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 A120511 a(n) = min{j>0 : A006949(j) = n}. 7
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 OFFSET 1,2 LINKS Antti Karttunen, Table of n, a(n) for n = 1..10000 C. Deugau and F. Ruskey, Complete k-ary Trees and Generalized Meta-Fibonacci Sequences C. Deugau and F. Ruskey, Complete k-ary Trees and Generalized Meta-Fibonacci Sequences, J. Integer Seq., Vol. 12. [This is a later version than that in the GenMetaFib.html link] B. Jackson and F. Ruskey, Meta-Fibonacci Sequences, Binary Trees and Extremal Compact Codes, Electronic Journal of Combinatorics, 13 (2006), R26. FORMULA G.f.: P(z) = z / (1-z) * (1 + sum(z^(2^m) * (1 + 1 / (1 - z^(2^m))), m=0..infinity)) It appears that a(n) = a(ceil(n/2)) + n. - Georgi Guninski, Sep 08 2009 Comments from Max Alekseyev, Sep 08 2009: (Start) This can be proved as follows. Let b=A006949. It is known that b(n) = b(n-1-b(n-1)) + b(n-2-b(n-2)) and b(n-1) <= b(n) <= b(n-1)+1. The following claims are trivial: Claim 1. For any n, b(a(n))=n. Claim 2. If m=a(n) for some n, then a(b(m))=m. Claim 3. Let m=a(n). Then b(m)=n and b(m-1)=n-1, implying that b(m+1) = b(m-b(m)) + b(m-1-b(m-1)) = 2*b(m-n) is an even number. Claim 4. Each even number in A006949 is repeated at least two times while each odd number in A006949 appears only once. Proof. If n is even, then for m=a(n), we have b(m)=n and b(m+1)=n (from Claim 3), i.e., n is repeated at least twice. If n is odd, then for m=a(n), we cannot have b(m+1)=n since by Claim 3 b(m+1) must be even. QED Consider two cases: 1) If n is odd, then b(m+1) = n+1 = 2*b(m-n), i.e., b(m-n) = (n+1)/2. Claim 4 also implies b(m-2)=n-1. Therefore n = b(m) = b(m-1-b(m-1)) + b(m-2-b(m-2)) = b(m-n) + b(m-n-1). Since n is odd, we have b(m-n-1) m), from 2n-1 <= A120511(n) <= m it follows that A006949(m) <= (m+1)/2. Similarly, from m < A120511(n+1) < 2(n+1) + log2(n+1) - 1 <= 2(n+1) + log2((m+1)/2+1) - 1, it follows that A006949(m) >= (m - log2(m+3)) / 2. Therefore | A006949(m) - m/2 | <= log2(m+3)/2, which gives an interval of just logarithmic length to search for the value of A006949(m). (End) Comment from Frank Ruskey, Sep 11 2009: From p. 25 of the revised version of the Deugau-Ruskey, we have p(n) = s*ceil(log_k n) + (kn-d-1)/(k-1) where d is the sum of the digits of the k-ary expression of n-1. In the present case s = 1 and k = 2. Comments from Antti Karttunen, Dec 12 2013: (Start) a(n) = 2n + A080791(n) - A007814(n) - A036987(n-1) [This is essentially Max Alekseyev's above formula represented with A-numbers]. a(n) = A005408(n-1) + A080791(n-1) = A233273(n-1)-1 [The above formula reduces to this, because A080791(n)-A080791(n+1) = 1-(A007814(n+1)+A036987(n)) and A080791(2n+1) = A080791(n)]. (End) a(n) = 2*n-1 + A023416(2*n-1). - Reinhard Zumkeller, Apr 17 2014 MAPLE p := proc(n) if n=1 then return 1; end if; for j from p(n-1)+1 to infinity do if A006949(j) = n then return j; fi; od; end proc; PROG (PARI) { A120511(n) = local(t, k); t=binary(n); k=valuation(n, 2); 2*n + #t - sum(i=1, #t, t[i]) - k - (n==2^k) } /* Max Alekseyev, Sep 18 2009 */ (Scheme) (define (A120511 n) (+ n n (A080791 n) (- (A007814 n)) (- (A036987 (- n 1))))) (define (A120511 n) (+ (A005408 (- n 1)) (A080791 (- n 1)))) ;; Based on above PARI-program and its further reduction, from Antti Karttunen, Dec 12 2013 (Haskell) import Data.List (elemIndex); import Data.Maybe (fromJust) a120511 = (+ 1) . fromJust . (`elemIndex` (tail a006949_list)) -- Reinhard Zumkeller, Apr 17 2014 CROSSREFS Cf. A006949, A120522, A007814, A023416, A036987, A080791, A005408. a(n) = one less than A233273(n-1). Cf. A241218. Sequence in context: A091067 A269177 A269178 * A176864 A022550 A237883 Adjacent sequences:  A120508 A120509 A120510 * A120512 A120513 A120514 KEYWORD nonn AUTHOR Frank Ruskey and Chris Deugau (deugaucj(AT)uvic.ca), Jun 20 2006 EXTENSIONS Edited by Max Alekseyev, Sep 16 2009 More terms from Max Alekseyev, Sep 18 2009 STATUS approved

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