A293445 A multiplicative encoding (base-2 compressed) for the exponents of 3 obtained when using Shevelev's algorithm for computing A053446.
2, 2, 3, 12, 36, 3, 12, 24, 6, 48, 12, 20736, 82944, 12, 18, 864, 248832, 6, 20, 19906560, 59719680, 80, 8640, 720, 25920, 34560, 5, 80, 103195607040, 240, 480, 622080, 137594142720, 138240, 20, 59440669655040, 138240, 20, 14929920, 29859840, 240, 59719680, 8640, 720, 414720, 8640, 540, 447897600, 960, 46080, 34560, 59719680, 295814814232058265600, 5, 80
Offset: 1
Keywords
Examples
A001651(5) = 7 as 7 is the fifth number not divisible by 3. According to the algorithm described in the comment of A053446 we have in the form of a "finite continued fraction"
1 + 14
------ + 7
3^1
---------- + 14
3^1
----------------- + 7
3^2
---------------------- = 1
3^2
Cumulatively multiplying (with A019565) together the prime-numbers corresponding to 1-bits in the binary expansions of the exponents of 3 in the denominators (that are 1, 1, 2, 2, in binary 1, 1, 10, 10, with 1's in bit-positions 0 and 1), yields prime(0+1) * prime(0+1) * prime(1+1) * prime(1+1) = 2^2 * 3^2 = 36, thus a(5) = 36.
(Adapted from _Vladimir Shevelev_'s explanation in A053446.)
Another example: A001651(19) = 28 as 28 is the 19th number not divisible by 3. (1 + 28) is not a multiple of 3, so we start with (1 + 2*28) = 1+56 = 57 and proceed as:
1 + 56
------ + 56 [that is, (57/3) + 56 = 75]
3^1
---------- + 56 [that is, (75/3) + 56 = 81]
3^1
----------------- = 1 [that is, (81/81) = 1]
3^4
So we obtained exponents 1, 1, 4 (in binary "1", "1" and "100") where the 1-bits are in positions 0, 0 and 2. We form a product prime(0+1) * prime(0+1) * prime(2+1) = 2*2*5, thus a(19) = 20.
Links
- Antti Karttunen, Table of n, a(n) for n = 1..1458
Crossrefs
Programs
-
Scheme
(define (A293445 n) (define (next_one k m) (if (zero? (modulo (+ k m) 3)) (+ k m) (+ k m m))) (let* ((u (A001651 n)) (x_init (next_one 1 u))) (let loop ((x x_init) (z (A019565 (A007949 x_init)))) (let ((r (A038502 x))) (if (= 1 r) z (let ((x_next (next_one r u))) (loop x_next (* z (A019565 (A007949 x_next)))))))))) (define (A001651 n) (let ((x (- n 1))) (if (even? x) (+ 1 (* 3 (/ x 2))) (- (* 3 (/ (+ x 1) 2)) 1)))) (define (A038500 n) (A000244 (A007949 n))) (define (A038502 n) (/ n (A038500 n)))
Comments