Displaying 1-10 of 11 results found.
a(n) = 3*n*a(n-1) + 1, a(0) = 1.
+10
12
1, 4, 25, 226, 2713, 40696, 732529, 15383110, 369194641, 9968255308, 299047659241, 9868572754954, 355268619178345, 13855476147955456, 581929998214129153, 26186849919635811886, 1256968796142518970529
COMMENTS
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
REFERENCES
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, Tenth Printing, 1972, p. 262.
FORMULA
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 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 a(n) = Integral_{x = 0..inf} (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 gives the continued fraction representation exp(1/3) = 1 + 1/(3 - 3/(7 - 6/(10 - ... - (3*n - 2)/((3*n + 1) - ... )))). Cf. A010844. (End)
EXAMPLE
1 + 4*x + 25*x^2 + 226*x^3 + 2713*x^4 + 40696*x^5 + 732529*x^6 + ...
MATHEMATICA
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 *)
PROG
(PARI) {a(n) = if( n<0, 0, n! * sum(k=0, n, 3^(n-k) / k!))} /* Michael Somos, Sep 04 2013 */
a(n) = 5*n*a(n-1) + 1 with a(0)=1.
+10
7
1, 6, 61, 916, 18321, 458026, 13740781, 480927336, 19237093441, 865669204846, 43283460242301, 2380590313326556, 142835418799593361, 9284302221973568466, 649901155538149792621, 48742586665361234446576
FORMULA
a(n) = floor(e^(1/5)*5^n*n!).
a(n) = n!*Sum_{k=0..n} 5^(n-k)/k!.
E.g.f.: exp(x)/(1 - 5*x). (End)
a(n) = Sum_{k=0..n} P(n, k)*5^k. - Ross La Haye, Aug 29 2005 a(n) = hypergeometric_U(1, n+2 , 1/5)/5. - Peter Luschny, Nov 26 2014 a(n) = Integral_{x >= 0} (5*x + 1)^n*exp(-x) dx.
The e.g.f. y = exp(x)/(1 - 5*x) satisfies the differential equation (1 - 5*x)*y' = (6 - 5*x)*y.
a(n) = (5*n + 1)*a(n-1) - 5*(n - 1)*a(n-2).
The sequence b(n) := 5^n*n! also satisfies the same recurrence with b(0) = 1, b(1) = 5. This leads to the continued fraction representation a(n) = 5^n*n!*( 1 + 1/(5 - 5/(11 - 10/(16 - ... - (5*n - 5)/(5*n + 1) )))) for n >= 2. Taking the limit gives the continued fraction representation exp(1/5) = 1 + 1/(5 - 5/(11 - 10/(16 - ... - (5*n - 5)/((5*n + 1) - ... )))). Cf. A010844. (End)
EXAMPLE
a(2) = 5*2*a(1) + 1 = 10*6 + 1 = 61.
a(n) = 6*n*a(n-1) + 1 with a(0)=1.
+10
6
1, 7, 85, 1531, 36745, 1102351, 39684637, 1666754755, 80004228241, 4320228325015, 259213699500901, 17108104167059467, 1231783500028281625, 96079113002205966751, 8070645492185301207085, 726358094296677108637651
FORMULA
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 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)
EXAMPLE
a(2) = 6*2*a(1) + 1 = 12*7 + 1 = 85.
MATHEMATICA
nxt[{n_, a_}]:={n+1, 6a(n+1)+1}; NestList[nxt, {0, 1}, 20][[;; , 2]] (* Harvey P. Dale, Jul 17 2024 *)
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).
+10
4
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
FORMULA
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 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)
EXAMPLE
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, ...
MAPLE
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:
MATHEMATICA
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
a(n) = (n!)^2 * Sum_{k=0..n} 4^(n-k) / (k!)^2.
+10
4
1, 5, 81, 2917, 186689, 18668901, 2688321745, 526911062021, 134889231877377, 43704111128270149, 17481644451308059601, 8461115914433100846885, 4873602766713466087805761, 3294555470298303075356694437, 2582931488713869611079648438609, 2324638339842482649971683594748101
FORMULA
Sum_{n>=0} a(n) * x^n / (n!)^2 = BesselI(0,2*sqrt(x)) / (1 - 4*x).
a(0) = 1; a(n) = 4 * n^2 * a(n-1) + 1.
MATHEMATICA
Table[n!^2 Sum[4^(n - k)/k!^2, {k, 0, n}], {n, 0, 15}]
nmax = 15; CoefficientList[Series[BesselI[0, 2 Sqrt[x]]/(1 - 4 x), {x, 0, nmax}], x] Range[0, nmax]!^2
Expansion of e.g.f. -log(1-4*x) * exp(x)/4.
+10
4
0, 1, 6, 47, 540, 8429, 166210, 3952955, 109981816, 3502905369, 125648153278, 5011458069639, 219987094389524, 10538817637744005, 547118005892177018, 30595552548140425747, 1833501625083035349488, 117219490267316310468913
FORMULA
a(n) = n! * Sum_{k=0..n-1} 4^(n-1-k) / ((n-k) * k!).
a(0) = 0, a(1) = 1, a(n) = (4 * n - 3) * a(n-1) - 4 * (n-1) * a(n-2) + 1.
PROG
(PARI) my(N=20, x='x+O('x^N)); concat(0, Vec(serlaplace(-log(1-4*x)*exp(x)/4)))
(PARI) a(n) = n!*sum(k=0, n-1, 4^(n-1-k)/((n-k)*k!));
(PARI) a_vector(n) = my(v=vector(n+1)); v[1]=0; v[2]=1; for(i=2, n, v[i+1]=(4*i-3)*v[i]-4*(i-1)*v[i-1]+1); v;
CROSSREFS
Essentially partial sums of A056545.
E.g.f.: exp(x) / (1 - 4 * x)^(1/4).
+10
3
1, 2, 8, 64, 800, 13376, 278272, 6914048, 199629824, 6566164480, 242327576576, 9915111636992, 445432721932288, 21795710738038784, 1153805878313615360, 65700181140859518976, 4004182878034473254912, 260071258357260225609728, 17932703649301871611346944
FORMULA
a(n) = Sum_{k=0..n} binomial(n,k) * A007696(k). a(n) ~ n! * exp(1/4) * 4^n / (Gamma(1/4) * n^(3/4)). - Vaclav Kotesovec, Aug 14 2021
MAPLE
g:= proc(n) option remember; `if`(n<2, 1, (4*n-3)*g(n-1)) end:
a:= n-> add(binomial(n, k)*g(k), k=0..n):
MATHEMATICA
nmax = 18; CoefficientList[Series[Exp[x]/(1 - 4 x)^(1/4), {x, 0, nmax}], x] Range[0, nmax]!
Table[Sum[Binomial[n, k] 4^k Pochhammer[1/4, k], {k, 0, n}], {n, 0, 18}]
Table[HypergeometricU[1/4, n + 5/4, 1/4]/Sqrt[2], {n, 0, 18}]
a(n) = (-i)^n * Integral_{x>=0} H_n(i*x) * exp(-x), where H_n(x) is n-th Hermite polynomial, i=sqrt(-1).
+10
2
1, 2, 10, 60, 492, 4920, 59160, 828240, 13253520, 238563360, 4771297440, 104968543680, 2519245713600, 65500388553600, 1834010896798080, 55020326903942400, 1760650461445075200, 59862115689132556800, 2155036164826415270400, 81891374263403780275200
FORMULA
a(n) = exp(1/4)*(-2*i)^n * n!*( cos(Pi*n/2)*Gamma(n/2 +1, 1/4)/Gamma(n/2 +1) + i*Gamma((n+1)/2, 1/4)*sin(Pi*n/2)/Gamma((n+1)/2) ).
a(n) = 2^n*(n!/floor(n/2)!)*Gamma(ceiling((n+1)/2),1/4)*exp(1/4).
The swinging factorial A056040(n) divides a(n). (End)
MATHEMATICA
FunctionExpand@Table[Exp[1/4] (-2 I)^n n! (Cos[Pi n/2] Gamma[n/2 + 1, 1/4]/Gamma[n/2 + 1] + I Gamma[(n + 1)/2, 1/4] Sin[Pi n/2]/Gamma[(n + 1)/2]), {n, 0, 20}]
FunctionExpand@Table[2^n (n!/Floor[n/2]!) Gamma[Ceiling[(n+1)/2], 1/4] Exp[1/4], {n, 0, 19}] (* Peter Luschny, Oct 19 2016 *)
PROG
(Sage)
def A():
yield 1
yield 2
a, h, f, g, n, b = 10, 5, 1, 2, 2, False
while True:
yield a
if b:
f = h
h = 4 * n * h + 1
n += 1
a = (a * h) // f
else:
g += 4
a *= g
b = not b
a = A(); print([next(a) for _ in range(20)]) # Peter Luschny, Oct 19 2016 (PARI) for(n=0, 30, print1(round(2^n*(n!/floor(n/2)!)* incgam(ceil( (n+1)/2), 1/4)*exp(1/4)), ", ")) \\ G. C. Greubel, Jul 12 2018
Expansion of e.g.f. exp(2x)/(1-5x).
+10
1
1, 7, 74, 1118, 22376, 559432, 16783024, 587405968, 23496238976, 1057330754432, 52866537722624, 2907659574746368, 174459574484786176, 11339872341511109632, 793791063905777690624, 59534329792933326829568
COMMENTS
Second binomial transform of n!*5^n.
FORMULA
a(n) = 5*n*a(n-1) + 2^n, n > 0, a(0)=1.
D-finite with recurrence a(n) +(-5*n-2)*a(n-1) +10*(n-1)*a(n-2)=0. - R. J. Mathar, Aug 20 2021 a(n) = 5^n * n! * Sum_{k = 0..n} (2/5)^k/k! = 5^n * exp(2/5) * gamma(n + 1, 2/5). - Gerry Martens, Nov 07 2022
MAPLE
f:= proc(n) option remember; 5*n*procname(n-1)+2^n end proc:
f(0):= 1:
PROG
(PARI) my(x='x + O('x^25)); Vec(serlaplace(exp(2*x)/(1-5*x))) \\ Michel Marcus, Nov 08 2022
Triangle read by rows: T(n, k) = n! * 4^k * hypergeom([-k], [-n], 1/4).
+10
1
1, 1, 5, 2, 9, 41, 6, 26, 113, 493, 24, 102, 434, 1849, 7889, 120, 504, 2118, 8906, 37473, 157781, 720, 3000, 12504, 52134, 217442, 907241, 3786745, 5040, 20880, 86520, 358584, 1486470, 6163322, 25560529, 106028861, 40320, 166320, 686160, 2831160, 11683224, 48219366, 199040786, 821723673, 3392923553
FORMULA
T(n, k) = Sum_{j=0..k} 4^(k - j)*binomial(k, k - j)*(n - j)!.
EXAMPLE
Triangle starts:
[0] 1;
[1] 1, 5;
[2] 2, 9, 41;
[3] 6, 26, 113, 493;
[4] 24, 102, 434, 1849, 7889;
[5] 120, 504, 2118, 8906, 37473, 157781;
[6] 720, 3000, 12504, 52134, 217442, 907241, 3786745;
[7] 5040, 20880, 86520, 358584, 1486470, 6163322, 25560529, 106028861;
...
MATHEMATICA
T[n_, k_] := Sum[4^(k - j)*Binomial[k, k - j]*(n - j)!, {j, 0, k}];
Table[T[n, k], {n, 0, 8}, {k, 0, n}] // Flatten
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