, 1, 18, 7, 4, 3, 47, 1, 4, 1, 7, 11, 4, 1, 3, 1, 4, 7, 2, 1, 18, 1, 11, 3, 7, 1

Offset: 1

Antti Karttunen, Jan 11 2017

nonn

Antti Karttunen, Table of n, a(n) for n = 1..15127

Cf. A000032, A054494, A280695, A280699.

Cf. A057854 (gives the positions n > 1 where this sequence and A280696 obtain equal values).

;; A stand-alone program:
(define (A280694 n) (let loop ((l1 1) (l2 3) (lpd 1)) (cond ((> l1 n) (if (and (= 1 lpd) (even? n)) 2 lpd)) ((zero? (modulo n l1)) (loop l2 (+ l1 l2) l1)) (else (loop l2 (+ l1 l2) lpd)))))

a(n) = n / A280695(n).
Other identities. For all n >= 1:
a(A000032(n)) = A000032(n).
a(A057854(n)) = A280696(A057854(n)).
a(A000045(n)) = A280699(n).

A280697 a(n) = n / A280696(n); n divided by its largest Lucas proper divisor, a(1) = 1.

Original entry on oeis.org

1, 2, 3, 2, 5, 2, 7, 2, 3, 5, 11, 3, 13, 2, 5, 4, 17, 6, 19, 5, 3, 2, 23, 6, 25, 13, 9, 4, 29, 10, 31, 8, 3, 17, 5, 2, 37, 19, 13, 10, 41, 6, 43, 4, 15, 23, 47, 12, 7, 25, 17, 13, 53, 3, 5, 8, 19, 2, 59, 15, 61, 31, 9, 16, 65, 6, 67, 17, 23, 10, 71, 4, 73, 37, 25, 19, 7, 26, 79, 20, 27, 41, 83, 12, 85, 43, 3, 8, 89, 5, 13, 23, 31, 2, 95, 24, 97, 14, 9, 25, 101

Offset: 1

Antti Karttunen, Jan 11 2017

nonn

For n > 1, a(n) = n divided by the greatest Lucas number (A000032) that divides n and is less than n.

Antti Karttunen, Table of n, a(n) for n = 1..15127

Cf. A000032, A280695, A280696, A280687.

Scheme
```
(define (A280697 n) (/ n (A280696 n)))
```

a(n) = n / A280696(n).

cp's OEIS Frontend

A280694 Largest Lucas number (A000032) dividing n.

Views

Author

Keywords

Links

Crossrefs

Programs

Scheme

Formula