cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A057514 Number of peaks in mountain ranges encoded by A014486, number of leaves in the corresponding rooted plane trees (the root node is never counted as a leaf).

Original entry on oeis.org

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

Views

Author

Antti Karttunen, Sep 03 2000

Keywords

Comments

Sum_{i=A014137(n)..(A014137(n+1)-1)} a(i) = A001700(n), i.e., A001700(n) gives the total number of leaves in all ordered trees with n + 1 edges.

Crossrefs

a(n)-1 gives the number of zeros in A071153(n) (for n>=1).

Programs

  • Python
    def a005811(n): return bin(n^(n>>1))[2:].count("1")
    def ok(n): # This function after Peter Luschny
        B=bin(n)[2:] if n!=0 else 0
        s=0
        for b in B:
            s+=1 if b=="1" else -1
            if s<0: return 0
        return s==0
    def A(n): return [0] + [i for i in range(1, n + 1) if ok(i)]
    l=A(200)
    print([a005811(l[i])//2 for i in range(len(l))]) # Indranil Ghosh, May 21 2017

Formula

a(n) = A005811(A014486(n))/2 = A000120(A003188(A014486(n)))/2.