A367066 a(n) = ((Sum_{i=1..n} A367065(i))-2)/(n+2).
0, 1, 1, 2, 3, 3, 4, 5, 5, 6, 6, 7, 8, 8, 9, 10, 10, 11, 11, 12, 13, 13, 14, 15, 15, 16, 16, 17, 18, 18, 19, 19, 20, 21, 21, 22, 23, 23, 24, 24, 25, 26, 26, 27, 28, 28, 29, 29, 30, 31, 31, 32, 32, 33, 34, 34, 35, 36, 36, 37, 37, 38, 39, 39, 40, 41, 41, 42, 42, 43
Offset: 1
Keywords
Links
- Muharem Avdispahić and Faruk Zejnulahi, An integer sequence with a divisibility property, Fibonacci Quarterly, Vol. 58:4 (2020), 321-333.
Crossrefs
Cf. A367065.
Programs
-
Mathematica
zlist={-1,2,4}; mlist={-1,0,1}; For[n=3,n<=101,n++,If[MemberQ[zlist,mlist[[n]]],AppendTo[mlist,mlist[[n]]+1]; AppendTo[zlist,mlist[[n+1]]+n+1];,AppendTo[mlist,mlist[[n]]];AppendTo[zlist,mlist[[n+1]]];];]; mlist=Drop[mlist,1];mlist
-
Python
z_list=[-1,2,4] m_list=[-1,0,1] n=2 for n in range(2, 100): if m_list[n] in z_list: m_list.append(m_list[n] + 1) z_list.append(m_list[n+1] + n+2) else: m_list.append(m_list[n]) z_list.append(m_list[n+1]) print(m_list[1:])
Formula
Conjecture: a(n) = floor(n/phi + 1/phi^3) - [n+2 = Fibonacci(2*j+1) for some j], where phi = (1+sqrt(5))/2 and [] is the Iverson bracket. - Jon E. Schoenfield, Nov 03 2023
Comments