A000004 -id:A000004 - cp's OEIS frontend

A360099 To get A(n,k), replace 0's in the binary expansion of n with (-1) and interpret the result in base k; square array A(n,k), n>=0, k>=0, read by antidiagonals.

0, 0, 1, 0, 1, -1, 0, 1, 0, 1, 0, 1, 1, 2, -1, 0, 1, 2, 3, -1, 1, 0, 1, 3, 4, 1, 1, -1, 0, 1, 4, 5, 5, 3, 1, 1, 0, 1, 5, 6, 11, 7, 5, 3, -1, 0, 1, 6, 7, 19, 13, 11, 7, -2, 1, 0, 1, 7, 8, 29, 21, 19, 13, 1, 0, -1, 0, 1, 8, 9, 41, 31, 29, 21, 14, 3, 0, 1, 0, 1, 9, 10, 55, 43, 41, 31, 43, 16, 5, 2, -1

Offset: 0

Alois P. Heinz, Jan 25 2023

The empty bit string is used as binary expansion of 0, so A(0,k) = 0.

			Square array A(n,k) begins:
   0,  0, 0,  0,  0,   0,   0,   0,   0,   0,   0, ...
   1,  1, 1,  1,  1,   1,   1,   1,   1,   1,   1, ...
  -1,  0, 1,  2,  3,   4,   5,   6,   7,   8,   9, ...
   1,  2, 3,  4,  5,   6,   7,   8,   9,  10,  11, ...
  -1, -1, 1,  5, 11,  19,  29,  41,  55,  71,  89, ...
   1,  1, 3,  7, 13,  21,  31,  43,  57,  73,  91, ...
  -1,  1, 5, 11, 19,  29,  41,  55,  71,  89, 109, ...
   1,  3, 7, 13, 21,  31,  43,  57,  73,  91, 111, ...
  -1, -2, 1, 14, 43,  94, 173, 286, 439, 638, 889, ...
   1,  0, 3, 16, 45,  96, 175, 288, 441, 640, 891, ...
  -1,  0, 5, 20, 51, 104, 185, 300, 455, 656, 909, ...

Alois P. Heinz, Antidiagonals n = 0..200, flattened

Columns k=0-6, 10 give: A062157, A145037, A006257, A147991, A147992, A153777, A147993, A359925.
Rows n=0-10 give: A000004, A000012, A023443, A000027(k+1), A165900, A002061, A165900(k+1), A002061(k+1), A083074, A152618, A062158.
Main diagonal gives A360096.
Cf. A030300, A057427.

A:= proc(n, k) option remember; local m;
     `if`(n=0, 0, k*A(iquo(n, 2, 'm'), k)+2*m-1)
    end:
seq(seq(A(n, d-n), n=0..d), d=0..12);
# second Maple program:
A:= (n, k)-> (l-> add((2*l[i]-1)*k^(i-1), i=1..nops(l)))(Bits[Split](n)):
seq(seq(A(n, d-n), n=0..d), d=0..12);

G.f. for column k satisfies g_k(x) = k*(x+1)*g_k(x^2) + x/(1+x).
A(n,k) = k*A(floor(n/2),k)+2*(n mod 2)-1 for n>0, A(0,k)=0.
A(n,k) mod 2 = A057427(n) if k is even.
A(n,k) mod 2 = A030300(n) if k is odd and n>=1.
A(2^(n+1),1) + n = 0.

A062160 Square array T(n,k) = (n^k - (-1)^k)/(n+1), n >= 0, k >= 0, read by falling antidiagonals.

Original entry on oeis.org

0, 1, 0, -1, 1, 0, 1, 0, 1, 0, -1, 1, 1, 1, 0, 1, 0, 3, 2, 1, 0, -1, 1, 5, 7, 3, 1, 0, 1, 0, 11, 20, 13, 4, 1, 0, -1, 1, 21, 61, 51, 21, 5, 1, 0, 1, 0, 43, 182, 205, 104, 31, 6, 1, 0, -1, 1, 85, 547, 819, 521, 185, 43, 7, 1, 0, 1, 0, 171, 1640, 3277, 2604, 1111, 300, 57, 8, 1, 0, -1, 1, 341, 4921, 13107, 13021, 6665, 2101, 455, 73, 9, 1, 0

Offset: 0

Henry Bottomley, Jun 08 2001

For n >= 1, T(n, k) equals the number of walks of length k between any two distinct vertices of the complete graph K_(n+1). - Peter Bala, May 30 2024

			From _Seiichi Manyama_, Apr 12 2019: (Start)
Square array begins:
   0, 1, -1,  1,  -1,    1,    -1,      1, ...
   0, 1,  0,  1,   0,    1,     0,      1, ...
   0, 1,  1,  3,   5,   11,    21,     43, ...
   0, 1,  2,  7,  20,   61,   182,    547, ...
   0, 1,  3, 13,  51,  205,   819,   3277, ...
   0, 1,  4, 21, 104,  521,  2604,  13021, ...
   0, 1,  5, 31, 185, 1111,  6665,  39991, ...
   0, 1,  6, 43, 300, 2101, 14706, 102943, ... (End)

Seiichi Manyama, Antidiagonals n = 0..139, flattened
M. Dukes and C. D. White, Web Matrices: Structural Properties and Generating Combinatorial Identities, arXiv:1603.01589 [math.CO], 2016.

Rows include A062157, A000035, A001045, A015518, A015521, A015531, A015540, A015552, A015565, A015577, A015585, A015592, A015609.
Columns include A000004, A000012, A023443, A002061, A062158, A060884, A062159, A060888.
Related to repunits in negative bases (cf. A055129 for positive bases).
Main diagonal gives A081216.
Cf. A109502.

seq(print(seq((n^k - (-1)^k)/(n+1), k = 0..10)), n = 0..10); # Peter Bala, May 31 2024

T[n_,k_]:=(n^k - (-1)^k)/(n+1); Join[{0},Table[Reverse[Table[T[n-k,k],{k,0,n}]],{n,12}]]//Flatten (* Stefano Spezia, Feb 20 2024 *)

T(n, k) = n^(k-1) - n^(k-2) + n^(k-3) - ... + (-1)^(k-1) = n^(k-1) - T(n, k-1) = n*T(n, k-1) - (-1)^k = (n - 1)*T(n, k-1) + n*T(n, k-2) = round[n^k/(n+1)] for n > 1.
T(n, k) = (-1)^(k+1) * resultant( n*x + 1, (x^k-1)/(x-1) ). - Max Alekseyev, Sep 28 2021
G.f. of row n: x/((1+x) * (1-n*x)). - Seiichi Manyama, Apr 12 2019
E.g.f. of row n: (exp(n*x) - exp(-x))/(n+1). - Stefano Spezia, Feb 20 2024
From Peter Bala, May 31 2024: (Start)
Binomial transform of the m-th row: Sum_{k = 0..n} binomial(n, k)*T(m, k) = (m + 1)^(n-1) for n >= 1.
Let R(m, x) denote the g.f. of the m-th row of the square array. Then R(m_1, x) o R(m_2, x) = R(m_1 + m_2 + m_1*m_2, x),  where o denotes the black diamond product of power series as defined by Dukes and White. Cf. A109502.
T(m_1 + m_2 + m_1*m_2, k) = Sum_{i = 0..k} Sum_{j = i..k} binomial(k, i)* binomial(k-i, j-i)*T(m_1, j)*T(m_2, k-i).  (End)

A065167 Table T(n,k) read by antidiagonals, where the k-th row gives the permutation t->t+k of Z, folded to N (k >= 0, n >= 1).

Original entry on oeis.org

1, 2, 2, 3, 4, 4, 4, 1, 6, 6, 5, 6, 2, 8, 8, 6, 3, 8, 4, 10, 10, 7, 8, 1, 10, 6, 12, 12, 8, 5, 10, 2, 12, 8, 14, 14, 9, 10, 3, 12, 4, 14, 10, 16, 16, 10, 7, 12, 1, 14, 6, 16, 12, 18, 18, 11, 12, 5, 14, 2, 16, 8, 18, 14, 20, 20, 12, 9, 14, 3, 16, 4, 18, 10, 20, 16, 22, 22, 13, 14, 7, 16, 1

Offset: 0

Antti Karttunen, Oct 19 2001

Simple periodic site swap permutations of natural numbers.

Row n of the table (starting from n=0) gives a permutation of natural numbers corresponding to the simple, infinite, periodic site swap pattern ...nnnnn...

			Table begins:
1 2 3 4 5 6 7 ...
2 4 1 6 3 8 5 ...
4 6 2 8 1 10 3 ...
6 8 4 10 2 12 1 ...

Joe Buhler and R. L. Graham, Juggling Drops and Descents, Amer. Math. Monthly, 101, (no. 6) 1994, 507 - 519.
Juggling Information Service, Site Swap FAQs

Successive rows and associated site swap sequences, starting from the zeroth row: (A000027, A000004), (A065164, A000012), (A065165, A007395), (A065166, A010701). Cf. also A065171, A065174, A065177. trinv given at A054425.

PerSS_table := (n) -> PerSS((((trinv(n)-1)*(((1/2)*trinv(n))+1))-n)+1, (n-((trinv(n)*(trinv(n)-1))/2))); PerSS := (n,c) -> Z2N(N2Z(n)+c);
N2Z := n -> ((-1)^n)*floor(n/2); Z2N := z -> 2*abs(z)+`if`((z < 1),1,0);
[seq(PerSS_table(j),j=0..119)];

Let f: Z -> N be given by f(z) = 2z if z>0 else 2|z|+1, with inverse g(z) = z/2 if z even else (1-z)/2. Then the n-th term of the k-th row is f(g(n)+k).

A092092 Back and Forth Summant S(n, 3): a(n) = Sum{i=0..floor(2n/3)} (n-3i).

Original entry on oeis.org

1, 1, 0, 3, 2, 0, 5, 3, 0, 7, 4, 0, 9, 5, 0, 11, 6, 0, 13, 7, 0, 15, 8, 0, 17, 9, 0, 19, 10, 0, 21, 11, 0, 23, 12, 0, 25, 13, 0, 27, 14, 0, 29, 15, 0, 31, 16, 0, 33, 17, 0, 35, 18, 0, 37, 19, 0, 39, 20, 0, 41, 21, 0, 43, 22, 0, 45, 23, 0, 47, 24, 0, 49, 25, 0, 51, 26, 0, 53, 27, 0, 55, 28

Offset: 1

Jahan Tuten (jahant(AT)indiainfo.com), Mar 29 2004

The terms for n>1 can also be defined by: a(n)=0 if n==0 (mod 3), and otherwise a(n) equals the inverse of 3 in Z/nZ*. - José María Grau Ribas, Jun 18 2013

The subsequence of nonzero terms is essentially the same as A026741. - Giovanni Resta, Jun 18 2013

F. Smarandache, Back and Forth Summants, Arizona State Univ., Special Collections, 1972.

Robert Israel, Table of n, a(n) for n = 1..10000
J. Dezert, editor, Smarandacheials, Mathematics Magazine, Aurora, Canada
Index entries for linear recurrences with constant coefficients, signature (0,0,2,0,0,-1).

Cf. A092094, A092397.
Other values of k: A000004 (k = 1, 2), A027656 (k = 4), A092093 (k = 5).
Cf. A226782 - A226787 for inverses of 4,5,6,.. in Z/nZ*.

f:= proc(n) local t;
t:= n mod 3;
if t = 0 then 0 elif t = 1 then 2/3*(n+1/2) else (n+1)/3 fi
end proc:
map(f, [$1..100]); # Robert Israel, May 19 2016

LinearRecurrence[{0, 0, 2, 0, 0, -1}, {1, 1, 0, 3, 2, 0}, 100] (* Jean-François Alcover, Jun 04 2020 *)

S(n, k=3) = local(s, x); s = n; x = n - k; while (x >= -n, s = s + x; x = x - k); s;

a(3n) = 0; a(3n+1) = 2n+1; a(3n+2) = n+1.
G.f.: x*(1+x+x^3) / ( (x-1)^2*(1+x+x^2)^2 ). - R. J. Mathar, Jun 26 2013
a(n) = Sum_{k=1..n} k*( floor((3k-1)/n)-floor((3k-2)/n) ). - Anthony Browne, May 17 2016

Edited and extended by David Wasserman, Dec 19 2005

A108306 Expansion of (3x+1)/(1-3x-3*x^2).

Original entry on oeis.org

1, 6, 21, 81, 306, 1161, 4401, 16686, 63261, 239841, 909306, 3447441, 13070241, 49553046, 187869861, 712268721, 2700415746, 10238053401, 38815407441, 147160382526, 557927369901, 2115263257281, 8019571881546, 30404505416481, 115272231894081

Offset: 0

Creighton Dement, Jul 24 2005

Binomial transform is A055271. May be seen as a ibasefor-transform of the zero-sequence A000004 with respect to the floretion given in the program code.
The sequence is the INVERT transform of (1, 5, 10, 20, 40, 80, 160, ...) and can be obtained by extracting the upper left terms of matrix powers of [(1,5); (1,2)]. These results are a case (a=5, b=2) of the conjecture:  The INVERT transform of a sequence starting (1, a, a*b, a*b^2, a*b^3, ...) is equivalent to extracting the upper left terms of powers of the  2x2 matrix [(1,a); (1,b)]. - Gary W. Adamson, Jul 31 2016
From Klaus Purath, Mar 09 2023: (Start)
For any terms (a(n), a(n+1)) = (x, y), -3*x^2 - 3*x*y + y^2 = 15*(-3)^n = A082111(2)*(-3)^n. This is valid in general for all recursive sequences (t) with constant coefficients (3,3) and t(0) = 1: -3*x^2 - 3*x*y + y^2 = A082111(t(1)-4)*(-3)^n.
By analogy to this, for three consecutive terms (x, y, z) of any sequence (t) of the form (3,3) with t(0) = 1: y^2 - x*z = A082111(t(1)-4)*(-3)^n. (End)

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Martin Burtscher, Igor Szczyrba and Rafał Szczyrba, Analytic Representations of the n-anacci Constants and Generalizations Thereof, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.5.
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (3,3).

Cf. A055271.

Cf. A084057.

I:=[1,6]; [n le 2 select I[n] else 3*Self(n-1)+3*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Aug 01 2016

seriestolist(series((3*x+1)/(1-3*x-3*x^2), x=0,25));

CoefficientList[Series[(3 x + 1) / (1 - 3 x - 3 x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Aug 01 2016 *)

Recurrence: a(0)=1; a(1)=6; a(n) = 3a(n-1) + 3a(n-2) - N-E. Fahssi, Apr 20 2008

A171522 Denominator of 1/n^2-1/(n+2)^2.

Original entry on oeis.org

0, 9, 16, 225, 144, 1225, 576, 3969, 1600, 9801, 3600, 20449, 7056, 38025, 12544, 65025, 20736, 104329, 32400, 159201, 48400, 233289, 69696, 330625, 97344, 455625, 132496, 613089, 176400, 808201, 230400, 1046529, 295936, 1334025, 374544, 1677025, 467856

Offset: 0

Paul Curtz, Dec 11 2009

This is the third column in the table of denominators of the hydrogenic spectra (the main diagonal A147560):
0,   0,   0,   0,   0,   0,   0,   0... A000004
1,   4,   9,  16,  25,  36,  49,  64... A000290
1,  36,  16, 100,   9, 196,  64, 324... A061038
1, 144, 225,  12, 441, 576,  81, 900... A061040
1, 400, 144, 784,  64,1296, 400,1936... A061042
1, 900 1225,1600,2025, 100,3025,3600... A061044
1,1764, 576, 324, 225,4356,  48,6084... A061046
1,3136,3969,4900,5929,7056,8281, 196... A061048.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,5,0,-10,0,10,0,-5,0,1).

Cf. A105371. Bisections: A060300, A069075.

A171522 := proc(n) if n = 0 then 0 else lcm(n+2,n) ; %^2 ; end if ; end:
seq(A171522(n),n=0..70) ; # R. J. Mathar, Dec 15 2009

a[n_] := If[EvenQ[n], (n*(n+2))^2/4, (n*(n+2))^2]; Table[a[n], {n, 0, 36}] (* Jean-François Alcover, Jun 13 2017 *)

concat(0, Vec(x*(x^8+4*x^6+16*x^5+190*x^4+64*x^3+180*x^2+16*x+9) / ((x-1)^5*-(x+1)^5) + O(x^100))) \\ Colin Barker, Nov 05 2014

a(n) = (A066830(n+1))^2.
a(n) = -((-5+3*(-1)^n)*n^2*(2+n)^2)/8. - Colin Barker, Nov 05 2014
G.f.: x*(x^8+4*x^6+16*x^5+190*x^4+64*x^3+180*x^2+16*x+9) / ((x-1)^5*-(x+1)^5). - Colin Barker, Nov 05 2014

Edited and extended by R. J. Mathar, Dec 15 2009

A199572 Number of round trips of length n on the cycle graph C_2 from any of the two vertices.

Original entry on oeis.org

1, 0, 4, 0, 16, 0, 64, 0, 256, 0, 1024, 0, 4096, 0, 16384, 0, 65536, 0, 262144, 0, 1048576, 0, 4194304, 0, 16777216, 0, 67108864, 0, 268435456, 0, 1073741824, 0, 4294967296, 0, 17179869184, 0, 68719476736, 0, 274877906944, 0

Offset: 0

Wolfdieter Lang, Nov 08 2011

See the array and triangle A199571 for the general cycle graph C_N counting.
This is A000302 and A000004 interleaved. - Omar E. Pol, Nov 09 2011
With offset = 1:  Number of ways to separate n distinguishable objects into an odd size pile and an even size pile.  For example:  a(3) = 4 because we have: {{1},{2,3}}; {{2},{1,3}}; {{3},{1,2}}; {{1,2,3},{}}. - Geoffrey Critzer, Jun 10 2013
Inverse Stirling transform of A065143. - Vladimir Reshetnikov, Nov 01 2015

			a(2) = 4 from starting with vertex no. 1, with edges e1 and e2 to vertex no. 2: e1e1, e2e2, e1e2 and e2e1.

R. J. Mathar, Counting Walks on Finite Graphs, Section 1.
Index entries for linear recurrences with constant coefficients, signature (0,4).

Cf. A000007 (N=1), A078008 (N=3). a(n) is second row of array w(N,L) A199571, and second column of the triangle a(K,N) A199571.

Cf. A065143 (Stirling transform).

nn = 39; Drop[Range[0, nn]! CoefficientList[Series[ Sinh[x] Cosh[x], {x, 0, nn}],x], 1] (* Geoffrey Critzer, Jun 10 2013 *)

vector(100, n, n--; (2^(n) +(-2)^n)/2) \\ Altug Alkan, Nov 02 2015

a(n) = (2^n + (-2)^n)/2 = 2^(n-1)*(1 + (-1)^n).
O.g.f.: 1/(1-(2*x)^2).
E.g.f.: cosh(2*x)=U(0) where U(k) = 1 + 2*x^2/((4*k+1)*(2*k+1) - x^2*(4*k+1)*(2*k+1)/(x^2 + (4*k+3)*(k+1)/U(k+1))); (continued fraction). - Sergei N. Gladkovskii, Oct 23 2012

A262221 a(n) = 25n(n + 1)/2 + 1.

Original entry on oeis.org

1, 26, 76, 151, 251, 376, 526, 701, 901, 1126, 1376, 1651, 1951, 2276, 2626, 3001, 3401, 3826, 4276, 4751, 5251, 5776, 6326, 6901, 7501, 8126, 8776, 9451, 10151, 10876, 11626, 12401, 13201, 14026, 14876, 15751, 16651, 17576, 18526, 19501, 20501, 21526, 22576, 23651

Offset: 0

Bruno Berselli, Sep 15 2015

Also centered 25-gonal (or icosipentagonal) numbers.
This is the case k=25 of the formula (k*n*(n+1) - (-1)^k + 1)/2. See table in Links section for similar sequences.
For k=2*n, the formula shown above gives A011379.
Primes in sequence: 151, 251, 701, 1951, 3001, 4751, 10151, 12401, ...

E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 51 (23rd row of the table).

Bruno Berselli, Table of n, a(n) for n = 0..1000
Bruno Berselli, Table of numbers of the form (k*n*(n+1)-(-1)^k+1)/2.
Eric Weisstein's World of Mathematics, Centered Polygonal Numbers.
Index entries for sequences related to centered polygonal numbers.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A008607 (first differences), A011379, A034856, A054254, A080956, A123296, A186349, A255184.
Cf. centered polygonal numbers listed in A069190.
Similar sequences of the form (k*n*(n+1) - (-1)^k + 1)/2 with -1 <= k <= 26: A000004, A000124, A002378, A005448, A005891, A028896, A033996, A035008, A046092, A049598, A060544, A064200, A069099, A069125, A069126, A069128, A069130, A069132, A069174, A069178, A080956, A124080, A163756, A163758, A163761, A164136, A173307.

Magma
```
[25*n*(n+1)/2+1: n in [0..50]];
```

Table[25 n (n + 1)/2 + 1, {n, 0, 50}]
25*Accumulate[Range[0,50]]+1 (* or *) LinearRecurrence[{3,-3,1},{1,26,76},50] (* Harvey P. Dale, Jan 29 2023 *)

PARI
```
vector(50, n, n--; 25*n*(n+1)/2+1)
```
Sage
```
[25*n*(n+1)/2+1 for n in (0..50)]
```

G.f.: (1 + 23*x + x^2)/(1 - x)^3.
a(n) = a(-n-1) = 3*a(n-1) - 3*a(n-2) + a(n-3).
a(n) = A123296(n) + 1.
a(n) = A000217(5*n+2) - 2.
a(n) = A034856(5*n+1).
a(n) = A186349(10*n+1).
a(n) = A054254(5*n+2) with n>0, a(0)=1.
a(n) = A000217(n+1) + 23*A000217(n) + A000217(n-1) with A000217(-1)=0.
Sum_{i>=0} 1/a(i) = 1.078209111... = 2*Pi*tan(Pi*sqrt(17)/10)/(5*sqrt(17)).
From Amiram Eldar, Jun 21 2020: (Start)
Sum_{n>=0} a(n)/n! = 77*e/2.
Sum_{n>=0} (-1)^(n+1) * a(n)/n! = 23/(2*e). (End)
E.g.f.: exp(x)*(2 + 50*x + 25*x^2)/2. - Elmo R. Oliveira, Dec 24 2024

A268833 Square array A(n, k) = A101080(k, A003188(n+A006068(k))), read by descending antidiagonals, where A003188 is the binary Gray code, A006068 is its inverse, and A101080(x,y) gives the Hamming distance between binary expansions of x and y.

Original entry on oeis.org

0, 0, 1, 0, 1, 2, 0, 1, 2, 1, 0, 1, 2, 3, 2, 0, 1, 2, 3, 2, 3, 0, 1, 2, 1, 2, 1, 2, 0, 1, 2, 3, 2, 3, 2, 1, 0, 1, 2, 1, 2, 3, 4, 3, 2, 0, 1, 2, 1, 2, 3, 4, 3, 2, 3, 0, 1, 2, 3, 2, 3, 4, 3, 2, 3, 4, 0, 1, 2, 3, 2, 3, 4, 3, 2, 1, 4, 3, 0, 1, 2, 1, 2, 1, 2, 3, 2, 3, 2, 3, 2, 0, 1, 2, 1, 2, 3, 2, 1, 2, 3, 2, 3, 2, 3, 0, 1, 2, 3, 2, 3, 4, 3, 2, 3, 4, 1, 2, 1, 2

Offset: 0

Antti Karttunen, Feb 15 2016

The entry at row n, column k, gives the Hamming distance between binary expansions of k and A003188(n+A006068(k)). When Gray code is viewed as a traversal of vertices of an infinite dimensional hypercube by bit-flipping (see the illustration "Visualized as a traversal of vertices of a tesseract" in the Wikipedia's "Gray code" article) the argument k is the "address" (the binary code given inside each vertex) of the starting vertex, and argument n tells how many edges forward along the Gray code path we should hop from it (to the direction that leads away from the vertex with code 0000...). A(n, k) gives then the Hamming distance between the starting and the ending vertex. For how this works with case n=3, see comments in A268676. - Antti Karttunen, Mar 11 2024

			The top left [0 .. 24] X [0 .. 24] section of the array:
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2
1, 3, 3, 1, 3, 1, 1, 3, 3, 1, 1, 3, 1, 3, 3, 1, 3, 1, 1, 3, 1, 3, 3, 1, 1
2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2
3, 1, 3, 3, 3, 3, 3, 1, 3, 3, 3, 1, 3, 1, 3, 3, 3, 3, 3, 1, 3, 1, 3, 3, 3
2, 2, 4, 4, 4, 4, 2, 2, 4, 4, 2, 2, 2, 2, 4, 4, 4, 4, 2, 2, 2, 2, 4, 4, 2
1, 3, 3, 3, 3, 3, 1, 3, 3, 3, 1, 3, 1, 3, 3, 3, 3, 3, 1, 3, 1, 3, 3, 3, 1
2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2
3, 3, 1, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 1, 3, 3, 3, 3, 3, 3, 3, 1, 3, 3
4, 4, 2, 2, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 2, 2, 4, 4, 4, 4, 4, 4, 2, 2, 4
3, 3, 3, 1, 5, 3, 3, 5, 5, 3, 3, 5, 3, 3, 3, 1, 5, 3, 3, 5, 3, 3, 3, 1, 3
2, 2, 2, 2, 4, 4, 4, 4, 4, 4, 4, 4, 2, 2, 2, 2, 4, 4, 4, 4, 2, 2, 2, 2, 2
3, 1, 3, 3, 3, 5, 5, 3, 3, 5, 5, 3, 3, 1, 3, 3, 3, 5, 5, 3, 3, 1, 3, 3, 3
2, 2, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 2, 2, 4, 4, 4, 4, 4, 4, 2, 2, 4, 4, 2
1, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 1, 3, 3, 3, 3, 3, 3, 3, 1, 3, 3, 3, 1
2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2
3, 3, 3, 3, 1, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3
4, 4, 4, 4, 2, 2, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4
3, 5, 5, 3, 3, 1, 3, 3, 5, 3, 3, 5, 3, 5, 5, 3, 5, 3, 3, 5, 3, 5, 5, 3, 3
4, 4, 4, 4, 2, 2, 2, 2, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4
5, 3, 3, 5, 3, 3, 3, 1, 5, 5, 5, 3, 5, 3, 5, 5, 5, 5, 5, 3, 5, 3, 5, 5, 5
4, 4, 4, 4, 4, 4, 2, 2, 6, 6, 4, 4, 4, 4, 6, 6, 6, 6, 4, 4, 4, 4, 6, 6, 4
3, 3, 3, 3, 3, 3, 1, 3, 5, 5, 3, 5, 3, 5, 5, 5, 5, 5, 3, 5, 3, 5, 5, 5, 3
2, 2, 2, 2, 2, 2, 2, 2, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 2

Antti Karttunen, Table of n, a(n) for n = 0..32895; the first 256 antidiagonals of array
Wikipedia, Gray code.

Transpose A268834.
Main diagonal: A268835.
Column 0: A005811.
Row 0: A000004, Row 1: A000012, Row 2: A007395, Row 3: A268676.
Cf. A003188, A006068.
Cf. also A268726, A268727.

A101080[n_, k_]:= DigitCount[BitXor[n, k], 2, 1];A003188[n_]:=BitXor[n, Floor[n/2]]; A006068[n_]:=If[n<2, n, Block[{m=A006068[Floor[n/2]]}, 2m + Mod[Mod[n,2] + Mod[m, 2], 2]]]; a[r_, 0]:= 0; a[0, c_]:=c; a[r_, c_]:= A003188[1 + A006068[a[r - 1, c - 1]]]; A[r_, c_]:=A101080[c, a[r, r + c]]; Table[A[c, r - c], {r, 0, 20}, {c, 0, r}] // Flatten (* Indranil Ghosh, Apr 02 2017 *)

b(n) = if(n<1, 0, b(n\2) + n%2);
A101080(n, k) = b(bitxor(n, k));
A003188(n) = bitxor(n, n\2);
A006068(n) = if(n<2, n, {my(m = A006068(n\2)); 2*m + (n%2 + m%2)%2});
A268820(r, c) = if(r==0, c, if(c==0, 0, A003188(1 + A006068(A268820(r - 1, c - 1)))));
A(r, c) = A101080(c, A268820(r, r + c));
for(r=0, 20, for(c=0, r, print1(A(c, r - c),", ");); print();) \\ Indranil Ghosh, Apr 02 2017

up_to = 32895; \\ = binomial(1+256,2)-1.\\ A003188 and A006068 as above.
A268833sq(n, k) = hammingweight(bitxor(n,A003188(k+A006068(n))));
A268833list(up_to) = { my(v = vector(up_to), i=0); for(a=0,oo, for(col=0,a, i++; if(i > up_to, return(v)); v[i] = A268833sq(a-col,col))); (v); };
v268833 = A268833list(1+up_to);
A268833(n) = v268833[1+n]; \\ Antti Karttunen, Mar 11 2024

def A101080(n, k): return bin(n^k)[2:].count("1")
def A003188(n): return n^(n//2)
def A006068(n):
    if n<2: return n
    else:
        m=A006068(n//2)
        return 2*m + (n%2 + m%2)%2
def A268820(r, c): return c if r<1 else 0 if c<1 else A003188(1 + A006068(A268820(r - 1, c - 1)))
def a(r, c): return A101080(c, A268820(r, r + c))
for r in range(21):
    print([a(c, r - c) for c in range(r + 1)]) # Indranil Ghosh, Apr 02 2017

(define (A268833 n) (A268833bi (A002262 n) (A025581 n)))
(define (A268833bi row col) (A101080bi col (A268820bi row (+ row col))))

A(row,col) = A101080(col, A268820(row, row+col)).

A(n, k) = A101080(k, A003188(n+A006068(k))). - Antti Karttunen, Mar 11 2024

Definition simplified by Antti Karttunen, Mar 11 2024

A305622 Triangle read by rows: T(n,k) is the number of chiral pairs of rows of n colors with exactly k different colors.

Original entry on oeis.org

0, 0, 1, 0, 2, 3, 0, 6, 18, 12, 0, 12, 72, 120, 60, 0, 28, 267, 780, 900, 360, 0, 56, 885, 4188, 8400, 7560, 2520, 0, 120, 2880, 20400, 63000, 95760, 70560, 20160, 0, 240, 9000, 93120, 417000, 952560, 1164240, 725760, 181440, 0, 496, 27915, 409140, 2551440, 8217720, 14817600, 15120000, 8164800, 1814400, 0, 992, 85233, 1748220, 14802900, 64614960, 161247240, 239500800, 209563200, 99792000, 19958400

Offset: 1

Robert A. Russell, Jun 06 2018

If the row is achiral, i.e., the same as its reverse, we ignore it. If different from its reverse, we count it and its reverse as a chiral pair.

			The triangle begins:
  0;
  0,   1;
  0,   2,     3;
  0,   6,    18,     12;
  0,  12,    72,    120,      60;
  0,  28,   267,    780,     900,     360;
  0,  56,   885,   4188,    8400,    7560,     2520;
  0, 120,  2880,  20400,   63000,   95760,    70560,    20160;
  0, 240,  9000,  93120,  417000,  952560,  1164240,   725760,  181440;
  ...
For T(3,2)=2, the chiral pairs are AAB-BAA and ABB-BBA.  For T(3,3)=3, the chiral pairs are ABC-CBA, ACB-BCA, and BAC-CAB.

Andrew Howroyd, Table of n, a(n) for n = 1..1275 (first 50 rows)

Columns 1-6 are A000004, A122746(n-2), A305623, A305624, A305625, and A305626.
Row sums are A327091.
Cf. A008277, A019538, A293500, A305621.

with(combinat):
a:=(n,k)->(factorial(k)/2)* (Stirling2(n,k)-Stirling2(ceil(n/2),k)): seq(seq(a(n,k),k=1..n),n=1..11); # Muniru A Asiru, Sep 27 2018

Table[(k!/2) (StirlingS2[n, k] - StirlingS2[Ceiling[n/2], k]), {n, 1, 15}, {k, 1, n}] // Flatten

T(n,k) = (k!/2) * (stirling(n,k,2) - stirling(ceil(n/2),k,2));
for (n=1, 10, for(k=1, n, print1(T(n,k), ", ")); print); \\ Michel Marcus, Sep 27 2018

T(n,k) = (k!/2) * (S2(n,k) - S2(ceiling(n/2),k)) where S2(n,k) is the Stirling subset number A008277.
T(n,k) = (A019538(n,k) - A019538(ceiling(n/2),k)) / 2.
T(n,k) = A019538(n,k) - A305621(n,k).
G.f. for column k: k! x^k / (2*Product_{i=1..k}(1-ix)) - k! (x^(2k-1)+x^(2k)) / (2*Product{i=1..k}(1-i x^2)). - Robert A. Russell, Sep 26 2018
T(n, k) = Sum_{i=0..k} (-1)^(k-i)*binomial(k,i)*A293500(n, i). - Andrew Howroyd, Sep 13 2019

cp's OEIS Frontend

A360099 To get A(n,k), replace 0's in the binary expansion of n with (-1) and interpret the result in base k; square array A(n,k), n>=0, k>=0, read by antidiagonals.

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A062160 Square array T(n,k) = (n^k - (-1)^k)/(n+1), n >= 0, k >= 0, read by falling antidiagonals.

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A065167 Table T(n,k) read by antidiagonals, where the k-th row gives the permutation t->t+k of Z, folded to N (k >= 0, n >= 1).

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A092092 Back and Forth Summant S(n, 3): a(n) = Sum{i=0..floor(2n/3)} (n-3i).

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A108306 Expansion of (3*x+1)/(1-3*x-3*x^2).

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A171522 Denominator of 1/n^2-1/(n+2)^2.

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A199572 Number of round trips of length n on the cycle graph C_2 from any of the two vertices.

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A108306 Expansion of (3x+1)/(1-3x-3*x^2).

A262221 a(n) = 25n(n + 1)/2 + 1.