This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
(2^0)-1 (0) is reached from (2^1)-1 (1) with one step by subtracting A000120(1) from 1. (2^1)-1 (1) is reached from (2^2)-1 (3) with one step by subtracting A000120(3) from 3. (2^2)-1 (3) is reached from (2^3)-1 (7) with two steps by first subtracting A000120(7) from 7 -> 4, and then subtracting A000120(4) from 4 -> 3. Thus a(0)=a(1)=1 and a(2)=2.
We consider 0 to have no nonleading zeros, so first we get to 0 -> 0+1+0 = 1, and 1 is odd, so we go one step down from the starting value a(0)=0, and thus a(1) = -1. 1 has no nonleading zeros, so we get 1 -> 1+1+0 = 2, and 2 is even, so we go one step up, and thus a(2) = 0. 2 has one nonleading zero in binary "10", so we get 2 -> 2+1+1 = 4, and 4 is also even, so we go one step up, and thus a(3) = 1. 4 has two nonleading zeros in binary "100", so we get 4 -> 4+2+1 = 7, 7 is odd, so we go one step down, and thus a(4) = 0.
A070939 = n->#binary(n)+!n; \\ From M. F. Hasler A080791 = n->if(n<1,0,(A070939(n)-hammingweight(n))); A233272 = n->(n + A080791(n) + 1); A257806_write_bfile(up_to_n) = { my(n,a_n=0,b_n=0); for(n=0, up_to_n, write("b257806.txt", n, " ", a_n); b_n = A233272(b_n); a_n += ((-1)^b_n)); }; A257806_write_bfile(8727);
def A257806_print_upto(n): a = 0 b = 0 for n in range(n): print(b, end=", ") ta = a c0 = 0 while ta>0: c0 += 1-(ta&1) ta >>= 1 a += 1 + c0 b += ((2*(1-(a&1))) - 1) # By Antti Karttunen after Alex Ratushnyak's Python-code for A216431.
(define (A257806 n) (- (A257808 n) (A257807 n)))
;; Using memoizing definec-macro. (definec (A257806 n) (if (zero? n) n (+ (expt -1 (A233271 n)) (A257806 (- n 1)))))
(2^0)-1 (0) is reached from (2^1)-1 (1) with one step by subtracting A000120(1) from 1. Zero is an even number, so a(0)=1. (2^1)-1 (1) is reached from (2^2)-1 (3) with one step by subtracting A000120(3) from 3. One is not an even number, so a(1)=0. (2^2)-1 (3) is reached from (2^3)-1 (7) with two steps by first subtracting A000120(7) from 7 -> 4, and then subtracting A000120(4) from 4 -> 3. Four is an even number, but three is not, so a(2)=1.
(1!)-1 (0) is reached from (2!)-1 (1) with one step by subtracting A034968(1) from 1. Zero is an even number, so a(1)=1. (2!)-1 (1) is reached from (3!)-1 (5) with two steps by first subtracting A034968(5) from 5 -> 2, and then subtracting A034968(2) from 2 -> 1. Two is an even number, but one is not, so a(2)=1. (3!)-1 (5) is reached from (4!)-1 (23) with five steps by repeatedly subtracting the sum of digits in factorial expansion as: 23 - 6 = 17, 17 - 5 = 12, 12 - 2 = 10, 10 - 3 = 7, 7 - 2 = 5. Of these only 12 and 10 are even numbers, so a(3)=2.
(definec (A219662 n) (if (< n 2) n (let loop ((i (- (A000142 (1+ n)) (A000217 n) 1)) (s 0)) (cond ((isA000142? (1+ i)) (+ s (- 1 (modulo i 2)))) (else (loop (A219651 i) (+ s (- 1 (modulo i 2))))))))) (define (isA000142? n) (and (> n 0) (let loop ((n n) (i 2)) (cond ((= 1 n) #t) ((not (zero? (modulo n i))) #f) (else (loop (/ n i) (1+ i)))))))
(1!)-1 (0) is reached from (2!)-1 (1) with one step by subtracting A034968(1) from 1. Zero is not an odd number, so a(1)=0. (2!)-1 (1) is reached from (3!)-1 (5) with two steps by first subtracting A034968(5) from 5 -> 2, and then subtracting A034968(2) from 2 -> 1. Two is not an odd number, but one is, so a(2)=1. (3!)-1 (5) is reached from (4!)-1 (23) with five steps by repeatedly subtracting the sum of digits in factorial expansion as: 23 - 6 = 17, 17 - 5 = 12, 12 - 2 = 10, 10 - 3 = 7, 7 - 2 = 5. Of these (after 23) only 17, 7 and 5 are odd numbers, so a(3)=3.
(definec (A219663 n) (if (< n 2) 0 (let loop ((i (- (A000142 (1+ n)) (A000217 n) 1)) (s 0)) (cond ((isA000142? (1+ i)) (+ s (modulo i 2))) (else (loop (A219651 i) (+ s (modulo i 2)))))))) (define (isA000142? n) (and (> n 0) (let loop ((n n) (i 2)) (cond ((= 1 n) #t) ((not (zero? (modulo n i))) #f) (else (loop (/ n i) (1+ i)))))))
For n=5 we start iterating with map m(n) = A236840(n) from the initial value (2^(5+1))-2 = 62. Thus we get m(62) = 60, m(60) = 58, m(58) = 54, m(54) = 50, m(50) = 46, m(46) = 42, m(42) = 36, m(36) = 32 and finally m(32) = 30, which is (2^5)-2. Of the nine numbers encountered, only 58, 54, 50, 46, 42 and 30 are of the form 4n+2, thus a(5) = 6. Note that the initial value 2^(n+1)-2 is not included in the cases, but the final (2^n) - 2 is.
\\ Use the PARI-code given in A255125.
(define (A255126 n) (if (zero? n) n (let loop ((i (- (expt 2 (+ 1 n)) 4)) (s 1)) (cond ((pow2? (+ 2 i)) s) (else (loop (- i (A005811 i)) (+ s (A021913 i)))))))) ;; Alternatively: (define (A255126 n) (add (COMPOSE A000035 A255057) (A255062 n) (A255061 (+ 1 n)))) (define (A255126 n) (add (COMPOSE A000035 A255067) (A255062 n) (A255061 (+ 1 n)))) (define (add intfun lowlim uplim) (let sumloop ((i lowlim) (res 0)) (cond ((> i uplim) res) (else (sumloop (1+ i) (+ res (intfun i)))))))
For n=0 we count the odious numbers (A000069) found in range A255056(0..0), and A255056(0) = 0 is not an odious number, thus a(0) = 0. For n=1 we count the odious numbers in range A255056(1..1), and A255056(1) = 2 is an odious number, thus a(1) = 1. For n=2 we look at the numbers in range A255056(2..3), i.e. 4 and 6 and while 4 is an odious number, 6 is not, thus a(2) = 1. For n=5 we look at the numbers in range A255056(12..20) which are (32, 36, 42, 46, 50, 54, 58, 60, 62), or if we take them in the order the come when iterating A236840 (as in A255066(12..20): 62, 60, 58, 54, 50, 46, 42, 36, 32), that is, we start iterating with map m(n) = A236840(n) from the initial value (2^(5+1))-2 = 62. Thus we get m(62) = 60, m(60) = 58, m(58) = 54, m(54) = 50, m(50) = 46, m(46) = 42, m(42) = 36 and finally m(36) = 32 which is (2^5). Of the nine numbers encountered, only 62, 50, 42 and 32 are odious numbers, thus a(5) = 4.
\\ Use the Pari-code given in A255063.
(define (A255064 n) (if (zero? n) n (let loop ((i (- (expt 2 (+ 1 n)) 4)) (s (modulo n 2))) (cond ((pow2? (+ 2 i)) s) (else (loop (- i (A005811 i)) (+ s (A010060 i))))))))
(define (A255064 n) (add A254114 (A255062 n) (A255061 (+ 1 n))))
(define (A255064 n) (add (COMPOSE A010060 A255066) (A255062 n) (A255061 (+ 1 n))))
Comments