A285064 -id:A285064 - cp's OEIS frontend

A126390 a(n) = Sum_{i=0..n} 2^iB(i)binomial(n,i) where B(n) = Bell numbers A000110(n).

1, 3, 13, 71, 457, 3355, 27509, 248127, 2434129, 25741939, 291397789, 3510328695, 44782460313, 602513988107, 8518757813637, 126179029108463, 1952609274344353, 31492811964616163, 528249539951292461, 9197240228562763687, 165923214676585626729

Offset: 0

N. J. A. Sloane, Aug 04 2007

nonn

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Toufik Mansour and Mark Shattuck, A recurrence related to the Bell numbers, INTEGERS 11 (2011), #A67.

Cf. A000110, A000296, A005493, A124311, A126617, A284859, A285064.

with(combstruct):seq(count(([S, {N=Union(Z, S, P), S=Set(Union(Z, P), card>=0), P=Set(Union(Z, Z), card>=1)}, labeled], size=n)), n=0..20); # Zerinvary Lajos, Mar 18 2008

Table[ Sum[ 2^k Binomial[n, k] BellB[k], {k, 0, n}], {n, 0, 30}] (* Karol A. Penson and Olivier Gérard, Oct 22 2007 *)

x='x+O('x^66); Vec(serlaplace((exp(exp(2*x)-1+x)))) \\ Joerg Arndt, May 13 2013

E.g.f.: exp(exp(2*x)-1+x). - Vladeta Jovovic, Aug 04 2007
a(n) = e^(-1)* 2^n * Sum_{k>=0} (k + 1/2)^n / k!. This is a Dobinski-type formula. - Karol A. Penson and Olivier Gérard, Oct 22 2007
G.f.: 1/Q(0), where Q(k)= 1 - (2*k+3)*x - 4*(k+1)*x^2/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, May 03 2013
G.f.: 1/Q(0), where Q(k)= 1 - x - 2*x/(1 - 2*x*(2*k+1)/(1 - x - 2*x/(1 - 2*x*(2*k+2)/Q(k+1)))); (continued fraction). - Sergei N. Gladkovskii, May 13 2013
a(0) = 1; a(n) = a(n-1) + Sum_{k=1..n} binomial(n-1,k-1) * 2^k * a(n-k). - Ilya Gutkovskiy, Jun 21 2022
From Vaclav Kotesovec, Jun 22 2022: (Start)
a(n) ~ Bell(n) * (2 + LambertW(n)/n)^n.
a(n) ~ Bell(n) * 2^n * sqrt(n) * log(n)^(-1/2 + 1/(2*log(n)) - 1/(2*log(n)^2)) * exp(log(log(n))^2/(4*log(n)^2)). (End)
a(n) ~ 2^n * n^(n + 1/2) * exp(n/LambertW(n) - n - 1) / (sqrt(1 + LambertW(n)) * LambertW(n)^(n + 1/2)). - Vaclav Kotesovec, Jun 27 2022

A285061 Sheffer triangle S2[4,1] = (exp(x), exp(4*x) - 1).

Original entry on oeis.org

1, 1, 4, 1, 24, 16, 1, 124, 240, 64, 1, 624, 2656, 1792, 256, 1, 3124, 26400, 33920, 11520, 1024, 1, 15624, 250096, 546560, 331520, 67584, 4096, 1, 78124, 2313360, 8105664, 7822080, 2745344, 372736, 16384, 1, 390624, 21132736, 114627072, 165398016, 88940544, 20299776, 1966080, 65536, 1, 1953124, 191757120, 1574682880, 3270274560, 2529343488, 863256576, 138215424, 10027008, 262144

Offset: 0

Wolfdieter Lang, Apr 13 2017

For Sheffer triangles (infinite lower triangular exponential convolution matrices) see the W. Lang link under A006232, with references.
This is a generalization of the Sheffer triangle Stirling2(n, m) = A048993(n, m) denoted by (exp(x), exp(x)-1), which could be named S2[1,0].
For the Sheffer triangle (exp(x), (1/4)*(exp(4*x) - 1)) see A111578, also the P. Bala link where this triangle, T(n, m)/4^m, is named S_{(4,0,1)}. The triangle T(n, m)*m! is given in A285066.
The a-sequence for this Sheffer triangle has e.g.f. 4*x/log(1+x) and is 4*A006232(n)/A006233(n) (Cauchy numbers of the first kind).
The z-sequence has e.g.f. (4/(log(1+x)))*(1 - 1/(1+x)^(1/4)) and is A285062(n)/A285063(n).
The main diagonal gives A000302.
The row sums give A285064. The alternating row sums give A285065.
The first column sequences are A000012, 4*A003463, 4^2*A016234. For the e.g.f.s and o.g.f.s see the formula section.
This triangle appears in the o.g.f. G(n, x) of the sequence {(1 + 4*m)^n}{m>=0}, as G(n, x) = Sum{m=0..n} T(n, m)*m!*x^m/(1-x)^(m+1), n >= 0. Hence the corresponding e.g.f. is, by the linear inverse Laplace transform, E(n, t) = Sum_{m >=0}(1 + 4*m)^n t^m/m! = exp(t)*Sum_{m=0..n} T(n, m)*t^m.
The corresponding Euler triangle with reversed rows is rEu(n, k) = Sum_{m=0..k} (-1)^(k-m)*binomial(n-m, k-m)*T(n, k)*k!, 0 <= k <= n. This is A225118 with row reversion.
In general the Sheffer triangle S2[d,a] appears in the reordering of the operator (a*1 + d*E_x) = Sum_{m=0..n} S2(d,a;n,m) x^m*(d_x)^m with the derivative d_x and the Euler operator E_x = x*d_x. For [d,a] = [1,0] this becomes the standard Stirling2 property.

			The triangle T(n,m) begins:
  n\m  0      1        2         3         4        6        7       8     9
  0:   1
  1:   1      4
  2:   1     24       16
  3:   1    124      240        64
  4:   1    624     2656      1792       256
  5:   1   3124    26400     33920     11520     1024
  6:   1  15624   250096    546560    331520    67584     4096
  7:   1  78124  2313360   8105664   7822080  2745344   372736   16384
  8:   1 390624 21132736 114627072 165398016 88940544 20299776 1966080 65536
  ...
Three term recurrence: T(4, 1) = 4*1 + (1 + 4*1)*124 = 624.
Recurrence for row polynomial R(3, x) (Meixner type): ((1 + 4*x) + 4*x*d_x)*(1 + 24*x + 16*x^2) = 1 + 124*x + 240*x^2 + 64*x^3.
Boas-Buck recurrence for column m = 2, and n = 4: T(4, 2) = (1/2)*(2*(2 + 4*2)*T(3, 2) + 2*6*(-4)^2*Bernoulli(2)*T(2, 2)) = (1/2)*(20*240 + 12*16*(1/6)*16) = 2656. - _Wolfdieter Lang_, Aug 11 2017

Michael De Vlieger, Table of n, a(n) for n = 0..11475, rows n = 0..150, flattened.
Paweł Hitczenko, A class of polynomial recurrences resulting in (n/log n, n/log^2 n)-asymptotic normality, arXiv:2403.03422 [math.CO], 2024. See p. 9.
Wolfdieter Lang, On Sums of Powers of Arithmetic Progressions, and Generalized Stirling, Eulerian and Bernoulli numbers, arXiv:1707.04451 [math.NT], 2017.

Cf. A000012, A000302, 4*A003463, A111578, A006232/A006233, 16*A016234, A282629, A285062/A285063, A285064, A285065, A285066.

Table[Sum[Binomial[n, k]*4^k*StirlingS2[k, m], {k, 0, n}], {n, 0, 20}, {m, 0, n}] // Flatten (* Indranil Ghosh, May 06 2017 *)

T(n, m) = sum(k=0, n, binomial(n, k)*4^k*stirling(k, m, 2));
for(n=0, 20, for(m=0, n, print1(T(n, m),", ");); print();) \\ Indranil Ghosh, May 06 2017

Three term recurrence (from the conversion property mentioned above, with [d,a] =[4,1]): T(n, -1) = 0, T(0, 0) = 1, T(n, m) = 0 if n < m. T(n, m) = 4*T(n-1, m-1) + (1 + 4*m)*T(n-1, m) for n >= 1, m = 0..n.
T(n, m) = Sum_{k=0..m} binomial(m,k)*(-1)^(k-m)*(1 + 4*k)^n/m!, 0 <= m <= n, satisfying the recurrence.
E.g.f. of the row polynomials R(n, x) = Sum_{m=0..n} T(n, m)*x^m: exp(z)*exp(x*(exp(4*z) - 1)). This is the e.g.f. of the triangle.
E.g.f. for the sequence of column m: exp(x)*((exp(3*x) - 1)^m)/m! (Sheffer property).
O.g.f. for sequence of column m: (4*x)^m/Product_{j=0..m} (1 - (1 + 4*j)*x) (by Laplace transform of the e.g.f.).
T(n, m) = Sum_{k=0..n} binomial(n, k)* 4^k*Stirling2(k, m), 0 <= m <= n, where Stirling2 is given in A048993.
A nontrivial recurrence for the column m=0 entries T(n, 0) = 4^n from the z-sequence given above: T(n,0) = n*Sum_{j=0..n-1} z(j)*T(n-1,j), n >= 1, T(0, 0) = 1.
Recurrence for column m >= 1 entries from the a-sequence given above: T(n, m) = (n/m)* Sum_{j=0..n-m} binomial(m-1+j, m-1)*a(j)*T(n-1, m-1+j), m >= 1.
Recurrence for row polynomials R(n, x) (Meixner type): R(n, x) = ((1 + 4*x) + 4*x*d_x)*R(n-1, x), with differentiation d_x, for n >= 1, with input R(0, x) = 1.
Boas-Buck recurrence for column sequence m: T(n, m) = (1/(n - m))*((n/2)*(2 + 4*m)*T(n-1, m) + m*Sum_{p=m..n-2} binomial(n, p)*(-4)^(n-p)*Bernoulli(n-p)*T(p, m)), for n > m >= 0, with input T(m,m) = 4^m. See a comment and references in A282629. An example is given below. - Wolfdieter Lang, Aug 11 2017

A307066 a(n) = exp(-1) * Sum_{k>=0} (n*k + 1)^n/k!.

Original entry on oeis.org

1, 2, 13, 199, 5329, 216151, 12211597, 909102342, 85761187393, 9957171535975, 1390946372509101, 229587693339867567, 44117901231194922193, 9748599124579281064294, 2451233017637221706477037, 695088863051920283838281851, 220558203335628758134165860609

Offset: 0

Ilya Gutkovskiy, Jun 24 2019

nonn

G. C. Greubel, Table of n, a(n) for n = 0..250

Cf. A000110, A126390, A134980, A284859, A285064, A292914, A307080.

A307066:= func< n | (&+[Binomial(n,k)*n^k*Bell(k): k in [0..n]]) >;
[A307066(n): n in [0..31]]; // G. C. Greubel, Jan 24 2024

Table[Exp[-1] Sum[(n k + 1)^n/k!, {k, 0, Infinity}], {n, 0, 16}]
Table[n! SeriesCoefficient[Exp[Exp[n x] + x - 1], {x, 0, n}], {n, 0, 16}]
Join[{1}, Table[Sum[Binomial[n, k] n^k BellB[k], {k, 0, n}], {n, 1, 16}]]

def A307066(n): return sum(binomial(n,k)*n^k*bell_number(k) for k in range(n+1))
[A307066(n) for n in range(31)] # G. C. Greubel, Jan 24 2024

a(n) = n! * [x^n] exp(exp(n*x) + x - 1).

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

A285065 Alternating row sums of Sheffer triangle S2[4,1] = A285061.

Original entry on oeis.org

1, -3, -7, 53, 497, -147, -44055, -437339, 971745, 90858205, 1254551513, -56188139, -361749699119, -7793811482035, -47717641321527, 2053219888651909, 77548473901557697, 1171383881442334141, -8155337883596701767

Offset: 0

Wolfdieter Lang, Apr 13 2017

See A285061 for details. This is a generalization of A000587.

Cf. A000587, A285061, A285064.

Table[Sum[Binomial[n, k]*BellB[k, -1]*4^k, {k, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Apr 19 2017 *)

from sympy import bell, binomial
def a(n): return sum([binomial(n, k)*bell(k, -1)*4**k for k in range(n + 1)]) # Indranil Ghosh, May 06 2017

a(n) = Sum_{m=0..n} (-1)^m*A285061(n, m), n >= 0.
E.g.f.: exp(x)*exp(1 - exp(4*x)).
a(n) = e*Sum_{m>=0} ((-1)^m / m!)*(1 + 4*m)^n, n >= 0, (Dobiński type formula).
a(n) = Sum_{k=0..n} binomial(n, k) * 4^k * A000587(k), n >= 0. - Vaclav Kotesovec, Apr 23 2017
a(0) = 1; a(n) = a(n-1) - Sum_{k=1..n} binomial(n-1,k-1) * 4^k * a(n-k). - Ilya Gutkovskiy, Nov 30 2023

A355167 a(n) = exp(-1/4) * Sum_{k>=0} (4k + 3)^n / (4^k k!).

Original entry on oeis.org

1, 4, 20, 128, 1008, 9280, 96704, 1120768, 14274816, 197833728, 2958521344, 47415508992, 809838505984, 14670950907904, 280760761434112, 5655835404271616, 119561580162646016, 2645030742360588288, 61087848487323959296, 1469652941137655103488, 36758243982057508175872, 954111239026567129595904

Offset: 0

Ilya Gutkovskiy, Jun 22 2022

nonn

Michael De Vlieger, Table of n, a(n) for n = 0..464
Adam Buck, Jennifer Elder, Azia A. Figueroa, Pamela E. Harris, Kimberly Harry, and Anthony Simpson, Flattened Stirling Permutations, arXiv:2306.13034 [math.CO], 2023. See p. 14.

Cf. A003576, A004213, A285064, A355164, A355165.

nmax = 21; CoefficientList[Series[Exp[3 x + (Exp[4 x] - 1)/4], {x, 0, nmax}], x] Range[0, nmax]!
a[0] = 1; a[n_] := a[n] = 3 a[n - 1] + Sum[Binomial[n - 1, k - 1] 4^(k - 1) a[n - k], {k, 1, n}]; Table[a[n], {n, 0, 21}]
Table[Sum[Binomial[n, k] 3^(n - k) 4^k BellB[k, 1/4], {k, 0, n}], {n, 0, 21}]

E.g.f.: exp(3*x + (exp(4*x) - 1) / 4).
a(0) = 1; a(n) = 3 * a(n-1) + Sum_{k=1..n} binomial(n-1,k-1) * 4^(k-1) * a(n-k).
a(n) = Sum_{k=0..n} binomial(n,k) * 3^(n-k) * A004213(k).
a(n) ~ 2^(2*n + 3/2) * n^(n + 3/4) * exp(n/LambertW(4*n) - n - 1/4) / (sqrt(1 + LambertW(4*n)) * LambertW(4*n)^(n + 3/4)). - Vaclav Kotesovec, Jun 27 2022

A367786 Expansion of e.g.f. exp(exp(4*x) - x - 1).

Original entry on oeis.org

1, 3, 25, 235, 2737, 36947, 563657, 9542715, 176920417, 3555369635, 76820077945, 1772943290763, 43469116126737, 1127040956393203, 30779951676185385, 882453651485815003, 26480355971228530369, 829522636694530362691, 27064267045022876869337, 917751849133986186857003

Offset: 0

Ilya Gutkovskiy, Nov 30 2023

nonn

Cf. A000110, A000296, A124311, A285064, A285065, A330605, A367785.

nmax = 19; CoefficientList[Series[Exp[Exp[4 x] - x - 1], {x, 0, nmax}], x] Range[0, nmax]!
a[0] = 1; a[n_] := a[n] = -a[n - 1] + Sum[Binomial[n - 1, k - 1] 4^k a[n - k], {k, 1, n}]; Table[a[n], {n, 0, 19}]
Table[Sum[(-1)^(n - k) Binomial[n, k] 4^k BellB[k], {k, 0, n}], {n, 0, 19}]

my(x='x+O('x^30)); Vec(serlaplace(exp(exp(4*x) - x - 1))) \\ Michel Marcus, Nov 30 2023

a(n) = exp(-1) * Sum_{k>=0} (4*k-1)^n / k!.
a(0) = 1; a(n) = -a(n-1) + Sum_{k=1..n} binomial(n-1,k-1) * 4^k * a(n-k).
a(n) = Sum_{k=0..n} (-1)^(n-k) * binomial(n,k) * 4^k * Bell(k).

A367837 Expansion of e.g.f. 1/(2 - x - exp(4*x)).

Original entry on oeis.org

1, 5, 66, 1294, 33752, 1100504, 43060176, 1965653232, 102548623744, 6018735869824, 392498702352128, 28155539333730560, 2203322337542003712, 186790304541786160128, 17053569926181643921408, 1668166923908523824576512, 174057374767036007615922176

Offset: 0

Seiichi Manyama, Dec 02 2023

nonn

Cf. A006155, A367835, A367836.

Cf. A285064, A343674, A355112, A367840.

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

a(0) = 1; a(n) = n * a(n-1) + Sum_{k=1..n} 4^k * binomial(n,k) * a(n-k).

A355162 a(n) = exp(-1) * Sum_{k>=0} (4*k + 2)^n / k!.

Original entry on oeis.org

1, 6, 52, 568, 7312, 107360, 1760576, 31760256, 623137024, 13179872768, 298391335936, 7189153167360, 183428957442048, 4935794590572544, 139571328018628608, 4134634425826115584, 127966201403431518208, 4127825849826169716736, 138477447400991610896384, 4822002684952714247929856

Offset: 0

Ilya Gutkovskiy, Jun 22 2022

nonn

Cf. A000110, A005493, A126390, A284859, A284864, A285064, A355163.

nmax = 19; CoefficientList[Series[Exp[Exp[4 x] + 2 x - 1], {x, 0, nmax}], x] Range[0, nmax]!
a[0] = 1; a[n_] := a[n] = 2 a[n - 1] + Sum[Binomial[n - 1, k - 1] 4^k a[n - k], {k, 1, n}]; Table[a[n], {n, 0, 19}]
Table[Sum[Binomial[n, k] 2^(n + k) BellB[k], {k, 0, n}], {n, 0, 19}]

E.g.f.: exp(exp(4*x) + 2 x - 1).
a(0) = 1; a(n) = 2 * a(n-1) + Sum_{k=1..n} binomial(n-1,k-1) * 4^k * a(n-k).
a(n) = Sum_{k=0..n} binomial(n,k) * 2^(n+k) * Bell(k).
a(n) = 2^n * A126390(n). - Vaclav Kotesovec, Jun 22 2022
a(n) ~ 4^n * n^(n + 1/2) * exp(n/LambertW(n) - n - 1) / (sqrt(1 + LambertW(n)) * LambertW(n)^(n + 1/2)). - Vaclav Kotesovec, Jun 27 2022

A355163 a(n) = exp(-1) * Sum_{k>=0} (4*k + 3)^n / k!.

Original entry on oeis.org

1, 7, 65, 743, 9921, 150151, 2526593, 46615783, 933072513, 20093861895, 462440842177, 11310514854375, 292627518129985, 7976748158144647, 228308400790500097, 6840702405678586343, 214000748166439723265, 6973447420429351808007, 236204029044752265931585, 8300724166287243795922151

Offset: 0

Ilya Gutkovskiy, Jun 22 2022

nonn

Cf. A000110, A005494, A126390, A284859, A284864, A285064, A355162.

nmax = 19; CoefficientList[Series[Exp[Exp[4 x] + 3 x - 1], {x, 0, nmax}], x] Range[0, nmax]!
a[0] = 1; a[n_] := a[n] = 3 a[n - 1] + Sum[Binomial[n - 1, k - 1] 4^k a[n - k], {k, 1, n}]; Table[a[n], {n, 0, 19}]
Table[Sum[Binomial[n, k] 3^(n - k) 4^k BellB[k], {k, 0, n}], {n, 0, 19}]

E.g.f.: exp(exp(4*x) + 3 x - 1).
a(0) = 1; a(n) = 3 * a(n-1) + Sum_{k=1..n} binomial(n-1,k-1) * 4^k * a(n-k).
a(n) = Sum_{k=0..n} binomial(n,k) * 3^(n-k) * 4^k * Bell(k).
a(n) ~ Bell(n) * (4 + 3*LambertW(n)/n)^n. - Vaclav Kotesovec, Jun 22 2022
a(n) ~ 4^n * n^(n + 3/4) * exp(n/LambertW(n) - n - 1) / (sqrt(1 + LambertW(n)) * LambertW(n)^(n + 3/4)). - Vaclav Kotesovec, Jun 27 2022

A355165 a(n) = exp(-1/4) * Sum_{k>=0} (4k + 2)^n / (4^k k!).

Original entry on oeis.org

1, 3, 13, 79, 601, 5339, 53861, 607527, 7560625, 102637235, 1506225085, 23726435583, 398852249097, 7120170905995, 134408217821205, 2673140092099543, 55832167947587425, 1221199519275467107, 27902127744298836845, 664446811342185649583, 16457968670922936733113, 423242969435491221774907

Offset: 0

Ilya Gutkovskiy, Jun 22 2022

nonn

Cf. A003576, A004213, A285064, A355164, A355167.

nmax = 21; CoefficientList[Series[Exp[2 x + (Exp[4 x] - 1)/4], {x, 0, nmax}], x] Range[0, nmax]!
a[0] = 1; a[n_] := a[n] = 2 a[n - 1] + Sum[Binomial[n - 1, k - 1] 4^(k - 1) a[n - k], {k, 1, n}]; Table[a[n], {n, 0, 21}]
Table[Sum[Binomial[n, k] 2^(n + k) BellB[k, 1/4], {k, 0, n}], {n, 0, 21}]

E.g.f.: exp(2*x + (exp(4*x) - 1) / 4).
a(0) = 1; a(n) = 2 * a(n-1) + Sum_{k=1..n} binomial(n-1,k-1) * 4^(k-1) * a(n-k).
a(n) = Sum_{k=0..n} binomial(n,k) * 2^(n-k) * A004213(k).
a(n) ~ 2^(2*n+1) * n^(n + 1/2) * exp(n/LambertW(4*n) - n - 1/4) / (sqrt(1 + LambertW(4*n)) * LambertW(4*n)^(n + 1/2)). - Vaclav Kotesovec, Jun 27 2022

cp's OEIS Frontend

A126390 a(n) = Sum_{i=0..n} 2^i*B(i)*binomial(n,i) where B(n) = Bell numbers A000110(n).

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A285061 Sheffer triangle S2[4,1] = (exp(x), exp(4*x) - 1).

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A307066 a(n) = exp(-1) * Sum_{k>=0} (n*k + 1)^n/k!.

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A285065 Alternating row sums of Sheffer triangle S2[4,1] = A285061.

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A355167 a(n) = exp(-1/4) * Sum_{k>=0} (4*k + 3)^n / (4^k * k!).

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A367786 Expansion of e.g.f. exp(exp(4*x) - x - 1).

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A367837 Expansion of e.g.f. 1/(2 - x - exp(4*x)).

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A355162 a(n) = exp(-1) * Sum_{k>=0} (4*k + 2)^n / k!.

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Mathematica

Formula

A355163 a(n) = exp(-1) * Sum_{k>=0} (4*k + 3)^n / k!.

A126390 a(n) = Sum_{i=0..n} 2^iB(i)binomial(n,i) where B(n) = Bell numbers A000110(n).

A355167 a(n) = exp(-1/4) * Sum_{k>=0} (4k + 3)^n / (4^k k!).

A355165 a(n) = exp(-1/4) * Sum_{k>=0} (4k + 2)^n / (4^k k!).