A026272 -id:A026272 - cp's OEIS frontend

A026274 Greatest k such that s(k) = n, where s = A026272.

3, 5, 8, 11, 13, 16, 18, 21, 24, 26, 29, 32, 34, 37, 39, 42, 45, 47, 50, 52, 55, 58, 60, 63, 66, 68, 71, 73, 76, 79, 81, 84, 87, 89, 92, 94, 97, 100, 102, 105, 107, 110, 113, 115, 118, 121, 123, 126, 128, 131, 134, 136, 139, 141, 144

Offset: 1

Clark Kimberling

This is the upper s-Wythoff sequence, where s(n)=n+1.
See comments at A026273.
Conjecture: This sequence consists precisely of those numbers without a 1 or 2 in their Zeckendorf representation. [In other words, numbers which are the sum of distinct nonconsecutive Fibonacci numbers greater than 2.] - Charles R Greathouse IV, Jan 28 2015
A Beatty sequence with complement A026273. - Robert G. Wilson v, Jan 30 2015
A035612(a(n)+1) = 1. - Reinhard Zumkeller, Jul 20 2015
From Michel Dekking, Mar 12 2018: (Start)
One has r*r*(n-2*r+3) = n*r^2 -2r^3+3*r^2 = (n+1)*r^2 -2, where r = (1+sqrt(5))/2.
So a(n) = floor((n+1)*r^2)-2, and we see that this sequence is simply the Beatty sequence of the square of the golden ratio, shifted spatially and temporally. In other words, if w = A001950 = 2,5,7,10,13,15,18,20,... is the upper Wythoff sequence, then a(n) = w(n+1) - 2.
(End)
From Michel Dekking, Apr 05 2020: (Start)
Proof of the conjecture by Charles R Greathouse IV.
Let Z(n) = d(L)...d(1)d(0) be the Zeckendorf expansion of n. Well-known is:
      d(0) = 1 if and only if n = floor(k*r^2) - 1
for some integer k (see A003622).
Then the same characterization holds for n with d(1)d(0) = 01, since 11 does not appear in a Zeckendorf expansion. But such an n has predecessor n-1 which always has an expansion with d(1)d(0) = 00. Combined with my comment from March 2018, this proves the conjecture (ignoring n = 0). (End)
It appears that these are the integers m for which A007895(m+1) > A007895(m) where A007895(m) is the number of terms in Zeckendorf representation of m. - Michel Marcus, Oct 30 2020
This follows directly from Theorem 4 in my paper "Points of increase of the sum of digits function of the base phi expansion". - Michel Dekking, Oct 31 2020

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Hung Viet Chu, Difference in the Number of Summands in the Zeckendorf Partitions of Consecutive Integers, arXiv:2010.15592 [math.NT], 2020.
Michel Dekking, Points of increase of the sum of digits function of the base phi expansion, arXiv:2003.14125 [math.CO], 2020.
F. Michel Dekking, The sum of digits functions of the Zeckendorf and the base phi expansions, Theoretical Computer Science (2021) Vol. 859, 70-79.

Cf. A184117, A026273, A001950, A003622.

a026274 n = a026274_list !! (n-1)
a026274_list = map (subtract 1) $ tail $ filter ((== 1) . a035612) [1..]
-- Reinhard Zumkeller, Jul 20 2015

r=(1+Sqrt[5])/2;
a[n_]:=Floor[r*r*(n+2r-3)];
Table[a[n],{n,200}]
Table[Floor[GoldenRatio^2 (n+2*GoldenRatio-3)],{n,60}] (* Harvey P. Dale, Dec 23 2022 *)

a(n)=my(w=quadgen(20),phi=(1+w)/2); phi^2*(n+2*phi-3)\1 \\ Charles R Greathouse IV, Nov 10 2021

from math import isqrt
def A026274(n): return (n+1+isqrt(5*(n+1)**2)>>1)+n-1 # Chai Wah Wu, Aug 17 2022

a(n) = floor(r*r*(n+2r-3)), where r = (1+sqrt(5))/2 = A001622. [Corrected by Tom Edgar, Jan 30 2015]
a(n) = 3*n - floor[(n+1)/(1+phi)], phi = (1+sqrt(5))/2. - Joshua Tobin (tobinrj(AT)tcd.ie), May 31 2008
a(n) = A003622(n+1) - 1 for n>1 (conjectured). - Michel Marcus, Oct 30 2020
This conjectured formula follows directly from the formula a(n) = floor((n+1)*r^2)-2 in my Mar 12 2018 comment above. - Michel Dekking, Oct 31 2020

Extended by Clark Kimberling, Jan 14 2011

A026273 a(n) = least k such that s(k) = n, where s = A026272.

Original entry on oeis.org

1, 2, 4, 6, 7, 9, 10, 12, 14, 15, 17, 19, 20, 22, 23, 25, 27, 28, 30, 31, 33, 35, 36, 38, 40, 41, 43, 44, 46, 48, 49, 51, 53, 54, 56, 57, 59, 61, 62, 64, 65, 67, 69, 70, 72, 74, 75, 77, 78, 80, 82, 83, 85, 86, 88, 90, 91, 93, 95, 96, 98, 99

Offset: 1

Clark Kimberling

nonn

This is the lower s-Wythoff sequence, where s(n)=n+1.
See A184117 for the definition of lower and upper s-Wythoff sequences.  The first few terms of a and its complement, b=A026274, are obtained generated as follows:
s=(2,3,4,5,6,...);
a=(1,2,4,6,7,...)=A026273;
b=(3,5,8,11,13,...)=A026274.
Briefly:  b=s+a, and a=mex="least missing".
From Michel Dekking, Mar 12 2018: (Start)
One has r*(n-2*r+3) = n*r-2r^2+3*r = (n+1)*r-2.
So  a(n) = (n+1)*r-2, and we see that this sequence is simply the Beatty sequence of the golden ratio, shifted spatially and temporally. In other words: if w = A000201 = 1,3,4,6,8,9,11,12,14,...  is the lower Wythoff sequence, then a(n) = w(n+2) - 2.
(N.B. As so often, there is the 'offset 0 vs 1 argument', w = A000201 has offset 1; it would have been better to give (a(n)) offset 1, too).
This observation also gives an answer to Lenormand's question, and a simple proof of Mathar's conjecture in A059426.
(End)

Jon Asier Bárcena-Petisco, Luis Martínez, María Merino, Juan Manuel Montoya, and Antonio Vera-López, Fibonacci-like partitions and their associated piecewise-defined permutations, arXiv:2503.19696 [math.CO], 2025. See p. 3.

Cf. A184117, A026274.

r=(1+Sqrt[5])/2;
a[n_]:=Floor[r*(n-2r+3)];
b[n_]:=Floor[r*r*(n+2r-3)];
Table[a[n],{n,200}]   (* A026273 *)
Table[b[n],{n,200}]   (* A026274 *)

a(n) = floor[r*(n-2*r+3)], where r=golden ratio.

b(n) = floor[(r^2)*(n+2*r-3)] = floor(n*A104457-A134972+1).

Extended by Clark Kimberling, Jan 14 2011

A026275 Sum of numbers between the two n's in A026272.

Original entry on oeis.org

2, 4, 11, 21, 28, 43, 53, 73, 96, 111, 139, 170, 190, 226, 249, 290, 334, 362, 411, 442, 496, 553, 589, 651, 716, 757, 827, 871, 946, 1024, 1073, 1156, 1242, 1296, 1387, 1444, 1540, 1639, 1701, 1805, 1870, 1979, 2091, 2161, 2278

Offset: 1

Clark Kimberling

nonn

A014552 Number of solutions to Langford (or Langford-Skolem) problem (up to reversal of the order).

Original entry on oeis.org

0, 0, 1, 1, 0, 0, 26, 150, 0, 0, 17792, 108144, 0, 0, 39809640, 326721800, 0, 0, 256814891280, 2636337861200, 0, 0, 3799455942515488, 46845158056515936, 0, 0, 111683611098764903232, 1607383260609382393152, 0, 0

Offset: 1

John E. Miller (john@timehaven.us), Eric W. Weisstein, N. J. A. Sloane

These are also called Langford pairings.
2*a(n) = A176127(n) gives the number of ways of arranging the numbers 1,1,2,2,...,n,n so that there is one number between the two 1's, two numbers between the two 2's, ..., n numbers between the two n's.
a(n) > 0 iff n == 0 or 3 (mod 4).

			Solutions for n=3 and 4: 312132 and 41312432.
Solution for n=16: 16, 14, 12, 10, 13, 5, 6, 4, 15, 11, 9, 5, 4, 6, 10, 12, 14, 16, 13, 8, 9, 11, 7, 1, 15, 1, 2, 3, 8, 2, 7, 3.

Jaromir Abrham, "Exponential lower bounds for the numbers of Skolem and extremal Langford sequences," Ars Combinatoria 22 (1986), 187-198.
M. Gardner, Mathematical Magic Show, New York: Vintage, pp. 70 and 77-78, 1978.
M. Gardner, Mathematical Magic Show, Revised edition published by Math. Assoc. Amer. in 1989. Contains a postscript on pp. 283-284 devoted to a discussion of early computations of the number of Langford sequences.
R. K. Guy, The unity of combinatorics, Proc. 25th Iranian Math. Conf, Tehran, (1994), Math. Appl 329 129-159, Kluwer Dordrecht 1995, Math. Rev. 96k:05001.
D. E. Knuth, The Art of Computer Programming, Vol. 4A, Section 7.1.1, p. 2.
M. Krajecki, Christophe Jaillet and Alain Bui, "Parallel tree search for combinatorial problems: A comparative study between OpenMP and MPI," Studia Informatica Universalis 4 (2005), 151-190.
Roselle, David P. Distributions of integers into s-tuples with given differences. Proceedings of the Manitoba Conference on Numerical Mathematics (Univ. Manitoba, Winnipeg, Man., 1971), pp. 31--42. Dept. Comput. Sci., Univ. Manitoba, Winnipeg, Man., 1971. MR0335429 (49 #211). - From N. J. A. Sloane, Jun 05 2012

Ali Assarpour, Amotz Bar-Noy, and Ou Liuo, Counting the Number of Langford Skolem Pairings, arXiv:1507.00315 [cs.DM], 2015.
Gheorghe Coserea, Solutions for n=7.
Gheorghe Coserea, Solutions for n=8.
Gheorghe Coserea, MiniZinc model for generating solutions.
R. O. Davies, On Langford's problem II, Math. Gaz., 1959, vol. 43, 253-255.
Elin Farnell, Puzzle Pedagogy: A Use of Riddles in Mathematics Education, PRIMUS, July 2016, pp. 202-211.
M. Krajecki, L(2,23)=3,799,455,942,515,488.
C. D. Langford, 2781. Parallelograms with Integral Sides and Diagonals, Math. Gaz., 1958, vol. 42, p. 228.
J. E. Miller, Langford's Problem
G. Nordh, Perfect Skolem sequences, arXiv:math/0506155 [math.CO], 2005.
Zan Pan, Conjectures on the number of Langford sequences, (2021).
Michael Penn, Why is this list "nice"? -- Langford's Problem, YouTube video, 2022.
C. J. Priday, On Langford's Problem I, Math. Gaz., 1959, vol. 43, 250-255.
W. Schneider, Langford's Problem
T. Skolem, On certain distributions of integers in pairs with given differences, Math. Scand., 1957, vol. 5, 57-68.
T. Saito and S. Hayasaka, Langford sequences: a progress report, Math. Gaz., 1979, vol. 63, #426, 261-262.
J. E. Simpson, Langford Sequences: perfect and hooked, Discrete Math., 1983, vol. 44, #1, 97-104.
Eric Weisstein's World of Mathematics, Langford's Problem.

See A050998 for further examples of solutions.
If the zeros are omitted we get A192289.
Cf. A059106, A059107, A059108, A125762, A026272.

a(n) = A176127(n)/2.

a(20) from Ron van Bruchem and Mike Godfrey, Feb 18 2002
a(21)-a(23) sent by John E. Miller (john@timehaven.us) and Pab Ter (pabrlos(AT)yahoo.com), May 26 2004. These values were found by a team at Université de Reims Champagne-Ardenne, headed by Michael Krajecki, using over 50 processors for 4 days.
a(24)=46845158056515936 was computed circa Apr 15 2005 by the Krajecki team. - Don Knuth, Feb 03 2007
Edited by Max Alekseyev, May 31 2011
a(27) from the J. E. Miller web page "Langford's problem"; thanks to Eric Desbiaux for reporting this. - N. J. A. Sloane, May 18 2015. However, it appears that the value was wrong. - N. J. A. Sloane, Feb 22 2016
Corrected and extended using results from the Assarpour et al. (2015) paper by N. J. A. Sloane, Feb 22 2016 at the suggestion of William Rex Marshall.

A026242 a(n) = j if n is L(j), else a(n) = k if n is U(k), where L = A000201, U = A001950 (lower and upper Wythoff sequences).

Original entry on oeis.org

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

Offset: 1

Clark Kimberling

Every positive integer occurs exactly twice. a(n) is the parent of n in the tree at A074049. - Clark Kimberling, Dec 24 2010
Apparently, if n=F(m) (a Fibonacci number), one of two circumstances arise:
I. a(n)=F(m-1) and a(n-1)=F(m-2). When this happens, a(n) occurs for the first time and a(n-1) occurs for the second time;
II. a(n)=F(m-2) and a(n-1)=F(m-1). When this happens, a(n) occurs for the second time and a(n-1) occurs for the first time. - Bob Selcoe, Sep 18 2014
These are the numerators when all fractions, j/r and k/r^2, are arranged in increasing order (where r = golden ratio and j,k are positive integers). - Clark Kimberling, Mar 02 2015

Robert G. Wilson v, Table of n, a(n) for n = 1..10000 (first 999 terms from M. F. Hasler)
S. Mneimneh, Fibonacci in The Curriculum: Not Just a Bad Recurrence, in Proceeding SIGCSE '15 Proceedings of the 46th ACM Technical Symposium on Computer Science Education, Pages 253-258. See Figure 2.
Jeffrey Shallit, Fibonacci automaton for a(n)

Cf. A026272, A074049, A089910.

Cf. A000045 (Fibonacci numbers).

mx = 100; gr = GoldenRatio; LW[n_] := Floor[n*gr]; UW[n_] := Floor[n*gr^2]; alw = Array[LW, Ceiling[mx/gr]]; auw = Array[UW, Ceiling[mx/gr^2]]; f[n_] := If[ MemberQ[alw, n], Position[alw, n][[1, 1]], Position[auw, n][[1, 1]]]; Array[f, mx] (* Robert G. Wilson v, Sep 17 2014 *)

my(A=vector(10^4),i,j=0); while(#A>=i=A000201(j++), A[i]=j; (i=A001950(j))>#A || A[i]=j); A026242=A \\ M. F. Hasler, Sep 16 2014 and Sep 18 2014

A026242=vector(#A002251,n,abs(A002251[n]-n)) \\ M. F. Hasler, Sep 17 2014

a(n) = a(m) if a(m) has already occurred exactly once and n = a(m) + m; otherwise, a(n) = least positive integer that has not yet occurred.
a(n) = abs(A002251(n) - n).
n = a(n) + a(n-1) unless n = A089910(m); if n = A089910(m), then n = a(n) + a(n-1) - m. - Bob Selcoe, Sep 20 2014
There is a 17-state automaton that accepts the Zeckendorf (Fibonacci) representation of n and a(n), in parallel.  See the file a026242.pdf. - Jeffrey Shallit, Dec 21 2023

A337226 Lexicographically earliest sequence of positive integers with the property that, for all k > 0, there is at most one j such that a(j) = a(j+k).

Original entry on oeis.org

1, 1, 2, 1, 3, 4, 2, 5, 1, 6, 3, 7, 8, 9, 4, 10, 2, 11, 5, 12, 1, 13, 6, 14, 15, 3, 16, 7, 17, 18, 8, 19, 20, 21, 22, 9, 23, 4, 24, 10, 25, 2, 26, 11, 27, 5, 28, 12, 29, 1, 30, 13, 31, 6, 32, 33, 14, 34, 15, 35, 36, 3, 37, 16, 38, 39, 40, 7, 41, 42, 17, 43, 18, 44, 45, 8

Offset: 1

Samuel B. Reid, Aug 19 2020

The sequence initially appears to be trivially fractal in that the removal of the first occurrence of each value seems to yield the original sequence. This pattern continues until a(121) where, if the sequence were fractal in this way, the value would be 72 or 1. The actual value is 13, so the pattern is broken.

Conjecture: For all k > 0, there is exactly one j such that a(j) = a(j+k). For 0 < k < 11911, this conjecture holds.

			  1 1 2 1 3 4 2
   (1)1 2 1 3 4   k = 1
      1(1)2 1 3   k = 2
       (1)1 2 1   k = 3
          1 1(2)  k = 4
            1 1   k = 5
              1   k = 6
Coincidences are circled. There can only be one coincidence per row.
a(3) cannot be 1 because that would result in two coincidences for k = 1.
a(5) cannot be 1 or 2 because those values would result in two coincidences for k = 1 and k = 2, respectively.
a(7) cannot be 1, 3, or 4 because those values would result in two coincidences for k = 3, k = 2, and k = 1, respectively. It can, however, be 2 because this results in no double coincidences.

Samuel B. Reid, Table of n, a(n) for n = 1..10000
Samuel B. Reid, Python program for A337226

Cf. A003602, A014552, A026272.

Python
```
# See Links section.
```

A363654 Lexicographically earliest sequence of positive integers such that the n-th pair of identical terms encloses exactly a(n) terms.

Original entry on oeis.org

1, 2, 1, 3, 2, 3, 1, 2, 4, 3, 4, 5, 3, 6, 7, 4, 3, 8, 6, 9, 7, 8, 4, 10, 3, 7, 4, 6, 8, 9, 11, 10, 12, 3, 9, 13, 7, 3, 11, 9, 14, 15, 13, 16, 17, 7, 18, 3, 19, 20, 11, 14, 9, 20, 17, 15, 19, 20, 18, 21, 11, 22, 3, 23, 24, 25, 26, 14, 17, 9, 27, 28, 29, 15, 26, 19

Offset: 1

Eric Angelini and Gavin Lupo, Jun 13 2023

nonn

Pairs are numbered according to the position of the second term.

			a(1) = 1. The 1st constructed pair encloses 1 term:  [1, 2, 1].
a(2) = 2. The 2nd constructed pair encloses 2 terms: [2, 1, 3, 2].
a(3) = 1. The 3rd constructed pair encloses 1 term:  [3, 2, 3].
a(4) = 3. The 4th constructed pair encloses 3 terms: [1, 3, 2, 3, 1].
a(5) = 2. The 5th constructed pair encloses 2 terms: [2, 3, 1, 2].
a(6) = 3. The 6th constructed pair encloses 3 terms: [3, 1, 2, 4, 3].
a(7) = 1. The 7th constructed pair encloses 1 term:  [4, 3, 4].
...

Neal Gersh Tolunsky, Table of n, a(n) for n = 1..10000
Eric Angelini, A seq with a strange behavior (or not)?
Neal Gersh Tolunsky, Scatterplot of first 100000 terms

Cf. A026272, A363708, A363757.

from itertools import count, islice
def agen(): # generator of terms
    a, indexes = [1], {1: 0}
    yield a[-1]
    for i in count(0):
        num = 1
        while True:
            if num in indexes:
                if (len(a) - indexes[num]) == (a[i]+1):
                    an = num; indexes[an] = len(a); a.append(an); yield an
                    break
                else:
                    num += 1
            else:
                an = max(a)+1; indexes[an] = len(a); a.append(an); yield an
                num = 1
print(list(islice(agen(), 100))) # Gavin Lupo and Michael S. Branicky, Jun 13 2023

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.

A026350 a(n) = a(m) if a(m) has already occurred exactly once and n = a(m)+m-1, else a(n) = least positive integer that has not yet occurred.

Original entry on oeis.org

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

Offset: 1

Clark Kimberling

nonn

Cf. A026272.

A026350=vector(#A026242+1,n,if(n>1,A026242[n-1])+1) \\ Assumes the vector A026242 is defined, cf. there for PARI code. - M. F. Hasler, Sep 17 2014

a(n+1)=A026242(n)+1, for all n>0. - M. F. Hasler, Sep 17 2014

A026354 a(n) = a(m) if a(m) has already occurred exactly once and n = a(m)+m-2, else a(n) = least positive integer that has not yet occurred.

Original entry on oeis.org

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

Offset: 1

Clark Kimberling

nonn

Jeffrey Shallit, Automaton for a(n)

Cf. A026272.

A026354=vector(2+#A026242,n,if(n>3,A026242[n-2]+2,n)) \\ Assumes the vector A026242 is defined, see there for PARI code. - M. F. Hasler, Sep 17 2014

a(n)=n for n<4, a(n)=A026242(n-2)+2 for all n>2, a(n)=A026272(n-4)+3 for n>4. - M. F. Hasler, Sep 17 2014

There is a 35-state automaton that accepts, in parallel, the Fibonacci (Zeckendorf) representation of n and a(n). See the file a026354.pdf. - Jeffrey Shallit, Dec 21 2023

cp's OEIS Frontend

A026274 Greatest k such that s(k) = n, where s = A026272.

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A026273 a(n) = least k such that s(k) = n, where s = A026272.

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A026275 Sum of numbers between the two n's in A026272.

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A014552 Number of solutions to Langford (or Langford-Skolem) problem (up to reversal of the order).

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A026242 a(n) = j if n is L(j), else a(n) = k if n is U(k), where L = A000201, U = A001950 (lower and upper Wythoff sequences).

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A337226 Lexicographically earliest sequence of positive integers with the property that, for all k > 0, there is at most one j such that a(j) = a(j+k).

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A363654 Lexicographically earliest sequence of positive integers such that the n-th pair of identical terms encloses exactly a(n) terms.

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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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