A047974 -id:A047974 - cp's OEIS frontend

A377955 a(n) = n! * Sum_{k=0..n} binomial(k+3,n-k) / k!.

1, 4, 15, 58, 241, 1056, 4879, 23710, 120033, 635356, 3478351, 19796514, 115988305, 703052728, 4372581711, 28022140486, 183804777409, 1238244635700, 8520907808143, 60061024788106, 431735704061361, 3171780156493264, 23730347517489295, 181115025566445678

Offset: 0

Seiichi Manyama, Nov 12 2024

Cf. A018191, A047974, A377954, A377956.

a(n) = n!*sum(k=0, n, binomial(k+3, n-k)/k!);

E.g.f.: (1 + x)^3 * exp(x + x^2).

a(n) = -(n-5)*a(n-1) + 3*(n-1)*a(n-2) + 2*(n-1)*(n-2)*a(n-3) for n > 2.

A381984 E.g.f. A(x) satisfies A(x) = exp(x) * B(x), where B(x) = 1 + x*B(x)^3 is the g.f. of A001764.

Original entry on oeis.org

1, 2, 9, 94, 1649, 40146, 1246057, 47004014, 2087644449, 106709890114, 6170322084041, 398219508589662, 28376096583546769, 2212797385807852754, 187441592012756668329, 17139223549605292448686, 1682551982313514625386817, 176505773149909540258262274, 19704960849698723062181296009

Offset: 0

Seiichi Manyama, Mar 11 2025

For each positive integer k, the sequence obtained by reducing a(n) modulo k is a periodic sequence with period dividing k. For example, modulo 6 the sequence becomes [1, 2, 3, 4, 5, 0, 1, 2, 3, 4, 5, 0, ...] with period 6. Cf. A047974. - Peter Bala, Mar 13 2025

Cf. A381985, A381986, A381987.

Cf. A001763, A001764.

seq(simplify(hypergeom([-n, 1/3, 2/3], [3/2], -27/4)), n = 0..18); # Peter Bala, Mar 13 2025

Table[HypergeometricPFQ[{-n, 1/3, 2/3}, {3/2}, -27/4], {n, 0, 20}] (* Vaclav Kotesovec, Mar 14 2025 *)

a(n) = n!*sum(k=0, n, binomial(3*k+1, k)/((3*k+1)*(n-k)!));

a(n) = n! * Sum_{k=0..n} A001764(k)/(n-k)!.
From Peter Bala, Mar 13 2025: (Start)
a(n) = hypergeom([-n, 1/3, 2/3], [3/2], -27/4).
2*(2*n + 1)*a(n) = (27*n^2 - 19*n + 4)*a(n-1) - 2*(n - 1)*(27*n - 25)*a(n-2) + 27*(n - 1)*(n - 2)*a(n-3) with a(0) = 0, a(1) = 2 and a(2) = 9. (End)
a(n) ~ 3^(3*n + 1/2) * n^(n + 1/2) / (2^(2*n + 3/2) * exp(n - 4/27) * n^(3/2)). - Vaclav Kotesovec, Mar 14 2025

A381987 E.g.f. A(x) satisfies A(x) = exp(x) * B(x), where B(x) = 1 + x*B(x)^4 is the g.f. of A002293.

Original entry on oeis.org

1, 2, 11, 160, 3941, 134486, 5851327, 309520436, 19283504585, 1382980764106, 112223497464371, 10165461405056552, 1016801830348902061, 111312715288354681310, 13237965546409421546471, 1699516550894276788156156, 234263144339070269872076177, 34507561203827621878485498386

Offset: 0

Seiichi Manyama, Mar 12 2025

For each positive integer k, the sequence obtained by reducing a(n) modulo k is a periodic sequence with period dividing k. For example, modulo 5 the sequence becomes [1, 2, 1, 0, 1, 1, 2, 1, 0, 1, ...] with period 5. In particular, a(5*n+3) == 0 (mod 5). Cf. A047974. - Peter Bala, Mar 13 2025

Cf. A381984, A381988, A381989.

Cf. A002293, A365340.

seq(simplify(hypergeom([-n, 1/2, 1/4, 3/4], [2/3, 4/3], -256/27)), n = 0..17); # Peter Bala, Mar 13 2025

Table[HypergeometricPFQ[{-n, 1/2, 1/4, 3/4}, {2/3, 4/3}, -256/27], {n, 0, 20}] (* Vaclav Kotesovec, Mar 14 2025 *)

a(n) = n!*sum(k=0, n, binomial(4*k+1, k)/((4*k+1)*(n-k)!));

a(n) = n! * Sum_{k=0..n} A002293(k)/(n-k)!.
From Peter Bala, Mar 13 2025: (Start)
a(n) = hypergeom([-n, 1/2, 1/4, 3/4], [2/3, 4/3], -256/27).
3*(3*n - 1)*(3*n + 1)*a(n) = n*(256*n^2 - 303*n + 95)*a(n-1) - 3*(n - 1)*(256*n^2 - 485*n + 245)*a(n-2) + 3*(256*n - 375)*(n - 1)*(n - 2)*a(n-3) - 256*(n - 1)*(n - 2)*(n - 3)*a(n-4) with a(0) = 1, a(1) = 2, a(2) = 11 and a(3) = 160. (End)
a(n) ~ 2^(8*n+1) * n^(n-1) / (3^(3*n + 3/2) * exp(n - 27/256)). - Vaclav Kotesovec, Mar 14 2025

A080833 E.g.f.: exp( x/(1 - x - x^2) ).

Original entry on oeis.org

1, 1, 3, 19, 145, 1401, 16051, 213403, 3223809, 54514225, 1019601091, 20890209891, 465156779473, 11181638663209, 288536019179955, 7953590111627371, 233211718410856321, 7246720953253750113, 237849724555558441219, 8221578401608012672435, 298505383888840158941841

Offset: 0

Emanuele Munarini, Mar 28 2003

From Peter Bala, Mar 25 2022: (Start)
The sequence terms are odd. 3 divides a(3*n-1), 9 divides a(9*n-1) and 27 divides a(27*n-1); 5 divides a(5*n+4), 25 divides a(25*n+9) and 125 divides a(125*n+34); 7 divides a(7*n+6), 49 divides a(49*n+34) and 343 divides a(343*n + 83); 15 divides a(15*n+14) and 17 divides a(17*n+13).
More generally, the congruence a(n+k) == a(n) (mod k) holds for all n and k. It follows that the sequence obtained by taking a(n) modulo a fixed positive integer k is periodic with exact period dividing k. For example, taken modulo 7 the sequence becomes [1, 1, 3, 5, 5, 1, 0, 1, 1, 3, 3, 5, 5, 1, 0, ...], a purely periodic sequence with period 7. (End)

Peter Bala, Integer sequences that become periodic on reduction modulo k for all k

Cf. A005443.

CoefficientList[Series[E^(x/(1-x-x^2)), {x, 0, 20}], x]* Range[0, 20]! (* Vaclav Kotesovec, Jun 27 2013 *)

my(x='x+O('x^30)); Vec(serlaplace(exp(x/(1-x-x^2)))) \\ Michel Marcus, Jun 07 2021

E.g.f.: exp( x/(1 - x - x^2) ).
a(n) = n!*sum{i=0..n, sum{j=0..n, C(i+j-1, j)*C(j, n-i-j)/i!}}. - Paul Barry, Aug 29 2005
E.g.f.: 1 + x*(E(0)-1)/(x+1) where E(k) =  1 + 1/(k+1)/(1-x-x^2)/(1-x/(x+1/E(k+1) )); (recursively defined continued fraction). - Sergei N. Gladkovskii, Jan 27 2013
Recurrence: a(n) = (2*n-1)*a(n-1) + (n-2)*(n-1)*a(n-2) - (n-2)*(n-1)*(2*n-7)*a(n-3) - (n-4)*(n-3)*(n-2)*(n-1)*a(n-4). - Vaclav Kotesovec, Jun 27 2013
a(n) ~ ((1+sqrt(5))/2)^n*exp(2*sqrt(n)/5^(1/4)-n-1/10)*n^(n-1/4)/(sqrt(2)*5^(1/8)). - Vaclav Kotesovec, Jun 27 2013
a(0) = 1; a(n) = Sum_{k=1..n} binomial(n-1,k-1) * k! * Fibonacci(k) * a(n-k). - Ilya Gutkovskiy, Jun 07 2021

A189423 Expansion of e.g.f. exp(log(1+x) + log(1+x)^2).

Original entry on oeis.org

1, 1, 2, 0, 10, -50, 368, -3052, 28740, -302220, 3508152, -44532048, 613399752, -9109006920, 145029146208, -2463935369040, 44482964644368, -850291412311152, 17153458120885152, -364163960169826944, 8114899768747511712, -189364681355153357088, 4617713773733245962240

Offset: 0

Vladimir Kruchinin, Apr 21 2011

sign

Seiichi Manyama, Table of n, a(n) for n = 0..447
Vladimir Kruchinin and D. V. Kruchinin, Composita and their properties, arXiv:1103.2582 [math.CO], 2011-2013.

Cf. A005444, A047974.

a(n):=sum(sum(k!*binomial(m,k-m)*stirling1(n,k),k,m,n)/m!,m,1,n);

my(N=30, x='x+O('x^N)); Vec(serlaplace(exp(log(1+x)*(1+log(1+x))))) \\ Seiichi Manyama, May 14 2022

a(n) = sum(k=0, n, k!*sum(j=0, k\2, 1/(j!*(k-2*j)!))*stirling(n, k, 1)); \\ Seiichi Manyama, May 14 2022

a(0) = 1; a(n) = Sum_{m=1..n} Sum_{k=m..n} k!*binomial(m,k-m)*stirling1(n,k)/m! for n>0.

a(n) = Sum_{k=0..n} A047974(k) * Stirling1(n,k). - Seiichi Manyama, May 14 2022

A294213 E.g.f.: exp(1/((1-x)*(1-x^2)) - 1).

Original entry on oeis.org

1, 1, 5, 25, 193, 1601, 16741, 190345, 2509025, 35825473, 569012581, 9716400761, 180303804385, 3569527588225, 75681964322693, 1700163418683241, 40499757023856961, 1016190431274596225, 26843084299482509125, 743180975111364212953

Offset: 0

Seiichi Manyama, Oct 25 2017

Seiichi Manyama, Table of n, a(n) for n = 0..435

Column k=2 of A294212.

nmax = 20; CoefficientList[Series[E^(1/((1-x)*(1-x^2)) - 1), {x, 0, nmax}], x] * Range[0, nmax]! (* Vaclav Kotesovec, Oct 26 2017 *)

N=66; x='x+O('x^N); Vec(serlaplace(exp(1/((1-x)*(1-x^2))-1)))

a(n) ~ exp(-61/96 + 3*n^(1/3)/4 + 3*n^(2/3)/2 - n) * n^(n - 1/6) / sqrt(3) * (1 + 25/(64*n^(1/3))). - Vaclav Kotesovec, Oct 26 2017
a(n) = n*a(n-1) + (n-1)*(2*n - 1)*a(n-2) - 2*(n-3)*(n-2)*(n-1)*a(n-3) - (n-4)*(n-3)*(n-2)*(n-1)*a(n-4) + (n-5)*(n-4)*(n-3)*(n-2)*(n-1)*a(n-5). - Vaclav Kotesovec, Dec 02 2021
From Peter Bala, Oct 17 2023: (Start)
a(n+k) == a(n) (mod k) for all n and k >= 1. Hence for each k, the sequence a(n) taken modulo k is a periodic sequence and the period divides k. Cf. A047974.
a(5*n + 2) == a(5*n + 3) == 0 (mod 5);
a(25*n + 3) == a(25*n + 8) == a(25*n + 13) == a(25*n + 17) == a(25*n + 18) == a(25*n + 23) == 0 (mod 5^2);
a(125*n + 18) == a(125*n + 67) == a(125*n + 93) == a(125*n + 118) == 0 (mod 5^3). (End)

A296618 Expansion of the e.g.f. exp(-x)/sqrt(1-4*x).

Original entry on oeis.org

1, 1, 9, 89, 1265, 22929, 506809, 13220521, 397585761, 13543386785, 515418398441, 21673889807481, 998003450868049, 49942515803293489, 2698849517019693465, 156631203355106962889, 9716434375682706344129, 641592631434102757993281

Offset: 0

Emanuele Munarini, Dec 17 2017

nonn

Peter Bala, Integer sequences that become periodic on reduction modulo k for all k

Cf. A052143.

A296618 := n -> (-1)^n*(I/2)*KummerU(1/2, n+3/2, -1/4):
seq(simplify(A296618(n)), n=0..17); # Peter Luschny, Dec 18 2017

Table[Sum[Binomial[n,k]Binomial[2k,k]k! (-1)^(n-k),{k,0,n}],{n,0,18}]
CoefficientList[Series[Exp[-x]/Sqrt[1-4x], {x,0,18}], x] Range[0,18]!

makelist(sum(binomial(n,k)*binomial(2*k,k)*k!*(-1)^(n-k),k,0,n),n,0,12);

x='x+O('x^99); Vec(serlaplace(exp(-x)/sqrt(1-4*x))) \\ Altug Alkan, Dec 17 2017

a(n) = Sum_{k=0..n} binomial(n,k)*binomial(2*k,k)*k!*(-1)^(n-k).
a(n) = (i/2)*(-1)^n*U(1/2,n+3/2,-1/4), where U denotes the Kummer U function.
D-finite with recurrence: a(n+2) - (4*n+5)*a(n+1) - 4*(n+1)*a(n) = 0.
Sum_{k=0..n} binomial(n,k)*a(k)*a(n-k) = Sum_{k=0..n} binomial(n,k)*(-1)^(n-k)*2^(n+k)*k!.
Conjectures:
1) a(n+1) == a(n) (mod n) for all n >= 1.
2) a(n+k) == (-1)^k*a(n) (mod k) for all n and k >= 1.
a(n) ~ 2^(2*n + 1/2) * n^n / exp(n + 1/4). - Vaclav Kotesovec, Dec 17 2017
From Peter Bala, Jun 19 2023: (Start)
a(n) == 1 (mod 4).
Conjecture 1) above follows immediately from the stated recurrence equation. In fact, a(n+1) == a(n) (mod 8*n) for n >= 1.
For a proof of Conjecture 2) see the Bala link (corollary to Theorem 1, p. 5). (End)

A361569 Expansion of e.g.f. exp(x^4/24 * (1+x)^4).

Original entry on oeis.org

1, 0, 0, 0, 1, 20, 180, 840, 1715, 2520, 88200, 1940400, 29111775, 303603300, 2188286100, 12549537000, 143029511625, 3397035642000, 71419225878000, 1170096883956000, 15075357741068625, 163540869094102500, 2025016641129982500, 40912918773391665000

Offset: 0

Seiichi Manyama, Mar 16 2023

nonn

Cf. A047974, A361567, A361568.

Cf. A361280.

my(N=30, x='x+O('x^N)); Vec(serlaplace(exp(x^4/24*(1+x)^4)))

a_vector(n) = my(v=vector(n+1)); v[1]=1; for(i=1, n, v[i+1]=(i-1)!/24*sum(j=4, i, j*binomial(4, j-4)*v[i-j+1]/(i-j)!)); v;

a(n) = n! * Sum_{k=0..floor(n/4)} binomial(4*k,n-4*k)/(24^k * k!).
a(0) = 1; a(n) = ((n-1)!/24) * Sum_{k=4..n} k * binomial(4,k-4) * a(n-k)/(n-k)!.
a(n) = (n-1)*(n-2)*(n-3)/24 * (4*a(n-4) + 20*(n-4)*a(n-5) + 36*(n-4)*(n-5)*a(n-6) + 28*(n-4)*(n-5)*(n-6)*a(n-7) + 8*(n-4)*(n-5)*(n-6)*(n-7)*a(n-8)). -Seiichi Manyama, Jun 16 2024

A361571 Expansion of e.g.f. exp( (x * (1+x))^3 ).

Original entry on oeis.org

1, 0, 0, 6, 72, 360, 1080, 15120, 302400, 3689280, 32659200, 359251200, 6965481600, 133880947200, 2070484416000, 30305353478400, 559684629504000, 12582442768896000, 271843009108070400, 5401042458152140800, 111578968350001152000, 2657164887872022528000

Offset: 0

Seiichi Manyama, Mar 16 2023

nonn

Harvey P. Dale, Table of n, a(n) for n = 0..476

Cf. A047974, A361570.

Cf. A361279.

With[{nn=30},CoefficientList[Series[Exp[(x(1+x))^3],{x,0,nn}],x] Range[0,nn]!] (* Harvey P. Dale, Jul 06 2025 *)

my(N=30, x='x+O('x^N)); Vec(serlaplace(exp((x*(1+x))^3)))

a_vector(n) = my(v=vector(n+1)); v[1]=1; for(i=1, n, v[i+1]=(i-1)!*sum(j=3, i, j*binomial(3, j-3)*v[i-j+1]/(i-j)!)); v;

a(n) = n! * Sum_{k=0..floor(n/3)} binomial(3*k,n-3*k)/k!.

a(0) = 1; a(n) = (n-1)! * Sum_{k=3..n} k * binomial(3,k-3) * a(n-k)/(n-k)!.

A376512 Expansion of e.g.f. exp(x^2 * (1 + x)).

Original entry on oeis.org

1, 0, 2, 6, 12, 120, 480, 2520, 21840, 120960, 937440, 8316000, 60540480, 570810240, 5465940480, 49037788800, 523588665600, 5504686387200, 57816850291200, 678823104960000, 7844848544332800, 93064133530368000, 1184800751111577600, 14967781957781452800

Offset: 0

Seiichi Manyama, Sep 25 2024

Cf. A047974, A376513.

Cf. A118589, A227937, A373740.

a(n) = n!*sum(k=0, n\2, binomial(k, n-2*k)/k!);

a(n) = n! * Sum_{k=0..floor(n/2)} binomial(k,n-2*k)/k!.
a(n) = (n-1) * (2*a(n-2) + 3*(n-2)*a(n-3)).
a(n) ~ 3^(n/3 - 1/2) * exp(4/81 - 2*3^(-7/3)*n^(1/3) + 3^(-2/3)*n^(2/3) - 2*n/3) * n^(2*n/3) * (1 + 223/(3^(20/3)*n^(1/3))). - Vaclav Kotesovec, Sep 26 2024

cp's OEIS Frontend

A377955 a(n) = n! * Sum_{k=0..n} binomial(k+3,n-k) / k!.

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A381984 E.g.f. A(x) satisfies A(x) = exp(x) * B(x), where B(x) = 1 + x*B(x)^3 is the g.f. of A001764.

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A381987 E.g.f. A(x) satisfies A(x) = exp(x) * B(x), where B(x) = 1 + x*B(x)^4 is the g.f. of A002293.

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A080833 E.g.f.: exp( x/(1 - x - x^2) ).

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A189423 Expansion of e.g.f. exp(log(1+x) + log(1+x)^2).

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A294213 E.g.f.: exp(1/((1-x)*(1-x^2)) - 1).

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A296618 Expansion of the e.g.f. exp(-x)/sqrt(1-4*x).

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A361569 Expansion of e.g.f. exp(x^4/24 * (1+x)^4).

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