A014601 -id:A014601 - cp's OEIS frontend

A050998 Inequivalent solutions to Langford (or Langford-Skolem) problem of arranging the numbers 1,1,2,2,3,3,...,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, listed by length and lexicographic order.

231213, 23421314, 14156742352637, 14167345236275, 15146735423627, 15163745326427, 15167245236473, 15173465324726, 16135743625427, 16172452634753, 17125623475364, 17126425374635, 23627345161475

Offset: 1

Eric W. Weisstein

Entries are indexed by numbers n == -1 or 0 mod 4 (A014601).
More precisely, for each given n = (3, 4, 7, 8, ...) in A014601, all of the A014552(n) inequivalent solutions are listed in lexicographic order. For example,  a(1), a(2) and a(3) correspond to n=3, 4 and 7, but a(4) is not the first solution for n=8 but the second solution for n=7. - M. F. Hasler, Nov 12 2015
"Inequivalent" means that for two solutions related by symmetry (reading the digits backwards), only the (lexicographic) smaller one is listed. - M. F. Hasler, Nov 15 2015
It is unclear how the sequence goes on after the first 1+1+26+150 terms, with the solutions for n >= 11. Will a solution s=(s[1],...,s[n]) be coded again by Sum_{i=1..n} s[i]*b^(n-i) in base b=10, or in some larger base b >= n+1? Maybe using as many decimal digits as needed, i.e., b=100 for 11 <= n <= 99? - M. F. Hasler, Nov 16 2015

			The first n which allows a solution (A014552(n) > 0; n in A014601) is n=3, the solutions are a(1) = 231213 and the same read backwards, 312132.
The next solutions are given for n=4, again there is only A014552(4)=1 solution a(2) = 23421314 up to reversal (41312432, not listed).
Then follow the A014552(7)=26 (inequivalent) solutions for n=7, viz. a(3)-a(28).

M. Gardner, Mathematical Magic Show, New York: Vintage, pp. 70 and 77-78, 1978.

Seiichi Manyama, Table of n, a(n) for n = 1..178
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.
C. D. Langford, Problem, Math. Gaz., 1958, vol. 42, p. 228.
J. Miller, Langford's problem.
Eric Weisstein's World of Mathematics, Langford's Problem.

See A014552 (the main entry for this problem) for number of solutions.

Definition clarified by M. F. Hasler, Nov 15 2015

A193867 Odd central polygonal numbers.

Original entry on oeis.org

1, 7, 11, 29, 37, 67, 79, 121, 137, 191, 211, 277, 301, 379, 407, 497, 529, 631, 667, 781, 821, 947, 991, 1129, 1177, 1327, 1379, 1541, 1597, 1771, 1831, 2017, 2081, 2279, 2347, 2557, 2629, 2851, 2927, 3161, 3241, 3487, 3571, 3829, 3917, 4187, 4279

Offset: 1

Omar E. Pol, Aug 15 2011

Even triangular numbers plus 1.

Union of A188135 and A185438 without repetitions (A188135 is a bisection of this sequence. Another bisection is A185438 but without its initial term).

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Cf. A000124, A014494, A014601, A185438, A188135, A193868.

Select[Accumulate[Range[0,100]],EvenQ]+1 (* or *) LinearRecurrence[{1,2,-2,-1,1},{1,7,11,29,37},50] (* Harvey P. Dale, Nov 29 2014 *)

Vec(-x*(x^2+1)*(x^2+6*x+1) / ((1+x)^2*(x-1)^3) + O(x^100)) \\ Colin Barker, Jan 27 2016

a(n) = A000124(A014601(n-1)).
a(n) = 1 + A014494(n-1).
G.f.: -x*(x^2+1)*(x^2+6*x+1) / ( (1+x)^2*(x-1)^3 ). - R. J. Mathar, Aug 25 2011
From Colin Barker, Jan 27 2016: (Start)
a(n) = (4*n^2+2*(-1)^n*n-4*n-(-1)^n+3)/2.
a(n) = 2*n^2-n+1 for n even.
a(n) = 2*n^2-3*n+2 for n odd. (End)
Sum_{n>=1} 1/a(n) = 2*Pi*sinh(sqrt(7)*Pi/4)/(sqrt(7)*(2*cosh(sqrt(7)*Pi/4) - sqrt(2))). - Amiram Eldar, May 11 2025

A274757 Numbers k such that 6*k+1 is a triangular number (A000217).

Original entry on oeis.org

0, 9, 15, 42, 54, 99, 117, 180, 204, 285, 315, 414, 450, 567, 609, 744, 792, 945, 999, 1170, 1230, 1419, 1485, 1692, 1764, 1989, 2067, 2310, 2394, 2655, 2745, 3024, 3120, 3417, 3519, 3834, 3942, 4275, 4389, 4740, 4860, 5229, 5355, 5742, 5874, 6279, 6417

Offset: 1

Colin Barker, Jul 04 2016

Numbers of the type floor(3*m*(m+1)/4) for which floor(3*m*(m+1)/4) = 3*floor(m*(m+1)/4). A014601 lists the values of m. - Bruno Berselli, Jan 13 2017

Numbers of the form 3*k*(4*k + 1) for k in Z. - Peter Bala, Nov 21 2024

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Cf. A000217, A014601.
Cf. A000096 (k+1), A074377 (2*k+1), A045943 (3*k+1), A274681 (4*k+1), A085787 (5*k+1).
Cf. similar sequences listed in A274830.

Table[3 (2 n - 1) (2 n + (-1)^n - 1)/4, {n, 1, 60}] (* Bruno Berselli, Jul 08 2016 *)
LinearRecurrence[{1,2,-2,-1,1},{0,9,15,42,54},50] (* Harvey P. Dale, Apr 13 2025 *)

PARI
```
isok(n) = ispolygonal(6*n+1, 3)
```

select(n->ispolygonal(6*n+1, 3), vector(7000, n, n-1))

concat(0, Vec(3*x^2*(3+2*x+3*x^2)/((1-x)^3*(1+x)^2) + O(x^60)))

G.f.: 3*x^2*(3 + 2*x + 3*x^2) / ((1 - x)^3*(1 + x)^2).
a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5) for n>5.
a(n) = 3*(2*n - 1)*(2*n + (-1)^n - 1)/4. Therefore:
a(n) = 3*n*(2*n - 1)/2 for n even,
a(n) = 3*(n-1)*(2*n - 1)/2 for n odd.

A281333 a(n) = 1 + floor(n/2) + floor(n^2/3).

Original entry on oeis.org

1, 1, 3, 5, 8, 11, 16, 20, 26, 32, 39, 46, 55, 63, 73, 83, 94, 105, 118, 130, 144, 158, 173, 188, 205, 221, 239, 257, 276, 295, 316, 336, 358, 380, 403, 426, 451, 475, 501, 527, 554, 581, 610, 638, 668, 698, 729, 760, 793, 825, 859, 893, 928, 963, 1000, 1036, 1074, 1112, 1151, 1190

Offset: 0

Bruno Berselli, Jan 20 2017

Bruno Berselli, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,1,0,-1,-1,1).

Subsequences: A033577, A244805 (numbers of the form 1 + k/2 + k^2/3), A212978 (second bisection).
Cf. A236771: n + floor(n/2) + floor(n^2/3).
Cf. A008619: 1 + floor(n/2); A087483: 1 + floor(n^2/3).
Cf. A000212, A004526.

[1 + n div 2 + n^2 div 3: n in [0..60]];

A281333:=n->1 + floor(n/2) + floor(n^2/3): seq(A281333(n), n=0..100); # Wesley Ivan Hurt, Feb 09 2017

Table[1 + Floor[n/2] + Floor[n^2/3], {n, 0, 60}]
LinearRecurrence[{1,1,0,-1,-1,1},{1,1,3,5,8,11},80] (* Harvey P. Dale, Sep 29 2024 *)

makelist(1+floor(n/2)+floor(n^2/3), n, 0, 60);

vector(60, n, n--; 1+floor(n/2)+floor(n^2/3))

[1+int(n/2)+int(n**2/3) for n in range(60)]

[1+floor(n/2)+floor(n^2/3) for n in range(60)]

G.f.: (1 + x^2 + x^3 + x^4)/((1 + x)*(1 + x + x^2)*(1 - x)^3).
a(n) = a(n-1) + a(n-2) - a(n-4) - a(n-5) + a(n-6).
a(n) = 1 + floor(n/2 + n^2/3).
a(n) = (12*n^2 + 18*n + 4*(-1)^(2*n/3) + 4*(-1)^(-2*n/3) + 9*(-1)^n + 19)/36.
a(n) - n = a(-n).
a(6*k+r) = 12*k^2 + (4*r+3)*k + a(r), where 0 <= r <= 5. Particular cases:
a(6*k) = A244805(k+1), a(6*k+1) = A033577(k).
a(n+2) -  a(n) =   A004773(n+2).
a(n+3) -  a(n) =   A014601(n+2).
a(n+4) -  a(n) =   A047480(n+3).
a(n) - a(-n+3) = 2*A001651(n-1).
a(n) + a(-n+3) = 2*A097922(n-1).
a(n) = 1 + A004526(n) + A000212(n) = A008619(n) + A000212(n). - Omar E. Pol, Dec 23 2020

A283394 a(n) = 3n(3*n + 7)/2 + 4.

Original entry on oeis.org

4, 19, 43, 76, 118, 169, 229, 298, 376, 463, 559, 664, 778, 901, 1033, 1174, 1324, 1483, 1651, 1828, 2014, 2209, 2413, 2626, 2848, 3079, 3319, 3568, 3826, 4093, 4369, 4654, 4948, 5251, 5563, 5884, 6214, 6553, 6901, 7258, 7624, 7999, 8383, 8776, 9178, 9589, 10009

Offset: 0

Bruno Berselli, Mar 23 2017

Sum_{k = 0..n} (3*k + r)^3 is divisible by 3*n*(3*n + 2*r + 3)/2 + r^2: the sequence corresponds to the case r = 2 of this formula (other cases are listed in Crossrefs section).
Also, Sum_{k = 0..n} (3*k + 2)^3 / a(n) gives 2, 7, 15, 26, 40, 57, 77, 100, 126, 155, 187, 222, ... (A005449).
a(n) is even if n belongs to A014601. No term is divisible by 3, 5, 7 and 11.

Bruno Berselli, Table of n, a(n) for n = 0..1000
Nickolas Arustamyan, Christopher Cox, Erik Lundberg, Sean Perry, and Zvi Rosen, On the Number of Equilibria Balancing Newtonian Point Masses with a Central Force, arXiv:2106.11416 [math-ph], 2021.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A005449, A034856, A095794, A101881, A165157, A186349.

Sequences with formula 3*n*(3*n + 2*r + 3)/2 + r^2: A038764 (r=-1), A027468 (r=0), A081271 (r=1), this sequence (r=2), A027468 (r=3; offset: -1), A080855 (r=4; offset: -2).

Magma
```
[3*n*(3*n+7)/2+4: n in [0..50]];
```

Table[3 n (3 n + 7)/2 + 4, {n, 0, 50}]
LinearRecurrence[{3,-3,1},{4,19,43},50] (* Harvey P. Dale, Mar 02 2019 *)

Maxima
```
makelist(3*n*(3*n+7)/2+4, n, 0, 50);
```

a(n) = 3*n*(3*n + 7)/2 + 4; \\ Indranil Ghosh, Mar 24 2017

Python
```
[3*n*(3*n+7)/2+4 for n in range(50)]
```
Sage
```
[3*n*(3*n+7)/2+4 for n in range(50)]
```

O.g.f.: (4 + 7*x - 2*x^2)/(1 - x)^3.
E.g.f.: (8 + 30*x + 9*x^2)*exp(x)/2.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
a(n) = A081271(-n-2).
a(n) = 3*A095794(n+1) + 1.
a(n) = A034856(3*n+2) = A101881(6*n+2) = A165157(6*n+3) = A186349(6*n+3).
The inverse binomial transform yields 4, 15, 9, 0 (0 continued), therefore:
a(n) = 4*binomial(n,0) + 15*binomial(n,1) + 9*binomial(n,2).

A365968 Irregular triangle read by rows: T(n,k) (0 <= n, 0 <= k < 2^n). An infinite binary tree with root node 0 in row n = 0. Each node then has left child (2j) - k - 1 and right child (2j) - k + 1, where j and k are the values of the parent and grandparent nodes respectively.

Original entry on oeis.org

0, -1, 1, -3, -1, 1, 3, -6, -4, -2, 0, 0, 2, 4, 6, -10, -8, -6, -4, -4, -2, 0, 2, -2, 0, 2, 4, 4, 6, 8, 10, -15, -13, -11, -9, -9, -7, -5, -3, -7, -5, -3, -1, -1, 1, 3, 5, -5, -3, -1, 1, 1, 3, 5, 7, 3, 5, 7, 9, 9, 11, 13, 15, -21, -19, -17, -15, -15, -13, -11, -9

Offset: 0

John Tyler Rascoe, Sep 23 2023

For n in A014601 row n will contain all even numbers from 0 to A000217(n).

For n in A042963 row n will contain all odd numbers from 1 to A000217(n).

			Triangle begins:
        k=0   1   2   3   4   5   6   7   8   9  10  11  12  13  14  15
  n=0:    0;
  n=1:   -1,  1;
  n=2:   -3, -1,  1,  3;
  n=3:   -6, -4, -2,  0,  0,  2,  4,  6;
  n=4:  -10, -8, -6, -4, -4, -2,  0,  2, -2,  0,  2,  4,  4,  6,  8, 10;
  ...
The binary tree starts with root 0 in row n = 0. For rows n < 2, k = 0.
In row n = 3, the parent node -3 has left child -6 = 2*(-3) - (-1) - 1.
The tree begins:
row
[n]
[0]                   ______0______
                     /             \
[1]              __-1__           __1__
                /      \         /     \
[2]           -3       -1       1       3
              / \      / \     / \     / \
[3]         -6  -4   -2   0   0   2   4   6
.

John Tyler Rascoe, Rows n = 0..12, flattened

Cf. A004718, A274575, A360173, A363718.

T(n,k) = sum(i=0,n-1, if(bittest(k,i), i+1, -(i+1))); \\ Kevin Ryde, Nov 14 2023

def A365968(n, k):
    b, x = bin(k)[2:].zfill(n), 0
    for i in range(0, n):
        x += (-1)**(int(b[n-(i+1)])+1)*(i+1)
    return(x) # John Tyler Rascoe, Nov 12 2023

T(n,k) = - Sum_{i=0..n-1} (i+1)*(-1)^b[i] where the binary expansion of k is k = Sum_{i=0..n-1} b[i]*2^i. - Kevin Ryde, Nov 14 2023

A056699 First differences are 2,1,-2,3 (repeated).

Original entry on oeis.org

1, 3, 4, 2, 5, 7, 8, 6, 9, 11, 12, 10, 13, 15, 16, 14, 17, 19, 20, 18, 21, 23, 24, 22, 25, 27, 28, 26, 29, 31, 32, 30, 33, 35, 36, 34, 37, 39, 40, 38, 41, 43, 44, 42, 45, 47, 48, 46, 49, 51, 52, 50, 53, 55, 56, 54, 57, 59, 60, 58, 61, 63, 64, 62, 65, 67, 68, 66

Offset: 1

Michael Knauth (knauth_jur(AT)yahoo.de), Nov 21 2003

Second quadrisection of natural numbers shifted right two places. - Ralf Stephan, Jun 10 2005

A permutation of the natural numbers partitioned into quadruples [4k-3,4k-1,4k,4k-2] for k > 0. Partition the natural number sequence into quadruples starting with (1,2,3,4); swap the second and third elements, then swap the third and fourth elements; repeat for all quadruples. - Guenther Schrack, Oct 18 2017

David A. Corneth, Table of n, a(n) for n = 1..10000
Index entries for sequences that are permutations of the natural numbers.
Index entries for linear recurrences with constant coefficients, signature (1,0,0,1,-1).

Inverse: A284307.
Sequence of fixed points: A016813(n-1) for n > 0.
Odd elements: A005408(n-1) for n > 0.
Indices of odd elements: A042963(n) for n > 0.
Even elements: 2*A103889(n) for n > 0.
Indices of even elements: A014601(n) for n > 0.
Cf. A067060, A092486, A116966, A132400, A292576, A263449, A042938 (bisection)

a = [1 3 4 2];
max = 10000;  % Generation of a b-file
for n := 5:max
   a(n) = a(n-4) + 4;
end;
% Guenther Schrack, Oct 18 2017

[Floor((n - ((-1)^n + (-1)^(n*(n-1)/2)*(2+(-1)^n)) / 2)): n in [1..100]]; // Vincenzo Librandi, Feb 05 2018

LinearRecurrence[{1,0,0,1,-1},{1,3,4,2,5},70] (* Harvey P. Dale, May 10 2014 *)
Table[Floor[(n - ((-1)^n + (-1)^(n (n - 1) / 2) (2 + (-1)^n)) / 2)], {n, 100}] (* Vincenzo Librandi, Feb 05 2018 *)

for(n=1, 10000, print1(n - ((-1)^n + (-1)^(n*(n-1)/2)*(2+(-1)^n))/2, ", ")) \\ Guenther Schrack, Oct 18 2017

G.f.: x*(2*x^4 - 2*x^3 + x^2 + 2*x + 1)/((x-1)^2*(x+1)*(x^2+1)). - Colin Barker, Nov 08 2012
From Guenther Schrack, Oct 18 2017: (Start)
a(n) = a(n-4) + 4 for n > 4.
a(n) = n + periodic[0,1,1,-2].
a(n) = A092486(A067060(n) - 1) for n > 0.
a(n) = A292576(n) - 2*((-1)^floor(n/2)) for n > 0.
a(A116966(n-1)) = A263449(n-1) for n > 0.
A263449(a(n) - 1) = A116966(n-1) for n > 0.
a(n+2) - a(n) = (-1)^floor(n^2/4)*A132400(n+1) for n > 0.
a(n) = a(n-1) + a(n-4) - a(n-5) for n > 5. (End)
a(n) = A298364(n-1) + 1 for n > 1. - Guenther Schrack, Feb 04 2018

A139131 Squarefree kernel of n*(n+1)/2.

Original entry on oeis.org

1, 3, 6, 10, 15, 21, 14, 6, 15, 55, 66, 78, 91, 105, 30, 34, 51, 57, 190, 210, 231, 253, 138, 30, 65, 39, 42, 406, 435, 465, 62, 66, 561, 595, 210, 222, 703, 741, 390, 410, 861, 903, 946, 330, 345, 1081, 282, 42, 35, 255, 1326, 1378, 159, 165, 770, 798, 1653, 1711

Offset: 1

Reinhard Zumkeller, Apr 10 2008

Reinhard Zumkeller, Table of n, a(n) for n = 1..1000

Cf. A000217, A007947, A014601, A014695, A078636.

Times@@(Transpose[FactorInteger[#]][[1]])&/@Accumulate[Range[60]] (* Harvey P. Dale, Jun 06 2013 *)

a(n) = vecprod(factor(n*(n+1)/2)[, 1]); \\ Amiram Eldar, May 12 2025

a(n) = A007947(A000217(n)).
a(A014601(n)) = A078636(A014601(n)).
a(n) = A078636(n) / A014695(n).

A191665 Dispersion of A042963 (numbers >1, congruent to 1 or 2 mod 4), by antidiagonals.

Original entry on oeis.org

1, 2, 3, 5, 6, 4, 10, 13, 9, 7, 21, 26, 18, 14, 8, 42, 53, 37, 29, 17, 11, 85, 106, 74, 58, 34, 22, 12, 170, 213, 149, 117, 69, 45, 25, 15, 341, 426, 298, 234, 138, 90, 50, 30, 16, 682, 853, 597, 469, 277, 181, 101, 61, 33, 19, 1365, 1706, 1194, 938, 554

Offset: 1

Clark Kimberling, Jun 11 2011

Row 1:  A000975
Row 2:  A081254
Row 3:  A081253
Row 4:  A052997
For a background discussion of dispersions, see A191426.
...
Each of the sequences (4n, n>2), (4n+1, n>0), (3n+2, n>=0), generates a dispersion.  Each complement (beginning with its first term >1) also generates a dispersion.  The six sequences and dispersions are listed here:
...
A191663=dispersion of A042948 (0 or 1 mod 4 and >1)
A054582=dispersion of A005843 (0 or 2 mod 4 and >1; evens)
A191664=dispersion of A014601 (0 or 3 mod 4 and >1)
A191665=dispersion of A042963 (1 or 2 mod 4 and >1)
A191448=dispersion of A005408 (1 or 3 mod 4 and >1, odds)
A191666=dispersion of A042964 (2 or 3 mod 4)
...
EXCEPT for at most 2 initial terms (so that column 1 always starts with 1):
A191663 has 1st col A042964, all else A042948
A054582 has 1st col A005408, all else A005843
A191664 has 1st col A042963, all else A014601
A191665 has 1st col A014601, all else A042963
A191448 has 1st col A005843, all else A005408
A191666 has 1st col A042948, all else A042964
...
There is a formula for sequences of the type "(a or b mod m)", (as in the Mathematica program below):
   If f(n)=(n mod 2), then (a,b,a,b,a,b,...) is given by
   a*f(n+1)+b*f(n), so that "(a or b mod m)" is given by
   a*f(n+1)+b*f(n)+m*floor((n-1)/2)), for n>=1.

			Northwest corner:
1...2...5....10...21
3...6...13...26...53
4...9...18...37...74
7...14..29...58...117
8...17..34...69...138

Ivan Neretin, Table of n, a(n) for n = 1..5050 (first 100 antidiagonals, flattened)

Cf. A042963, A014601, A191426, A191664.

(* Program generates the dispersion array T of the increasing sequence f[n] *)
r = 40; r1 = 12;  c = 40; c1 = 12;
a = 2; b = 5; m[n_] := If[Mod[n, 2] == 0, 1, 0];
f[n_] := a*m[n + 1] + b*m[n] + 4*Floor[(n - 1)/2]
Table[f[n], {n, 1, 30}]  (* A042963: (2+4k,5+4k) *)
mex[list_] := NestWhile[#1 + 1 &, 1, Union[list][[#1]] <= #1 &, 1, Length[Union[list]]]
rows = {NestList[f, 1, c]};
Do[rows = Append[rows, NestList[f, mex[Flatten[rows]], r]], {r}];
t[i_, j_] := rows[[i, j]];
TableForm[Table[t[i, j], {i, 1, 10}, {j, 1, 10}]]
(* A191665 *)
Flatten[Table[t[k, n - k + 1], {n, 1, c1}, {k, 1, n}]]
(* A191665  *)

A191666 Dispersion of A042964 (numbers congruent to 2 or 3 mod 4), by antidiagonals.

Original entry on oeis.org

1, 2, 4, 3, 7, 5, 6, 14, 10, 8, 11, 27, 19, 15, 9, 22, 54, 38, 30, 18, 12, 43, 107, 75, 59, 35, 23, 13, 86, 214, 150, 118, 70, 46, 26, 16, 171, 427, 299, 235, 139, 91, 51, 31, 17, 342, 854, 598, 470, 278, 182, 102, 62, 34, 20, 683, 1707, 1195, 939, 555, 363

Offset: 1

Clark Kimberling, Jun 11 2011

Row 1:  A005578
Row 2:  A160113
For a background discussion of dispersions, see A191426.
...
Each of the sequences (4n, n>2), (4n+1, n>0), (3n+2, n>=0), generates a dispersion.  Each complement (beginning with its first term >1) also generates a dispersion.  The six sequences and dispersions are listed here:
...
A191663=dispersion of A042948 (0 or 1 mod 4 and >1)
A054582=dispersion of A005843 (0 or 2 mod 4 and >1; evens)
A191664=dispersion of A014601 (0 or 3 mod 4 and >1)
A191665=dispersion of A042963 (1 or 2 mod 4 and >1)
A191448=dispersion of A005408 (1 or 3 mod 4 and >1, odds)
A191666=dispersion of A042964 (2 or 3 mod 4)
...
EXCEPT for at most 2 initial terms (so that column 1 always starts with 1):
A191663 has 1st col A042964, all else A042948
A054582 has 1st col A005408, all else A005843
A191664 has 1st col A042963, all else A014601
A191665 has 1st col A014601, all else A042963
A191448 has 1st col A005843, all else A005408
A191666 has 1st col A042948, all else A042964
...
There is a formula for sequences of the type "(a or b mod m)", (as in the Mathematica program below):
   If f(n)=(n mod 2), then (a,b,a,b,a,b,...) is given by
   a*f(n+1)+b*f(n), so that "(a or b mod m)" is given by
   a*f(n+1)+b*f(n)+m*floor((n-1)/2)), for n>=1.

			Northwest corner:
1...2...3....6...11
4...7...14....27...54
5...10...19...38...75
8...15..30...59...118
8...18..35...70...139

Ivan Neretin, Table of n, a(n) for n = 1..5050 (first 100 antidiagonals, flattened)

Cf. A042963, A014601, A191426, A191663.

(* Program generates the dispersion array T of the increasing sequence f[n] *)
r = 40; r1 = 12;  c = 40; c1 = 12;
a = 2; b = 3; m[n_] := If[Mod[n, 2] == 0, 1, 0];
f[n_] := a*m[n + 1] + b*m[n] + 4*Floor[(n - 1)/2]
Table[f[n], {n, 1, 30}]  (* A042964: (2+4k,3+4k) *)
mex[list_] := NestWhile[#1 + 1 &, 1, Union[list][[#1]] <= #1 &, 1, Length[Union[list]]]
rows = {NestList[f, 1, c]};
Do[rows = Append[rows, NestList[f, mex[Flatten[rows]], r]], {r}];
t[i_, j_] := rows[[i, j]];
TableForm[Table[t[i, j], {i, 1, 10}, {j, 1, 10}]] (* A191666 *)
Flatten[Table[t[k, n - k + 1], {n, 1, c1}, {k, 1, n}]] (* A191666  *)

cp's OEIS Frontend

A050998 Inequivalent solutions to Langford (or Langford-Skolem) problem of arranging the numbers 1,1,2,2,3,3,...,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, listed by length and lexicographic order.

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A193867 Odd central polygonal numbers.

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Mathematica

PARI

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A274757 Numbers k such that 6*k+1 is a triangular number (A000217).

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A281333 a(n) = 1 + floor(n/2) + floor(n^2/3).

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A283394 a(n) = 3*n*(3*n + 7)/2 + 4.

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A365968 Irregular triangle read by rows: T(n,k) (0 <= n, 0 <= k < 2^n). An infinite binary tree with root node 0 in row n = 0. Each node then has left child (2*j) - k - 1 and right child (2*j) - k + 1, where j and k are the values of the parent and grandparent nodes respectively.

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A056699 First differences are 2,1,-2,3 (repeated).

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A283394 a(n) = 3n(3*n + 7)/2 + 4.

A365968 Irregular triangle read by rows: T(n,k) (0 <= n, 0 <= k < 2^n). An infinite binary tree with root node 0 in row n = 0. Each node then has left child (2j) - k - 1 and right child (2j) - k + 1, where j and k are the values of the parent and grandparent nodes respectively.