A327022 -id:A327022 - cp's OEIS frontend

A241171 Triangle read by rows: Joffe's central differences of zero, T(n,k), 1 <= k <= n.

1, 1, 6, 1, 30, 90, 1, 126, 1260, 2520, 1, 510, 13230, 75600, 113400, 1, 2046, 126720, 1580040, 6237000, 7484400, 1, 8190, 1171170, 28828800, 227026800, 681080400, 681080400, 1, 32766, 10663380, 494053560, 6972966000, 39502663200, 95351256000, 81729648000, 1, 131070, 96461910, 8203431600, 196556560200, 1882311631200, 8266953895200, 16672848192000, 12504636144000

Offset: 1

N. J. A. Sloane, Apr 22 2014

T(n,k) gives the number of ordered set partitions of the set {1,2,...,2*n} into k even sized blocks. An example is given below. Cf. A019538 and A156289. - Peter Bala, Aug 20 2014

			Triangle begins:
1,
1, 6,
1, 30, 90,
1, 126, 1260, 2520,
1, 510, 13230, 75600, 113400,
1, 2046, 126720, 1580040, 6237000, 7484400,
1, 8190, 1171170, 28828800, 227026800, 681080400, 681080400,
1, 32766, 10663380, 494053560, 6972966000, 39502663200, 95351256000, 81729648000,
...
From _Peter Bala_, Aug 20 2014: (Start)
Row 2: [1,6]
k  Ordered set partitions of {1,2,3,4} into k blocks    Number
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
1   {1,2,3,4}                                             1
2   {1,2}{3,4}, {3,4}{1,2}, {1,3}{2,4}, {2,4}{1,3},       6
    {1,4}{2,3}, {2,3}{1,4}
(End)

H. T. Davis, Tables of the Mathematical Functions. Vols. 1 and 2, 2nd ed., 1963, Vol. 3 (with V. J. Fisher), 1962; Principia Press of Trinity Univ., San Antonio, TX, Vol. 2, p. 283.
S. A. Joffe, Calculation of the first thirty-two Eulerian numbers from central differences of zero, Quart. J. Pure Appl. Math. 47 (1914), 103-126.
S. A. Joffe, Calculation of eighteen more, fifty in all, Eulerian numbers from central differences of zero, Quart. J. Pure Appl. Math. 48 (1917-1920), 193-271.

Muniru A Asiru, Rows n = 1..50 of triangle, flattened

Case m=2 of the polynomials defined in A278073.
Cf. A000680 (diagonal), A094088 (row sums), A000364 (alternating row sums), A281478 (central terms), A327022 (refinement).
Diagonals give A002446, A213455, A241172, A002456.
Cf. A019538, A036969, A156289.

Flat(List([1..10],n->List([1..n],k->1/(2^(k-1))*Sum([1..k],j->(-1)^(k-j)*Binomial(2*k,k-j)*j^(2*n))))); # Muniru A Asiru, Feb 27 2019

T := proc(n,k) option remember;
if k > n then 0
elif k=0 then k^n
elif k=1 then 1
else k*(2*k-1)*T(n-1,k-1)+k^2*T(n-1,k); fi;
end: # Minor edit to make it also work in the (0,0)-offset case. Peter Luschny, Sep 03 2022
for n from 1 to 12 do lprint([seq(T(n,k), k=1..n)]); od:

T[n_, k_] /; 1 <= k <= n := T[n, k] = k(2k-1) T[n-1, k-1] + k^2 T[n-1, k]; T[, 1] = 1; T[, ] = 0; Table[T[n, k], {n, 1, 9}, {k, 1, n}] (* _Jean-François Alcover, Jul 03 2019 *)

@cached_function
def A241171(n, k):
    if n == 0 and k == 0: return 1
    if k < 0 or k > n: return 0
    return (2*k^2 - k)*A241171(n - 1, k - 1) + k^2*A241171(n - 1, k)
for n in (1..6): print([A241171(n, k) for k in (1..n)]) # Peter Luschny, Sep 06 2017

T(n,k) = 0 if k <= 0 or k > n, = 1 if k=1, otherwise T(n,k) = k*(2*k-1)*T(n-1,k-1) + k^2*T(n-1,k).
Related to Euler numbers A000364 by A000364(n) = (-1)^n*Sum_{k=1..n} (-1)^k*T(n,k). For example, A000364(3) = 61 = 90 - 30 + 1.
From Peter Bala, Aug 20 2014: (Start)
T(n,k) = 1/(2^(k-1))*Sum_{j = 1..k} (-1)^(k-j)*binomial(2*k,k-j)*j^(2*n).
T(n,k) = k!*A156289(n,k) = k!*(2*k-1)!!*A036969.
E.g.f.: A(t,z) := 1/( 1 - t*(cosh(z) - 1) ) = 1 + t*z^2/2! + (t + 6*t^2)*z^4/4! + (t + 30*t^2 + 90*t^3)*z^6/6! + ... satisfies the partial differential equation d^2/dz^2(A) = D(A), where D = t^2*(2*t + 1)*d^2/dt^2 + t*(5*t + 1)*d/dt + t.
Hence the row polynomials R(n,t) satisfy the differential equation R(n+1,t) = t^2*(2*t + 1)*R''(n,t) + t*(5*t + 1)*R'(n,t) + t*R(n,t) with R(0,t) = 1, where ' indicates differentiation w.r.t. t. This is equivalent to the above recurrence equation.
Recurrence for row polynomials: R(n,t) = t*( Sum_{k = 1..n} binomial(2*n,2*k)*R(n-k,t) ) with R(0,t) := 1.
Row sums equal A094088(n) for n >= 1.
A100872(n) = (1/2)*R(n,2). (End)

A327023 Ordered set partitions of the set {1, 2, ..., 3*n} with all block sizes divisible by 3, irregular triangle T(n, k) for n >= 0 and 0 <= k < A000041(n), read by rows.

Original entry on oeis.org

1, 1, 1, 20, 1, 168, 1680, 1, 440, 924, 55440, 369600, 1, 910, 10010, 300300, 1261260, 33633600, 168168000, 1, 1632, 37128, 48620, 1113840, 24504480, 17153136, 326726400, 2058376320, 34306272000, 137225088000

Offset: 0

Peter Luschny, Aug 27 2019

T_{m}(n, k) gives the number of ordered set partitions of the set {1, 2, ..., m*n} into sized blocks of shape m*P(n, k), where P(n, k) is the k-th integer partition of n in the 'canonical' order A080577. Here we assume the rows of A080577 to be 0-based and m*[a, b, c,..., h] = [m*a, m*b, m*c,..., m*h]. Here is case m = 3. For instance 3*P(4, .) = [[12], [9, 3], [6, 6], [6, 3, 3], [3, 3, 3, 3]].

			Triangle starts (note the subdivisions by ';' (A072233)):
[0] [1]
[1] [1]
[2] [1;   20]
[3] [1;  168;  1680]
[4] [1;  440,   924;  55440;  369600]
[5] [1;  910, 10010; 300300, 1261260; 33633600; 168168000]
[6] [1; 1632, 37128,  48620; 1113840, 24504480,  17153136; 326726400, 2058376320;
     34306272000; 137225088000]
.
T(4, 1) = 440 because [9, 3] is the integer partition 3*P(4, 1) in the canonical order and there are 220 set partitions which have the shape [9, 3]. Finally, since the order of the sets is taken into account, one gets 2!*220 = 440.

Row sums: A243664, alternating row sums: A002115, main diagonal: A014606, central column A281479, by length: A278073.
Cf. A178803 (m=0), A133314 (m=1), A327022 (m=2), this sequence (m=3), A327024 (m=4).
Cf. A080577, A000041, A072233.

# uses[GenOrdSetPart from A327022]
def A327023row(n): return GenOrdSetPart(3, n)
for n in (0..6): print(A327023row(n))

A327024 Ordered set partitions of the set {1, 2, ..., 4*n} with all block sizes divisible by 4, irregular triangle T(n, k) for n >= 0 and 0 <= k < A000041(n), read by rows.

Original entry on oeis.org

1, 1, 1, 70, 1, 990, 34650, 1, 3640, 12870, 2702700, 63063000, 1, 9690, 251940, 26453700, 187065450, 17459442000, 305540235000, 1, 21252, 1470942, 2704156, 154448910, 8031343320, 9465511770, 374796021600, 3975514943400, 231905038365000, 3246670537110000

Offset: 0

Peter Luschny, Aug 27 2019

T_{m}(n, k) gives the number of ordered set partitions of the set {1, 2, ..., m*n} into sized blocks of shape m*P(n, k), where P(n, k) is the k-th integer partition of n in the 'canonical' order A080577. Here we assume the rows of A080577 to be 0-based and m*[a, b, c,..., h] = [m*a, m*b, m*c,..., m*h]. Here is case m = 4. For instance 4*P(4, .) = [[16], [12, 4], [8, 8], [8, 4, 4], [4, 4, 4, 4]].

			Triangle starts (note the subdivisions by ';' (A072233)):
[0] [1]
[1] [1]
[2] [1;    70]
[3] [1;   990;   34650]
[4] [1;  3640,   12870;  2702700;  63063000]
[5] [1;  9690,  251940; 26453700, 187065450; 17459442000; 305540235000]
[6] [1; 21252, 1470942,  2704156; 154448910,  8031343320,   9465511770;
     374796021600, 3975514943400; 231905038365000; 3246670537110000]
.
T(4, 1) = 3640 because [12, 4] is the integer partition 4*P(4, 1) in the canonical order and there are 1820 set partitions which have the shape [12, 4]. Finally, since the order of the sets is taken into account, one gets 2!*1820 = 3640.

Row sums: A243665, alternating row sums: A211212, main diagonal: A014608, central column: A281480, by length: A278074.
Cf. A178803 (m=0), A133314 (m=1), A327022 (m=2), A327023 (m=3), this sequence (m=4).
Cf. A080577, A000041, A072233.

# uses[GenOrdSetPart from A327022]
def A327024row(n): return GenOrdSetPart(4, n)
for n in (0..6): print(A327024row(n))

A356900 a(n) = P(n, 1/2) where P(n, x) = x^(-n)Sum_{k=0..n} A241171(n, k)x^k.

Original entry on oeis.org

1, 1, 8, 154, 5552, 321616, 27325088, 3200979664, 494474723072, 97390246272256, 23820397371219968, 7083386168647642624, 2516691244849530785792, 1052914814802404260765696, 512347915163742179541659648, 286902390859642414913802102784, 183187476890368376930869730803712

Offset: 0

Peter Luschny, Sep 03 2022

nonn

Other special values of this Euler type polynomials are: P(n, -1) = A000364(n); P(n, -1/2) = A002105(n); P(n, 1) = A094088(n), where we always make the assumption that the offset of the sequences is 0. A partition refinement of Joffe's triangle A241171 is A327022.

Cf. A241171, A327022, A000364, A002105, A094088, A269941.

a := n -> 2^n*add(A241171(n, k)*(1/2)^k, k = 0..n):
seq(a(n), n = 0..16);

# Using function PtransMatrix from A269941.
def E(n, v):
    eulr = lambda n: 1 / ((2 * n - 1) * (2 * n))
    norm = lambda n, k: (1 / v)^n * factorial(2 * n)
    P = PtransMatrix(n, eulr, norm)
    return [(-1)^j * sum([v^k * P[j][k] for k in range(j + 1)]) for j in range(n)]
A356900List = lambda n: E(n, -1/2); print(A356900List(17))
# A002105List = lambda n: E(n, 1/2) returns the reduced tangent numbers A002105.

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A241171 Triangle read by rows: Joffe's central differences of zero, T(n,k), 1 <= k <= n.

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A327023 Ordered set partitions of the set {1, 2, ..., 3*n} with all block sizes divisible by 3, irregular triangle T(n, k) for n >= 0 and 0 <= k < A000041(n), read by rows.

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A327024 Ordered set partitions of the set {1, 2, ..., 4*n} with all block sizes divisible by 4, irregular triangle T(n, k) for n >= 0 and 0 <= k < A000041(n), read by rows.

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A356900 a(n) = P(n, 1/2) where P(n, x) = x^(-n)*Sum_{k=0..n} A241171(n, k)*x^k.

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A356900 a(n) = P(n, 1/2) where P(n, x) = x^(-n)Sum_{k=0..n} A241171(n, k)x^k.