A213714 Inverse function for injection A005187.

0, 1, 0, 2, 3, 0, 0, 4, 5, 0, 6, 7, 0, 0, 0, 8, 9, 0, 10, 11, 0, 0, 12, 13, 0, 14, 15, 0, 0, 0, 0, 16, 17, 0, 18, 19, 0, 0, 20, 21, 0, 22, 23, 0, 0, 0, 24, 25, 0, 26, 27, 0, 0, 28, 29, 0, 30, 31, 0, 0, 0, 0, 0, 32, 33, 0, 34, 35, 0, 0, 36, 37, 0, 38, 39, 0, 0, 0, 40, 41, 0, 42, 43, 0, 0, 44, 45, 0, 46, 47, 0

Offset: 0

Author

Antti Karttunen, Oct 26 2012

Keywords

• nonn

Comments

a(0)=0; thereafter if n occurs as a term of A005187, a(n)=its position in A005187, otherwise zero. This works as an "inverse" function for A005187 in a sense that a(A005187(n)) = n for all n.

a(n)*A234017(n) = 0 for all n.

Links

• Antti Karttunen, Table of n, a(n) for n = 0..8192

Crossrefs

Can be used when computing A213715, A213723, A213724, A233275, A233277. Cf. A005187, A046699, A079559, A234017, A230414.

Programs

Haskell

import Data.List (genericIndex)
a213714 n = genericIndex a213714_list n
a213714_list = f [0..] a005187_list 0 where
   f (x:xs) ys'@(y:ys) i | x == y    = i : f xs ys (i+1)
                         | otherwise = 0 : f xs ys' i
-- Reinhard Zumkeller, May 01 2015

Python

from sympy import factorial
def a046699(n):
    if n<3: return 1
    s=1
    while factorial(2*s)%(2**(n - 1))>0: s+=1
    return s
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))
def a(n): return 0 if n==0 else a079559(n)*(a046699(n + 2) - 1) # Indranil Ghosh, Jun 11 2017

Formula

a(0)=0, for n>0, a(n) = A079559(n) * (A046699(n+2)-1) [With A046699's October 2012 starting offset. Incorrect indexing shown in this formula corrected by Antti Karttunen, Dec 18 2013]