A350297 -id:A350297 - cp's OEIS frontend

A350149 Triangle read by rows: T(n, k) = n^(n-k)*k!.

1, 1, 1, 4, 2, 2, 27, 9, 6, 6, 256, 64, 32, 24, 24, 3125, 625, 250, 150, 120, 120, 46656, 7776, 2592, 1296, 864, 720, 720, 823543, 117649, 33614, 14406, 8232, 5880, 5040, 5040, 16777216, 2097152, 524288, 196608, 98304, 61440, 46080, 40320, 40320

Offset: 0

Robert B Fowler, Dec 27 2021

T(n,k) are the denominators in a double summation power series for the definite integral of x^x. First expand x^x = exp(x*log(x)) = Sum_{n>=0} (x*log(x))^n/n!, then integrate each of the terms to get the double summation for F(x) = Integral_{t=0..x} t^t = Sum_{n>=1} (Sum_{k=0..n-1} (-1)^(n+k+1)*x^n*(log(x))^k/T(n,k)).
This is a definite integral, because lim {x->0} F(x) = 0.
The value of F(1) = 0.78343... = A083648 is known humorously as the Sophomore's Dream (see Borwein et al.).

			Triangle T(n,k) begins:
--------------------------------------------------------------------------
n/k         0        1       2       3      4      5      6      7      8
--------------------------------------------------------------------------
0  |        1,
1  |        1,       1,
2  |        4,       2,      2,
3  |       27,       9,      6,      6,
4  |      256,      64,     32,     24,    24,
5  |     3125,     625,    250,    150,   120,   120,
6  |    46656,    7776,   2592,   1296,   864,   720,   720,
7  |   823543,  117649,  33614,  14406,  8232,  5880,  5040,  5040,
8  | 16777216, 2097152, 524288, 196608, 98304, 61440, 46080, 40320, 40320.
...

Borwein, J., Bailey, D. and Girgensohn, R., Experimentation in Mathematics: Computational Paths to Discovery, A. K. Peters 2004.
William Dunham, The Calculus Gallery, Masterpieces from Newton to Lebesgue, Princeton University Press, Princeton NJ 2005.

G. C. Greubel, Rows n = 0..50 of the triangle, flattened
Eric Weisstein's World of Mathematics, Sophomore's dream
Wikipedia, Sophomore's dream

Cf. A000312 (first column), A000169 (2nd column), A003308 (3rd column excluding first term), A000142 (main diagonal), A000142 (2nd diagonal excluding first term), A112541 (row sums).
Values of the integral: A083648, A073009.
Cf. A061711, A350297.

A350149:= func< n,k | n^(n-k)*Factorial(k) >;
[A350149(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Jul 31 2022

T := (n, k) -> n^(n - k)*k!:
seq(seq(T(n, k), k = 0..n), n = 0..9); # Peter Luschny, Jan 07 2022

T[n_, k_]:= n^(n-k)*k!; Table[T[n, k], {n, 0,12}, {k,0,n}]//Flatten (* Amiram Eldar, Dec 27 2021 *)

def A350149(n,k): return n^(n-k)*factorial(k)
flatten([[A350149(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Jul 31 2022

T(n, 0) = A000312(n).
T(n, 1) = A000169(n).
T(n, 2) = A003308(n), n >= 2.
Sum_{k=0..n} T(n, k) = A112541(n).
T(n, n) = A000142(n).
T(n, n-1) = A000142(n), n >= 1.
T(n,k) = A061711(n) * (n+1) / A350297(n+1,k). - Robert B Fowler, Jan 11 2022

A350358 Value of -F(0), where F(x) is the indefinite integral of x^(1/x).

Original entry on oeis.org

4, 2, 0, 3, 6, 9, 5, 8, 8, 7, 8, 3, 2, 0, 2, 2, 9, 8, 1, 3, 2, 4, 3, 9, 3, 8, 1, 8, 1, 8, 0, 8, 8, 1, 8, 9, 9, 8, 1, 5, 4, 9, 5, 5, 3, 2, 8, 7, 1, 2, 2, 3, 9, 5, 1, 4, 5, 3, 5, 4, 0, 7, 3, 5, 4, 3, 6, 4, 0, 1, 2, 0, 8, 1, 2, 1, 8, 7, 0, 6, 2, 2, 7, 3, 1, 5, 1, 4

Offset: 0

Robert B Fowler, Dec 26 2021

The indefinite integral of x^(1/x) can be derived by expanding
    x^(1/x) = exp(log(x)/x) = Sum_{n>=0} (log(x)/x)^n/n!,
then integrating term-by-term to get the double summation
    F(x) = x + (1/2)*(log(x))^2
             - (Sum_{n>=2} (n-1)^(-n-1) * x^(-n+1)/n!
             * (Sum_{k=0..n} A350297(n,k)*(log(x))^k)).
We assume the constant of integration in F(x) is zero.
To compute definite integrals of x^(1/x) for ranges of x starting at x=0 or x=1, we would need the values
    F(0) = lim {x->0} F(x) = -0.4203695887832022981324...
    F(1) = 1 - Sum_{n>=2} (n-1)^(-n-1) = -0.06687278808178024266...
Note that the definite Integral_{t=0..1} x^(1/x) = F(1)-F(0) = A175999.
The calculation of F(0) requires some care. See the FORMULA below.
Since x^(1/x) is the inverse of the infinite exponentiation function E(y) = y^(y^(y^(...))) = x, the definite integrals of these two functions are related by
    (Integral_{t=0..y} E(t) dt) + (Integral_{t=0..x} t^(1/t) dt) = x*y.
Note that the repeated exponentiation in E(y) converges for 0 <= y < e^(1/e), but diverges for y > e^(1/e).

			0.4203695887832022981324...

Peter Luschny, Table of n, a(n) for n = 0..999
Jonathan Sondow and Diego Marques, Algebraic and transcendental solutions of some exponential equations, Annales Mathematicae et Informaticae 37 (2010) 151-164; see Figure 5.
Eric Weisstein's World of Mathematics, Steiner's Problem.

Cf. A350297, A175999, A176000, A073229.

# The chosen parameters give about 100 exact decimal places.
using Nemo
RR = RealField(1100)
function F(b::Int, x::arb)
    lnx = log(x)
    s = sum(gamma_regularized(RR(n+2), RR(n)*lnx) * RR(n)^(-n-2) for n in 1:b)
    x + (lnx * lnx) / RR(2) - s
end
println( F(1400, RR(0.015)) ) # Peter Luschny, Dec 26 2021

# The chosen parameters give about 100 exact decimal places.
Digits := 400: F := proc(b, x) local s, lnx; lnx := log(x);
   s := add(add((n*lnx)^k / k!, k = 0..n+1) / (n*x)^(n+2), n = 1..b);
   x - x^2 * s + lnx^2 / 2 end:
F(2000, 0.01); # Peter Luschny, Dec 27 2021

RealDigits[N[Sum[1/n^(n+2), {n, 1, 100}] + Integrate[x^(1/x), {x, 0, 1}] - 1, 110]][[1]] (* Amiram Eldar, Dec 29 2021 *)

The calculation of F(0) requires some care, because terms in the formula for F(x) diverge for x=0, but converge for all x > 0, although convergence is progressively slower as x approaches zero. To calculate F(0), first choose a desired precision d (absolute error). Then choose any x such that 0 < x^(1/x) < d, and evaluate F(x) as defined above.
Since F(x)-F(0) < x^(1/x) < d, F(0)=F(x) to the desired precision.
For small x, the main summation terms initially increase in absolute value (but with alternating signs), reach a maximum of about 1/d at n = -log(x)/x, then decrease at an accelerating rate and reach a value of around d at n = -3.5911*log(x)/x, at which point the large terms have mostly cancelled out, and further terms are below precision level.
For example, for a final precision of d = 10^-50, the calculation must allow for intermediate terms and sums as large as 1/d = 10^50, so these terms must be evaluated to at least 100 digits. For 50 digits, F(.03) is a suitable choice, because .03^(1/.03) = 1.7273...*10^-51 < 10^-50. A few extra digits should also be allowed for round off error.

More terms from Hugo Pfoertner, Dec 26 2021

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A350149 Triangle read by rows: T(n, k) = n^(n-k)*k!.

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