Mikhail Kurkov - cp's OEIS frontend

A386381 Main diagonal of A386363.

1, 1, 1, 2, 8, 56, 640, 10960, 264640, 8581760, 360331520, 19031302400, 1235451750400, 96722377139200, 8988790940876800, 978442125179648000, 123324448870740377600, 17820979140159760793600, 2926936219425738642227200, 542215853077506417192140800, 112527512540808439576566169600

Offset: 0

Mikhail Kurkov, Jul 20 2025

nonn

Cf. A000010, A000111, A060818, A386363.

a := proc(n) local T; T := proc(n, k) option remember; ifelse(k = 0, 0^n, ifelse(k = 1, T(n-1, n-1), T(n, k-1) + (n - 2)*T(n-1, n-k))) end: T(n, n) end:
seq(a(n), n = 0..20);  # Peter Luschny, Jul 21 2025

upto(n) = {my(v1, v2, v3);
v1 = vector(n+1, i, 0); v1[1] = 1;
v2 = vector(n+1, i, 0); v2[1] = 1;
for(i=1, n, v3 = v1; v1[1] = 0; v1[2] = v3[i];
for(j=2, i, v1[j+1] = v1[j] + (i-2)*v3[i-j+1]);
v2[i+1] = v1[i+1]); v2}

This sequence has surprising divisibility properties. Let n and m be any natural numbers, E(n) = A000111(n), and phi(n) = A000010(n).
Conjecture 1: m divides a(m*n + k) <=> m divides a(k). In particular, a(k) divides a(a(k)*n + k).
Conjecture 2: m > 2 divides E(phi(m)*n + k) <=> m divides E(k). In particular, E(k) divides E(phi(E(k))*n + k) for k > 3.
From Peter Luschny, Jul 20 2025: (Start)
Conjecture: 2*a(n) is divisible by A060818(n). (End)

A386393 Triangle T(n, k) (1 <= k <= n) read by rows: T(n, k) is the numerator of R(n, k) where R(n, k) = R(n, k-1)/2 + R(n-1, k-1) for 1 < k <= n with R(n, 1) = R(n-1, n-1) for n > 1, R(1, 1) = 1.

Original entry on oeis.org

1, 1, 3, 3, 7, 19, 19, 43, 99, 251, 251, 555, 1243, 2827, 6843, 6843, 14875, 32635, 72411, 162875, 381851, 381851, 819803, 1771803, 3860443, 8494747, 18918747, 43357211, 43357211, 92234139, 197168923, 423959707, 918096411, 2005424027, 4427023643, 9976746651

Offset: 1

Mikhail Kurkov, Jul 20 2025

Denominator of R(n, k) is 2^((n-1)*(n-2)/2+k-1).

			Triangle begins:
       1;
       1,      3;
       3,      7,      19;
      19,     43,      99,     251;
     251,    555,    1243,    2827,    6843;
    6843,  14875,   32635,   72411,  162875,   381851;
  381851, 819803, 1771803, 3860443, 8494747, 18918747, 43357211;

Cf. A305562.

rows_upto(n) = {my(A, v1, v2, v3);
v1 = vector(n, i, 0); v1[1] = 1;
v2 = vector(n, i, 0); v2[1] = [1];
for(i=2, n, v3 = v1; v1[1] = v3[i-1];
for(j=2, i, v1[j] = v1[j-1]/2 + v3[j-1]);
A = 2^((i-1)*(i-2)/2);
v2[i] = vector(i, j, A*2^(j-1)*v1[j])); v2}

A386363 Variation of triangle of Entringer numbers (A008281) read by rows: T(n, k) = T(n, k-1) + (n-2)*T(n-1, n-k) for 1 < k <= n, T(n, 1) = T(n-1, n-1) for n > 0, T(n, 0) = 0^n.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 1, 2, 2, 0, 2, 6, 8, 8, 0, 8, 32, 50, 56, 56, 0, 56, 280, 480, 608, 640, 640, 0, 640, 3840, 6880, 9280, 10680, 10960, 10960, 0, 10960, 76720, 140800, 196480, 237760, 260800, 264640, 264640, 0, 264640, 2117120, 3942720, 5607040, 6982400, 7968000, 8505040, 8581760, 8581760

Offset: 0

Mikhail Kurkov, Jul 19 2025

			Triangle begins:
  1;
  0,     1;
  0,     1,     1;
  0,     1,     2,      2;
  0,     2,     6,      8,      8;
  0,     8,    32,     50,     56,     56;
  0,    56,   280,    480,    608,    640,    640;
  0,   640,  3840,   6880,   9280,  10680,  10960,  10960;
  0, 10960, 76720, 140800, 196480, 237760, 260800, 264640, 264640;

Paolo Xausa, Table of n, a(n) for n = 0..11475 (rows 0..150 of triangle, flattened).

Cf. A008281, A386381 (main diagonal).

T := proc(n, k) option remember; ifelse(k = 0, 0^n, ifelse(k = 1, T(n-1, n-1), T(n, k-1) + (n - 2)*T(n-1, n-k))) end: seq(seq(T(n, k), k = 0..n), n = 0..9); # Peter Luschny, Jul 19 2025

A386363[n_, k_] := A386363[n, k] = Switch[k, 0, Boole[n == 0], 1, A386363[n-1, n-1], _, A386363[n, k-1] + (n-2)*A386363[n-1, n-k]];
Table[A386363[n, k], {n, 0, 10}, {k, 0, n}] (* Paolo Xausa, Jul 21 2025 *)

rows_upto(n) = {my(v1, v2, v3);
v1 = vector(n+1, i, 0); v1[1] = 1;
v2 = vector(n+1, i, 0); v2[1] = [1];
for(i=1, n, v3 = v1; v1[1] = 0; v1[2] = v3[i];
for(j=2, i, v1[j+1] = v1[j] + (i-2)*v3[i-j+1]);
v2[i+1] = vector(i+1, j, v1[j])); v2}

from functools import cache
@cache
def seidel(n: int)-> list[int]:
    if n == 0: return [1]
    rowA = seidel(n - 1)
    row = [0] + seidel(n - 1)
    row[1] = row[n]
    for k in range(2, n + 1):
        row[k] = row[k - 1] + (n - 2) * rowA[n - k]
    return row
def A386363row(n: int) -> list[int]: return seidel(n)
for n in range(10): print(A386363row(n))  # Peter Luschny, Jul 20 2025

A381810 Array read by downward antidiagonals: A(n,k) is a generalization of odd columns of A125790 defined in Comments for n > 0, k >= 0.

Original entry on oeis.org

2, 4, 4, 6, 16, 6, 8, 36, 20, 10, 10, 64, 42, 84, 14, 12, 100, 72, 286, 100, 20, 14, 144, 110, 680, 322, 120, 26, 16, 196, 156, 1330, 744, 364, 140, 36, 18, 256, 210, 2300, 1430, 816, 406, 656, 46, 20, 324, 272, 3654, 2444, 1540, 888, 3396, 740, 60, 22, 400, 342, 5456, 3850, 2600, 1650, 10816, 3682, 840, 74

Offset: 1

Mikhail Kurkov, May 05 2025

This is generalization in the sense that first column of A125790 is A000123(2^(n-1)) while in this square array column zero is conjecturally A000123(n).
A(n,k) = v_{A001511(n)} where we start with vector v of fixed length L(n) = A070939(n) with elements v_i = A125790(i,2*k+1), pre-calculate A078121 up to L(n)-th row, reserve t as an empty vector of fixed length L(n) and for i=1..A119387(n+1), for j=1..L(n)-i+1 apply t := v (at the beginning of each cycle for i) and also apply v_j := Sum_{k=1..j+1} A078121(j,k-1)*t_k if R(n,L(n)-i) = 1, otherwise v_j := Sum_{k=1..j+1} A078121(j,k-1)*t_k*(-1)^(j+k+1). Here R(n,k) = floor(n/(2^k)) mod 2 is the (k+1)-th bit in the binary expansion of n.
Conjecture: sequence A(n,k) for fixed n is a polynomial of degree A070939(n).

			Array begins:
===========================================================
n\k|  0    1     2      3      4      5       6       7 ...
---+-------------------------------------------------------
1  |  2,   4,    6,     8,    10,    12,     14,     16 ...
2  |  4,  16,   36,    64,   100,   144,    196,    256 ...
3  |  6,  20,   42,    72,   110,   156,    210,    272 ...
4  | 10,  84,  286,   680,  1330,  2300,   3654,   5456 ...
5  | 14, 100,  322,   744,  1430,  2444,   3850,   5712 ...
6  | 20, 120,  364,   816,  1540,  2600,   4060,   5984 ...
7  | 26, 140,  406,   888,  1650,  2756,   4270,   6256 ...
8  | 36, 656, 3396, 10816, 26500, 55056, 102116, 174336 ...
  ...

Cf. A000123, A001511, A007814, A053645, A062383, A070939, A078121, A106400, A119387, A125790, A236206.

upto1(n) = my(v1); v1 = vector(n+1, i, vector(i, j, j==1 || j==i)); for(i=2, n, for(j=1, i-1, v1[i+1][j+1] = sum(k=j-1, i-1, v1[i][k+1]*v1[k+1][j]))); v1
A(n,m) = my(L = logint(n,2), A = valuation(n,2), B = logint(n>>A,2), v1, v2, v3); v1 = upto1(L+2); v2 = vector(L+2, i, vecsum(v1[i])); for(i=1, 2*m, v2 = vector(L+2, i, sum(j=1, i, v1[i][j]*v2[j]))); for(i=1, B, v3 = v2; for(j=1, L-i+1, v2[j+1] = sum(k=1, j+1, v1[j+1][k]*v3[k+1]*if(!bittest(n,L-i+1), (-1)^(j+k+1), 1)))); v2[A+2]

upto1(n) = my(v1); v1 = vector(n+1, i, vector(i, j, j==1 || j==i)); for(i=2, n, for(j=1, i-1, v1[i+1][j+1] = sum(k=j-1, i-1, v1[i][k+1]*v1[k+1][j]))); v1
upto2(n,m) = my(L = logint(n,2), A = valuation(n,2), B = logint(n>>A,2), v1, v2, v3, v4, v5); v1 = upto1(L+2); v2 = vector(L+2, i, 1); v3 = vector(m+1, i, 0); for(s=0, m, for(i=1, min(s+1,2), v2 = vector(L+2, i, sum(j=1, i, v1[i][j]*v2[j]))); v4 = v2; for(i=1, B, v5 = v4; for(j=1, L-i+1, v4[j+1] = sum(k=1, j+1, v1[j+1][k]*v5[k+1]*if(!bittest(n,L-i+1), (-1)^(j+k+1), 1)))); v3[s+1] = v4[A+2]); v3 \\ slightly modified version of the first program, some kind of memoization; generates A(n,k) for k=0..m

A(2^(n-1),k) = A125790(n,2*k+1) for n > 0, k >= 0.
Conjectured formulas: (Start)
A(n,0) = A000123(n) for n > 0.
A(n,k) = Sum_{j=0..k} A000123(A062383(n)*j+n)*A106400(k-j) for n > 0, k >= 0.
If we change v_i = A125790(i,2*k+1) to v_i = A125790(i,2*k) to get similar generalization of even columns, then for resulting array B(n,k) we have B(n,k) = Sum_{j=0..k} A000123(A062383(n)*j+A053645(n))*A106400(k-j) for n > 0, k >= 0.
2*(k+1) divides A(n,k) for n > 0 if (k+1) is a term of A236206.
G.f. for n-th row is f(A070939(n)+1,n) for n > 0 where f(n,k) = (Sum_{(c_0 + c_1 + ... + c_{n-1}) == 2*k (mod 2^n), 0 <= c_i < 2^n, 2^i divides c_i} x^((c_0 + c_1 + ... + c_{n-1} - 2*k)/2^n))/(1-x)^n for n > 0, k >= 0. Similarly, g.f. for n-th row of B(n,k) is f(A070939(n)+1,A053645(n)).
G.f. for n-th row is (Sum_{i=0..L(n)-1} x^i * Sum_{j=0..i} binomial(L(n)+1,j)*A(n,i-j)*(-1)^j)/(1-x)^(L(n)+1) for n > 0 where L(n) = A070939(n).
s(4*n+1) = 1 for n >= 0, s(4*n) = s(4*n+2) = 1 if A010060(n) = 1 for n > 0 where s(n) = A007814(Sum_{k=0..n-1} A(k+1,n-k-1)). (End)

A383410 Array read by downward antidiagonals: A(n,k) = Sum_{i=0..n-1} Sum_{j=0..k+1} binomial(n-1,i)binomial(k+1,j)A(i,j) with A(0,k) = 1, n >= 0, k >= 0.

Original entry on oeis.org

1, 1, 2, 1, 4, 8, 1, 8, 22, 44, 1, 16, 62, 154, 308, 1, 32, 178, 554, 1306, 2612, 1, 64, 518, 2038, 5690, 12994, 25988, 1, 128, 1522, 7634, 25366, 66338, 148282, 296564, 1, 256, 4502, 29014, 115298, 346366, 867002, 1908274, 3816548, 1, 512, 13378, 111554, 532726, 1844042, 5179798, 12564434, 27333706, 54667412

Offset: 0

Mikhail Kurkov, Apr 26 2025

			Array begins:
==================================================================
n\k|     0      1      2       3        4         5          6 ...
---+--------------------------------------------------------------
0  |     1      1      1       1        1         1          1 ...
1  |     2      4      8      16       32        64        128 ...
2  |     8     22     62     178      518      1522       4502 ...
3  |    44    154    554    2038     7634     29014     111554 ...
4  |   308   1306   5690   25366   115298    532726    2495570 ...
5  |  2612  12994  66338  346366  1844042   9985054   54865658 ...
6  | 25988 148282 867002 5179798 31540898 195320182 1227693842 ...
  ...

Cf. A005649.

A(m, n=m)={my(r=vectorv(m+1), v=vector(m+1, j, vector(n+m-j+2, k, (j==1)))); r[1] = v[1][1..n+1];
for(i=1, m, v[i+1] = vector(#v[i+1], k, sum(j=1, i, sum(q=1, k+1, binomial(i-1,j-1)*binomial(k,q-1)*v[j][q]))); r[1+i] = v[i+1][1..n+1]); Mat(r)}
{ A(6) }

Conjecture: A(n,0) = A005649(n).

A383019 Triangle T(n,k) read by rows (n >= 0, k >= 0) with g.f. 1/(1 - f(0)x - xy/(1 - f(1)x - xy/(1 - f(2)x - xy/(1 - f(3)x - xy/(1 - f(4)x - xy/(1 - ...)))))) where f(n) = n + 1 for n >= 0.

Original entry on oeis.org

1, 1, 1, 1, 4, 2, 1, 11, 16, 5, 1, 26, 80, 64, 14, 1, 57, 324, 490, 256, 42, 1, 120, 1170, 2944, 2730, 1024, 132, 1, 247, 3948, 15403, 22400, 14322, 4096, 429, 1, 502, 12776, 73960, 157564, 152064, 72072, 16384, 1430, 1, 1013, 40264, 335856, 1004400, 1368796, 953344, 351780, 65536, 4862

Offset: 0

Mikhail Kurkov, Apr 12 2025

Conjecture: T(n,k) is the number of paths of length n + k steps from (0,0) to (2*n,0) on or above the x-axis with steps U = (1,1), D = (1,-1), and L = (2,0), where the level steps L at height m have f(m) = m + 1 colors for m >= 0. The same seems to work for any f(n). Also, it looks like that if we modify g.f. to 1/(1 - f(0)*x - g(0)*x*y/(1 - f(1)*x - g(1)*x*y/(1 - f(2)*x - g(2)*x*y/(1 - f(3)*x - g(3)*x*y/(1 - f(4)*x - g(4)*x*y/(1 - ...)))))), then steps U at height m have g(m) colors for m >= 0.

			Triangle begins:
  1;
  1,    1;
  1,    4,     2;
  1,   11,    16,     5;
  1,   26,    80,    64,     14;
  1,   57,   324,   490,    256,     42;
  1,  120,  1170,  2944,   2730,   1024,   132;
  1,  247,  3948, 15403,  22400,  14322,  4096,   429;
  1,  502, 12776, 73960, 157564, 152064, 72072, 16384, 1430;
  ...

Cf. A302285 (row sums).

rows_upto(n) = my(A = 1, x = 'x, y = 'y, v1); forstep(j=n, 1, -1, A = 1 - j*x - x*y/A + x*O(x^n)); v1 = Vec(1/A); v1 = vector(n+1, i, Vecrev(v1[i]))

Conjecture: row polynomials are R(n,0,x) where R(n,k,x) = R(n-1,k+1,x) + x*Sum_{i=0..n-1} Sum_{j=0..k} binomial(k,j)*R(n-i-1,j,x)*R(i,k-j,x) for n > 0, k >= 0 with R(0,k,x) = 1 for k >= 0.

Conjecture: using modified g.f. given in the comments, the n-th row polynomial is v_n for n > 0 where we start with vector v of fixed length m with elements v_i = Sum_{j=0..i-1} f(j) + g(j)*x, reserve t as an empty vector of fixed length m and for i=1..m-1, for j=i+1..m, apply t := v (at the beginning of each cycle for i) and also apply v_j := [j = (i+1)]*f(0)*v_{j-1} + [j > (i+1)]*(v_{j-1} + f(j-i-1)*t_{j-1}) + g(j-i-1)*x*v_j.

A380971 Irregular triangle T(n, k), n >= 0, k > 0, read by rows with row polynomials R(n, x) such that R(2n+1, x) = xR(n, x) for n >= 0, R(2n,x) = wt(n)x*((x+1)^wt(n) - x^wt(n)) + Sum_{k=1..wt(n)} kx^kT(n,k) for n > 0 with R(0,x) = 0 where wt(n) = A000120(n).

Original entry on oeis.org

1, 2, 0, 1, 2, 4, 3, 0, 2, 2, 6, 0, 0, 1, 4, 12, 0, 2, 4, 3, 9, 9, 4, 0, 3, 2, 8, 0, 0, 2, 4, 16, 0, 2, 6, 3, 9, 12, 0, 0, 0, 1, 6, 28, 0, 4, 12, 3, 13, 21, 0, 0, 2, 4, 6, 27, 36, 0, 3, 9, 9, 4, 16, 24, 16, 5, 0, 4, 2, 10, 0, 0, 3, 4, 20, 0, 2, 8, 3, 9, 15, 0

Offset: 0

Mikhail Kurkov, Feb 10 2025

Row n length is A000120(n) except for n = 2^k - 1 which are empty rows.

			Irregular triangle begins:
  -
  -
  1;
  -
  2;
  0,  1;
  2,  4;
  -
  3;
  0,  2;
  2,  6;
  0,  0, 1;
  4, 12;
  0,  2, 4;
  3,  9, 9;
  -

Cf. A000120, A373183, A380179, A380944.

row(n) = if(n==0, [], my(x = 'x, A = 0, B = 0); forstep(i=logint(n, 2), 0, -1, A = if(bittest(n, i), B++; x*A, B*x*((x+1)^B - x^B) + sum(k=1, B, k*x^k*polcoeff(A, k, x)))); Vecrev(A/x))

Conjectures: (Start)
b(2^m*(2k+1)) = b(2^m*(2^wt(k)-1)) + Sum_{i=1..wt(k)} (i+1)^m*T(k,i)*(-1)^(wt(k)-i) for m >= 0, k >= 0 where b(n) = A380944(n) and where wt(n) = A000120(n). Note that this formula is recursive for k != 2^q - 1. We can also use b(2^m*(2^n-1)) = (n+1)^m - n*n!*c(m,n+1) for n >= 0, m >= 0 where c(n,k) = Sum_{i=0..n-k} Stirling2(k+i,k) for n >= 0, k >= 0.
A380179(n,k) = Sum_{i=0..2^(n+1)-1, [wt(i)<=(k+1)]*T(i,k+1)*(-1)^(wt(i)-k+1) for 0 <= k < n. (End)

A380944 a(n) = b(n,A000120(n)) for n >= 0 where b(n,k) is defined in Comments.

Original entry on oeis.org

1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 9, 3, 4, 1, 5, 4, 7, 3, 13, 5, 7, 2, 23, 9, 12, 3, 16, 4, 5, 1, 6, 5, 9, 4, 17, 7, 10, 3, 31, 13, 18, 5, 25, 7, 9, 2, 53, 23, 32, 9, 44, 12, 15, 3, 64, 16, 20, 4, 25, 5, 6, 1, 7, 6, 11, 5, 21, 9, 13, 4, 39, 17, 24, 7, 34, 10

Offset: 0

Mikhail Kurkov, Feb 09 2025

Here b(2n+1,k) = b(n,k) + b(n,k-1) for n >= 0, k > 0, b(2n,k) = (A000120(n)-k+1)*b(2n+1,k) for n > 0, k > 0 with T(n, 0) = 1 for n >= 0, T(0, k) = 0 for k > 0 (see A379817, A379819 for similar recurrence).

Cf. A000120, A379817, A379819.

b(n,k) = if(k==0, 1, if(n==0, 0, if(n%2, b((n-1)/2,k) + b((n-1)/2,k-1), (hammingweight(n/2)-k+1)*b(n+1,k))))
a(n) = b(n, hammingweight(n))

Conjecture: a(2^m*(2^n-1)) = (n+1)^m - n*n!*c(m,n+1) for n >= 0, m >= 0 where c(n,k) = Sum_{i=0..n-k} Stirling2(k+i,k) for n >= 0, k >= 0.

A380179 Triangle T(n,k) read by rows: T(n,k) = -binomial(n+1,k) + Sum_{i=0..k} Sum_{j=0..i+1} (i+1)^(n-i+j)(-1)^(k-i)/(j!(k-i)!) for 0 <= k <= n.

Original entry on oeis.org

1, 1, 1, 1, 5, 1, 1, 14, 14, 1, 1, 33, 68, 30, 1, 1, 72, 257, 218, 55, 1, 1, 151, 873, 1189, 553, 91, 1, 1, 310, 2812, 5734, 4094, 1204, 140, 1, 1, 629, 8802, 25916, 26484, 11598, 2352, 204, 1, 1, 1268, 27107, 112718, 158840, 96702, 28566, 4236, 285, 1

Offset: 0

Mikhail Kurkov, Jan 14 2025

			Triangle begins:
  1;
  1,    1;
  1,    5,     1;
  1,   14,    14,      1;
  1,   33,    68,     30,      1;
  1,   72,   257,    218,     55,     1;
  1,  151,   873,   1189,    553,    91,     1;
  1,  310,  2812,   5734,   4094,  1204,   140,    1;
  1,  629,  8802,  25916,  26484, 11598,  2352,  204,   1;
  1, 1268, 27107, 112718, 158840, 96702, 28566, 4236, 285, 1;

Cf. A347420.

T(n,k) = if(k >= 0 && n >= k, -binomial(n+1, k) + sum(i=0, k, sum(j=0, i+1, (i+1)^(n-i+j)*(-1)^(k-i)/(j!*(k-i)!))))

Conjecture: A347420(n) = 2^n + Sum_{k=1..n-1} T(n-1, k) for n >= 0.

A379819 Irregular table T(n, k), n >= 0, k >= 0, read by rows such that T(n,k) = f(n,k)/f(2^k-1,k) where f(n,k) is defined in Comments.

Original entry on oeis.org

1, 1, 1, 4, 2, 1, 3, 1, 10, 4, 4, 8, 2, 10, 13, 3, 1, 7, 6, 1, 22, 8, 10, 18, 4, 28, 30, 6, 4, 20, 14, 2, 49, 47, 9, 10, 36, 22, 3, 22, 56, 31, 4, 1, 15, 25, 10, 1, 46, 16, 22, 38, 8, 64, 64, 12, 10, 46, 30, 4, 118, 102, 18, 28, 88, 48, 6, 64, 138, 68, 8, 4

Offset: 0

Mikhail Kurkov, Jan 03 2025

Here f(n,k) = b(2^k*(2n+1)) - Sum_{j=1..k} b(2^(j-1)*(2n+1))*R(k,j) for n >= 0, k >= 0 where b(n) = A379818(n) and where R(k,j) is the unique solution to b(2^k*(2^i-1)) = Sum_{j=1..k} b(2^(j-1)*(2^i-1))*R(k,j) for k > 0, 1 <= i <= k.

Row n length is A000120(n) + 1.

			Irregular table begins:
   1;
   1,  1;
   4,  2;
   1,  3,  1;
  10,  4;
   4,  8,  2;
  10, 13,  3;
   1,  7,  6,  1;
  22,  8;
  10, 18,  4;
  28, 30,  6;
   4, 20, 14,  2;
  49, 47,  9;
  10, 36, 22,  3;
  22, 56, 31,  4;
   1, 15, 25, 10, 1;

Cf. A000120, A341392, A379818.

upto(n) = my(A, v1); v1 = vector(n+1, i, 0); v1[1] = 1; for(i=1, n, v1[i+1] = v1[i\2+1] + if(i%2, 0, A = 1 << valuation(i/2, 2); v1[i/2-A+1] + v1[i-A+1] + v1[i\(4*A)+1])); v1 \\ from A379818
R(k) = my(v1, M1, M2); v1 = upto(2^k*(2^k-1)); M1 = matrix(k, k, i, j, v1[2^(j-1)*(2^i-1)+1]); M2 = matrix(k, 1, i, j, v1[2^k*(2^i-1)+1]); M1 = matsolve(M1, M2)
row(n) = my(A = hammingweight(n), v1, v2, v3); v1 = upto(2^A*(2*n+1)); v2 = vector(A, i, R(i)); v3 = vector(A, i, (v1[2^i*(2*n+1)+1] - sum(j=1, i, v1[2^(j-1)*(2*n+1)+1]*v2[i][j,1]))/(v1[2^i*(2*(2^i-1)+1)+1] - sum(j=1, i, v1[2^(j-1)*(2*(2^i-1)+1)+1]*v2[i][j,1]))); concat(v1[n+1], v3)

Conjectures: (Start)
f(2^k-1,k) = (k+1)*A130032(k+1) for k >= 0.
T(2^n-1, k) = Stirling2(n+1, k+1) for n >= 0, 0 <= k <= n.
T(n,k) = c(n,wt(n)-k) for n >= 0, 0 <= k <= wt(n) where c(2n+1,k) = c(n,k) + (wt(n)-k+2)*c(n,k-1), c(2n,k) = (wt(n)-k+1)*c(2n+1,k) + [(wt(n)-k+1) > 0]*c(n,k-1) for n > 0, k > 0 with c(n,0) = A341392(n) for n >= 0, c(0,k) = 0 for k > 0 and where wt(n) = A000120(n). (End)

cp's OEIS Frontend

User: Mikhail Kurkov

A386381 Main diagonal of A386363.

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A386393 Triangle T(n, k) (1 <= k <= n) read by rows: T(n, k) is the numerator of R(n, k) where R(n, k) = R(n, k-1)/2 + R(n-1, k-1) for 1 < k <= n with R(n, 1) = R(n-1, n-1) for n > 1, R(1, 1) = 1.

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A386363 Variation of triangle of Entringer numbers (A008281) read by rows: T(n, k) = T(n, k-1) + (n-2)*T(n-1, n-k) for 1 < k <= n, T(n, 1) = T(n-1, n-1) for n > 0, T(n, 0) = 0^n.

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A381810 Array read by downward antidiagonals: A(n,k) is a generalization of odd columns of A125790 defined in Comments for n > 0, k >= 0.

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A383410 Array read by downward antidiagonals: A(n,k) = Sum_{i=0..n-1} Sum_{j=0..k+1} binomial(n-1,i)*binomial(k+1,j)*A(i,j) with A(0,k) = 1, n >= 0, k >= 0.

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A383019 Triangle T(n,k) read by rows (n >= 0, k >= 0) with g.f. 1/(1 - f(0)*x - x*y/(1 - f(1)*x - x*y/(1 - f(2)*x - x*y/(1 - f(3)*x - x*y/(1 - f(4)*x - x*y/(1 - ...)))))) where f(n) = n + 1 for n >= 0.

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A380971 Irregular triangle T(n, k), n >= 0, k > 0, read by rows with row polynomials R(n, x) such that R(2n+1, x) = x*R(n, x) for n >= 0, R(2n,x) = wt(n)*x*((x+1)^wt(n) - x^wt(n)) + Sum_{k=1..wt(n)} k*x^k*T(n,k) for n > 0 with R(0,x) = 0 where wt(n) = A000120(n).

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A380944 a(n) = b(n,A000120(n)) for n >= 0 where b(n,k) is defined in Comments.

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A383410 Array read by downward antidiagonals: A(n,k) = Sum_{i=0..n-1} Sum_{j=0..k+1} binomial(n-1,i)binomial(k+1,j)A(i,j) with A(0,k) = 1, n >= 0, k >= 0.

A383019 Triangle T(n,k) read by rows (n >= 0, k >= 0) with g.f. 1/(1 - f(0)x - xy/(1 - f(1)x - xy/(1 - f(2)x - xy/(1 - f(3)x - xy/(1 - f(4)x - xy/(1 - ...)))))) where f(n) = n + 1 for n >= 0.

A380971 Irregular triangle T(n, k), n >= 0, k > 0, read by rows with row polynomials R(n, x) such that R(2n+1, x) = xR(n, x) for n >= 0, R(2n,x) = wt(n)x*((x+1)^wt(n) - x^wt(n)) + Sum_{k=1..wt(n)} kx^kT(n,k) for n > 0 with R(0,x) = 0 where wt(n) = A000120(n).

A380179 Triangle T(n,k) read by rows: T(n,k) = -binomial(n+1,k) + Sum_{i=0..k} Sum_{j=0..i+1} (i+1)^(n-i+j)(-1)^(k-i)/(j!(k-i)!) for 0 <= k <= n.