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

Author

Antti Karttunen, Apr 15 2015

Keywords

• nonn

Comments

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.

Links

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

Crossrefs

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

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

Programs

Mathematica

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 *)

Scheme

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

Formula

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).