A078943 -id:A078943 - cp's OEIS frontend

A356080 Variation on Recamán's sequence (A005132) that is intended to be a permutation of the nonnegative integers, essentially as envisaged by the original comments in A078943. See comments below for details.

0, 1, 3, 6, 2, 7, 13, 20, 28, 37, 27, 16, 4, 17, 31, 46, 30, 47, 29, 48, 68, 89, 111, 134, 158, 133, 159, 132, 160, 131, 101, 70, 38, 5, 39, 74, 110, 147, 109, 148, 108, 149, 107, 150, 106, 151, 105, 58, 10, 59, 9, 60, 8, 61, 115, 170, 226, 283, 341, 400, 460, 399, 337, 274, 210, 145, 79, 12, 80, 11

Offset: 1

Peter Munn, Jul 25 2022

nonn

a(1) = 0; for n >= 1, a(n+1) = a(n)-n or a(n)+n; and no two terms are equal.
Subject to the pairwise constraints above, the sequence describes a path between (nodes labeled with) nonnegative integers that is extended in stages so that the extended path ends with the least missing number from the unextended path.
We denote the number at the end of the unextended path as k, the path from 0 to k as P_k and the least number absent from P_k as m_k. We call a path that starts at k an extension route if it extends P_k and satisfies the pairwise constraints. [clarification added, Peter Munn, Aug 07 2024]
An extension route R_k from P_k is suitable if (1) R_k ends in m_k and (2) R_k is the start of a "look-ahead" extension route (R_k + R') that ends in a record number (i.e., greater than the largest in P_k + R_k). If there is no suitable R_k the sequence finishes at k. Otherwise we denote the lexicographically earliest shortest such R_k as E_m_k, and extend P_k as P_m_k = P_k + E_m_k.
Note that any E_m_k ends in m_k (i.e., without any look-ahead R' being included) and so E_k ends in k and beware the sequence might end there. Nevertheless, once we determine that the sequence continues and includes m_k (perhaps because at least 1 suitable R_k has been found) 1 or more terms of E_m_k will be known, and this is the point in time we may include terms after k in the published sequence.
Conjecture (Peter Munn): the sequence is infinite and so a permutation of the nonnegative integers.

			a(2) = a(1) + 1 = 1, since a(1) - 1 = -1 is a negative integer.
We now find the lexicographically earliest shortest route to the least missing number, 2. Any extension route has a(3) = a(2) + 2 = 3 since a(2) - 2 = -1 is a negative integer. Any extension route has a(4) = a(3) + 3 = 6, since a(3) - 3 = 0 is already in the sequence. So a(3) = 3, a(4) = 6, a(5) = a(4) - 4 = 2 is the only way to reach 2 by a(5); no shorter route exists. Lastly, we must check an onward route exists to a new record term (greater than 6). This is provided by a(6) = a(5) + 5 = 7, so we have determined a(3) = 3, a(4) = 6, a(5) = 2.

Peter Munn, Table of n, a(n) for n = 1..424
Peter Munn and Ali Sada, Selection of relevant SeqFan posts, July 2022.
Peter Munn, Initial validation program (LibreOffice Basic)

Cf. A005132, A078943.

|a(n) - a(n+1)| = n.

If a(n) = a(m) then n = m.

A063733 A variant of Recamán's sequence: a(0)=1, a(n) = a(n-1)-(n-1) if positive and new, else a(n) = a(n-1)+(n-1).

Original entry on oeis.org

1, 1, 2, 4, 7, 3, 8, 14, 21, 13, 22, 12, 23, 11, 24, 10, 25, 9, 26, 44, 63, 43, 64, 42, 19, 43, 18, 44, 17, 45, 16, 46, 15, 47, 80, 114, 79, 115, 78, 40, 79, 39, 80, 38, 81, 37, 82, 36, 83, 35, 84, 34, 85, 33, 86, 32, 87, 31, 88, 30, 89, 29, 90, 28, 91, 27

Offset: 0

N. J. A. Sloane, Sep 05 2001

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
Index entries for sequences related to Recamán's sequence

See A005132, which is the main entry for this sequence. A063753 = 1 + A005132. A row of A066201. Also a row of A066202.
See also A078943.
Cf. A141126.

a063733 n = a063733_list !! n
a063733_list = 1 : f 0 [1] where
   f x ys@(y:_) | u > 0 && u `notElem` ys = u : f (x + 1) (u : ys)
                | otherwise               = v : f (x + 1) (v : ys)
                where u = y - x; v = x + y
-- Reinhard Zumkeller, Jul 02 2015

a[0] = 1; a[n_] := a[n] = If[an = a[n-1] - (n-1); an > 0 && FreeQ[Array[a, n-1], an], an, a[n-1] + (n-1)]; Table[a[n], {n, 0, 65}] (* Jean-François Alcover, Feb 18 2018 *)

l=[1]
for n in range(1, 101):
    x = l[n - 1] - (n - 1)
    if x > 0 and x not in l:
        l.append(x)
    else:
        l.append(l[n - 1] + (n - 1))
print(l) # Indranil Ghosh, Jun 02 2017

a(n+1) = A005132(n) - 1; A005132 is Recamán's sequence. - Franklin T. Adams-Watters, Mar 12 2010

A174162 a(1) = 2. Let k >= 1 be the minimal integer such that 2ka(n-1) + 1 has at least one prime divisor which is not already in the sequence. Then a(n) is the smallest such divisor.

Original entry on oeis.org

2, 5, 11, 23, 47, 19, 3, 7, 29, 59, 17, 103, 619, 2477, 991, 661, 3967, 2267, 907, 191, 383, 13, 53, 107, 43, 173, 347, 139, 31, 83, 167, 67, 269, 359, 719, 1439, 2879, 443, 887, 71, 61, 41, 137, 823, 37, 149, 199, 797, 1063, 709, 2837, 227, 101, 607, 3643, 21859

Offset: 1

Vladimir Shevelev, Mar 10 2010

nonn

Conjectures: 1) The sequence is a permutation of prime numbers; 2) k = k(n) runs all positive integers.

Cf. A005132, A063733, A078943, A174161.

a = {2}; Do[k = 1; While[(d = Complement[FactorInteger[2 k a[[-1]] + 1][[All, 1]], a]) == {}, k++]; AppendTo[a, Min@d], {n, 50}]; a (* Jinyuan Wang, Feb 26 2020 *)

More terms from Jinyuan Wang, Feb 26 2020

A174161 a(1) = 2. Let k >= 1 be the minimal integer such that 2ka(n-1) - 1 has at least one prime divisor which is not already in the sequence. Then a(n) is the smallest such divisor.

Original entry on oeis.org

2, 3, 5, 19, 37, 73, 29, 23, 7, 13, 17, 11, 43, 257, 79, 157, 313, 139, 277, 41, 163, 31, 61, 487, 59, 47, 281, 1123, 449, 359, 239, 53, 211, 421, 101, 67, 89, 71, 283, 113, 677, 2707, 5413, 433, 173, 691, 1381, 251, 167, 1669, 5563, 7417, 44501, 431, 1723, 2297

Offset: 1

Vladimir Shevelev, Mar 10 2010

nonn

Conjectures: 1) The sequence is a permutation of prime numbers; 2) k = k(n) runs all positive integers.

Ivan Neretin, Table of n, a(n) for n = 1..10000

Cf. A005132, A063733, A078943, A174162.

a = {2}; Do[k = 1; While[(d = Complement[FactorInteger[2 k a[[-1]] - 1][[All, 1]], a]) == {}, k++]; AppendTo[a, Min@d], {n, 50}]; a (* Ivan Neretin, Dec 04 2018 *)

Terms from a(22) onwards corrected by Ivan Neretin, Dec 04 2018

cp's OEIS Frontend

A356080 Variation on Recamán's sequence (A005132) that is intended to be a permutation of the nonnegative integers, essentially as envisaged by the original comments in A078943. See comments below for details.

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A063733 A variant of Recamán's sequence: a(0)=1, a(n) = a(n-1)-(n-1) if positive and new, else a(n) = a(n-1)+(n-1).

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A174162 a(1) = 2. Let k >= 1 be the minimal integer such that 2ka(n-1) + 1 has at least one prime divisor which is not already in the sequence. Then a(n) is the smallest such divisor.

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Mathematica

Extensions

A174161 a(1) = 2. Let k >= 1 be the minimal integer such that 2ka(n-1) - 1 has at least one prime divisor which is not already in the sequence. Then a(n) is the smallest such divisor.

Views

Author

Keywords

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Programs

Mathematica

Extensions

cp's OEIS Frontend

A356080 Variation on Recamán's sequence (A005132) that is intended to be a permutation of the nonnegative integers, essentially as envisaged by the original comments in A078943. See comments below for details.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Formula

A063733 A variant of Recamán's sequence: a(0)=1, a(n) = a(n-1)-(n-1) if positive and new, else a(n) = a(n-1)+(n-1).

Views

Author

Keywords

Links

Crossrefs

Programs

Haskell

Mathematica

Python

Formula

A174162 a(1) = 2. Let k >= 1 be the minimal integer such that 2*k*a(n-1) + 1 has at least one prime divisor which is not already in the sequence. Then a(n) is the smallest such divisor.

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Author

Keywords

Comments

Crossrefs

Programs

Mathematica

Extensions

A174161 a(1) = 2. Let k >= 1 be the minimal integer such that 2*k*a(n-1) - 1 has at least one prime divisor which is not already in the sequence. Then a(n) is the smallest such divisor.

Views

Author

Keywords

Comments

Links

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Programs

Mathematica

Extensions

A174162 a(1) = 2. Let k >= 1 be the minimal integer such that 2ka(n-1) + 1 has at least one prime divisor which is not already in the sequence. Then a(n) is the smallest such divisor.

A174161 a(1) = 2. Let k >= 1 be the minimal integer such that 2ka(n-1) - 1 has at least one prime divisor which is not already in the sequence. Then a(n) is the smallest such divisor.