A080036 -id:A080036 - cp's OEIS frontend

A080580 a(1)=1; for n>1, a(n)=a(n-1)+2 if n is already in the sequence, a(n)=a(n-1)+4 otherwise.

1, 5, 9, 13, 15, 19, 23, 27, 29, 33, 37, 41, 43, 47, 49, 53, 57, 61, 63, 67, 71, 75, 77, 81, 85, 89, 91, 95, 97, 101, 105, 109, 111, 115, 119, 123, 125, 129, 133, 137, 139, 143, 145, 149, 153, 157, 159, 163, 165, 169, 173, 177, 179, 183, 187, 191, 193

Offset: 1

N. J. A. Sloane, Mar 23 2003

nonn

Also, positions of 0 in A284939; complement of A284941. - R. J. Mathar_, Apr 24 2017
Proof that this is the same as the positions of 0 in A284939, from Joe Slater, Apr 26 2017: (Start)
The sequence A284939 consists of a concatenation of the words 01 and 1101, each word representing a single term from the previous iteration of the transformation taken in order.
Because the sequence has reached its fixed point we don't need to compare these words to the terms of the previous iteration: we can actually relate the present iteration to itself.
Thus the following table relates terms of A284939 to its words. We would end up with the same sequence by concatenating either the terms or the words:
   0 -> 01
   1 -> 1101
   1 -> 1101
   1 -> 1101
   0 -> 01
   1 -> 1101
   ... ...
Consider any arbitrary term A284939(n) and its successor A284939(n+1) .  They relate to the n-th and (n+1)-st words, and since neither of the possible words have sequential zeros there are only three possibilities for the two terms (0,1; 1,0; and 1,1) and therefore three possibilities for the words: 01 1101, 1101 01, or 1101 1101.
We can see that the only time there is a gap of three 1's between the first and second zeros will be when the two terms A284939(n) and A284939(n+1) are (1,0). Therefore, if the n-th zero of the sequence (corresponding to the term A284939(n)) is at position k, it will be followed by a zero at position k+2 if the term A284939(n+1)=0, but otherwise it will be followed by a zero at position k+4.
Let's relate this to the present sequence A080580:
We know sequence A284939 starts with a zero in position 1 (i.e., A284939(1)=0), so we can make a sequence S (say) listing the position of the zeros with S(1)=1. From our earlier discussion we know that the n-th zero of A284939 relates to the n-th term of A284939. Therefore, if S(n)=k, the following term S(n+1) will be +4 if A284939(n+1)=1 and +2 if A284939(n+1)=0.
But we don't actually need to refer to A284939 at all! When we come to S(n+1) we can just see whether n+1 appears in our sequence. If the number n+1 already appears in S then we know that A284939(n+1)=0, so S(n+1)=S(n)+2. If the number n+1 does not appear in S then we know that A284939(n+1)=1, so S(n+1)=S(n)+4.
This rule is exactly the rule of A080580, which means that A080580 and S are identical, and A080580 manages to predict the positions of the zeros in A284939 without ever referring to the sequence itself.  QED
(End)
In the Fokkink-Joshi paper, this sequence is the Cloitre (0,1,2,4)-hiccup sequence. - Michael De Vlieger, Jul 29 2025

Clark Kimberling, Table of n, a(n) for n = 1..10000
Benoit Cloitre, A study of a family of self-referential sequences, arXiv:2506.18103 [math.GM], 2025. See p. 15.
Benoit Cloitre, N. J. A. Sloane and M. J. Vandermast, Numerical analogues of Aronson's sequence, J. Integer Seqs., Vol. 6 (2003), #03.2.2.
Benoit Cloitre, N. J. A. Sloane and M. J. Vandermast, Numerical analogues of Aronson's sequence, arXiv:math/0305308 [math.NT], 2003.
Robbert Fokkink and Gandhar Joshi, On Cloitre's hiccup sequences, arXiv:2507.16956 [math.CO], 2025. See pp. 3, 5.

Cf. A080455-A080458, A080036, A080037.

A080580 := proc(n)
    option remember;
    if n = 1 then
        1;
    else
        known := false ;
        for i from 1 to n-1 do
            if procname(i) = n then
                known := true;
                break;
            end if;
        end do:
        if known then
            procname(n-1)+2 ;
        else
            procname(n-1)+4 ;
        end if;
    end if;
end proc:
seq(A080580(n),n=1..100) ; # R. J. Mathar, Apr 25 2017

a[1] = 1; a[n_] := a[n] = If[MemberQ[Array[a, n-1], n], a[n-1]+2, a[n-1]+4 ];
Array[a, 60] (* Jean-François Alcover, Nov 23 2017 *)

Edited by N. J. A. Sloane, Apr 27 2017

A363757 Lexicographically earliest sequence of positive integers such that the n-th pair of consecutive equal values are separated by a(n) distinct terms, with pairs numbered according to the position of the second term in the pair.

Original entry on oeis.org

1, 2, 1, 3, 2, 3, 4, 1, 3, 2, 5, 4, 5, 3, 4, 6, 1, 5, 2, 6, 4, 7, 3, 7, 5, 3, 1, 4, 8, 2, 1, 6, 3, 2, 3, 8, 9, 7, 8, 7, 1, 9, 7, 8, 5, 10, 4, 3, 2, 9, 2, 6, 8, 7, 3, 11, 1, 8, 3, 1, 10, 3, 6, 9, 7, 3, 12, 5, 12, 8, 3, 8, 2, 12, 9, 1, 7, 12, 13, 4, 9, 11, 8, 4, 2, 8, 10, 1, 10, 13, 6

Offset: 1

Neal Gersh Tolunsky, Jun 23 2023

nonn

The word 'distinct' differentiates this sequence from A363654.
A000124 gives the index of the first occurrence of n, and A080036 gives the indices of the remaining terms. A record high term occurs when its corresponding pair number would be the previous record high, since that would have to use all terms between the enclosing pair, which is impossible.
A083920(n) gives the number of pairs in the first n terms of this sequence.
If pairs are numbered according to the position of the first term in the pair (rather than second), this becomes A001511 (the ruler function).

			The 1st pair (1,2,1) encloses 1 term because a(1)=1.
The 2nd pair (2,1,3,2) encloses 2 distinct terms because a(2)=2.
The 3rd pair (3,2,3) encloses 1 term because a(3)=1.
The 4th pair (1,3,2,3,4,1) encloses 3 distinct terms because a(4)=3.
a(4)=3 since if we place a 1 or a 2 (creating the second pair), this would enclose less than a(2)=2 distinct terms, so a(4) must be the smallest unused number, which is 3.

Neal Gersh Tolunsky, Table of n, a(n) for n = 1..10000
Neal Gersh Tolunsky, Graph of first 100000 terms

Cf. A363654, A363708, A026272, A330896.

A080353 a(1)=5; for n>1, a(n)=a(n-1)+1 if n is already in the sequence, a(n)=a(n-1)+2 otherwise.

Original entry on oeis.org

5, 7, 9, 11, 12, 14, 15, 17, 18, 20, 21, 22, 24, 25, 26, 28, 29, 30, 32, 33, 34, 35, 37, 38, 39, 40, 42, 43, 44, 45, 47, 48, 49, 50, 51, 53, 54, 55, 56, 57, 59, 60, 61, 62, 63, 65, 66, 67, 68, 69, 70, 72, 73, 74, 75, 76, 77, 79, 80, 81, 82, 83, 84, 86, 87, 88, 89

Offset: 1

N. J. A. Sloane, Mar 20 2003

nonn

Conjecture: This sequence is equivalent to N \ A008478. - Michael De Vlieger, Jul 02 2025

Michael De Vlieger, Table of n, a(n) for n = 1..10000
Benoit Cloitre, A study of a family of self-referential sequences, arXiv:2506.18103 [math.GM], 2025. See p. 13.
Benoit Cloitre, N. J. A. Sloane and M. J. Vandermast, Numerical analogues of Aronson's sequence, J. Integer Seqs., Vol. 6 (2003), #03.2.2.
Benoit Cloitre, N. J. A. Sloane and M. J. Vandermast, Numerical analogues of Aronson's sequence (math.NT/0305308)

Cf. A080036, A080037. Differences give A080354.

a[1] = 5; a[n_] := a[n] = If[MemberQ[Array[a, n-1], n], a[n-1]+1, a[n-1]+2]; Array[a, 67] (* Jean-François Alcover, Oct 08 2018 *)

a(n) = n + floor(sqrt(6*n)) + O(1).

A080590 a(1)=1; for n>1, a(n)=a(n-1)+3 if n is already in the sequence, a(n)=a(n-1)+4 otherwise.

Original entry on oeis.org

1, 5, 9, 13, 16, 20, 24, 28, 31, 35, 39, 43, 46, 50, 54, 57, 61, 65, 69, 72, 76, 80, 84, 87, 91, 95, 99, 102, 106, 110, 113, 117, 121, 125, 128, 132, 136, 140, 143, 147, 151, 155, 158, 162, 166, 169, 173, 177, 181, 184, 188, 192, 196, 199, 203, 207

Offset: 1

N. J. A. Sloane, Mar 23 2003

nonn

In the Fokkink-Joshi paper, this sequence is the Cloitre (0,1,3,4)-hiccup sequence. - Michael De Vlieger, Jul 29 2025

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Benoit Cloitre, A study of a family of self-referential sequences, arXiv:2506.18103 [math.GM], 2025. See p. 9.
Benoit Cloitre, N. J. A. Sloane and M. J. Vandermast, Numerical analogues of Aronson's sequence, J. Integer Seqs., Vol. 6 (2003), #03.2.2.
Benoit Cloitre, N. J. A. Sloane and M. J. Vandermast, Numerical analogues of Aronson's sequence, arXiv:math/0305308 [math.NT], 2003.
Robbert Fokkink and Gandhar Joshi, On Cloitre's hiccup sequences, arXiv:2507.16956 [math.CO], 2025. See p. 3.

Cf. A003151, A080455-A080458, A080036, A080037.

Cf. A081834.

a080590 n = a080590_list !! (n-1)
a080590_list = 1 : f 2 [1] where
   f x zs@(z:_) = y : f (x + 1) (y : zs) where
     y = z + (if x `elem` zs then 3 else 4)
-- Reinhard Zumkeller, Sep 26 2014

a[n_] := a[n] = If[n == 1, 1, If[MemberQ[Array[a, n-1], n], a[n-1]+3, a[n-1]+4]]; Array[a, 56] (* Jean-François Alcover, Oct 05 2018 *)

A080600 a(n) = ceiling(n*(3 + sqrt(13))/2).

Original entry on oeis.org

0, 4, 7, 10, 14, 17, 20, 24, 27, 30, 34, 37, 40, 43, 47, 50, 53, 57, 60, 63, 67, 70, 73, 76, 80, 83, 86, 90, 93, 96, 100, 103, 106, 109, 113, 116, 119, 123, 126, 129, 133, 136, 139, 143, 146, 149, 152, 156, 159, 162, 166, 169, 172, 176, 179, 182, 185, 189

Offset: 0

N. J. A. Sloane, Mar 23 2003

nonn

a(0)=0, a(1)=4; for n > 1, a(n) = a(n-1) + 4 if n is already in the sequence, a(n) = a(n-1) + 3 otherwise.

In the Fokkink-Joshi paper, this sequence is the Cloitre (0,4,4,3)-hiccup sequence. - Michael De Vlieger, Jul 29 2025

Michael De Vlieger, Table of n, a(n) for n = 0..10000
Benoit Cloitre, A study of a family of self-referential sequences, arXiv:2506.18103 [math.GM], 2025. See p. 7.
Benoit Cloitre, N. J. A. Sloane and M. J. Vandermast, Numerical analogues of Aronson's sequence, J. Integer Seqs., Vol. 6 (2003), #03.2.2.
Benoit Cloitre, N. J. A. Sloane and M. J. Vandermast, Numerical analogues of Aronson's sequence, arXiv:math/0305308 [math.NT], 2003.
Robbert Fokkink and Gandhar Joshi, On Cloitre's hiccup sequences, arXiv:2507.16956 [math.CO], 2025. See p. 3.

Equals A080081 + 1 for n > 0. Cf. A080455-A080458, A080036, A080037.

With[{c=(3+Sqrt[13])/2},Table[Ceiling[c*n],{n,0,60}]] (* Harvey P. Dale, Oct 30 2021 *)

A164845 a(n) = (6 + 10n + 5n^2 + n^3)/2.

Original entry on oeis.org

3, 11, 27, 54, 95, 153, 231, 332, 459, 615, 803, 1026, 1287, 1589, 1935, 2328, 2771, 3267, 3819, 4430, 5103, 5841, 6647, 7524, 8475, 9503, 10611, 11802, 13079, 14445, 15903, 17456, 19107, 20859, 22715, 24678, 26751, 28937, 31239, 33660, 36203, 38871

Offset: 0

Paul Curtz, Aug 28 2009

Row sums of the triangle defined by non-interrupted runs in A080036.
If the sequence of integers is split at positions defined by A000124 we obtain A080036. Its runs of consecutive integers can be placed into rows of a triangle:
    3;
    5,  6;
    8,  9, 10;
   12, 13, 14, 15;
   17, 18, 19, 20, 21;
   ...
The a(n) are the row sums of this triangle.
The a(n) are also the binomial transform of the quasi-finite sequence 3, 8, 8, 3, 0 (0 continued).
An associated integer sequence could be defined by a(n)/A026741(n+1) = 3, 11, 9, 27, ...

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A135278.

[3+5*n+5*n^2/2+n^3/2: n in [0..50]]; // Vincenzo Librandi, Aug 07 2011

Table[(6 + 10*n + 5*n^2 + n^3)/2, {n,0,50}] (* or *) LinearRecurrence[{4, -6, 4, -1}, {3, 11, 27, 54}, 50] (* G. C. Greubel, Apr 21 2018 *)

for(n=0, 50, print1((6+10*n+5*n^2+n^3)/2, ", ")) \\ G. C. Greubel, Apr 21 2018

a(n) = A162607(n+3) + n.
First differences: a(n+1) - a(n) = A104249(n+2), i.e., a(n) = a(n-1) + 3*n^2/2 + 7*n/2 +3.
Second differences: a(n+2) - 2*a(n+1) + a(n) = A016789(n+2).
a(n) = 2*a(n-1) - a(n-2) + 3*n + 5, n>1.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + 3, n>2.
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4), n>3.
G.f.: (3-x+x^2)/(x-1)^4.
E.g.f.: (6 + 16*x + 8*x^2 + x^3)*exp(x)/2. - G. C. Greubel, Apr 21 2018

Edited and extended by R. J. Mathar, Aug 31 2009

Corrected typo in recurrence, observed by Paul Curtz - R. J. Mathar, Sep 25 2009

A361684 Ramsey core number rc(n,n).

Original entry on oeis.org

2, 5, 8, 11, 15, 18, 22, 25, 28, 32, 35, 39, 42, 45, 49, 52, 56, 59, 62, 66, 69, 73, 76, 80, 83, 86, 90, 93, 97, 100, 103, 107, 110, 114, 117, 121, 124, 127, 131, 134, 138, 141, 144, 148, 151, 155, 158, 161, 165, 168, 172, 175, 179, 182, 185, 189, 192, 196, 199, 202

Offset: 1

Allan Bickle, Mar 28 2023

The Ramsey core number rc(s,t) is the smallest n such that for all edge 2-colorings of K_n, either the factor induced by the first color contains an s-core or the second factor contains a t-core.  (A k-core is a subgraph with minimum degree at least k.)
The beginning of the square array is:
   2,  3,  4,  5,  6,  7,  8,  9, 10, 11, 12, ...
   3,  5,  6,  8,  9, 10, 12, 13, 14, 15, 17, ...
   4,  6,  8, 10, 11, 13, 14, 15, 17, 18, 19, ...
   5,  8, 10, 11, 13, 15, 16, 18, 19, 20, 22, ...
   6,  9, 11, 13, 15, 16, 18, 20, 21, 23, 24, ...
   7, 10, 13, 15, 16, 18, 20, 21, 23, 25, 26, ...
   8, 12, 14, 16, 18, 20, 22, 23, 25, 26, 28, ...

			For order 5, one of the two factors has at least 5 edges, and so contains a cycle.  For order 4, K_4 decomposes into two paths.  Thus rc(2,2)=5.

R. Klein and J. Schönheim, Decomposition of K_{n} into degenerate graphs, In Combinatorics and Graph Theory Hefei 6-27, April 1992. World Scientific. Singapore, New Jersey, London, Hong Kong, 141-1.

Paolo Xausa, Table of n, a(n) for n = 1..10000
Allan Bickle, The k-Cores of a Graph, Ph.D. Dissertation, Western Michigan University, 2010.
Allan Bickle, Structural results on maximal k-degenerate graphs, Discuss. Math. Graph Theory 32 4 (2012), 659-676.
Allan Bickle, A Survey of Maximal k-degenerate Graphs and k-Trees, Theory and Applications of Graphs 0 1 (2024) Article 5.
Sascha Stoll, On Subgraphs With Minimum Degree Restrictions, Master’s Thesis, Karlsruhe Institute of Technology, 2019.

Cf. A361261 (array of rc(s,t)), A080036 (rc(2,n)).

A361684[n_]:=Ceiling[2n-3/2+Sqrt[2(n-1)^2+9/4]];
Array[A361684,100] (* Paolo Xausa, Dec 01 2023 *)

a(n) = rc(n,n) = ceiling(2*n - 3/2 + sqrt(2*(n-1)^2 + 9/4)).

A361261 Array of Ramsey core number rc(s,t) read by antidiagonals.

Original entry on oeis.org

2, 3, 3, 4, 5, 4, 5, 6, 6, 5, 6, 8, 8, 8, 6, 7, 9, 10, 10, 9, 7, 8, 10, 11, 11, 11, 10, 8, 9, 12, 13, 13, 13, 13, 12, 9, 10, 13, 14, 15, 15, 15, 14, 13, 10, 11, 14, 15, 16, 16, 16, 16, 15, 14, 11, 12, 15, 17, 18, 18, 18, 18, 18, 17, 15, 12, 13, 17, 18, 19, 20, 20, 20, 20, 19, 18, 17, 13

Offset: 1

Allan Bickle, Mar 28 2023

The Ramsey core number rc(s,t) is the smallest n such that for all edge 2-colorings of K_n, either the factor induced by the first color contains an s-core or the second factor contains a t-core. (A k-core is a subgraph with minimum degree at least k.)

			For order 5, one of the two factors has at least 5 edges, and so contains a cycle. For order 4, K_4 decomposes into two paths. Thus rc(2,2) = 5.
The square array begins:
   2,  3,  4,  5,  6,  7,  8,  9, 10, 11, 12, ...
   3,  5,  6,  8,  9, 10, 12, 13, 14, 15, 17, ...
   4,  6,  8, 10, 11, 13, 14, 15, 17, 18, 19, ...
   5,  8, 10, 11, 13, 15, 16, 18, 19, 20, 22, ...
   6,  9, 11, 13, 15, 16, 18, 20, 21, 23, 24, ...
   7, 10, 13, 15, 16, 18, 20, 21, 23, 25, 26, ...
   8, 12, 14, 16, 18, 20, 22, 23, 25, 26, 28, ...

R. Klein and J. Schönheim, Decomposition of K_{n} into degenerate graphs, In Combinatorics and Graph Theory Hefei 6-27, April 1992. World Scientific. Singapore, New Jersey, London, Hong Kong, 141-155.

Allan Bickle, The k-Cores of a Graph, Ph.D. Dissertation, Western Michigan University, 2010.
Allan Bickle, Structural results on maximal k-degenerate graphs, Discuss. Math. Graph Theory 32 4 (2012), 659-676.
Allan Bickle, A Survey of Maximal k-degenerate Graphs and k-Trees, Theory and Applications of Graphs 0 1 (2024) Article 5.
Sascha Stoll, On Subgraphs With Minimum Degree Restrictions, Master's Thesis, Karlsruhe Institute of Technology, 2019.

Cf. A361684 (rc(n,n)), A080036 (rc(2,n)).

rc[s_,t_]:=Ceiling[s+t-3/2+Sqrt[2(s-1)(t-1)+9/4]]; Flatten[Table[rc[s-t+1,t],{s,12},{t,s}]] (* Stefano Spezia, Apr 03 2023 *)

rc(s,t) = ceiling(s + t - 3/2 + sqrt(2*(s-1)*(t-1) + 9/4)).

A080579 a(1)=1; for n>1, a(n)=a(n-1)+1 if n is already in the sequence, a(n)=a(n-1)+4 otherwise.

Original entry on oeis.org

1, 5, 9, 13, 14, 18, 22, 26, 27, 31, 35, 39, 40, 41, 45, 49, 53, 54, 58, 62, 66, 67, 71, 75, 79, 80, 81, 85, 89, 93, 94, 98, 102, 106, 107, 111, 115, 119, 120, 121, 122, 126, 130, 134, 135, 139, 143, 147, 148, 152, 156, 160, 161, 162, 166, 170, 174, 175

Offset: 1

N. J. A. Sloane, Mar 23 2003

nonn

In the Fokkink-Joshi paper, this sequence is the Cloitre (0,1,1,4)-hiccup sequence. - Michael De Vlieger, Jul 29 2025

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Benoit Cloitre, N. J. A. Sloane and M. J. Vandermast, Numerical analogues of Aronson's sequence, J. Integer Seqs., Vol. 6 (2003), #03.2.2.
Benoit Cloitre, N. J. A. Sloane and M. J. Vandermast, Numerical analogues of Aronson's sequence (math.NT/0305308)
Robbert Fokkink and Gandhar Joshi, On Cloitre's hiccup sequences, arXiv:2507.16956 [math.CO], 2025. See pp. 3, 5.

Cf. A080455-A080458, A080036, A080037.

a080579 n = a080579_list !! (n-1)
a080579_list = 1 : f 2 [1] where
   f x zs@(z:_) = y : f (x + 1) (y : zs) where
     y = if x `elem` zs then z + 1 else z + 4
-- Reinhard Zumkeller, Sep 26 2014

a[1] = 1; a[n_] := a[n] = If[MemberQ[Array[a, n-1], n], a[n-1]+1, a[n-1]+4]; Array[a, 60] (* Jean-François Alcover, Oct 08 2018 *)

A080667 a(1)=3; for n>1, a(n)=a(n-1)+4 if n is already in the sequence, a(n)=a(n-1)+3 otherwise.

Original entry on oeis.org

3, 6, 10, 13, 16, 20, 23, 26, 29, 33, 36, 39, 43, 46, 49, 53, 56, 59, 62, 66, 69, 72, 76, 79, 82, 86, 89, 92, 96, 99, 102, 105, 109, 112, 115, 119, 122, 125, 129, 132, 135, 138, 142, 145, 148, 152, 155, 158, 162, 165, 168, 171, 175, 178, 181, 185, 188

Offset: 1

N. J. A. Sloane, Mar 23 2003

nonn

In the Fokkink-Joshi paper, this sequence is the Cloitre (0,3,4,3)-hiccup sequence. - Michael De Vlieger, Jul 29 2025

Benoit Cloitre, N. J. A. Sloane and M. J. Vandermast, Numerical analogues of Aronson's sequence, J. Integer Seqs., Vol. 6 (2003), #03.2.2.
Benoit Cloitre, N. J. A. Sloane and M. J. Vandermast, Numerical analogues of Aronson's sequence, arXiv:math/0305308 [math.NT], 2003.
Robbert Fokkink and Gandhar Joshi, On Cloitre's hiccup sequences, arXiv:2507.16956 [math.CO], 2025. See p. 3.

Cf. A080455, A080456, A080457, A080458, A080036, A080037.

a[1] = 3; a[n_] := a[n] = a[n-1] + If[MemberQ[Array[a, n-1], n], 4, 3];
Array[a, 60] (* Jean-François Alcover, Nov 25 2018 *)

a(n) = floor(n*r + 1/(1+r)) where r = (3 + sqrt(13))/2.

cp's OEIS Frontend

A080580 a(1)=1; for n>1, a(n)=a(n-1)+2 if n is already in the sequence, a(n)=a(n-1)+4 otherwise.

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A363757 Lexicographically earliest sequence of positive integers such that the n-th pair of consecutive equal values are separated by a(n) distinct terms, with pairs numbered according to the position of the second term in the pair.

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A080353 a(1)=5; for n>1, a(n)=a(n-1)+1 if n is already in the sequence, a(n)=a(n-1)+2 otherwise.

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A080590 a(1)=1; for n>1, a(n)=a(n-1)+3 if n is already in the sequence, a(n)=a(n-1)+4 otherwise.

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A080600 a(n) = ceiling(n*(3 + sqrt(13))/2).

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A164845 a(n) = (6 + 10*n + 5*n^2 + n^3)/2.

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A361684 Ramsey core number rc(n,n).

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A361261 Array of Ramsey core number rc(s,t) read by antidiagonals.

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A164845 a(n) = (6 + 10n + 5n^2 + n^3)/2.