A213714 -id:A213714 - cp's OEIS frontend

A256992 Position of n in either of the complementary sequences, A005187 or A055938: a(n) = A213714(n) + A234017(n).

1, 1, 2, 3, 2, 3, 4, 5, 4, 6, 7, 5, 6, 7, 8, 9, 8, 10, 11, 9, 10, 12, 13, 11, 14, 15, 12, 13, 14, 15, 16, 17, 16, 18, 19, 17, 18, 20, 21, 19, 22, 23, 20, 21, 22, 24, 25, 23, 26, 27, 24, 25, 28, 29, 26, 30, 31, 27, 28, 29, 30, 31, 32, 33, 32, 34, 35, 33, 34, 36, 37, 35, 38, 39, 36, 37, 38, 40, 41, 39, 42, 43, 40, 41, 44, 45, 42

Offset: 1

Antti Karttunen, Apr 15 2015

nonn

In other words, if n = A005187(k) for some k >= 1, then a(n) = k, otherwise it must be that n = A055938(h) for some h, and then a(n) = h.
Each n occurs exactly twice, first at a(A005187(n)), then at a(A055938(n)). Cf. also A257126.
When iterating a(n), a(a(n)), a(a(a(n))), etc, A256993(n) gives the number of steps to reach one, from any starting value n >= 1.

Antti Karttunen, Table of n, a(n) for n = 1..16384

Cf. A005187, A055938, A079559, A213714, A234017.

Cf. also A256991 (variant), A256993, A257126.

With[{nn = 92}, Function[{g, h}, Flatten@ Table[If[MemberQ[g, n], First@ Position[g, n] - 1, First@ Position[h, n]], {n, Min[Length /@ {g, h}]}]] @@ {Table[2 n - DigitCount[2 n, 2, 1], {n, 0, nn}], Complement[Range@ Last@ #, #] &@ Table[IntegerExponent[(2 n)!, 2], {n, 0, nn}]} ] (* Michael De Vlieger, Dec 12 2016, after Harvey P. Dale at A005187 and Jean-François Alcover at A055938 *)

(define (A256992 n) (+ (A213714 n) (A234017 n)))
(define (A256992 n) (if (not (zero? (A079559 n))) (A213714 n) (A234017 n)))

a(n) = A213714(n) + A234017(n).
a(n) = A256991(n) + A079559(n).
If A079559(n) = 1, a(n) = A213714(n), otherwise a(n) = A234017(n).

A256991 If A079559(n) = 1, a(n) = A213714(n) - 1, otherwise a(n) = A234017(n).

Original entry on oeis.org

0, 1, 1, 2, 2, 3, 3, 4, 4, 5, 6, 5, 6, 7, 7, 8, 8, 9, 10, 9, 10, 11, 12, 11, 13, 14, 12, 13, 14, 15, 15, 16, 16, 17, 18, 17, 18, 19, 20, 19, 21, 22, 20, 21, 22, 23, 24, 23, 25, 26, 24, 25, 27, 28, 26, 29, 30, 27, 28, 29, 30, 31, 31, 32, 32, 33, 34, 33, 34, 35, 36, 35, 37, 38, 36, 37, 38, 39, 40, 39, 41, 42, 40, 41, 43, 44, 42

Offset: 1

Antti Karttunen, Apr 15 2015

nonn

In other words, if n = A005187(k) for some k >= 1, then a(n) = k-1, otherwise it must be that n = A055938(h) for some h, and then a(n) = h.
In binary trees like A233276 and A233278, a(n) gives the contents at the parent node of node containing n, for any n >= 1.
When iterating a(n), a(a(n)), a(a(a(n))), and so on, A070939(n) = A256478(n) + A256479(n) = A257248(n) + A257249(n) gives the number of steps needed to reach zero, from any starting value n >= 1.

Antti Karttunen, Table of n, a(n) for n = 1..16384

Cf. A005187, A055938, A079559, A213714, A234017.

Cf. also A256992 (variant), A256478, A256479, A233276, A233278, A257248, A257249.

(define (A256991 n) (if (not (zero? (A079559 n))) (+ -1 (A213714 n)) (A234017 n)))

If A079559(n) = 1, a(n) = A213714(n) - 1, otherwise a(n) = A234017(n).

a(n) = A256992(n) - A079559(n) = A213714(n) + A234017(n) - A079559(n).

A254112 Row index of n in A254105: If A234017(n) = 0, then a(n) = A213714(n), otherwise a(n) = a(A234017(n)).

Original entry on oeis.org

1, 1, 2, 3, 1, 2, 4, 5, 3, 6, 7, 1, 2, 4, 8, 9, 5, 10, 11, 3, 6, 12, 13, 7, 14, 15, 1, 2, 4, 8, 16, 17, 9, 18, 19, 5, 10, 20, 21, 11, 22, 23, 3, 6, 12, 24, 25, 13, 26, 27, 7, 14, 28, 29, 15, 30, 31, 1, 2, 4, 8, 16, 32, 33, 17, 34, 35, 9, 18, 36, 37, 19, 38, 39, 5, 10, 20, 40, 41, 21, 42, 43, 11, 22, 44, 45, 23, 46, 47, 3, 6, 12, 24, 48, 49, 25, 50

Offset: 1

Antti Karttunen, Jan 27 2015

nonn

Antti Karttunen, Table of n, a(n) for n = 1..8192

Row index of n in A254105. Column index for A254107.
Cf. A254111 (corresponding column index).
Cf. A213714, A234017.

If A234017(n) = 0, then a(n) = A213714(n), otherwise a(n) = a(A234017(n)).

A256478 a(0) = 0; and for n >= 1, if A079559(n) = 1, then a(n) = 1 + a(A213714(n)-1), otherwise a(n) = a(A234017(n)).

Original entry on oeis.org

0, 1, 1, 2, 2, 1, 2, 3, 3, 2, 2, 3, 1, 2, 3, 4, 4, 3, 3, 3, 2, 2, 4, 2, 3, 3, 4, 1, 2, 3, 4, 5, 5, 4, 4, 4, 3, 3, 4, 3, 3, 3, 5, 2, 2, 4, 3, 4, 2, 4, 5, 3, 3, 2, 3, 4, 4, 5, 1, 2, 3, 4, 5, 6, 6, 5, 5, 5, 4, 4, 5, 4, 4, 4, 5, 3, 3, 4, 4, 4, 3, 4, 6, 3, 3, 3, 3, 5, 5, 4, 2, 2, 4, 3, 5, 3, 4, 5, 6, 2, 4, 4, 4, 5, 3, 4, 3, 3, 2, 5, 5, 3, 6, 2, 4, 4, 3, 4, 5, 5, 6, 1, 2, 3, 4, 5, 6, 7, 7

Offset: 0

Antti Karttunen, Apr 15 2015

nonn

a(n) tells how many nonzero terms of A005187 are encountered when traversing toward the root of binary tree A233276, starting from the node containing n. This count includes both n (in case it is a term of A005187) and 1 (but not 0). See also comments in A256479 and A256991.

The 1's (seem to) occur at positions given by A000325.

Antti Karttunen, Table of n, a(n) for n = 0..16384

One more than A257248.

Cf. A000120, A000325, A005187, A070939, A079559, A080791, A213714, A234017, A233275, A233276, A233277, A255559, A256479, A256991.

a(0) = 0; and for n >= 1, if A079559(n) = 1, then a(n) = 1 + a(A213714(n)-1), otherwise a(n) = a(A234017(n)).
a(n) = A000120(A233277(n)). [Binary weight of A233277(n).]
Other identities and observations. For all n >= 1:
a(n) = 1 + A257248(n) = 1 + A080791(A233275(n)).
a(n) = A070939(n) - A256479(n).
a(n) >= A255559(n).

A256479 a(1) = 0, and for n > 1, if A079559(n) = 0, then a(n) = 1 + a(A234017(n)), otherwise a(n) = a(A213714(n)-1).

Original entry on oeis.org

0, 1, 0, 1, 2, 1, 0, 1, 2, 2, 1, 3, 2, 1, 0, 1, 2, 2, 2, 3, 3, 1, 3, 2, 2, 1, 4, 3, 2, 1, 0, 1, 2, 2, 2, 3, 3, 2, 3, 3, 3, 1, 4, 4, 2, 3, 2, 4, 2, 1, 3, 3, 4, 3, 2, 2, 1, 5, 4, 3, 2, 1, 0, 1, 2, 2, 2, 3, 3, 2, 3, 3, 3, 2, 4, 4, 3, 3, 3, 4, 3, 1, 4, 4, 4, 4, 2, 2, 3, 5, 5, 3, 4, 2, 4, 3, 2, 1, 5, 3, 3, 3, 2, 4, 3, 4, 4, 5, 2, 2, 4, 1, 5, 3, 3, 4, 3, 2, 2, 1, 6, 5, 4, 3, 2, 1, 0, 1

Offset: 1

Antti Karttunen, Apr 15 2015

nonn

a(n) tells how many terms of A055938 are encountered when traversing toward the root of binary tree A233276, starting from the node containing n. This count includes also n in case it itself is a term of A055938. See also comments in A256478 and A256991.

Antti Karttunen, Table of n, a(n) for n = 1..16384

One less than A257249.
Cf. A000120, A055938, A070939, A079559, A080791, A213714, A234017, A233275, A233276, A233277, A256478, A256991.
Cf. also A000225 (gives the positions zeros).

a(1) = 0, and for n > 1, if A079559(n) = 0, then a(n) = 1 + a(A234017(n)), otherwise a(n) = a(A213714(n)-1).
a(n) = A080791(A233277(n)). [Number of nonleading zeros in the binary representation of A233277(n).]
Other identities. For all n >= 1:
a(n) = A257249(n) - 1 = A000120(A233275(n)) - 1.
a(n) = A070939(n) - A256478(n).
a(A000225(n)) = 0.

A257248 a(1) = 0; and for n > 1, if A079559(n) = 1, then a(n) = 1 + a(A213714(n)-1), otherwise a(n) = a(A234017(n)).

Original entry on oeis.org

0, 0, 1, 1, 0, 1, 2, 2, 1, 1, 2, 0, 1, 2, 3, 3, 2, 2, 2, 1, 1, 3, 1, 2, 2, 3, 0, 1, 2, 3, 4, 4, 3, 3, 3, 2, 2, 3, 2, 2, 2, 4, 1, 1, 3, 2, 3, 1, 3, 4, 2, 2, 1, 2, 3, 3, 4, 0, 1, 2, 3, 4, 5, 5, 4, 4, 4, 3, 3, 4, 3, 3, 3, 4, 2, 2, 3, 3, 3, 2, 3, 5, 2, 2, 2, 2, 4, 4, 3, 1, 1, 3, 2, 4, 2, 3, 4, 5, 1, 3, 3, 3, 4, 2, 3, 2, 2, 1, 4, 4, 2, 5, 1, 3, 3, 2, 3, 4, 4, 5, 0, 1, 2, 3, 4, 5, 6, 6

Offset: 1

Antti Karttunen, Apr 19 2015

nonn

a(n) tells how many nonzero terms of A005187 are encountered when traversing toward the root of binary tree A233276, starting from the node containing n and before 1 is reached. This count includes both n (in case it is a term of A005187) but excludes the 1 and 0 at the root. See also comments in A257249, A256478 and A256991.

Antti Karttunen, Table of n, a(n) for n = 1..8192

One less than A256478.

Cf. A000120, A000325, A005187, A070939, A079559, A080791, A213714, A234017, A233275, A233276, A233277, A255559, A257249, A256991.

a(1) = 0; and for n > 1, if A079559(n) = 1, then a(n) = 1 + a(A213714(n)-1), otherwise a(n) = a(A234017(n)).
a(n) = A080791(A233275(n)). [Number of nonleading zeros in the binary representation of A233275(n).]
Other identities. For all n >= 1:
a(n) = A256478(n)-1 = A000120(A233277(n))-1.
a(n) = A070939(n) - A257249(n).

A257249 a(0) = 1, and for n >= 1, if A079559(n) = 0, then a(n) = 1 + a(A234017(n)), otherwise a(n) = a(A213714(n)-1).

Original entry on oeis.org

1, 1, 2, 1, 2, 3, 2, 1, 2, 3, 3, 2, 4, 3, 2, 1, 2, 3, 3, 3, 4, 4, 2, 4, 3, 3, 2, 5, 4, 3, 2, 1, 2, 3, 3, 3, 4, 4, 3, 4, 4, 4, 2, 5, 5, 3, 4, 3, 5, 3, 2, 4, 4, 5, 4, 3, 3, 2, 6, 5, 4, 3, 2, 1, 2, 3, 3, 3, 4, 4, 3, 4, 4, 4, 3, 5, 5, 4, 4, 4, 5, 4, 2, 5, 5, 5, 5, 3, 3, 4, 6, 6, 4, 5, 3, 5, 4, 3, 2, 6, 4, 4, 4, 3, 5, 4, 5, 5, 6, 3, 3, 5, 2, 6, 4, 4, 5, 4, 3, 3, 2, 7, 6, 5, 4, 3, 2, 1, 2

Offset: 0

Antti Karttunen, Apr 19 2015

nonn

Because A233275(n) = A003188(n) for n = 1 .. 9, a(n) = A005811(n) for n = 1 .. 9.

Antti Karttunen, Table of n, a(n) for n = 0..8191

One more than A256479.

Cf. A000120, A003188, A005811, A055938, A070939, A079559, A080791, A213714, A234017, A233275, A233277, A233278, A257248, A256991.

a(0) = 1, and for n >= 1, if A079559(n) = 0, then a(n) = 1 + a(A234017(n)), otherwise a(n) = a(A213714(n)-1).
Other identities. For all n >= 1:
a(n) = A070939(n) - A257248(n).
a(n) = A000120(A233275(n)). [Binary weight of A233275(n).]
a(n) = 1 + A256479(n) = 1 + A080791(A233277(n)).

A279336 Permutation of natural numbers: a(1) = 1; for n > 1, if A079559(n) = 0, a(n) = 2*A234016(n), otherwise a(n) = A003961(a(A213714(n))).

Original entry on oeis.org

1, 2, 3, 5, 4, 6, 7, 9, 8, 15, 11, 10, 12, 14, 25, 27, 16, 35, 13, 18, 20, 21, 45, 22, 33, 49, 24, 26, 28, 30, 125, 81, 32, 77, 17, 34, 36, 75, 63, 38, 55, 175, 40, 42, 44, 39, 65, 46, 121, 135, 48, 50, 51, 99, 52, 105, 343, 54, 56, 58, 60, 62, 625, 243, 64, 143, 19, 66, 68, 57, 225, 70, 245, 275, 72, 74, 76, 69, 91, 78, 539, 189, 80

Offset: 1

Antti Karttunen, Dec 10 2016

nonn

For n > 1, a(n) = the number which is in the same position of array A246278 where n is located in array A256997.

Antti Karttunen, Table of n, a(n) for n = 1..10000
Index entries for sequences that are permutations of the natural numbers

Inverse permutation: A279337.
Cf. A003961, A005187, A079559, A213714, A234016, A246278, A256997, A256998.
Cf. also A278501, A279338 (a variant).

(definec (A279336 n) (cond ((= 1 n) n) ((zero? (A079559 n)) (* 2 (A234016 n))) (else (A003961 (A279336 (A213714 n))))))

a(1) = 1, for n > 1, if A079559(n) = 0 [when n is a term of A055938], a(n) = 2*A234016(n), otherwise a(n) = A003961(a(A213714(n))).
Other identities:
For all n >= 2, a(n) = A246278(A256998(n)).

A055938 Integers not generated by b(n) = b(floor(n/2)) + n (complement of A005187).

Original entry on oeis.org

2, 5, 6, 9, 12, 13, 14, 17, 20, 21, 24, 27, 28, 29, 30, 33, 36, 37, 40, 43, 44, 45, 48, 51, 52, 55, 58, 59, 60, 61, 62, 65, 68, 69, 72, 75, 76, 77, 80, 83, 84, 87, 90, 91, 92, 93, 96, 99, 100, 103, 106, 107, 108, 111, 114, 115, 118, 121, 122, 123, 124, 125, 126, 129

Offset: 1

Alford Arnold, Jul 21 2000

Note that the lengths of the consecutive runs in a(n) form sequence A001511.
Integers that are not a sum of distinct integers of the form 2^k-1. - Vladeta Jovovic, Jan 24 2003
Also n! never ends in this many 0's in base 2 - Carl R. White, Jan 21 2008
A079559(a(n)) = 0. - Reinhard Zumkeller, Mar 18 2009
These numbers are dead-end points when trying to apply the iterated process depicted in A071542 in reverse, i.e. these are positive integers i such that there does not exist k with A000120(i+k)=k. See also comments at A179016. - Antti Karttunen, Oct 26 2012
Conjecture: a(n)=b(n) defined as b(1)=2, for n>1, b(n+1)=b(n)+1 if n is already in the sequence, b(n+1)=b(n)+3 otherwise. If so, then see Cloitre comment in A080578. - Ralf Stephan, Dec 27 2013
Numbers n for which A257265(m) = 0. - Reinhard Zumkeller, May 06 2015. Typo corrected by Antti Karttunen, Aug 08 2015
Numbers which have a 2 in their skew-binary representation (cf. A169683). - Allan C. Wechsler, Feb 28 2025

			Since A005187 begins 0 1 3 4 7 8 10 11 15 16 18 19 22 23 25 26 31... this sequence begins 2 5 6 9 12 13 14 17 20 21

Complement of A005187. Setwise difference of A213713 and A213717.
Row 1 of arrays A257264, A256997 and also of A255557 (when prepended with 1). Equally: column 1 of A256995 and A255555.
Cf. also arrays A254105, A254107 and permutations A233276, A233278.
Left inverses: A234017, A256992.
Cf. A001511, A046699, A079559, A080578, A086343, A227359, A227408, A234016.
Gives positions of zeros in A213714, A213723, A213724, A213731, A257265, positions of ones in A213725-A213727 and A256989, positions of nonzeros in A254110.
Cf. also A010061 (integers that are not a sum of distinct integers of the form 2^k+1).
Analogous sequence for factorial base number system: A219658, for Fibonacci number system: A219638, for base-3: A096346. Cf. also A136767-A136774.
Cf. A257508, A257509, A257126.

a055938 n = a055938_list !! (n-1)
a055938_list = concat $
   zipWith (\u v -> [u+1..v-1]) a005187_list $ tail a005187_list
-- Reinhard Zumkeller, Nov 07 2011

a[0] = 0; a[1] = 1; a[n_Integer] := a[Floor[n/2]] + n; b = {}; Do[ b = Append[b, a[n]], {n, 0, 105}]; c =Table[n, {n, 0, 200}]; Complement[c, b]
(* Second program: *)
t = Table[IntegerExponent[(2n)!, 2], {n, 0, 100}]; Complement[Range[t // Last], t] (* Jean-François Alcover, Nov 15 2016 *)

L=listcreate();for(n=1,1000,for(k=2*n-hammingweight(n)+1,2*n+1-hammingweight(n+1),listput(L,k)));Vec(L) \\ Ralf Stephan, Dec 27 2013

def a053644(n): return 0 if n==0 else 2**(len(bin(n)[2:]) - 1)
def a043545(n):
    x=bin(n)[2:]
    return int(max(x)) - int(min(x))
def a079559(n): return 1 if n==0 else a043545(n + 1)*a079559(n + 1 - a053644(n + 1))
print([n for n in range(1, 201) if a079559(n)==0]) # Indranil Ghosh, Jun 11 2017, after the comment by Reinhard Zumkeller

;; utilizing COMPLEMENT-macro from Antti Karttunen's IntSeq-library)
(define A055938 (COMPLEMENT 1 A005187))
;; Antti Karttunen, Aug 08 2015

a(n) = A080578(n+1) - 2 = A080468(n+1) + 2*n (conjectured). - Ralf Stephan, Dec 27 2013
From Antti Karttunen, Aug 08 2015: (Start)
Other identities. For all n >= 1:
A234017(a(n)) = n.
A256992(a(n)) = n.
A257126(n) = a(n) - A005187(n).
(End)

More terms from Robert G. Wilson v, Jul 24 2000

A046699 a(1) = a(2) = 1, a(n) = a(n - a(n-1)) + a(n-1 - a(n-2)) if n > 2.

Original entry on oeis.org

1, 1, 2, 2, 3, 4, 4, 4, 5, 6, 6, 7, 8, 8, 8, 8, 9, 10, 10, 11, 12, 12, 12, 13, 14, 14, 15, 16, 16, 16, 16, 16, 17, 18, 18, 19, 20, 20, 20, 21, 22, 22, 23, 24, 24, 24, 24, 25, 26, 26, 27, 28, 28, 28, 29, 30, 30, 31, 32, 32, 32, 32, 32, 32, 33, 34, 34, 35, 36, 36, 36, 37

Offset: 1

R. K. Guy

Ignoring first term, this is the meta-Fibonacci sequence for s=0. - Frank Ruskey and Chris Deugau (deugaucj(AT)uvic.ca)

Except for the first term, n occurs A001511(n) times. - Franklin T. Adams-Watters, Oct 22 2006

Sequence was proposed by Reg Allenby.
B. W. Conolly, "Meta-Fibonacci sequences," in S. Vajda, editor, Fibonacci and Lucas Numbers and the Golden Section. Halstead Press, NY, 1989, pp. 127-138. See Eq. (2).
Michael Doob, The Canadian Mathematical Olympiad & L'Olympiade Mathématique du Canada 1969-1993, Canadian Mathematical Society & Société Mathématique du Canada, Problem 5, 1990, pp. 212-213, 1993.
S. Vajda, Fibonacci and Lucas Numbers and the Golden Section, Wiley, 1989, see p. 129.
S. Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 129.

N. J. A. Sloane, Table of n, a(n) for n = 1..20000
Altug Alkan, On a Generalization of Hofstadter's Q-Sequence: A Family of Chaotic Generational Structures, Complexity (2018) Article ID 8517125.
Joerg Arndt, Matters Computational (The Fxtbook)
Stephen M. Buckley, Wiring switches to more light bulbs, arXiv:2410.02460 [math.CO], 2024. See pp. 3, 12, 22.
Joseph Callaghan, John J. Chew III, and Stephen M. Tanny, On the behavior of a family of meta-Fibonacci sequences, SIAM Journal on Discrete Mathematics 18.4 (2005): 794-824. See p. 794. - _N. J. A. Sloane_, Apr 16 2014
M. Celaya and F. Ruskey, Morphic Words and Nested Recurrence Relations, arXiv preprint arXiv:1307.0153 [math.CO], 2013.
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]
C. Deugau and F. Ruskey, Complete k-ary Trees and Generalized Meta-Fibonacci Sequences
A. Erickson, A. Isgur, B. W. Jackson, F. Ruskey and S. M. Tanny, Nested recurrence relations with Conolly-like solutions, See Table 2.
Nathan Fox, A Slow Relative of Hofstadter's Q-Sequence, arXiv:1611.08244 [math.NT], 2016.
Nathan Fox, Trees, Fibonacci Numbers, and Nested Recurrences, Rutgers University Experimental Math Seminar, Mar 07, 2019.
Nathan Fox, Connecting Slow Solutions to Nested Recurrences with Linear Recurrent Sequences, arXiv:2203.09340 [math.CO], 2022.
The IMO Compendium, Problem 5, 22nd Canadian Mathematical Olympiad 1990.
Abraham Isgur, Mustazee Rahman, and Stephen Tanny, Solving non-homogeneous nested recursions using trees, Annals of Combinatorics 17.4 (2013): 695-710. See p. 695. - _N. J. A. Sloane_, Apr 16 2014
A. Isgur, R. Lech, S. Moore, S. Tanny, Y. Verberne, and Y. Zhang, Constructing New Families of Nested Recursions with Slow Solutions, SIAM J. Discrete Math., 30(2), 2016, 1128-1147. (20 pages); DOI:10.1137/15M1040505.
B. Jackson and F. Ruskey, Meta-Fibonacci Sequences, Binary Trees and Extremal Compact Codes
B. Jackson and F. Ruskey, Meta-Fibonacci Sequences, Binary Trees and Extremal Compact Codes, Electronic Journal of Combinatorics, 13 (2006), #R26, 13 pages.
Oliver Kullmann and Xishun Zhao, Parameters for minimal unsatisfiability: Smarandache primitive numbers and full clauses, arXiv preprint, arXiv:1505.02318 [cs.DM], 2015.
Thomas M. Lewis and Fabian Salinas, Optimal pebbling of complete binary trees and a meta-Fibonacci sequence, arXiv:2109.07328 [math.CO], 2021.
Ramin Naimi and Eric Sundberg, A Combinatorial Problem Solved by a Meta-Fibonacci Recurrence Relation, arXiv:1902.02929 [math.CO], 2019.
Index entries for Hofstadter-type sequences.
Index to sequences related to Olympiads.

Cf. A001511, A005185, A005187, A007843, A055938, A079559, A080578, A101925, A182105, A213714, A226222, A234016, A275363, A324473, A324475, A324477.

Callaghan et al. (2005)'s sequences T_{0,k}(n) for k=1 through 7 are A000012, A046699, A046702, A240835, A241154, A241155, A240830.

a046699 n = a046699_list !! (n-1)
a046699_list = 1 : 1 : zipWith (+) zs (tail zs) where
   zs = map a046699 $ zipWith (-) [2..] a046699_list
-- Reinhard Zumkeller, Jan 02 2012

[ n le 2 select 1 else Self(n - Self(n-1)) + Self(n-1 -Self(n-2)):n in [1..80]]; // Marius A. Burtea, Oct 17 2019

a := proc(n) option remember; if n <= 1 then return 1 end if; if n <= 2 then return 2 end if; return add(a(n - i + 1 - a(n - i)), i = 1 .. 2) end proc # Frank Ruskey and Chris Deugau (deugaucj(AT)uvic.ca)
a := proc(n) option remember; if n <= 2 then 1 else a(n - a(n-1)) + a(n-1 - a(n-2)); fi; end; # N. J. A. Sloane, Apr 16 2014

a[n_] := (k = 1; While[ !Divisible[(2*++k)!, 2^(n-1)]]; k); a[1] = a[2] = 1; Table[a[n], {n, 1, 72}] (* Jean-François Alcover, Oct 06 2011, after Benoit Cloitre *)
CoefficientList[ Series[1 + x/(1 - x)*Product[1 + x^(2^n - 1), {n, 6}], {x, 0, 80}], x] (* or *)
a[1] = a[2] = 1; a[n_] := a[n] = a[n - a[n - 1]] + a[n - 1 - a[n - 2]]; Array[a, 80] (* Robert G. Wilson v, Sep 08 2014 *)

a[1]:1$
a[2]:1$
a[n]:=a[n-a[n-1]]+a[n-1-a[n-2]]$
makelist(a[n],n,2,60); /* Martin Ettl, Oct 29 2012 */

a(n)=if(n<0,1,s=1;while((2*s)!%2^(n-1)>0,s++);s) \\ Benoit Cloitre, Jan 19 2007

from sympy import factorial
def a(n):
    if n<3: return 1
    s=1
    while factorial(2*s)%(2**(n - 1))>0: s+=1
    return s
print([a(n) for n in range(1, 101)]) # Indranil Ghosh, Jun 11 2017, after Benoit Cloitre

First differences seem to be A079559. - Vladeta Jovovic, Nov 30 2003. This is correct and not too hard to prove, giving the generating function x + x^2(1+x)(1+x^3)(1+x^7)(1+x^15).../(1-x). - Paul Boddington, Jul 30 2004
G.f.: x + x^2/(1-x) * Product_{n=1}^{infinity} (1 + x^(2^n-1)). - Frank Ruskey and Chris Deugau (deugaucj(AT)uvic.ca)
For n>=1, a(n)=w(n-1) where w(n) is the least k such that 2^n divides (2k)!. - Benoit Cloitre, Jan 19 2007
Conjecture: a(n+1) = a(n) + A215530(a(n) + n) for all n > 0. - Velin Yanev, Oct 17 2019
From Bernard Schott, Dec 03 2021: (Start)
a(n) <= a(n+1) <=  a(n) +1.
For n > 1, if a(n) is odd, then a(n+1) = a(n) + 1.
a(2^n+1) = 2^(n-1) + 1 for n > 0.
Results coming from the 5th problem proposed during the 22nd Canadian Mathematical Olympiad in 1990 (link IMO Compendium and Doob reference). (End)

cp's OEIS Frontend

A256992 Position of n in either of the complementary sequences, A005187 or A055938: a(n) = A213714(n) + A234017(n).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Scheme

Formula

A256991 If A079559(n) = 1, a(n) = A213714(n) - 1, otherwise a(n) = A234017(n).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Scheme

Formula

A254112 Row index of n in A254105: If A234017(n) = 0, then a(n) = A213714(n), otherwise a(n) = a(A234017(n)).

Views

Author

Keywords

Links

Crossrefs

Formula

A256478 a(0) = 0; and for n >= 1, if A079559(n) = 1, then a(n) = 1 + a(A213714(n)-1), otherwise a(n) = a(A234017(n)).

Views

Author

Keywords

Comments

Links

Crossrefs

Formula

A256479 a(1) = 0, and for n > 1, if A079559(n) = 0, then a(n) = 1 + a(A234017(n)), otherwise a(n) = a(A213714(n)-1).

Views

Author

Keywords

Comments

Links

Crossrefs

Formula

A257248 a(1) = 0; and for n > 1, if A079559(n) = 1, then a(n) = 1 + a(A213714(n)-1), otherwise a(n) = a(A234017(n)).

Views

Author

Keywords

Comments

Links

Crossrefs

Formula

A257249 a(0) = 1, and for n >= 1, if A079559(n) = 0, then a(n) = 1 + a(A234017(n)), otherwise a(n) = a(A213714(n)-1).

Views

Author

Keywords

Comments

Links

Crossrefs

Formula

A279336 Permutation of natural numbers: a(1) = 1; for n > 1, if A079559(n) = 0, a(n) = 2*A234016(n), otherwise a(n) = A003961(a(A213714(n))).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Scheme

Formula

A055938 Integers not generated by b(n) = b(floor(n/2)) + n (complement of A005187).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs