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A255063
Number of times an evil number is encountered when iterating from 2^(n+1)-2 to (2^n)-2 with the map x -> x - (number of runs in binary representation of x).
7
1, 0, 1, 2, 2, 5, 7, 14, 24, 52, 84, 173, 290, 586, 1038, 2025, 3740, 7177, 13498, 25832, 49027, 93918, 179291, 344128, 660058, 1270590, 2447944, 4728357, 9145214, 17718039, 34365068, 66717630, 129619518, 251953756, 489964171, 953141850, 1854911347
OFFSET
0,4
FORMULA
a(n) = Sum_{k = A255062(n) .. A255061(n+1)} A254113(k).
a(n) = Sum_{k = A255062(n) .. A255061(n+1)} A010059(A255066(k)).
Other identities. For all n >= 1:
a(n) = A255071(n) - A255064(n).
EXAMPLE
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.
PROG
(PARI)
\\ 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));
(Scheme, different versions)
(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)))))
;; Alternatively:
(define (A255063 n) (add A254113 (A255062 n) (A255061 (+ 1 n))))
(define (A255063 n) (add (COMPOSE A010059 A255066) (A255062 n) (A255061 (+ 1 n))))
CROSSREFS
KEYWORD
nonn
AUTHOR
Antti Karttunen, Feb 14 2015
STATUS
approved