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.
A236840 := proc(n) local i, b; if n=0 then 0 else b := convert(n, base, 2); select(i -> (b[i-1]<>b[i]), [$2..nops(b)]); n-1-nops(%) fi end: seq(A236840(i), i=0..69); # Peter Luschny, Apr 19 2014
a[n_] := n - Length@ Split[IntegerDigits[n, 2]]; a[0] = 0; Array[a, 100, 0] (* Amiram Eldar, Jul 16 2023 *)
(define (A236840 n) (- n (A005811 n)))
A005811(n) = hammingweight(bitxor(n,n\2)); A255071(n) = { my(k, i); k = (2^(n+1))-2; i = 1; while(1, k = k - A005811(k); if(!bitand(k+1,k+2),return(i),i++)); }; for(n=1, 48, write("b255071.txt", n, " ", A255071(n)));
(define (A255071 n) (- (A255072 (- (expt 2 (+ n 1)) 2)) (A255072 (- (expt 2 n) 2)))) (define (A255071shifted n) (add (COMPOSE A079944off2 A255056) (A255062 n) (A255061 (+ 1 n)))) (define (A079944off2 n) (A000035 (floor->exact (/ n (A072376 n))))) ;; Cf. A079944. ;; Shifted variant gives: (map A255071shifted (iota 16)) --> (0 1 2 3 5 9 16 29 53 97 178 328 608 1134 2126 4001)
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 60, 36 and 32 are multiples of four, thus a(5) = 3.
A005811(n) = hammingweight(bitxor(n, n\2)); write_A255125_and_A255126_and_A255071(n) = { my(k, i, s25, s26); k = (2^(n+1))-2; i = 1; s25 = 0; s26 = 0; while(1, if((k%4),s26++,s25++); k = k - A005811(k); if(!bitand(k+1, k+2), break, i++)); write("b255125.txt", n, " ", s25); write("b255126.txt", n, " ", s26); write("b255071.txt", n, " ", i); }; for(n=1,42,write_A255125_and_A255126_and_A255071(n));
(define (A255125 n) (if (zero? n) 1 (let loop ((i (- (expt 2 (+ 1 n)) 4)) (s 0)) (cond ((pow2? (+ 2 i)) s) (else (loop (- i (A005811 i)) (+ s (A133872 i)))))))) ;; Alternatively: (define (A255125 n) (add (COMPOSE A059841 A255057) (A255062 n) (A255061 (+ 1 n)))) (define (A255125 n) (add (COMPOSE A059841 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=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 evil numbers (A001969) found in range A255056(0..0), and A255056(0) = 0 is an evil number, thus a(0) = 1. For n=1 we count the evil numbers in range A255056(1..1), and A255056(1) = 2 is not an evil number, thus a(1) = 0. For n=2 we look at the numbers in range A255056(2..3), i.e. 4 and 6 and while 4 is not an evil number, 6 is, 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 we take them in the order they 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 60, 58, 54, 46 and 36 are evil numbers, thus a(5) = 5.
\\ Compute sequences A255063, A255064 and A255071 at the same time, starting from n=1: A005811(n) = hammingweight(bitxor(n, n\2)); write_A255063_and_A255064_and_A255071(n) = { my(k, i, s63, s64); k = (2^(n+1))-2; i = 1; s63 = 0; s64 = 0; while(1, if((hammingweight(k)%2),s64++,s63++); k = k - A005811(k); if(!bitand(k+1, k+2), break, i++)); write("b255063.txt", n, " ", s63); write("b255064.txt", n, " ", s64); write("b255071.txt", n, " ", i); }; for(n=1,36,write_A255063_and_A255064_and_A255071(n));
(define (A255063 n) (if (zero? n) 1 (let loop ((i (- (expt 2 (+ 1 n)) 4)) (s (modulo (+ 1 n) 2))) (cond ((pow2? (+ 2 i)) s) (else (loop (- i (A005811 i)) (+ s (A010059 i)))))))) (define (pow2? n) (and (> n 0) (zero? (A004198bi n (- n 1)))))
(define (A255063 n) (add A254113 (A255062 n) (A255061 (+ 1 n))))
(define (A255063 n) (add (COMPOSE A010059 A255066) (A255062 n) (A255061 (+ 1 n))))
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