A266117 Lexicographically first injection of positive integers beginning with a(1) such that a(n)*a(n+1) is a term of A265349, i.e., has no multiple occurrences of any nonzero digit when viewed in factorial base (A007623).

1, 2, 3, 4, 5, 10, 11, 6, 7, 12, 8, 9, 24, 13, 18, 19, 26, 14, 17, 22, 21, 16, 15, 32, 30, 20, 23, 29, 42, 28, 25, 48, 34, 53, 41, 54, 27, 40, 33, 36, 37, 61, 65, 44, 49, 72, 39, 38, 51, 52, 55, 59, 47, 46, 58, 50, 60, 62, 31, 66, 56, 57, 64, 45, 80, 63, 74, 69, 68, 35, 79, 100, 73, 43, 67, 70, 71, 78, 84, 87, 92, 90, 76, 96, 75

Offset

1,2

Comments

After a(1) = 1, always choose for a(n+1) the least unused k such that in factorial base representation (A007623), the product a(n)*a(n+1) will not show any of the nonzero digits present twice (or more times), regardless of the positions of the digits.

Links

Antti Karttunen, Table of n, a(n) for n = 1..14641

Eric Angelini, a(n)*a(n+1) shows at least twice the same digit, Posting on SeqFan-list Dec 21 2015. [Source of inspiration for this sequence.]

Index entries for sequences related to factorial base representation

Index entries for sequences that are permutations of the natural numbers

Example

For n = 6, we start searching from the least not yet used number in range a(1) .. a(5) [which is 6, because all the previous terms are fixed] for the first number whose product with a(5) = 5 results a number in A265349.
Multiplying 5 (in factorial base "21") with 6 (in factorial base "100") results 30, which in factorial base is "1100", containing digit "1" twice, thus 6 is disqualified.
Similarly, products 5*7, 5*8 and 5*9 result 35 = "1121", 40 = "1220" and 45 = "1311", where in all cases one of the nonzero digits occur more than once, so 7, 8 and 9 are also all disqualified.
But 5*10 = 50, which has a factorial base representation ("2010") that matches the criterion, thus a(6) = 10.

Prog

(Scheme, with defineperm1-macro from Antti Karttunen's IntSeq-library)
;; Reference-implementation, quite sub-optimal:
(defineperm1 (A266117 n) (if (= 1 n) n (let ((prev (A266117 (- n 1)))) (let loop ((k 1)) (cond ((and (not-lte? (A266118 k) (- n 1)) (= (A264990 (* k prev)) 1)) k) (else (loop (+ 1 k))))))))
;; We consider a > b (i.e. not less than b) also in case a is #f.
;; (Because of the stateful caching system used by defineperm1-macro):
(define (not-lte? a b) (cond ((not (number? a)) #t) (else (> a b))))

Crossrefs

Left inverse: A266118 (also the right inverse if this sequence is a permutation of the positive integers).

Cf. A264990, A265349.

Cf. also A266121, A266191 and A266195 for similar permutations.

Sequence in context: A370893 A072701 A261039 * A037473 A007092 A241213

Adjacent sequences: A266114 A266115 A266116 * A266118 A266119 A266120

Keyword

nonn,base

Author

Antti Karttunen, Dec 22 2015

Status

approved