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%I #21 Jun 03 2018 02:04:42
%S 1,0,1,2,2,5,7,14,24,52,84,173,290,586,1038,2025,3740,7177,13498,
%T 25832,49027,93918,179291,344128,660058,1270590,2447944,4728357,
%U 9145214,17718039,34365068,66717630,129619518,251953756,489964171,953141850,1854911347
%N 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).
%F a(n) = Sum_{k = A255062(n) .. A255061(n+1)} A254113(k).
%F a(n) = Sum_{k = A255062(n) .. A255061(n+1)} A010059(A255066(k)).
%F Other identities. For all n >= 1:
%F a(n) = A255071(n) - A255064(n).
%e 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.
%e 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.
%e 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.
%e 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.
%o (PARI)
%o \\ Compute sequences A255063, A255064 and A255071 at the same time, starting from n=1:
%o A005811(n) = hammingweight(bitxor(n, n\2));
%o 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); };
%o for(n=1,36,write_A255063_and_A255064_and_A255071(n));
%o (Scheme, different versions)
%o (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))))))))
%o (define (pow2? n) (and (> n 0) (zero? (A004198bi n (- n 1)))))
%o ;; Alternatively:
%o (define (A255063 n) (add A254113 (A255062 n) (A255061 (+ 1 n))))
%o (define (A255063 n) (add (COMPOSE A010059 A255066) (A255062 n) (A255061 (+ 1 n))))
%Y Cf. A001969, A005811, A010059, A236840, A254113, A255056, A255064, A255066, A255071.
%Y Similar sequences: A255125, A218542.
%K nonn
%O 0,4
%A _Antti Karttunen_, Feb 14 2015
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