A056546 -id:A056546 - cp's OEIS frontend

A010845 a(n) = 3na(n-1) + 1, a(0) = 1.

1, 4, 25, 226, 2713, 40696, 732529, 15383110, 369194641, 9968255308, 299047659241, 9868572754954, 355268619178345, 13855476147955456, 581929998214129153, 26186849919635811886, 1256968796142518970529

Offset: 0

Simon Plouffe

a(n)/(A000142*A000244) is an increasingly good approximation to cube root of e.
Related to Incomplete Gamma Function at 1/3. - Michael Somos, Mar 26 1999
For positive n, a(n) equals 3^n times the permanent of the n X n matrix with (4/3)'s along the main diagonal, and 1's everywhere else. - John M. Campbell, Jul 10 2011

			1 + 4*x + 25*x^2 + 226*x^3 + 2713*x^4 + 40696*x^5 + 732529*x^6 + ...

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, Tenth Printing, 1972, p. 262.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, Tenth Printing, 1972, p. 262.
Roland Bacher, Counting Packings of Generic Subsets in Finite Groups, Electr. J. Combinatorics, 19 (2012), #P7. - From _N. J. A. Sloane_, Feb 06 2013
M. Z. Spivey and L. L. Steil, The k-Binomial Transforms and the Hankel Transform, J. Integ. Seqs. Vol. 9 (2006), #06.1.1.

Cf. A000522, A010844, A056545, A056546, A056547 for analogs.

Table[ Gamma[ n, 1/3 ]*Exp[ 1/3 ]*3^(n-1), {n, 1, 24} ]
a[ n_] := If[ n<0, 0, Floor[ n! E^(1/3) 3^n ]] (* Michael Somos, Sep 04 2013 *)
Range[0, 20]! CoefficientList[Series[Exp[x]/(1 - 3 x), {x, 0, 20}], x] (* Vincenzo Librandi, Feb 17 2014 *)

{a(n) = if( n<0, 0, n! * sum(k=0, n, 3^(n-k) / k!))} /* Michael Somos, Sep 04 2013 */

E.g.f.: exp(x)/(1-3*x).
a(n) = floor( n!*e^(1/3)*3^n ) = n! * (Sum_{k=0..n} 3^(n-k) / k!) = n! * (e^(1/3) * 3^n - Sum_{k>n} 3^(n-k) / k!). - Michael Somos, Mar 26 1999
a(n) = Sum_{k=0..n} P(n, k)*3^k. - Ross La Haye, Aug 29 2005
Binomial transform of A032031. - Carl Najafi, Sep 11 2011
Conjecture: a(n) +(-3*n-1)*a(n-1) +3*(n-1)*a(n-2)=0. - R. J. Mathar, Feb 16 2014
a(n) = hypergeometric_U(1,n+2,1/3)/3. - Peter Luschny, Nov 26 2014
From Peter Bala, Mar 01 2017: (Start)
a(n) = Integral_{x >= 0} (3*x + 1)^n*exp(-x) dx.
The e.g.f. y = exp(x)/(1 - 3*x) satisfies the differential equation (1 - 3*x)*y' = (4 - 3*x)*y. Mathar's recurrence above follows easily from this.
The sequence b(n) := (3^n)*n! also satisfies Mathar's recurrence with b(0) = 1, b(1) = 3. This leads to the continued fraction representation a(n) = (3^n)*n!*( 1 + 1/(3 - 3/(7 - 6/(10 - ... - (3*n - 3)/(3*n + 1) )))) for n >= 2. Taking the limit as n -> oo gives the continued fraction representation exp(1/3) = 1 + 1/(3 - 3/(7 - 6/(10 - ... - (3*n - 3)/((3*n + 1) - ... )))). Cf. A010844. (End)

Better description and formulas from Michael Somos

More terms from James Sellers, Jul 04 2000

A056545 a(n) = 4na(n-1) + 1 with a(0)=1.

Original entry on oeis.org

1, 5, 41, 493, 7889, 157781, 3786745, 106028861, 3392923553, 122145247909, 4885809916361, 214975636319885, 10318830543354481, 536579188254433013, 30048434542248248729, 1802906072534894923741, 115385988642233275119425

Offset: 0

Henry Bottomley, Jun 20 2000

For positive n, a(n) equals 4^n times the permanent of the n X n matrix with (5/4)'s along the main diagonal and 1's everywhere else. - John M. Campbell, Jul 10 2011

			a(2) = 4*2*a(1) + 1 = 8*5 + 1 = 41.

Harvey P. Dale, Table of n, a(n) for n = 0..365
Michael Z. Spivey and Laura L. Steil, The k-Binomial Transforms and the Hankel Transform, Journal of Integer Sequences, Vol. 9 (2006), Article 06.1.1.

Cf. A000522, A010844, A010845, A056546, A056547, A001907 for analogs. A056545/(A000142*A000302) is an increasingly good approximation to 4th root of e.

Round@Table[Exp[1/4] 4^n Gamma[n + 1, 1/4], {n, 0, 20}] (* Round is equivalent to FullSimplify here, but is much faster; Vladimir Reshetnikov, Oct 14 2016 *)
nxt[{n_,a_}]:={n+1,4a(n+1)+1}; NestList[nxt,{0,1},20][[All,2]] (* Harvey P. Dale, Mar 19 2019 *)

a(n) = floor(e^(1/4)*4^n*n!).
From Philippe Deléham, Mar 14 2004: (Start)
a(n) = n!*Sum_{k=0..n} 4^(n-k)/k!.
E.g.f.: exp(x)/(1 - 4*x). (End)
a(n) = Sum_{k=0..n} P(n, k)*4^k. - Ross La Haye, Aug 29 2005
a(n) = hypergeometric_U(1, n+2 , 1/4)/4. - Peter Luschny, Nov 26 2014
a(n) = exp(1/4)*4^n*Gamma(n+1, 1/4). a(n) ~ sqrt(2*Pi)*4^n*n^(n+1/2)*exp(1/4-n). - Vladimir Reshetnikov, Oct 14 2016
From Peter Bala, Mar 01 2017: (Start)
a(n) = Integral_{x = 0..inf} (4*x + 1)^n*exp(-x) dx.
The e.g.f. y = exp(x)/(1 - 4*x) satisfies the differential equation (1 - 4*x)*y' = (5 - 4*x)*y.
a(n) = (4*n + 1)*a(n-1) - 4*(n - 1)*a(n-2).
The sequence b(n) := 4^n*n! also satisfies the same recurrence with b(0) = 1, b(1) = 4. This leads to the continued fraction representation a(n) = 4^n*n!*( 1 + 1/(4 - 4/(9 - 8/(13 - ... - (4*n - 4)/(4*n + 1) )))) for n >= 2. Taking the limit gives the continued fraction representation exp(1/4) = 1 + 1/(4 - 4/(9 - 8/(13 - ... - (4*n - 4)/((4*n + 1) - ... )))). Cf. A010844. (End)

More terms from James Sellers, Jul 04 2000

A056547 a(n) = 6na(n-1) + 1 with a(0)=1.

Original entry on oeis.org

1, 7, 85, 1531, 36745, 1102351, 39684637, 1666754755, 80004228241, 4320228325015, 259213699500901, 17108104167059467, 1231783500028281625, 96079113002205966751, 8070645492185301207085, 726358094296677108637651

Offset: 0

Henry Bottomley, Jun 20 2000

			a(2) = 6*2*a(1) + 1 = 12*7 + 1 = 85.

Harvey P. Dale, Table of n, a(n) for n = 0..346

Cf. A000522, A010844, A010845, A056545, A056546 for analogs. A056547/(A000142*A000400) is an increasingly good approximation to 6th root of e.

nxt[{n_,a_}]:={n+1,6a(n+1)+1}; NestList[nxt,{0,1},20][[;;,2]] (* Harvey P. Dale, Jul 17 2024 *)

a(n) = floor(e^(1/6)*6^n*n!).
a(n) = n!*Sum_{k=0..n} 6^(n-k)/k!. E.g.f.: exp(x)/(1 - 6*x). - Philippe Deléham, Mar 14 2004
From Peter Bala, Mar 01 2017: (Start)
a(n) = Integral_{x = 0..inf} (6*x + 1)^n*exp(-x) dx.
The e.g.f. y = exp(x)/(1 - 6*x) satisfies the differential equation (1 - 6*x)*y' = (7 - 6*x)*y.
a(n) = (6*n + 1)*a(n-1) - 6*(n - 1)*a(n-2).
The sequence b(n) := 6^n*n! also satisfies the same recurrence with b(0) = 1, b(1) = 6. This leads to the continued fraction representation a(n) = 6^n*n!*( 1 + 1/(6 - 6/(13 - 12/(19 - ... - (6*n - 6)/(6*n + 1) )))) for n >= 2. Taking the limit gives the continued fraction representation exp(1/6) = 1 + 1/(6 - 6/(13 - 12/(19 - ... - (6*n - 6)/((6*n + 1) - ... )))). Cf. A010844. (End)

More terms from James Sellers, Jul 04 2000

A320031 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of the e.g.f. exp(x)/(1 - k*x).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 5, 1, 1, 4, 13, 16, 1, 1, 5, 25, 79, 65, 1, 1, 6, 41, 226, 633, 326, 1, 1, 7, 61, 493, 2713, 6331, 1957, 1, 1, 8, 85, 916, 7889, 40696, 75973, 13700, 1, 1, 9, 113, 1531, 18321, 157781, 732529, 1063623, 109601, 1, 1, 10, 145, 2374, 36745, 458026, 3786745, 15383110, 17017969, 986410, 1

Offset: 0

Ilya Gutkovskiy, Oct 03 2018

			E.g.f. of column k: A_k(x) = 1 + (k + 1)*x/1! + (2*k^2 + 2*k + 1)*x^2/2! + (6*k^3 + 6*k^2 + 3*k + 1)*x^3/3! + (24*k^4 + 24*k^3 + 12*k^2 + 4*k + 1)*x^4/4! + ...
Square array begins:
  1,    1,     1,      1,       1,       1,  ...
  1,    2,     3,      4,       5,       6,  ...
  1,    5,    13,     25,      41,      61,  ...
  1,   16,    79,    226,     493,     916,  ...
  1,   65,   633,   2713,    7889,   18321,  ...
  1,  326,  6331,  40696,  157781,  458026,  ...

Alois P. Heinz, Antidiagonals n = 0..140, flattened

Columns k=0..6 give A000012, A000522, A010844, A010845, A056545, A056546, A056547.
Main diagonal gives A277452.
Cf. A007318, A320032, A371898.

A := (n, k) -> simplify(hypergeom([1, -n], [], -k)):
for n from 0 to 5 do seq(A(n, k), k=0..8) od; # Peter Luschny, Oct 03 2018
# second Maple program:
A:= proc(n, k) option remember;
      1 + `if`(n>0, k*n*A(n-1, k), 0)
    end:
seq(seq(A(n, d-n), n=0..d), d=0..12);  # Alois P. Heinz, May 09 2020

Table[Function[k, n! SeriesCoefficient[Exp[x]/(1 - k x), {x, 0, n}]][j - n], {j, 0, 10}, {n, 0, j}] // Flatten
Table[Function[k, HypergeometricPFQ[{1, -n}, {}, -k]][j - n], {j, 0, 10}, {n, 0, j}] // Flatten

E.g.f. of column k: exp(x)/(1 - k*x).
A(n,k) = Sum_{j=0..n} binomial(n,j)*j!*k^j.
A(n,k) = hypergeom_2F0([1, -n], [], -k).
A(n,k) = 1 + [n > 0] * k * n * A(n-1,k). - Alois P. Heinz, May 09 2020
A(n,k) = floor(n!*k^n*exp(1/k)), k > 0, n + k > 1. - Peter McNair, Dec 20 2021
From Werner Schulte, Apr 14 2024: (Start)
The LU decomposition of this array is given by the upper triangular matrix U which is the transpose of A007318 and the lower triangular matrix L = A371898, i.e., A(n, k) = Sum_{i=0..k} binomial(k, i) * A371898(n, i).
Conjecture: E.g.f. of row n is exp(x) * (Sum_{k=0..n} A371898(n, k) * x^k / k!). (End)

A336808 a(n) = (n!)^2 * Sum_{k=0..n} 5^(n-k) / (k!)^2.

Original entry on oeis.org

1, 6, 121, 5446, 435681, 54460126, 9802822681, 2401691556846, 768541298190721, 311259225767242006, 155629612883621003001, 94155915794590706815606, 67792259372105308907236321, 57284459169428986026614691246, 56138769986040406306082397421081, 63156116234295457094342697098716126

Offset: 0

Ilya Gutkovskiy, Jan 27 2021

nonn

Cf. A006040, A056546, A336804, A336805, A336807.

Table[n!^2 Sum[5^(n - k)/k!^2, {k, 0, n}], {n, 0, 15}]
nmax = 15; CoefficientList[Series[BesselI[0, 2 Sqrt[x]]/(1 - 5 x), {x, 0, nmax}], x] Range[0, nmax]!^2

Sum_{n>=0} a(n) * x^n / (n!)^2 = BesselI(0,2*sqrt(x)) / (1 - 5*x).

a(0) = 1; a(n) = 5 * n^2 * a(n-1) + 1.

A347013 E.g.f.: exp(x) / (1 - 5 * x)^(1/5).

Original entry on oeis.org

1, 2, 9, 88, 1361, 28182, 726889, 22414988, 803913441, 32867765002, 1508608850249, 76804271962848, 4294870015118641, 261673684619584862, 17252970318529474089, 1223896705010751194068, 92946073511938131386561, 7523666291578066678172562, 646658551118777059833155209

Offset: 0

Ilya Gutkovskiy, Aug 10 2021

nonn

Binomial transform of A008548.

Cf. A000522, A008548, A056546, A084262, A088992, A346258, A347012, A347014.

g:= proc(n) option remember; `if`(n<2, 1, (5*n-4)*g(n-1)) end:
a:= n-> add(binomial(n, k)*g(k), k=0..n):
seq(a(n), n=0..18);  # Alois P. Heinz, Aug 10 2021

nmax = 18; CoefficientList[Series[Exp[x]/(1 - 5 x)^(1/5), {x, 0, nmax}], x] Range[0, nmax]!
Table[Sum[Binomial[n, k] 5^k Pochhammer[1/5, k], {k, 0, n}], {n, 0, 18}]
Table[HypergeometricU[1/5, n + 6/5, 1/5]/5^(1/5), {n, 0, 18}]

a(n) = Sum_{k=0..n} binomial(n,k) * A008548(k).

a(n) ~ n! * exp(1/5) * 5^n / (Gamma(1/5) * n^(4/5)). - Vaclav Kotesovec, Aug 14 2021

cp's OEIS Frontend

A010845 a(n) = 3*n*a(n-1) + 1, a(0) = 1.

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A056545 a(n) = 4*n*a(n-1) + 1 with a(0)=1.

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A056547 a(n) = 6*n*a(n-1) + 1 with a(0)=1.

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A320031 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of the e.g.f. exp(x)/(1 - k*x).

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A336808 a(n) = (n!)^2 * Sum_{k=0..n} 5^(n-k) / (k!)^2.

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A347013 E.g.f.: exp(x) / (1 - 5 * x)^(1/5).

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A010845 a(n) = 3na(n-1) + 1, a(0) = 1.

A056545 a(n) = 4na(n-1) + 1 with a(0)=1.

A056547 a(n) = 6na(n-1) + 1 with a(0)=1.