A024167 -id:A024167 - cp's OEIS frontend

A190782 Triangle T(n,k), read by rows, of the coefficients of x^k in the expansion of Sum_(m=0..n) binomial(x,m) = (a(k)*x^k)/n!, n >= 0, 0 <= k <= n.

1, 1, 1, 2, 1, 1, 6, 5, 0, 1, 24, 14, 11, -2, 1, 120, 94, 5, 25, -5, 1, 720, 444, 304, -75, 55, -9, 1, 5040, 3828, 364, 1099, -350, 112, -14, 1, 40320, 25584, 15980, -4340, 3969, -1064, 210, -20, 1

Offset: 0

Mokhtar Mohamed, Dec 29 2012

There is a strong relation between this triangle and triangle A048994 which deals with the binomial (x,n), this triangle being dealing with the summation of this binomial.

Apparently A054651 with reversed rows. - Mathew Englander, May 17 2014

			Triangle begins:
n\k     0       1       2       3       4       5       6      7     8
0       1
1       1       1
2       2       1       1
3       6       5       0        1
4      24      14      11       -2      1
5     120      94       5       25     -5       1
6     720     444     304      -75     55      -9       1
7    5040    3828     364     1099   -350     112     -14      1
8   40320   25584   15980    -4340   3969   -1064     210    -20     1
...

Seiichi Manyama, Rows n = 0..139, flattened

T(2*n,n) gives A347987.
Column 0-5 give A000142, A024167, A348063, A348064, A348065, A348068.
Cf. A048994, A054651, A132393.

row[n_] := CoefficientList[ Series[ Sum[ Binomial[x, m], {m, 0, n}], {x, 0, n}], x]*n!; Table[row[n], {n, 0, 8}] // Flatten (* Jean-François Alcover, Jan 04 2013 *)

T(n,k) = T(n-1,k)+ T(n-1,k-1)- T(n-2,k-1)*(n-1)+ T(n-2,k)*(n-1)^2, T(n,n)=1, T(n,0)= n! for n >= 0.
T(n,k) = T(n-1,k)*n + (A048994(n,k)), T(n,n)= 1, T(n,0)= n! for n>= 0.
E.g.f. of column k: (log(1 + x))^k/(k! * (1 - x)). - Seiichi Manyama, Sep 26 2021
T(n, k) = Sum_{i=0..n-k} Stirling1(i+k, k)*n!/(i+k)!. - Igor Victorovich Statsenko, May 27 2024

A075830 Let u(1) = x and u(n+1) = (n^2/u(n)) + 1 for n >= 1; then a(n) is such that u(n) = (b(n)x + c(n))/(a(n)x + d(n)) (in lowest terms) and a(n), b(n), c(n), d(n) are positive integers.

Original entry on oeis.org

0, 1, 1, 5, 7, 47, 37, 319, 533, 1879, 1627, 20417, 18107, 263111, 237371, 52279, 95549, 1768477, 1632341, 33464927, 155685007, 166770367, 156188887, 3825136961, 3602044091, 19081066231, 18051406831, 57128792093, 7751493599

Offset: 1

Benoit Cloitre, Oct 14 2002

nonn

For x real <> 1 - 1/log(2), Lim_{n -> infinity} abs(u(n)-n) = abs((x - 1)/(1 + (x - 1)*log(2))). [Corrected by Petros Hadjicostas, May 18 2020]
From Petros Hadjicostas, May 05 2020: (Start)
Given x > 0, u(n) = (A075827(n)*x + A075828(n))/(a(n)*x + A075829(n)) = (b(n)*x + c(n))/(a(n)*x + d(n)) with gcd(gcd(b(n), c(n)), gcd(a(n), d(n))) = 1 for each n >= 1.
Conjecture 1: Define the sequences (A(n): n >= 1) and (B(n): n >= 1) by A(n+1) = n^2/A(n) + 1 for n >= 2 with A(1) = infinity and A(2) = 1, and B(n+1) = n^2/B(n) + 1 for n >= 3 with B(1) = 0, B(2) = infinity, and B(3) = 1. Then a(n) = denominator(A(n)), b(n) = numerator(A(n)), c(n) = numerator(B(n)), and d(n) = denominator(B(n)) (assuming infinity = 1/0). Also, gcd(a(n), d(n)) = 1.
In 2002, Michael Somos claimed that d(n) = A024168(n-1)/gcd(A024168(n-1), A024168(n)) for n >= 2. In 2006, N. J. A. Sloane claimed that a(n) = A058313(n-1) for n >= 2 while Alexander Adamchuk claimed that d(n) = A058312(n-1) - A058313(n-1) for n >= 2.
Conjecture 2: a(n) = A024167(n-1)/gcd(A024167(n-1), A024167(n)).
Conjecture 3: b(p) = a(p+1) for p = 1 or prime. In general, it seems that b(n) = A048671(n)*a(n+1) for all n for which A048671(n) < n.
Conjecture 4: c(n) = n*(a(n) + d(n)) - b(n) for n >= 1. (End)
All conjectures are proved in the link below except for the second part of Conjecture 3. - Petros Hadjicostas, May 21 2020

Petros Hadjicostas, Proofs of various results about the sequence u(n), 2020.

Apart from the leading term, same as A058313.
Cf. A075827 (= b), A075828 (= c), A075829 (= d).
Cf. A024167, A024168, A048671, A058312.

u(n)=if(n<2,x,(n-1)^2/u(n-1)+1);
a(n)=polcoeff(denominator(u(n)),1,x);

Name edited by Petros Hadjicostas, May 04 2020

A334958 GCD of consecutive terms of the factorial times the alternating harmonic series.

Original entry on oeis.org

1, 1, 1, 2, 2, 12, 12, 48, 144, 1440, 1440, 17280, 17280, 241920, 18144000, 145152000, 145152000, 2612736000, 2612736000, 10450944000, 219469824000, 4828336128000, 4828336128000, 115880067072000, 579400335360000, 15064408719360000, 135579678474240000, 26573616980951040000, 26573616980951040000

Offset: 1

Petros Hadjicostas, May 17 2020

nonn

For n = 1..14, we have a(n) = A025527(n), but a(15) = 18144000 <> 3628800 = A025527(15).
It appears that A025527(n) | a(n) for all n >= 1 and A025527(n) = a(n) for infinitely many n. In addition, it seems that a(n)/a(n-1) = A048671(n) for infinitely many n >= 2. However, I have not established these claims.
This sequence appears in formulas for sequences A075827, A075828, A075829, and A075830 (the first one of which was established in 2002 by Michael Somos).
Conjecture: a(n) = n! * Product_{p <= n} p^min(0, v_p(H'(n))), where the product ranges over primes p, H'(n) = Sum_{k=1..n} (-1)^(k+1)/k, and v_p(r) is the p-adic valuation of rational r (checked for n < 1100).

			A024167(4) = 4!*(1 - 1/2 + 1/3 - 1/4) = 14, A024167(5) = 5!*(1 - 1/2 + 1/3 - 1/4 + 1/5) = 94, A024168(4) = 4!*(1/2 - 1/3 + 1/4) = 10, and A024168(5) = 5!*(1/2 - 1/3 + 1/4 - 1/5) = 26. Then a(4) = gcd(14, 94) = gcd(10, 26) = gcd(14, 4!) = gcd(10, 4!) = gcd(14, 10) = 2.

Cf. A000142, A024167, A024168, A025527, A048671, A058312, A058313, A075827, A075828, A075829, A075830.

Cf. A056612 (similar sequence for the harmonic series).

b:= proc(n) b(n):= (-(-1)^n/n +`if`(n=1, 0, b(n-1))) end:
a:= n-> (f-> igcd(b(n)*f, f))(n!):
seq(a(n), n=1..30);  # Alois P. Heinz, May 18 2020

b[n_] := b[n] = -(-1)^n/n + If[n == 1, 0, b[n-1]];
a[n_] := GCD[b[n] #, #]&[n!];
Array[a, 30] (* Jean-François Alcover, Oct 27 2020, after Alois P. Heinz *)

def A():
    a, b, n = 1, 1, 2
    while True:
        yield gcd(a, b)
        b, a = a, a + b * n * n
        n += 1
a = A(); print([next(a) for  in range(29)]) # _Peter Luschny, May 19 2020

a(n) = gcd(A024167(n+1), A024167(n)) = gcd(A024168(n+1), A024168(n)) = gcd(A024167(n), n!) = gcd(A024168(n), n!) = gcd(A024167(n), A024168(n)).

A142980 a(1) = 1, a(2) = 5, a(n+2) = 5a(n+1) + (n + 1)^2a(n).

Original entry on oeis.org

1, 5, 29, 190, 1414, 11820, 110004, 1129200, 12686256, 154896480, 2043108000, 28958014080, 438997622400, 7088892491520, 121487996448000, 2202440792832000, 42113131054848000, 847071044402688000, 17880009683784192000, 395192695448291328000, 9127967350755133440000

Offset: 1

Peter Bala, Jul 17 2008

This is the case m = 2 of the more general recurrence a(1) = 1, a(2) = 2*m + 1, a(n+2) = (2*m + 1)*a(n+1) + (n + 1)^2*a(n), which arises when accelerating the convergence of Mercator's series for the constant log(2). See A142979 for remarks on the general case.

Seiichi Manyama, Table of n, a(n) for n = 1..448

Cf. A024167, A135148, A142979, A142981, A142982.

p := n -> 2*n^2+ 2*n+1: a := n -> n!*p(n)*sum ((-1)^(k+1)/(k*p(k-1)*p(k)), k = 1..n): seq(a(n), n = 1..20)

Module[{a, n}, RecurrenceTable[{a[n+2] == 5*a[n+1] + (n+1)^2*a[n], a[1] == 1, a[2] == 5}, a, {n, 25}]] (* Paolo Xausa, Dec 12 2024 *)

a(n) = n!*p(n)*Sum_{k = 1..n} (-1)^(k+1)/(k*p(k-1)*p(k)), where p(n) = 2*n^2 + 2*n + 1 = A001844(n) is the Ehrhart polynomial for the 2-dimensional cross polytope (a square).
Recurrence: a(1) = 1, a(2) = 5, a(n+2) = 5*a(n+1) + (n+1)^2*a(n).
The sequence b(n) := n!*p(n) satisfies the same recurrence with b(1) = 5, b(2) = 26.
Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(5 + 1^2/(5 + 2^2/(5 + 3^2/(5 + ... + (n-1)^2/5)))), for n >= 2.
The behavior of a(n) for large n is given by lim_{n -> oo} a(n)/b(n) = 1/(5 + 1^2/(5 + 2^2/(5 + 3^2/(5 + ... + n^2/(5 + ...))))) = Sum_{k >= 1} (-1)^(k+1)/(k*(4*k^4 + 1)) = log(2) - (1 - 1/2); the final equality is a result of Glaisher.
Thus a(n) ~ c*n^2*n! as n -> oo, where c = 2*log(2) - 1.
From Peter Bala, Dec 09 2024: (Start)
E.g.f.: A(x) = ((1 + x)^2 *log(1 + x) - 2*x^2)/(1 - x)^3 satisfies the differential equation 1 + (x + 5)*A(x) + (x^2 - 1)*A(x)' with A(0) = 0.
Sum_{k = 1..n} Stirling2(n, k)*a(k) = A135148(n+1). (End)

A281964 Real part of n!*Sum_{k=1..n} i^(k-1)/k, where i is sqrt(-1).

Original entry on oeis.org

1, 2, 4, 16, 104, 624, 3648, 29184, 302976, 3029760, 29698560, 356382720, 5111976960, 71567677440, 986336870400, 15781389926400, 289206418636800, 5205715535462400, 92506221468057600, 1850124429361152000, 41285515024760832000, 908281330544738304000

Offset: 1

Daniel Suteu, Feb 06 2017

nonn

			For n=5, a(5) = 104, which is the real part of 5!*(1/1 + i/2 - 1/3 - i/4 + 1/5) = 104+30*i.

Iain Fox, Table of n, a(n) for n = 1..450 (first 100 terms from Daniel Suteu)

The corresponding imaginary part is A282132.

Cf. A003881, A024167.

a(n) = real(n!*sum(k=1, n, I^(k-1)/k));

first(n) = x='x+O('x^(n+1)); Vec(serlaplace(atan(x)/(1 - x))) \\ Iain Fox, Dec 19 2017

a(n) ~ Pi/4 * n!.
a(1) = 1, a(n+1) = a(n)*(n+1) + n!*cos(Pi*n/2).
E.g.f.: arctan(x)/(1 - x). - Ilya Gutkovskiy, Dec 19 2017

A075827 Let u(1) = x and u(n+1) = (n^2/u(n)) + 1 for n >= 1; then a(n) is such that u(n) =(a(n)x + b(n))/(c(n)x + d(n)) (in lowest terms) and a(n), b(n), c(n), d(n) are positive integers.

Original entry on oeis.org

1, 1, 5, 14, 47, 222, 319, 2132, 5637, 16270, 20417, 217284, 263111, 3323194, 3920925, 764392, 1768477, 29382138, 33464927, 622740028, 3502177707, 3436155514, 3825136961, 86449058184, 95405331155, 469336577606, 514159128837, 1519292745404, 236266661971, 6755272778730, 7313175618421

Offset: 1

Benoit Cloitre, Oct 14 2002

nonn

For x real <> 1 - 1/log(2), Lim_{n -> infinity} abs(u(n) - n) = abs((x - 1)/(1 + (x - 1)*log(2))). [Corrected by Petros Hadjicostas, May 18 2020]

Petros Hadjicostas, Proofs of various results about the sequence u(n), 2020.

Cf. A075828 (= b), A075829 (= d), A075830 (= c).

Cf. A024167, A024168, A334958.

u(n) = if(n<2, x, (n-1)^2/u(n-1)+1);
a(n) = polcoeff(numerator(u(n)), 1, x);
for(n=1, 30, print1(a(n)", ")) \\ Petros Hadjicostas, May 06 2020

From Petros Hadjicostas, May 18 2020: (Start)
a(n) = A024167(n)/gcd(A024167(n), A024167(n-1)) = A024167(n)/A334958(n-1) for n >= 2. (Cf. Michael Somos's result for d = A075829 using A024168.)
u(n) = (A024167(n)*x + A024168(n))/(A024167(n-1)*x + A024168(n-1)) for n >= 2. (End)

Name edited by Petros Hadjicostas, May 06 2020

More terms from Michel Marcus, Aug 01 2025

A075828 Let u(1) = x and u(n+1) = (n^2/u(n)) + 1 for n >= 1; then a(n) is such that u(n) =(b(n)x + a(n))/(c(n)x + d(n)) (in lowest terms) and a(n), b(n), c(n), d(n) are positive integers.

Original entry on oeis.org

0, 1, 1, 10, 13, 138, 101, 1228, 1923, 8930, 7303, 115356, 97249, 1721846, 1484475, 388760, 681971, 14725926, 13093585, 308430212, 1386466053, 1685280806, 1529091919, 42052434936, 38450390845, 226713176794, 208661769963

Offset: 1

Benoit Cloitre, Oct 14 2002

nonn

For x real <> 1 - 1/log(2), Lim_{n -> infinity} abs(u(n) - n) = abs((x - 1)/(1 + (x - 1)*log(2))). [Corrected by Petros Hadjicostas, May 18 2020]

Petros Hadjicostas, Proofs of various results about the sequence u(n), 2020.

Cf. A075827 (= b), A075829 (= d), A075830 (= c).

Cf. A024167, A024168, A334958.

u(n) = if(n<2, x, (n-1)^2/u(n-1)+1);
a(n) = polcoeff(numerator(u(n)), 0 ,x)

From Petros Hadjicostas, May 18 2020: (Start)
a(n) = A024168(n)/gcd(A024168(n), A024168(n-1)) = A024168(n)/A334958(n) for n >= 2. (Cf. Michael Somos's claim for d = A075829 using A024168.)
u(n) = (A024167(n)*x + A024168(n))/(A024167(n-1)*x + A024168(n-1)) for n >= 2. (End)

Name edited by Petros Hadjicostas, May 06 2020

A142981 a(1) = 1, a(2) = 7, a(n+2) = 7a(n+1) + (n+1)^2a(n).

Original entry on oeis.org

1, 7, 53, 434, 3886, 38052, 406260, 4708368, 58959216, 794092320, 11454567840, 176267145600, 2883327788160, 49972442123520, 914939341344000, 17648374867200000, 357763095454464000, 7604722004802048000, 169148296960860672000, 3929342722459564032000, 95164717841561217024000

Offset: 1

Peter Bala, Jul 17 2008

This is the case m = 3 of the more general recurrence a(1) = 1, a(2) = 2*m + 1, a(n+2) = (2*m + 1)*a(n+1) + (n + 1)^2*a(n), which arises when accelerating the convergence of Mercator's series for the constant log(2). See A142979 for remarks on the general case.

Bruce C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag.

Seiichi Manyama, Table of n, a(n) for n = 1..447

Cf. A001845, A024167, A142979, A142980, A142982.

p := n -> (4*n^3+6*n^2+8*n+3)/3: a := n -> n!*p(n)*sum ((-1)^(k+1)/(k*p(k-1)*p(k)), k = 1..n): seq(a(n), n = 1..20)

Module[{a, n}, RecurrenceTable[{a[n+2] == 7*a[n+1] + (n+1)^2*a[n], a[1] == 1, a[2] == 7}, a, {n, 25}]] (* Paolo Xausa, Dec 12 2024 *)

a(n) = n!*p(n)*Sum_{k = 1..n} (-1)^(k+1)/(k*p(k-1)*p(k)), where p(n) = (4*n^3 + 6*n^2 + 8*n + 3)/3 = A001845(n) is the Ehrhart polynomial for the 3-dimensional cross polytope (the octahedron).
Recurrence: a(1) = 1, a(2) = 7, a(n+2) = 7*a(n+1) + (n + 1)^2*a(n). The sequence b(n):= n!*p(n) satisfies the same recurrence with b(1) = 7, b(2) = 50.
Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(7 + 1^2/(7 + 2^2/(7 + 3^2/(7 + ... + (n-1)^2/7)))), for n >= 2.
The behavior of a(n) for large n is given by limit_{n -> oo} a(n)/b(n) = Sum_{k >= 1} (-1)^(k+1)/(k*p(k-1)*p(k)) = 1/(7 + 1^2/(7 + 2^2/(7 + 3^2/(7 + ... + n^2/(7 + ...))))) = (1 - 1/2 + 1/3) - log(2); the final equality follows by a result of Ramanujan (see [Berndt, Chapter 12, Entry 29]). Thus a(n) ~ c*n^3*n! as n -> oo, where c = (10 - 12*log(2))/9.
E.g.f.: A(x) = (2*x*(4*x^2 + 3*x + 3) - 3*(x + 1)^3*log(1 + x) )/(3*(1 - x)^4) satisfies the differential equation 1 + (x + 7)*A(x) + (x^2 - 1)*A'(x) = 0 with A(0) = 0. - Peter Bala, Dec 09 2024

A142982 a(1) = 1, a(2) = 9, a(n+2) = 9a(n+1) + (n + 1)^2a(n).

Original entry on oeis.org

1, 9, 85, 846, 8974, 101916, 1240308, 16156656, 224789616, 3331795680, 52465122720, 875333381760, 15432978107520, 286828144485120, 5606317009440000, 114993185594112000, 2470155824763648000, 55464433059571200000, 1299510384759562752000, 31718253797341267968000

Offset: 1

Peter Bala, Jul 17 2008

This is the case m = 4 of the more general recurrence a(1) = 1, a(2) = 2*m + 1, a(n+2) = (2*m + 1)*a(n+1) + (n + 1)^2*a(n), which arises when accelerating the convergence of Mercator's series for the constant log(2). See A142979 for remarks on the general case.

Bruce C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag.

Seiichi Manyama, Table of n, a(n) for n = 1..446

Cf. A024167, A142979, A142980, A142981.

p := n -> (2*n^4+4*n^3+10*n^2+8*n+3)/3: a := n -> n!*p(n)*sum ((-1)^(k+1)/(k*p(k-1)*p(k)), k = 1..n): seq(a(n), n = 1..20);

Module[{a, n}, RecurrenceTable[{a[n+2] == 9*a[n+1] + (n+1)^2*a[n], a[1] == 1, a[2] == 9}, a, {n, 25}]] (* Paolo Xausa, Dec 12 2024 *)

a(n) = n!*p(n)*Sum_{k = 1..n} (-1)^(k+1)/(k*p(k-1)*p(k)), where p(n) = (2*n^4 + 4*n^3 + 10*n^2 + 8*n + 3)/3 = A001846(n) is the Ehrhart polynomial for the 4-dimensional cross polytope (the 16-cell).
Recurrence: a(1) = 1, a(2) = 9, a(n+2) = 9*a(n+1) + (n + 1)^2*a(n).
The sequence b(n) := n!*p(n) satisfies the same recurrence with b(1) = 9, b(2) = 82.
Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(9 + 1^2/(9 + 2^2/(9 + 3^2/(9 + ... + (n-1)^2/9)))), for n >= 2.
The behavior of a(n) for large n is given by limit_{n -> oo} a(n)/b(n) = Sum_{k >= 1} (-1)^(k+1)/(k*p(k-1)*p(k)) = 1/(9 + 1^2/(9 + 2^2/(9 + 3^2/(9 + ... + n^2/(9 + ...))))) = log(2) - (1 - 1/2 + 1/3 - 1/4); the final equality follows by a result of Ramanujan (see [Berndt, Chapter 12, Entry 29]).
Thus a(n) ~ c*n^4*n! as n -> oo, where c = (12*log(2) - 7)/18.
E.g.f.: A(x) = (3*(x + 1)^4*log(1 + x) - 4*x^2*(2*x^2 + 2*x + 3))/(3*(1 - x)^5) satisfies the differential equation 1 + (x + 9)*A(x) + (x^2 - 1)*A'(x) = 0 with A(0) = 0. - Peter Bala, Dec 09 2024

A080959 Square array of coefficients of binomial polynomials, read by antidiagonals.

Original entry on oeis.org

1, 2, 1, 3, 1, 5, 4, 0, 11, 14, 5, -2, 20, 14, 94, 6, -5, 34, -10, 214, 444, 7, -9, 55, -74, 454, 444, 3828, 8, -14, 85, -200, 974, -636, 8868, 25584, 9, -20, 126, -416, 2024, -4236, 21468, 25584, 270576, 10, -27, 180, -756, 3968, -13056, 56748, -55056, 633456, 2342880, 11, -35, 249, -1260, 7308, -31632, 146208, -377616, 1722096, 2342880, 29400480

Offset: 1

Paul Barry, Mar 01 2003

			Array, A(n, k), begin as:
   1,   1,   5,    14,    94,     444,    3828,      25584,     270576, ... A024167;
   2,   1,  11,    14,   214,     444,    8868,      25584,     633456, ... A080958;
   3,   0,  20,   -10,   454,    -636,   21468,     -55056,    1722096, ... ;
   4,  -2,  34,   -74,   974,   -4236,   56748,    -377616,    5471856, ... ;
   5,  -5,  55,  -200,  2024,  -13056,  146208,   -1325136,   16902576, ... ;
   6,  -9,  85,  -416,  3968,  -31632,  348816,   -3695952,   47457072, ... ;
   7, -14, 126,  -756,  7308,  -67032,  766296,   -9004752,  120758832, ... ;
   8, -20, 180, -1260, 12708, -129672, 1563336,  -19925712,  281929392, ... ;
   9, -27, 249, -1974, 21018, -234252, 2993436,  -40917312,  611923392, ... ;
  10, -35, 335, -2950, 33298, -400812, 5431116,  -79073472, 1248697152, ... ;
  11, -44, 440, -4246, 50842, -655908, 9411204, -145250688, 2417424768, ... ;
Antidiagonals, T(n, k), begin as:
   1;
   2,   1;
   3,   1,   5;
   4,   0,  11,    14;
   5,  -2,  20,    14,   94;
   6,  -5,  34,   -10,  214,    444;
   7,  -9,  55,   -74,  454,    444,   3828;
   8, -14,  85,  -200,  974,   -636,   8868,   25584;
   9, -20, 126,  -416, 2024,  -4236,  21468,   25584,  270576;
  10, -27, 180,  -756, 3968, -13056,  56748,  -55056,  633456, 2342880;
  11, -35, 249, -1260, 7308, -31632, 146208, -377616, 1722096, 2342880, 29400480;

G. C. Greubel, Antidiagonals n = 1..50, flattened

Cf. A024167, A080958.

Columns: A000027 (k=1), A080956 (k=2), A080957 (k=3).

A:= func< n,k | Factorial(k)*(&+[(-1)^(j+1)*Binomial(n+j,j)/j: j in [1..k]]) >;
A080959:= func< n,k | A(n-k,k) >;
[A080959(n,k): k in [1..n], n in [0..12]]; // G. C. Greubel, May 11 2025

A[n_, k_]:= k!*Sum[(-1)^(j+1)*Binomial[n+j,j]/j, {j,k}];
A080959[n_, k_]:= A[n-k, k];
Table[A080959[n,k], {n,0,12}, {k,n}]//Flatten (* G. C. Greubel, May 11 2025 *)

def A(n,k): return factorial(k)*sum((-1)^(j+1)*binomial(n+j,j)/j for j in range(1,k+1))
def A080959(n,k): return A(n-k,k)
print(flatten([[A080959(n,k) for k in range(1,n+1)] for n in range(13)])) # G. C. Greubel, May 11 2025

A(n, k) = k!*Sum_{j=1..k} (-1)^(j+1)*binomial(n+j, j)/j (array).

T(n, k) = A(n-k, k) (antidiagonals).

cp's OEIS Frontend

A190782 Triangle T(n,k), read by rows, of the coefficients of x^k in the expansion of Sum_(m=0..n) binomial(x,m) = (a(k)*x^k)/n!, n >= 0, 0 <= k <= n.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

A075830 Let u(1) = x and u(n+1) = (n^2/u(n)) + 1 for n >= 1; then a(n) is such that u(n) = (b(n)*x + c(n))/(a(n)*x + d(n)) (in lowest terms) and a(n), b(n), c(n), d(n) are positive integers.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

PARI

Extensions

A334958 GCD of consecutive terms of the factorial times the alternating harmonic series.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Maple

Mathematica

SageMath

Formula

A142980 a(1) = 1, a(2) = 5, a(n+2) = 5*a(n+1) + (n + 1)^2*a(n).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Mathematica

Formula

A281964 Real part of n!*Sum_{k=1..n} i^(k-1)/k, where i is sqrt(-1).

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

PARI

PARI

Formula

A075827 Let u(1) = x and u(n+1) = (n^2/u(n)) + 1 for n >= 1; then a(n) is such that u(n) =(a(n)*x + b(n))/(c(n)*x + d(n)) (in lowest terms) and a(n), b(n), c(n), d(n) are positive integers.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

PARI

Formula

Extensions

A075828 Let u(1) = x and u(n+1) = (n^2/u(n)) + 1 for n >= 1; then a(n) is such that u(n) =(b(n)*x + a(n))/(c(n)*x + d(n)) (in lowest terms) and a(n), b(n), c(n), d(n) are positive integers.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

PARI

Formula

Extensions

A142981 a(1) = 1, a(2) = 7, a(n+2) = 7*a(n+1) + (n+1)^2*a(n).

Views

A075830 Let u(1) = x and u(n+1) = (n^2/u(n)) + 1 for n >= 1; then a(n) is such that u(n) = (b(n)x + c(n))/(a(n)x + d(n)) (in lowest terms) and a(n), b(n), c(n), d(n) are positive integers.

A142980 a(1) = 1, a(2) = 5, a(n+2) = 5a(n+1) + (n + 1)^2a(n).

A075827 Let u(1) = x and u(n+1) = (n^2/u(n)) + 1 for n >= 1; then a(n) is such that u(n) =(a(n)x + b(n))/(c(n)x + d(n)) (in lowest terms) and a(n), b(n), c(n), d(n) are positive integers.

A075828 Let u(1) = x and u(n+1) = (n^2/u(n)) + 1 for n >= 1; then a(n) is such that u(n) =(b(n)x + a(n))/(c(n)x + d(n)) (in lowest terms) and a(n), b(n), c(n), d(n) are positive integers.

A142981 a(1) = 1, a(2) = 7, a(n+2) = 7a(n+1) + (n+1)^2a(n).

A142982 a(1) = 1, a(2) = 9, a(n+2) = 9a(n+1) + (n + 1)^2a(n).