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In the binary expansion of n replace each run of k 0's (or 1's) with 2k-1 0's (or 1's), except in the most significant run where we double the number of 0's (or 1's).
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%I #9 Dec 20 2023 17:42:00

%S 0,3,6,15,24,13,30,63,96,49,26,55,120,61,126,255,384,193,98,199,104,

%T 53,110,223,480,241,122,247,504,253,510,1023,1536,769,386,775,392,197,

%U 398,799,416,209,106,215,440,221,446,895,1920,961,482,967,488,245,494

%N In the binary expansion of n replace each run of k 0's (or 1's) with 2k-1 0's (or 1's), except in the most significant run where we double the number of 0's (or 1's).

%e a(1) = 3, as 1 is 1 in binary and doubling the number of 1's (in the only run) gives binary 11, 3 in decimal. a(9) = 49, as 9 is 1001 in binary and replacing the most significant run '1' with '11' and the center run '00' with '000' and the least significant run '1' with '1', we get 110001 in binary, 49 in decimal.

%o (MIT/GNU Scheme)

%o ;;; binexp->runcount1list and runcount1list->binexp given in A129594.

%o (define (A129602 n) (if (zero? n) n (let ((rl (binexp->runcount1list n))) (runcount1list->binexp (cons (* 2 (car rl)) (map (lambda (i) (- (* 2 i) 1)) (cdr rl)))))))

%Y Central diagonal of array A129600, a(n) = A129600bi(n, n). Cf. A129594. For n > 0, a(n) = A004760(A129603(n)+1).

%K nonn,base

%O 0,2

%A _Antti Karttunen_, May 01 2007

%E Edited definition. - _N. J. A. Sloane_, Dec 20 2023