A007526 -id:A007526 - cp's OEIS frontend

A097656 Binomial transform of A038507.

2, 4, 9, 24, 81, 358, 2021, 13828, 109857, 986922, 9865125, 108507160, 1302065441, 16926805678, 236975181189, 3554627504844, 56874039618753, 966858672535762, 17403456103546565, 330665665962928288, 6613313319249128577

Offset: 0

Ross La Haye, Sep 20 2004

nonn

			a(2) = 9 because P(2,0) = 1, P(2,1) = 2, P(2,2) = 2 while C(2,0) = 1, C(2,1) = 2, C(2,2) = 1 and 1 + 1 + 2 + 2 + 2 + 1 = 9.

Cf. A038507, A097204.

f[n_] := Sum[n!(k! + 1)/(k!(n - k)!), {k, 0, n}]; Table[ f[n], {n, 0, 20}] (* Robert G. Wilson v, Sep 24 2004 *)

a(n) = Sum_{k=0..n} n!*(k!+1) / (k!*(n-k)!) = Sum_{k=0..n} (P(n, k) + C(n, k)) = Sum_{k=0..n} P(n, k) + 2^n = A007526(n) + A000079(n). - Ross La Haye, Aug 24 2006

A097967 a(n) = Sum_{k=1..n} (P(n,k) + C(n,k)).

Original entry on oeis.org

0, 2, 7, 22, 79, 356, 2019, 13826, 109855, 986920, 9865123, 108507158, 1302065439, 16926805676, 236975181187, 3554627504842, 56874039618751, 966858672535760, 17403456103546563, 330665665962928286, 6613313319249128575

Offset: 0

Ross La Haye, Sep 21 2004

nonn

			a(2) = 7 because P(2,1) = 2, P(2,2) = 2 while C(2,1)= 2, C(2,2) = 1 and 2 + 2 + 2 + 1 = 7.

Cf. A097656, A007526, A000225.

A097967 := proc(n)
    add(n!*(k!+1)/k!/(n-k)!,k=1..n) ;
end proc: # R. J. Mathar, May 29 2013

f[n_] := Sum[n!(k! + 1)/(k!(n - k)!), {k, n}]; Table[ f[n], {n, 0, 20}] (* Robert G. Wilson v, Sep 24 2004 *)

a(n) = Sum_{k=1..n} n!(k!+1) / k!(n-k)! = Sum_{k=1..n} P(n, k)+2^n-1 = A007526(n) - A000225(n) - 1 = A097656(n) - 2.

Conjecture: a(n) +(-n-6)*a(n-1) +(6*n+7)*a(n-2) +(-13*n+14)*a(n-3) +4*(3*n-8)*a(n-4) +4*(-n+4)*a(n-5)=0. - R. J. Mathar, May 29 2013

Edited by Robert G. Wilson v, Sep 24 2004

A154987 Triangle read by rows: T(n,k) = t(n,k) + t(n,n-k), where t(n,k) = 2(n!/k!)(2*(n + k) - 1).

Original entry on oeis.org

-2, 4, 4, 13, 20, 13, 41, 69, 69, 41, 183, 268, 264, 268, 183, 1099, 1405, 1080, 1080, 1405, 1099, 7943, 9486, 5970, 4080, 5970, 9486, 7943, 65547, 75775, 43806, 20370, 20370, 43806, 75775, 65547, 604831, 685672, 384552, 149520, 77280, 149520, 384552, 685672, 604831

Offset: 0

Roger L. Bagula, Jan 18 2009

			     -2;
      4,     4;
     13,    20,    13;
     41,    69,    69,    41;
    183,   268,   264,   268,   183;
   1099,  1405,  1080,  1080,  1405,  1099;
   7943,  9486,  5970,  4080,  5970,  9486,  7943;
  65547, 75775, 43806, 20370, 20370, 43806, 75775, 65547;
  ...

G. C. Greubel, Rows n = 0..100 of the triangle, flattened

t:= proc(n,k) option remember; ## simplified t;
2*(n+k-1/2)*(n!/k!);
end proc:
A154987:= proc(n,k) ## n >= 0 and k = 0 .. n
t(n,k) + t(n,n-k)
end proc: # Yu-Sheng Chang, Apr 13 2020

(* First program *)
t[n_, k_]:= 2*n!*Gamma[n+k+1/2]/(k!*Gamma[n+k-1/2]);
T[n_, k_]:= t[n, k] + t[n,n-k];
Table[T[n,k], {n,0,10}, {k,0,n}]//Flatten
(* Second Program *)
T[n_, k_]:= Binomial[n, k]*((n-k)!*(2*n+2*k-1) + k!*(4*n-2*k-1));
Table[T[n, k], {n,0,10}, {k,0,n}]//Flatten (* G. C. Greubel, May 28 2020 *)

def T(n, k): return binomial(n, k)*(factorial(n-k)*(2*n+2*k-1) + factorial(k)*(4*n-2*k-1))
[[T(n, k) for k in (0..n)] for n in (0..10)] # G. C. Greubel, May 28 2020

T(n,k) = t(n,k) + t(n,n-k), where t(n,k) = 2*n!*Gamma(n + k + 1/2)/(k!*Gamma(n + k - 1/2)).
T(n,k) = t(n,k) + t(n,n-k), where t(n,k) = 2*(n+k-1/2)*(n!/k!). - Yu-Sheng Chang, Apr 13 2020
From G. C. Greubel, May 28 2020: (Start)
T(n,k) = binomial(n,k)*( (2*n+2*k-1)*(n-k)! + (4*n-2*k-1)*k! ).
T(n,n-k) = T(n,k), for k >= 0.
Sum_{k=0..n} T(n,k) = 2*( 2*n-1 + (2*n+1)*n!*e_{n-1}(1) ), where e_{n}(x) is the finite exponential function = Sum_{k=0..n} x^k/k!.
Sum_{k=0..n} T(n,k) = 2*( 2*n-1 + (2*n+1)*A007526(n) ).
T(n,0) = A175925(n-1) + 2*n.
T(n,1) = A007680(n) + A001107(n). (End)

Partially edited by Andrew Howroyd, Mar 26 2020

Additionally edited by G. C. Greubel, May 28 2020

A328290 Table T(b,n) = #{ k > 0 | nk has only distinct and nonzero digits in base b }, b >= 2, 1 <= n <= A051846(b).

Original entry on oeis.org

1, 4, 1, 0, 0, 1, 0, 1, 15, 5, 9, 0, 2, 3, 2, 0, 4, 1, 1, 0, 2, 1, 2, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 64, 42, 21, 9, 0, 14, 8, 4, 7, 0, 6, 4, 3, 6, 0, 3, 2, 5, 2, 0, 4, 5, 3, 2, 0, 0, 2, 1, 2, 0, 2, 1, 2, 1, 0, 1, 2, 1, 2, 0, 1, 3, 1, 1, 0, 1, 1, 2, 1, 0, 0, 0, 0, 2

Offset: 2

M. F. Hasler, Oct 11 2019

The table could also be considered as an infinite square array with T(b,n) = 0 for n > A051846(b) = the largest pandigital number in base b.

Can anyone find a simple formula for the index of the last terms > 1 in each row b?

			The table reads:  (column n >= 2 corresponds to the base)
   B \ n = 1      2      3      4      5      6      7     8      9      10  ...
   2       1     (0 ...)
   3       4      1      0      0      1      0      1    (0 ...)
   4      15      5      9      0      2      3      2     0      4       1  ...
   5      64     42     21      9      0     14      8     4      7       0  ...
   6     325    130     65     65    161      0     48    23     32      66  ...
   7    1956    651   1140    319    386    221      0   156    362     128  ...
   8   13699   5871   4506   1957   2748   1944   6277     0   1470    1189  ...
   9  109600  73588  27400  56930  21973  18397  15641  8305      0   14826  ...
  10  986409 438404 572175 219202 109601 255752 140515 109601 432645     0   ...
  (...)
In base 2, 1 is the only number with distinct nonzero digits, so T(2,1) = 1, T(2,n) = 0 for n > 1.
In base 3, {1, 2, 12_3 = 5, 21_3 = 7} are the only numbers with distinct nonzero digits, so T(3,1) = 4, T(3,2) = T(3,7) = T(3,7) = 1, T(3,n) = 0 for n > 7.
In base 4, {1, 2, 3, 12_4 = 6, 13_4 = 7, 21_4 = 9, ..., 321_4 = 57} are the only numbers with distinct nonzero digits, so T(4,n) = 0 for n > 57.

Cf. A328287 (row 10), A328288, A328277.

Column 1 is A007526 (number of nonnull variations of n distinct objects).

T(B,n)={my(S,T,U); for(L=1,B-1,T=vectorv(L,k,B^(k-1)); forperm(L,p, U=vecextract(T,p); forvec(D=vector(L,i,[1,B-1]),D*U%n||S++,2)));S}

T(b,b) = 0, since any multiple of b has a trailing digit 0 in base b.

T(b,A051846(b)) = 1 and T(b,n) = 0 for n > A051846(b) = (b-1)(b-2)..21 in base b.

A331798 E.g.f.: -log(1 - x) / ((1 - x) * (1 + log(1 - x))).

Original entry on oeis.org

0, 1, 5, 29, 204, 1714, 16862, 190826, 2447512, 35136696, 558727872, 9754239648, 185546362416, 3820734689472, 84687887312688, 2010622152615504, 50908186083448320, 1369376758488222336, 38998680958184088960, 1172297572938013827456, 37092793335394301708544

Offset: 0

Ilya Gutkovskiy, Jan 26 2020

nonn

Cf. A000254, A001008, A002805, A003713, A007526, A007840, A215916, A331797.

nmax = 20; CoefficientList[Series[-Log[1 - x]/((1 - x) (1 + Log[1 - x])), {x, 0, nmax}], x] Range[0, nmax]!
A007526[n_] := n! Sum[1/k!, {k, 0, n - 1}]; a[n_] := Sum[Abs[StirlingS1[n, k]] A007526[k], {k, 0, n}]; Table[a[n], {n, 0, 20}]
A007840[n_] := Sum[Abs[StirlingS1[n, k]] k!, {k, 0, n}]; a[n_] := Sum[Binomial[n, k] k! HarmonicNumber[k] A007840[n - k], {k, 1, n}]; Table[a[n], {n, 0, 20}]

a(n) = Sum_{k=0..n} = |Stirling1(n,k)| * A007526(k).
a(n) = Sum_{k=1..n} binomial(n,k) * k! * H(k) * A007840(n-k), where H(k) is the k-th harmonic number.
a(n) ~ n! / (1 - exp(-1))^(n+1). - Vaclav Kotesovec, Jan 26 2020

A348312 a(n) = n! * Sum_{k=0..n-1} 3^k / k!.

Original entry on oeis.org

0, 1, 8, 51, 312, 1965, 13248, 97839, 800208, 7260921, 72806040, 801515979, 9620317512, 125071036389, 1751016829968, 26265324194055, 420245416687392, 7144172815479921, 128595113003161512, 2443307154421058019, 48866143111666389720, 1026189005418216656541, 22576158119430894214368

Offset: 0

Ilya Gutkovskiy, Oct 11 2021

nonn

Cf. A007526, A053486, A066534, A348314.

Table[n! Sum[3^k/k!, {k, 0, n - 1}], {n, 0, 22}]
nmax = 22; CoefficientList[Series[x Exp[3 x]/(1 - x), {x, 0, nmax}], x] Range[0, nmax]!

a(n) = n!*sum(k=0, n-1, 3^k/k!); \\ Michel Marcus, Oct 11 2021

E.g.f.: x * exp(3*x) / (1 - x).
a(0) = 0; a(n) = n * (a(n-1) + 3^(n-1)).
a(n) ~ exp(3)*n!. - Stefano Spezia, Oct 11 2021

A348314 a(n) = n! * Sum_{k=0..n-1} 4^k / k!.

Original entry on oeis.org

0, 1, 10, 78, 568, 4120, 30864, 244720, 2088832, 19389312, 196514560, 2173194496, 26128665600, 339890756608, 4759410116608, 71395178280960, 1142340032364544, 19419853564641280, 349557673401188352, 6641597100292636672, 132831947503410872320, 2789470920661372502016

Offset: 0

Ilya Gutkovskiy, Oct 11 2021

nonn

Cf. A007526, A053487, A066534, A348312.

Table[n! Sum[4^k/k!, {k, 0, n - 1}], {n, 0, 21}]
nmax = 21; CoefficientList[Series[x Exp[4 x]/(1 - x), {x, 0, nmax}], x] Range[0, nmax]!

a(n) = n!*sum(k=0, n-1, 4^k/k!); \\ Michel Marcus, Oct 11 2021

E.g.f.: x * exp(4*x) / (1 - x).
a(0) = 0; a(n) = n * (a(n-1) + 4^(n-1)).
a(n) ~ exp(4)*n!. - Stefano Spezia, Oct 11 2021

A367962 Triangle read by rows. T(n, k) = Sum_{j=0..k} (n!/j!).

Original entry on oeis.org

1, 1, 2, 2, 4, 5, 6, 12, 15, 16, 24, 48, 60, 64, 65, 120, 240, 300, 320, 325, 326, 720, 1440, 1800, 1920, 1950, 1956, 1957, 5040, 10080, 12600, 13440, 13650, 13692, 13699, 13700, 40320, 80640, 100800, 107520, 109200, 109536, 109592, 109600, 109601

Offset: 0

Peter Luschny, Dec 06 2023

			  [0]   1;
  [1]   1,    2;
  [2]   2,    4,    5;
  [3]   6,   12,   15,   16;
  [4]  24,   48,   60,   64,   65;
  [5] 120,  240,  300,  320,  325,  326;
  [6] 720, 1440, 1800, 1920, 1950, 1956, 1957;

Cf. A094587, A000142 (T(n, 0)), A052849 (T(n, 1)), A000522 (T(n, n)), A007526 (T(n,n-1)), A038154 (T(n, n-2)), A355268 (T(n, n/2)), A367963(n) (T(2*n, n)/n!).

Cf. A001339 (row sums), A087208 (alternating row sums), A082030 (accumulated sums), A053482, A331689.

T := (n, k) -> add(n!/j!, j = 0..k):
seq(seq(T(n, k), k = 0..n), n = 0..9);

Module[{n=1},NestList[Append[n#,1+Last[#]n++]&,{1},10]] (* or *)
Table[Sum[n!/j!,{j,0,k}],{n,0,10},{k,0,n}] (* Paolo Xausa, Dec 07 2023 *)

from functools import cache
@cache
def a_row(n: int) -> list[int]:
    if n == 0: return [1]
    row = a_row(n - 1) + [0]
    for k in range(n): row[k] *= n
    row[n] = row[n - 1] + 1
    return row

def T(n, k): return sum(falling_factorial(n, n - j) for j in range(k + 1))
for n in range(9): print([T(n, k) for k in range(n + 1)])

T(n, k) = A094587(n, k) * A000522(k).
T(n, k) = e * (n! / k!) * Gamma(k + 1, 1).
Sum_{k=0..n} T(n, k) * 2^(n - k) = A053482(n).
Sum_{k=0..n} T(n, k) * binomial(n, k) = A331689(n).
Recurrence: T(n, n) = T(n, n-1) + 1 starting with T(0, 0) = 1.
For k <> n: T(n, k) = n * T(n-1, k).

A381888 Triangle read by rows: T(n, k) = (n + 1) * Sum_{j=k..n} binomial(n, j) * Eulerian1(j, j - k).

Original entry on oeis.org

1, 2, 2, 3, 9, 3, 4, 28, 28, 4, 5, 75, 165, 75, 5, 6, 186, 786, 786, 186, 6, 7, 441, 3311, 6181, 3311, 441, 7, 8, 1016, 12888, 40888, 40888, 12888, 1016, 8, 9, 2295, 47529, 241191, 404361, 241191, 47529, 2295, 9, 10, 5110, 168670, 1312750, 3445510, 3445510, 1312750, 168670, 5110, 10

Offset: 0

Peter Luschny, Mar 11 2025

Consider A381706, the number of permutations of k chosen numbers in [n] with i-1 descents, as a sequence of squares of size 1x1, 2x2, 3x3, ..., as displayed in the example section of A381706. Conjecture: T(n, k) is the sum of column k+1 of the (n+1)th square; in other words: T(n, k) = Sum_{j=0..n} b(n+1, j+1, k+1).

			Triangle starts:
  [0] 1;
  [1] 2,    2;
  [2] 3,    9,     3;
  [3] 4,   28,    28,      4;
  [4] 5,   75,   165,     75,      5;
  [5] 6,  186,   786,    786,    186,      6;
  [6] 7,  441,  3311,   6181,   3311,    441,     7;
  [7] 8, 1016, 12888,  40888,  40888,  12888,  1016,    8;
  [8] 9, 2295, 47529, 241191, 404361, 241191, 47529, 2295, 9;

Cf. A046802, A173018 (Eulerian1), A122045 (Euler), A058877 (column 1), A007526 (row sums), A381706 (generalized Eulerian).

T := (n, k) -> (n + 1)*add(binomial(n, j)*combinat:-eulerian1(j, j - k), j = k .. n):
for n from 0 to 8 do seq(T(n, k), k=0..n) od;
# Using the e.g.f.:
egf := ((y - 1)*(y*exp(x*y)*(x*y + 1) - (x + 1)*exp(x*(2*y - 1))))/(exp(x*(y - 1)) - y)^2:
ser := simplify(series(egf, x, 10)):
seq(seq(n!*coeff(coeff(ser, x, n), y, k), k = 0..n), n = 0..9);

# Using function eulerian1 from A173018.
def T(n: int, k: int) -> int:
    return (n + 1) * sum(binomial(n, j) * eulerian1(j, j-k) for j in (k..n))
def Trow(n) -> list[int]: return [T(n, k) for k in (0..n)]
for n in (0..8): print(f"{n}: ", Trow(n))

T(n, k) = n! * [y^k] [x^n] ((y - 1)*(y*exp(x*y)*(x*y + 1) - (x + 1)*exp(x*(2*y - 1)))) / (exp(x*(y - 1)) - y)^2.
Sum_{k=0..n} (-1)^k * T(n, k) = (-1)^n * (n + 1) * Euler(n).
T(n, k) = (n + 1) * A046802(n, k).

A081114 Triangle read by rows: T(n,k) = n*T(n-1,k) + n - k starting at T(n,n)=0.

Original entry on oeis.org

0, 1, 0, 4, 1, 0, 15, 5, 1, 0, 64, 23, 6, 1, 0, 325, 119, 33, 7, 1, 0, 1956, 719, 202, 45, 8, 1, 0, 13699, 5039, 1419, 319, 59, 9, 1, 0, 109600, 40319, 11358, 2557, 476, 75, 10, 1, 0, 986409, 362879, 102229, 23019, 4289, 679, 93, 11, 1, 0, 9864100, 3628799, 1022298, 230197, 42896, 6795, 934, 113, 12, 1, 0

Offset: 0

Henry Bottomley, Apr 16 2003

Taking the triangle into negative values of n and k would produce results close to (k+1)*e*n! - 1, i.e., one less than multiples of A000522 for nonnegative n.

			Triangle begins
    0;
    1,   0;
    4,   1,  0;
   15,   5,  1, 0;
   64,  23,  6, 1, 0;
  325, 119, 33, 7, 1, 0;

Columns include A007526 and A033312.

T(n,k) = if (k==n, 0, n*T(n-1,k) + n - k);
tabl(nn) = {for (n=0, nn, for (k=0, n, print1(T(n, k), ", ");); print(););} \\ Michel Marcus, Jun 16 2019

For k > 0, T(n, k) = ceiling((A001339(k-1)/(k-1)! - (k-1)*e) *n! - 1) where A001339(k-1) = ceiling((k-1)!*(k-1)*e) for k > 1.

T(n, 0) = floor(e*n! - 1) for n > 0; T(n, 1) = n! - 1. T(n, n)=0; T(n, n-1) = n+2; T(n, n-2) = n^2 + 3*n + 5 = A027688(n+1).

More terms from Michel Marcus, Jun 16 2019

cp's OEIS Frontend

A097656 Binomial transform of A038507.

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A097967 a(n) = Sum_{k=1..n} (P(n,k) + C(n,k)).

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A154987 Triangle read by rows: T(n,k) = t(n,k) + t(n,n-k), where t(n,k) = 2*(n!/k!)*(2*(n + k) - 1).

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A328290 Table T(b,n) = #{ k > 0 | nk has only distinct and nonzero digits in base b }, b >= 2, 1 <= n <= A051846(b).

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A331798 E.g.f.: -log(1 - x) / ((1 - x) * (1 + log(1 - x))).

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Mathematica

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A348312 a(n) = n! * Sum_{k=0..n-1} 3^k / k!.

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A348314 a(n) = n! * Sum_{k=0..n-1} 4^k / k!.

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A367962 Triangle read by rows. T(n, k) = Sum_{j=0..k} (n!/j!).

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A154987 Triangle read by rows: T(n,k) = t(n,k) + t(n,n-k), where t(n,k) = 2(n!/k!)(2*(n + k) - 1).