A062253 -id:A062253 - cp's OEIS frontend

A062254 3rd level triangle related to Eulerian numbers and binomial transforms (A062253 is second level, triangle of Eulerian numbers is first level and triangle with Z(0,0)=1 and Z(n,k)=0 otherwise is 0th level).

1, 6, 0, 25, 10, 0, 90, 120, 15, 0, 301, 896, 406, 21, 0, 966, 5376, 5586, 1176, 28, 0, 3025, 28470, 55560, 27910, 3123, 36, 0, 9330, 139320, 456525, 437100, 122520, 7860, 45, 0, 28501, 646492, 3312078, 5339719, 2912833, 494802, 19096, 55, 0

Offset: 0

Henry Bottomley, Jun 14 2001

Binomial transform of n^3*k^n is ((kn)^3 + 3(kn)^2 + (1 - k)(kn))*(k + 1)^(n - 3); of n^4*k^n is ((kn)^4 + 6(kn)^3 + (7 - 4k)(kn)^2 + (1 - 4k + k^2)(kn))*(k + 1)^(n - 4); of n^5*k^n is ((kn)^5 + 10(kn)^4 + (25 - 10k)(kn)^3 + (15 - 30k + 5k^2)(kn)^2 + (1 - 11k + 11k^2 - k^3)(kn))*(k + 1)^(n - 5); of n^6*k^n is ((kn)^6 + 15(kn)^5 + (65 - 20k)(kn)^4 + (90 - 120k + 15k^2)(kn)^3 + (31 - 146k + 91k^2 - 6k^3)(kn)^2 + (1 - 26k + 66k^2 - 26k^3 + k^4)(kn))*(k + 1)^(n - 6). This sequence gives the (unsigned) polynomial coefficients of (kn)^3.

			Rows start:
 (1),
 (6,0),
 (25,10,0),
 (90,120,15,0),
 ...

First column is A000392. Diagonals include A000007 and all but the start of A000217. Row sums are A000399.
Taking all the levels together to create a pyramid, one face would be A010054 as a triangle with a parallel face which is Pascal's triangle (A007318) with two columns removed, another face would be a triangle of Stirling numbers of the second kind (A008277) and a third face would be A000007 as a triangle, (cont.)
(cont.) with a triangle of Eulerian numbers (A008292), A062253, A062254 and A062255 as faces parallel to it. The row sums of this last group would provide a triangle of unsigned Stirling numbers of the first kind (A008275).

E(n, k) = if ((n<0) || (k<0), 0, if ((n==0) && (k==0), 1, (k+1)*E(n-1, k)+(n-k)*E(n-1, k-1)));
A2(n, k) = if ((n<0) || (k<0), 0, (k+2)*A2(n-1, k)+(n-k)*A2(n-1, k-1)+E(n, k));
A3(n, k) = if ((n<0) || (k<0), 0, (k+3)*A3(n-1, k)+(n-k)*A3(n-1, k-1) + A2(n, k));
row3(n) = vector(n+1, k, A3(n,k-1)); \\ Michel Marcus, Jan 27 2025

A(n, k) = (k+3)*A(n-1, k) + (n-k)*A(n-1, k-1) + A062253(n, k).

A062255 4th level triangle related to Eulerian numbers and binomial transforms (A062254 is third level, A062253 is second level, triangle of Eulerian numbers is first level and triangle with Z(0,0)=1 and Z(n,k)=0 otherwise is 0th level).

Original entry on oeis.org

1, 10, 0, 65, 20, 0, 350, 350, 35, 0, 1701, 3696, 1316, 56, 0, 7770, 30660, 24570, 4200, 84, 0, 34105, 220620, 325620, 131020, 12195, 120, 0, 145750, 1447050, 3513345, 2656720, 613140, 33330, 165, 0, 611501, 8901992, 33074448, 41503484, 18444833, 2634192, 87406, 220, 0

Offset: 0

Henry Bottomley, Jun 14 2001

Binomial transform of n^4*k^n is ((kn)^4 + 6(kn)^3 + (7 - 4k)(kn)^2 + (1 - 4k + k^2)(kn))*(k + 1)^(n - 4); of n^5*k^n is ((kn)^5 + 10(kn)^4 + (25 - 10k)(kn)^3 + (15 - 30k + 5k^2)(kn)^2 + (1 - 11k + 11k^2 - k^3)(kn))*(k + 1)^(n - 5); of n^6*k^n is ((kn)^6 + 15(kn)^5 + (65 - 20k)(kn)^4 + (90 - 120k + 15k^2)(kn)^3 + (31 - 146k + 91k^2 - 6k^3)(kn)^2 + (1 - 26k + 66k^2 - 26k^3 + k^4)(kn))*(k + 1)^(n - 6). This sequence gives the (unsigned) polynomial coefficients of (kn)^4.

			Rows start:
 (1),
 (10,0),
 (65,20,0),
 (350,350,35,0), etc.

First column is A000453. Diagonals include A000007 and all but the start of A000292. Row sums are A000454. Taking all the levels together to create a pyramid, one face would be A010054 as a triangle with a parallel face which is Pascal's triangle (A007318) with two columns removed, another face would be a triangle of Stirling numbers of the second kind (A008277) and a third face would be A000007 as a triangle, with a triangle of Eulerian numbers (A008292), A062253, A062254 and A062255 as faces parallel to it. The row sums of this last group would provide a triangle of unsigned Stirling numbers of the first kind (A008275).

E(n, k) = if ((n<0) || (k<0), 0, if ((n==0) && (k==0), 1, (k+1)*E(n-1, k)+(n-k)*E(n-1, k-1)));
A2(n, k) = if ((n<0) || (k<0), 0, (k+2)*A2(n-1, k)+(n-k)*A2(n-1, k-1)+E(n, k));
A3(n, k) = if ((n<0) || (k<0), 0, (k+3)*A3(n-1, k)+(n-k)*A3(n-1, k-1) + A2(n, k));
A4(n, k) = if ((n<0) || (k<0), 0, (k+4)*A4(n-1, k)+(n-k)*A4(n-1, k-1)+ A3(n, k));
row4(n) = vector(n+1, k, A4(n,k-1)); \\ Michel Marcus, Jan 27 2025

A(n, k) = (k+4)*A(n-1, k)+(n-k)*A(n-1, k-1) + A062254(n, k).

More terms from Michel Marcus, Jan 27 2025

A329369 Number of permutations of {1,2,...,m} with excedance set constructed by taking m-i (0 < i < m) if b(i-1) = 1 where b(k)b(k-1)...b(1)b(0) (0 <= k < m-1) is the binary expansion of n.

Original entry on oeis.org

1, 1, 3, 1, 7, 3, 7, 1, 15, 7, 17, 3, 31, 7, 15, 1, 31, 15, 37, 7, 69, 17, 37, 3, 115, 31, 69, 7, 115, 15, 31, 1, 63, 31, 77, 15, 145, 37, 81, 7, 245, 69, 155, 17, 261, 37, 77, 3, 391, 115, 261, 31, 445, 69, 145, 7, 675, 115, 245, 15, 391, 31, 63, 1, 127, 63

Offset: 0

Mikhail Kurkov, Nov 12 2019

Another version of A152884.
The excedance set of a permutation p of {1,2,...,m} is the set of indices i such that p(i) > i; it is a subset of {1,2,...,m-1}.
Great work on this subject was done by R. Ehrenborg and E. Steingrimsson, so most of the formulas given below are just their results translated into the language of the sequences which are related to the binary expansion of n.
Conjecture 1: equivalently, number of open tours by a biased rook on a specific f(n) X 1 board, which ends on a white cell, where f(n) = A070941(n) = floor(log_2(2n)) + 1 and cells are colored white or black according to the binary representation of 2n. A cell is colored white if the binary digit is 0 and a cell is colored black if the binary digit is 1. A biased rook on a white cell moves only to the left and otherwise moves only to the right. - Mikhail Kurkov, May 18 2021
Conjecture 2: this sequence is an inverse modulo 2 binomial transform of A284005. - Mikhail Kurkov, Dec 15 2021

			a(1) = 1 because the 1st excedance set is {m-1} and the permutations of {1,2,...,m} with such excedance set are 21, 132, 1243, 12354 and so on, i.e., for a given m we always have 1 permutation.
a(2) = 3 because the 2nd excedance set is {m-2} and the permutations of {1,2,...,m} with such excedance set are 213, 312, 321, 1324, 1423, 1432, 12435, 12534, 12543 and so on, i.e., for a given m we always have 3 permutations.
a(3) = 1 because the 3rd excedance set is {m-2, m-1} and the permutations of {1,2,...,m} with such excedance set are 231, 1342, 12453 and so on, i.e., for a given m we always have 1 permutation.

Mikhail Kurkov, Table of n, a(n) for n = 0..8191
Max Alekseyev, Recursion for the sum with Stirling numbers of both kinds, answer to question on MathOverflow (2024).
R. Ehrenborg and E. Steingrimsson, The excedance set of a permutation, Advances in Appl. Math., 24, 284-299, 2000.
Peter J. Taylor, Correctness of the algorithm for the A329369, A347205 and related sequences, answer to question on MathOverflow (2024).

Cf. A000120, A000523, A001511, A007814, A008292, A023416, A036987, A053645, A062253, A062254, A062255, A063250, A080791, A091344, A091347, A091348, A110501, A136126, A152884 (lexicographic order), A173018, A329718, A344902, A344947, A357990, A358612, A373183, A379817.
Similar recurrences: A006014, A124758, A217924, A243499, A284005, A290615, A341392, A347204, A347205.
Cf. A360288.

g:= proc(n) option remember;  2^padic[ordp](n, 2) end:
a:= proc(n) option remember; `if`(n=0, 1, (h-> a(h)+
     `if`(n::odd, 0, (t-> a(h-t)+a(n-t))(g(h))))(iquo(n, 2)))
    end:
seq(a(n), n=0..100);  # Alois P. Heinz, Jan 30 2023

a[n_] := a[n] = Which[n == 0, 1, OddQ[n], a[(n-1)/2], True, a[n/2] + a[n/2 - 2^IntegerExponent[n/2, 2]] + a[n - 2^IntegerExponent[n/2, 2]]];
a /@ Range[0, 65] (* Jean-François Alcover, Feb 13 2020 *)

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 \\ Mikhail Kurkov, Jun 06 2024

a(2n+1) = a(n) for n >= 0.
a(2n) = a(n) + a(n - 2^f(n)) + a(2n - 2^f(n)) for n > 0 with a(0) = 1 where f(n) = A007814(n) (equivalent to proposition 2.1 at the page 286, see R. Ehrenborg and E. Steingrimsson link).
a(2^m*(2n+1)) = Sum_{k=0..m} binomial(m+1,k) a(2^k*n) = a(2^m*n) + a(2^(m-1)*(2n+1)) + a(2^(m-1)*(4n+1)) for m > 0, n >= 0 (equivalent to proposition 2.5 at the page 287, see R. Ehrenborg and E. Steingrimsson link).
a(2n) = a(2*g(n)) + a(2n - 2^h(n)) + a(2*g(n) + 2^h(n)) for n > 0 with a(0) = 1 where g(n) = A053645(n), h(n) = A063250(n) (equivalent to proposition 2.1 at the page 286, see R. Ehrenborg and E. Steingrimsson link).
a(2n) = 2*a(n + g(n)) + a(2*g(n)) for n > 0, floor(n/3) < 2^(floor(log_2(n))-1) (in other words, for 2^m + k where 0 <= k < 2^(m-1), m > 0) with a(0) = 1 (just a special case of the previous formula, because for 2^m + k where 0 <= k < 2^(m-1), m > 0 we have 2^h(n) = n - g(n)).
a(2n) = a(f(n,-1)) + a(f(n,0)) + a(f(n,1)) for n > 0 with a(0) = 1 where f(n,k) = 2*(f(floor(n/2),k) + n mod 2) + k*A036987(n) for n > 1 with f(1,k) = abs(k) (equivalent to a(2n) = a(2*g(n)) + a(2n - 2^h(n)) + a(2*g(n) + 2^h(n))).
a(n) = Sum_{j=0..2^wt(n) - 1} (-1)^(wt(n) - wt(j)) Product_{k=0..wt(n) - 1} (1 + wt(floor(j/2^k)))^T(n,k) for n > 0 with a(0) = 1 where wt(n) = A000120(n), T(n,k) = T(floor(n/2), k - n mod 2) for k > 0 with T(n,0) = A001511(n) (equivalent to theorem 6.3 at page 296, see R. Ehrenborg and E. Steingrimsson link). Here T(n, k) - 1 for k > 0 is the length of the run of zeros between k-th pair of ones from the right side in the binary expansion of n. Conjecture 1: this formula is equivalent to inverse modulo 2 binomial transform of A284005.
Sum_{k=0..2^n-1} a(k) = (n+1)! for n >= 0.
a((4^n-1)/3) = A110501(n+1) for n >= 0.
a(2^2*(2^n-1)) = A091344(n+1),
a(2^3*(2^n-1)) = A091347(n+1),
a(2^4*(2^n-1)) = A091348(n+1).
More generally, a(2^m*(2^n-1)) = a(2^n*(2^m-1)) = S(n+1,m) for n >= 0, m >= 0 where S(n,m) = Sum_{k=1..n} k!*k^m*Stirling2(n,k)*(-1)^(n-k) (equivalent to proposition 6.5 at the page 297, see R. Ehrenborg and E. Steingrimsson link).
Conjecture 2: a(n) = (1 + A023416(n))*a(g(n)) + Sum_{k=0..floor(log_2(n))-1} (1-R(n,k))*a(g(n) + 2^k*(1 - R(n,k))) for n > 1 with a(0) = 1, a(1) = 1, where g(n) = A053645(n) and where R(n,k) = floor(n/2^k) mod 2 (at this moment this is the only formula here, which is not related to R. Ehrenborg's and E. Steingrimsson's work and arises from another definition given above, exactly conjectured definition with a biased rook). Here R(n,k) is the (k+1)-th bit from the right side in the binary expansion of n. - Mikhail Kurkov, Jun 23 2021
From Mikhail Kurkov, Jan 23 2023: (Start)
The formulas below are not related to R. Ehrenborg's and E. Steingrimsson's work.
Conjecture 3: a(n) = A357990(n, 1) for n >= 0.
Conjecture 4: a(2^m*(2k+1)) = Sum_{i=1..wt(k) + 2} i!*i^m*A358612(k, i)*(-1)^(wt(k) - i) for m >= 0, k >= 0 where wt(n) = A000120(n).
Conjecture 5: a(2^m*(2^n - 2^p - 1)) = Sum_{i=1..n} i!*i^m*(-1)^(n - i)*((i - p + 1)*Stirling2(n, i) - Stirling2(n - p, i - p) + Sum_{j=0..p-2} (p - j - 1)*Stirling2(n - p, i - j)/j! Sum_{k=0..j} (i - k)^p*binomial(j, k)*(-1)^k) for n > 2, m >= 0, 0 < p < n - 1. Here we consider that Stirling2(n, k) = 0 for n >= 0, k < 0. (End)
Conjecture 6: a(2^m*n + q) = Sum_{i=A001511(n+1)..A000120(n)+1} A373183(n, i)*a(2^m*(2^(i-1)-1) + q) for n >= 0, m >= 0, q >= 0. Note that this formula is recursive for n != 2^k - 1. Also, it is not related to R. Ehrenborg's and E. Steingrimsson's work. - Mikhail Kurkov, Jun 05 2024
From Mikhail Kurkov, Jul 10 2024: (Start)
a(2^m*(2^n*(2k+1) - 1)) = Sum_{i=1..m+1} a(2^i*k)*(-1)^(m-i+1)*Sum_{j=i..m+1} j^n*Stirling1(j, i)*Stirling2(m+1, j) for m >= 0, n >= 0, k >= 0 with a(0) = 1.
Proof: start with a(2^m*(2n+1)) = Sum_{k=0..m} binomial(m+1,k) a(2^k*n) given above and rewrite it as a(2^m*(2^n*(2k+1) - 1)) = Sum_{i=0..m} binomial(m+1, i) a(2^i*(2^(n-1)*(2k+1) - 1)).
Then conjecture that a(2^m*(2^n*(2k+1) - 1)) = Sum_{i=1..m+1} a(2^i*k)*f(n, m, i). From that it is obvious that f(0, m, i) = [i = (m+1)].
After that use a(2^m*(2^n*(2k+1) - 1)) = Sum_{i=0..m} binomial(m+1, i) Sum_{j=1..i+1} a(2^j*k)*f(n-1, i, j) = Sum_{i=1..m+1} a(2^i*k) Sum_{j=i-1..m} binomial(m+1, j)*f(n-1, j, i). From that it is obvious that f(n, m, i) = Sum_{j=i-1..m} binomial(m+1, j)*f(n-1, j, i).
Finally, all we need is to show that basic conditions and recurrence for f(n, m, i) gives f(n, m, i) = (-1)^(m-i+1)*Sum_{j=i..m+1} j^n*Stirling1(j, i)*Stirling2(m+1, j) (see Max Alekseyev link).
a(2^m*(2k+1)) = a(2^(m-1)*k) + (m+1)*a(2^m*k) + Sum_{i=1..m-1} a(2^m*k + 2^i) for m > 0, k >= 0.
Proof: start with a(2^(m+1)*(2k+1)) = a(2^m*k) + (m+2)*a(2^(m+1)*k) + Sum_{i=1..m} a(2^(m+1)*k + 2^i).
Then use a(2^m*(4k+1)) = a(2^m*k) + (m+1)*a(2^(m+1)*k) + Sum_{i=1..m-1} a(2^(m+1)*k + 2^i).
From that we get a(2^(m+1)*(2k+1)) - a(2^m*k) - (m+2)*a(2^(m+1)*k) - a(2^(m+1)*k + 2^m) = a(2^m*(4k+1)) - a(2^m*k) - (m+1)*a(2^(m+1)*k).
Finally, a(2^(m+1)*(2k+1)) = a(2^(m+1)*k) + a(2^m*(2*k+1)) + a(2^m*(4k+1)) which agrees with the a(2^m*(2n+1)) = a(2^m*n) + a(2^(m-1)*(2n+1)) + a(2^(m-1)*(4n+1)) given above.
This formula can be considered as an alternative to a(2^m*(2n+1)) = Sum_{k=0..m} binomial(m+1,k) a(2^k*n). There are algorithms for both these formulas that allow you to calculate them without recursion. However, even though it is necessary to calculate binomial coefficients in the mentioned formula, the triple-looped algorithm for it still works faster (see Peter J. Taylor link).
It looks like you can also change v2 in the mentioned algorithm to vector with elements a(2^m*(2^(i+A007814(n+1)-1)-1) + q) to get a(2^m*n + q) instead of a(n). This may have common causes with formula that uses A373183 given above. (End)
From Mikhail Kurkov, Jan 27 2025: (Start)
The formulas below are not related to R. Ehrenborg's and E. Steingrimsson's work.
Conjecture 7: A008292(n+1,k+1) = Sum_{i=0..2^n-1} [A000120(i) = k]*a(i) for n >= 0, k >= 0.
Conjecture 8: a(2^m*(2^n*(2k+1)-1)) = Sum_{i=0..m} Sum_{j=0..m-i} Sum_{q=0..i} binomial(m-i,j)*(m-j+1)^n*a(2^(q+1)*k)*L(m,i,q)*(-1)^j for m >= 0, n > 0, k >= 0 where L(n,k,m) = W(n-m,k-m,m+1) for n > 0, 0 <= k < n, 0 <= m <= k and where W(n,k,m) = (k+m)*W(n-1,k,m) + (n-k)*W(n-1,k-1,m) + [m > 1]*W(n,k,m-1) for 0 <= k < n, m > 0 with W(0,0,m) = 1, W(n,k,m) = 0 for n < 0 or k < 0.
In particular, W(n, k, 1) = A173018(n, k), W(n, k, 2) = A062253(n, k), W(n, k, 3) = A062254(n, k) and W(n, k, 4) = A062255(n, k).
Conjecture 9: a(n) = b(n,wt(n)) for n >= 0 where b(2n+1,k) = b(n,k) + (wt(n)-k+2)*b(n,k-1), b(2n,k) = (wt(n)-k+1)*b(2n+1,k) for n > 0, k > 0 with b(n,0) = A341392(n) for n >= 0, b(0,k) = 0 for k > 0 and where wt(n) = A000120(n) (see A379817).
More generally, a(2^m*(2k+1)) = ((m+1)!)^2*b(k,wt(k)-m) - Sum_{j=1..m} Stirling1(m+2,j+1)*a(2^(j-1)*(2k+1)) for m >= 0, k >= 0. Here we also consider that b(n,k) = 0 for k < 0. (End)
Conjecture 10: if we change b(n,0) = A341392(n) given above to b(n,0) = A341392(n)*x^n, then nonzero terms of the resulting polynomials for b(n,wt(n)) form c(n,k) such that a(n) = Sum_{k=0..A080791(n)} c(n,k) for n >= 0 where c(n,k) = (Product_{i=0..k-1} (1 + 1/A000120(floor(n/2^(A000523(n)-i))))) * Sum_{j=max{0,k-A080791(n)+A080791(A053645(n))}..A080791(A053645(n))} c(A053645(n),j) for n > 0, k >= 0 with c(0,0) = 1, c(0,k) = 0 for k > 0. - Mikhail Kurkov, Jun 19 2025

A293616 Array of generalized Eulerian number triangles read by ascending antidiagonals, with m >= 0, n >= 0 and 0 <= k <= n.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 3, 0, 0, 1, 6, 0, 1, 0, 1, 10, 0, 7, 1, 0, 1, 15, 0, 25, 4, 0, 0, 1, 21, 0, 65, 10, 0, 1, 0, 1, 28, 0, 140, 20, 0, 15, 4, 0, 1, 36, 0, 266, 35, 0, 90, 30, 1, 0, 1, 45, 0, 462, 56, 0, 350, 120, 5, 0, 0, 1, 55, 0, 750, 84, 0, 1050, 350, 15, 0, 1, 0

Offset: 0

Peter Luschny, Oct 14 2017

			Array starts:
m\j| 0   1  2     3    4  5       6       7    8  9      10      11      12
---|----------------------------------------------------------------------------
m=0| 1,  0, 0,    0,   0, 0,      0,      0,   0, 0,      0,      0,      0, ...
m=1| 1,  1, 0,    1,   1, 0,      1,      4,   1, 0,      1,     11,     11, ...
m=2| 1,  3, 0,    7,   4, 0,     15,     30,   5, 0,     31,    146,     91, ...
m=3| 1,  6, 0,   25,  10, 0,     90,    120,  15, 0,    301,    896,    406, ...
m=4| 1, 10, 0,   65,  20, 0,    350,    350,  35, 0,   1701,   3696,   1316, ...
m=5| 1, 15, 0,  140,  35, 0,   1050,    840,  70, 0,   6951,  11886,   3486, ...
m=6| 1, 21, 0,  266,  56, 0,   2646,   1764, 126, 0,  22827,  32172,   8022, ...
m=7| 1, 28, 0,  462,  84, 0,   5880,   3360, 210, 0,  63987,  76692,  16632, ...
m=8| 1, 36, 0,  750, 120, 0,  11880,   5940, 330, 0, 159027, 165792,  31812, ...
m=9| 1, 45, 0, 1155, 165, 0,  22275,   9900, 495, 0, 359502, 331617,  57057, ...
   A000217, A001296,A000292,A001297,A027789,A000332,A001298,A293610,A293611, ...
.
m\j| ...    13  14      15       16       17      18      19 20
---|----------------------------------------------------------------
m=0| ...,    0, 0,       0,       0,       0,      0,      0, 0, ...  [A000007]
m=1| ...,    1, 0,       1,      26,      66,     26,      1, 0, ...  [A173018]
m=2| ...,    6, 0,      63,     588,     868,    238,      7, 0, ...  [A062253]
m=3| ...,   21, 0,     966,    5376,    5586,   1176,     28, 0, ...  [A062254]
m=4| ...,   56, 0,    7770,   30660,   24570,   4200,     84, 0, ...  [A062255]
m=5| ...,  126, 0,   42525,  129780,   84630,  12180,    210, 0, ...
m=6| ...,  252, 0,  179487,  446292,  245322,  30492,    462, 0, ...
m=7| ...,  462, 0,  627396, 1315776,  625086,  68376,    924, 0, ...
m=8| ...,  792, 0, 1899612, 3444012, 1440582, 140712,   1716, 0, ...
m=9| ..., 1287, 0, 5135130, 8198190, 3063060, 270270,   3003, 0, ...
          A000389, A112494, A293612, A293613,A293614,A000579.
.
The parameter m runs over the triangles and j indexes the triangles by reading them by rows. Let T(m, n) denote the row [T(m, n, k) for 0 <= k <= n] and T(m) denote the triangle [T(m, n) for n >= 0]. Then for instance T(2) is the triangle A062253, T(4, 2) is row 2 of A062255 (which is [65, 20, 0]) and T(4, 2, 1) = 20.

Eric Weisstein's World of Mathematics, Nielsen Generalized Polylogarithm.

A000217(n) = T(n, 1, 0), A001296(n) = T(n, 2, 0), A000292(n) = T(n, 2, 1),
A001297(n) = T(n, 3, 0), A027789(n) = T(n, 3, 1), A000332(n) = T(n, 3, 2),
A001298(n) = T(n, 4, 0), A293610(n) = T(n, 4, 1), A293611(n) = T(n, 4, 2),
A000389(n) = T(n, 4, 3), A112494(n) = T(n, 5, 0), A293612(n) = T(n, 5, 1),
A293613(n) = T(n, 5, 2), A293614(n) = T(n, 5, 3), A000579(n) = T(n, 5, 4),
A144969(n) = T(n, 6, 0), A000580(n) = T(n, 6, 5), A000295(n) = T(1, n, 1),
A000460(n) = T(1, n, 2), A000498(n) = T(1, n, 3), A000505(n) = T(1, n, 4),
A000514(n) = T(1, n, 5), A001243(n) = T(1, n, 6), A001244(n) = T(1, n, 7),
A126646(n) = T(2, n, 0), A007820(n) = T(n, n, 0).
Cf. A173018, A062253, A062254, A062255, A293609.

A293616 := proc(m, n, k) option remember:
if m = 0 then m^n elif k < 0 or k > n then 0 elif n = 0 then 1 else
(k+m)*A293616(m,n-1,k) + (n-k)*A293616(m,n-1,k-1) + A293616(m-1,n,k) fi end:
for m in [$0..4] do for n in [$0..6] do print(seq(A293616(m, n, k), k=0..n)) od od;
# Sample uses:
A001298 := n -> A293616(n, 4, 0): A293614 := n -> A293616(n, 5, 3):
# Flatten:
a := proc(n) local w; w := proc(k) local t, s; t := 1; s := 1;
while t <= k do s := s + 1; t := t + s od; [s - 1, s - t + k] end:
seq(A293616(n - k, w(k)[1], w(k)[2]), k=0..n) end: seq(a(n), n = 0..11);

GenEulerianRow[0, n_] := Table[If[n==0 && j==0,1,0], {j,0,n}];
GenEulerianRow[m_, n_] := If[n==0,{1},Join[CoefficientList[x^(-m) (1 - x)^(n+m)
    PolyLog[-n-m, m, x] /. Log[1-x] -> 0, x], {0}]];
(* Sample use: *)
A173018Row[n_] := GenEulerianRow[1, n]; Table[A173018Row[n], {n, 0, 6}]

T(m, n, k) = (k + m)*T(m, n-1, k) + (n - k)*T(m, n-1, k-1) + T(m-1, n, k) with boundary conditions T(0, n, k) = 0^n; T(m, n, k) = 0 if k < 0 or k > n; and T(m, 0, k) = 0^k.

Let h(m, n) = x^(-m)*(1 - x)^(n + m)*PolyLog(-n - m, m, x) and p(m, n) the polynomial given by the expansion of h(m, n) after replacing log(1 - x) by 0. Then T(m, n, k) is the k-th coefficient of p(m, n) for 0 <= k < n.

A373183 Irregular table 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) = x(R(n, x+1) - R(n, x)) for n > 0 with R(0, x) = x.

Original entry on oeis.org

1, 0, 1, 1, 2, 0, 0, 1, 3, 4, 0, 1, 2, 1, 3, 3, 0, 0, 0, 1, 7, 8, 0, 3, 4, 3, 8, 6, 0, 0, 1, 2, 7, 15, 9, 0, 1, 3, 3, 1, 4, 6, 4, 0, 0, 0, 0, 1, 15, 16, 0, 7, 8, 7, 18, 12, 0, 0, 3, 4, 17, 34, 18, 0, 3, 8, 6, 3, 11, 15, 8, 0, 0, 0, 1, 2, 31, 57, 27, 0, 7, 15

Offset: 0

Mikhail Kurkov, May 27 2024

Row n length is A000120(n) + 1.

			Irregular table begins:
  1;
  0,  1;
  1,  2;
  0,  0, 1;
  3,  4;
  0,  1, 2;
  1,  3, 3;
  0,  0, 0, 1;
  7,  8;
  0,  3, 4;
  3,  8, 6
  0,  0, 1, 2
  7, 15, 9;
  0,  1, 3, 3;
  1,  4, 6, 4;
  0,  0, 0, 0, 1;

Cf. A000120, A001511, A007814, A053645, A062253, A062254, A062255, A062383, A087734, A090996, A173018, A279209, A290255, A329369, A341392, A357990, A358612.

row(n) = my(x = 'x, A = x); forstep(i=if(n == 0, -1, logint(n, 2)), 0, -1, A = if(bittest(n, i), x*A, x*(subst(A, x, x+1) - A))); Vecrev(A/x)

Conjectured formulas: (Start)
R(2n, x) = R(n, x) + R(n - 2^f(n), x) + R(2n - 2^f(n), x) where f(n) = A007814(n) (see A329369).
b(2^m*n + q) = Sum_{i=A001511(n+1)..A000120(n)+1} T(n, i)*b(2^m*(2^(i-1)-1) + q) for n >= 0, m >= 0, q >= 0 where b(n) = A329369(n). Note that this formula is recursive for n != 2^k - 1.
R(n, x) = c(n, x)
where c(2^k - 1, x) = x^(k+1) for k >= 0,
c(n, x) = Sum_{i=0..s(n)} p(n, s(n)-i)*Sum_{j=0..i} (s(n)-j+1)^A279209(n)*binomial(i, j)*(-1)^j,
p(n, k) = Sum_{i=0..k} c(t(n) + (2^i - 1)*A062383(t(n)), x)*L(s(n), k, i) for 0 <= k < s(n) with p(n, s(n)) = c(t(n) + (2^s(n) - 1)*A062383(t(n)), x),
s(n) = A090996(n), t(n) = A087734(n),
L(n, k, m) are some integer coefficients defined for n > 0, 0 <= k < n, 0 <= m <= k that can be represented as W(n-m, k-m, m+1)
and where W(n, k, m) = (k+m)*W(n-1, k, m) + (n-k)*W(n-1, k-1, m) + [m > 1]*W(n, k, m-1) for 0 <= k < n, m > 0 with W(0, 0, m) = 1, W(n, k, m) = 0 for n < 0 or k < 0.
In particular, W(n, k, 1) = A173018(n, k), W(n, k, 2) = A062253(n, k), W(n, k, 3) = A062254(n, k) and W(n, k, 4) = A062255(n, k).
Here s(n), t(n) and A279209(n) are unique integer sequences such that n can be represented as t(n) + (2^s(n) - 1)*A062383(t(n))*2^A279209(n) where t(n) is minimal. (End)
Conjectures from Mikhail Kurkov, Jun 19 2024: (Start)
T(n, k) = d(n, 1, A000120(n) - k + 2) where d(n, m, k) = (m+1)^g(n)*d(h(n), m+1, k) - m^(g(n)+1)*d(h(n), m, k-1) for n > 0, m > 0, k > 0 with d(n, m, 0) = 0 for n >= 0, m > 0, d(0, m, k) = [k <= m]*abs(Stirling1(m, m-k+1)) for m > 0, k > 0, g(n) = A290255(n) and where h(n) = A053645(n). In particular, d(n, 1, 1) = A341392(n).
Sum_{i=A001511(n+1)..wt(n)+k} d(n, k, wt(n)-i+k+1)*A329369(2^m*(2^(i-1)-1) + q) = k!*A357990(2^m*n + q, k) for n >= 0, k > 0, m >= 0, q >= 0 where wt(n) = A000120(n).
If we change R(0, x) to Product_{i=0..m-1} (x+i), then for resulting irregular table U(n, k, m) we have U(n, k, m) = d(n, m, A000120(n) - k + m + 1).
T(n, k) = (-1)^(wt(n)-k+1)*Sum_{i=1..wt(n)-k+3} Stirling1(wt(n)-i+3, k+1)*A358612(n, wt(n)-i+3) for n >= 0, k > 0 where wt(n) = A000120(n). (End)
Conjecture: T(2^m*(2k+1), q) = (-1)^(wt(k)-q)*Sum_{i=q..wt(k)+2} Stirling1(i,q)*A358612(k,i)*i^m for m >= 0, k >= 0, q > 0 where wt(n) = A000120(n). - Mikhail Kurkov, Jan 17 2025

cp's OEIS Frontend

A062254 3rd level triangle related to Eulerian numbers and binomial transforms (A062253 is second level, triangle of Eulerian numbers is first level and triangle with Z(0,0)=1 and Z(n,k)=0 otherwise is 0th level).

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A062255 4th level triangle related to Eulerian numbers and binomial transforms (A062254 is third level, A062253 is second level, triangle of Eulerian numbers is first level and triangle with Z(0,0)=1 and Z(n,k)=0 otherwise is 0th level).

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Extensions

A329369 Number of permutations of {1,2,...,m} with excedance set constructed by taking m-i (0 < i < m) if b(i-1) = 1 where b(k)b(k-1)...b(1)b(0) (0 <= k < m-1) is the binary expansion of n.

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A293616 Array of generalized Eulerian number triangles read by ascending antidiagonals, with m >= 0, n >= 0 and 0 <= k <= n.

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A373183 Irregular table 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) = x*(R(n, x+1) - R(n, x)) for n > 0 with R(0, x) = x.

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A293609 Triangle read by rows, a refinement of the central Stirling numbers of the first kind A187646, T(n, k) for n >= 0 and 0 <= k <= n.

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A373183 Irregular table 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) = x(R(n, x+1) - R(n, x)) for n > 0 with R(0, x) = x.