Zenan Sabanac - cp's OEIS frontend

A367069 a(n) = ((Sum_{i=1..n} A367067(i))-3)/(n+3).

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

Offset: 1

Zenan Sabanac, Dec 17 2023

nonn

For a positive integer k define the Avdispahić-Zejnulahi sequence AZ(k) by b(1)=k, and thereafter b(n) is the least positive integer not yet in the sequence such that Sum_{i=1..n} b(i) == k (mod n+k). Define the Avdispahić-Zejnulahi means sequence AZM(k) by a(n) = ((Sum_{i=1..n} b(i))-k)/(n+k). This is the AZM(3) sequence.

Muharem Avdispahić and Faruk Zejnulahi, An integer sequence with a divisibility property, Fibonacci Quarterly, Vol. 58:4 (2020), 321-333.

Cf. A367067.

Cf. A073869 (AZM(0)), A367068 (AZM(1)), A367066 (AZM(2)).

zlist = {-1, 3, 5};
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 + 2];,
    AppendTo[mlist, mlist[[n]]]; AppendTo[zlist, mlist[[n + 1]]];];];
mlist = Drop[mlist, 1]; mlist

z_list=[-1, 3, 5]
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+3)
    else:
        m_list.append(m_list[n])
        z_list.append(m_list[n+1])
print(m_list[1:])

A367068 a(n) = ((Sum_{i=1..n} A340510(i))-1)/(n+1).

Original entry on oeis.org

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

Offset: 1

Zenan Sabanac, Dec 17 2023

nonn

For a positive integer k define the Avdispahić-Zejnulahi sequence AZ(k) by b(1)=k, and thereafter b(n) is the least positive integer not yet in the sequence such that Sum_{i=1..n} b(i) == k (mod n+k).
Define the Avdispahić-Zejnulahi means sequence AZM(k) by a(n) = ((Sum_{i=1..n} b(i))-k)/(n+k). This is the AZM(1) sequence.
Is this a duplicate of A005379? For n<=1300 at least a(n)=A005379(n). - R. J. Mathar, Jan 30 2024

Muharem Avdispahić and Faruk Zejnulahi, An integer sequence with a divisibility property, Fibonacci Quarterly, Vol. 58:4 (2020), 321-333.

Cf. A340510.

Cf. A073869 (AZM(0)), A367066 (AZM(2)).

A367068 := proc(n)
    add(A340510(i),i=1..n)-1 ;
    %/(n+1) ;
end proc:
seq(A367068(n),n=1..50) ; # R. J. Mathar, Jan 30 2024

zlist = {-1, 1, 3};
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];, AppendTo[mlist, mlist[[n]]];
     AppendTo[zlist, mlist[[n + 1]]];];];
mlist = Drop[mlist, 1]; mlist

z_list=[-1, 1, 3]
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+1)
    else:
        m_list.append(m_list[n])
        z_list.append(m_list[n+1])
print(m_list[1:])

For n>2, a(n) = a(n-1) if a(n-1) <> A340510(k) (for k=1..n-1) and a(n) = a(n-1)+1=A340510(n)-n otherwise. (See Proposition 3.1. of Avdispahić and Zejnulahi in the link above).

A367067 a(1)=3, thereafter a(n) is the least positive integer not yet in the sequence such that Sum_{i=1..n} a(i) == 3 (mod n+3).

Original entry on oeis.org

3, 5, 1, 8, 2, 11, 13, 4, 16, 18, 6, 21, 7, 24, 26, 9, 29, 10, 32, 34, 12, 37, 39, 14, 42, 15, 45, 47, 17, 50, 52, 19, 55, 20, 58, 60, 22, 63, 23, 66, 68, 25, 71, 73, 27, 76, 28, 79, 81, 30, 84, 31, 87, 89, 33, 92, 94, 35, 97, 36, 100, 102, 38, 105

Offset: 1

Zenan Sabanac, Nov 03 2023

nonn

This is the Avdispahić-Zejnulahi sequence AZ(3).

Note that AZ(3) is the third term in a sequence of permutations of the set of positive integers defined by a specific divisibility property (see Links section and Crossrefs for details).

Muharem Avdispahić and Faruk Zejnulahi, An integer sequence with a divisibility property, Fibonacci Quarterly, Vol. 58:4 (2020), 321-333.
Jeffrey Shallit, Proving properties of some greedily-defined integer recurrences via automata theory, arXiv:2308.06544 [cs.DM], 2023.

Cf. A340510, A367065.

lst = {3};
f[s_List] := Block[{k = 1, len = 4 + Length@lst, t = Plus @@ lst},
  While[MemberQ[s, k] || Mod[k + t, len] != 3, k++];
  AppendTo[lst, k]]; Nest[f, lst, 100]

z_list=[-1,3,5]
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+3)
    else:
        m_list.append(m_list[n])
        z_list.append(m_list[n+1])
print(z_list[1:])

A367066 a(n) = ((Sum_{i=1..n} A367065(i))-2)/(n+2).

Original entry on oeis.org

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

Zenan Sabanac, Nov 03 2023

nonn

For a positive integer k define the Avdispahić-Zejnulahi sequence AZ(k) by b(1)=k, and thereafter b(n) is the least positive integer not yet in the sequence such that Sum_{i=1..n} b(i) == k (mod n+k). Define the Avdispahić-Zejnulahi means sequence AZM(k) by a(n) = ((Sum_{i=1..n} b(i))-k)/(n+k). This is the AZM(2) sequence.

Muharem Avdispahić and Faruk Zejnulahi, An integer sequence with a divisibility property, Fibonacci Quarterly, Vol. 58:4 (2020), 321-333.

Cf. A367065.

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

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:])

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

A367065 a(1)=2, thereafter a(n) is the least positive integer not yet in the sequence such that Sum_{i=1..n} a(i) == 2 (mod n+2).

Original entry on oeis.org

2, 4, 1, 7, 9, 3, 12, 14, 5, 17, 6, 20, 22, 8, 25, 27, 10, 30, 11, 33, 35, 13, 38, 40, 15, 43, 16, 46, 48, 18, 51, 19, 54, 56, 21, 59, 61, 23, 64, 24, 67, 69, 26, 72, 74, 28, 77, 29, 80, 82, 31, 85, 32, 88, 90, 34, 93, 95, 36, 98, 37, 101, 103, 39, 106, 108, 41, 111, 42, 114

Offset: 1

Zenan Sabanac, Nov 03 2023

nonn

This is the Avdispahić-Zejnulahi sequence AZ(2). For a positive integer k, the Avdispahić-Zejnulahi sequence AZ(k) is given by: a(1)=k, thereafter a(n) is the least positive integer not yet in the sequence such that Sum_{i=1..n} a(i) == k (mod n+k). It is interesting to note that (AZ(k)) represents a sequence of permutations of the set of positive integers. (See Links section for details concerning AZ(1).)

Muharem Avdispahić and Faruk Zejnulahi, An integer sequence with a divisibility property, Fibonacci Quarterly, Vol. 58:4 (2020), 321-333.
Jeffrey Shallit, Proving properties of some greedily-defined integer recurrences via automata theory, arXiv:2308.06544 [cs.DM], 2023.

A340510 is the AZ(1) sequence. A002251 is the AZ(0) sequence.

lst={2}; f[s_List]:=Block[{k=1,len=3+Length@lst,t=Plus@@lst},While[MemberQ[s,k]||Mod[k+t,len]!=2,k++]; AppendTo[lst,k]]; Nest[f,lst,100]

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(z_list[1:])

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User: Zenan Sabanac

A367069 a(n) = ((Sum_{i=1..n} A367067(i))-3)/(n+3).

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A367068 a(n) = ((Sum_{i=1..n} A340510(i))-1)/(n+1).

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A367067 a(1)=3, thereafter a(n) is the least positive integer not yet in the sequence such that Sum_{i=1..n} a(i) == 3 (mod n+3).

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A367066 a(n) = ((Sum_{i=1..n} A367065(i))-2)/(n+2).

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A367065 a(1)=2, thereafter a(n) is the least positive integer not yet in the sequence such that Sum_{i=1..n} a(i) == 2 (mod n+2).

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Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Python