Hywel Normington - cp's OEIS frontend

A360809 Decimal expansion of the area under the curve of the reciprocal of the Luschny factorial function from zero to infinity.

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

Offset: 1

Hywel Normington, Feb 21 2023

			2.58670505978680822777810687294696021357309627424893612446708242258594...

Peter Luschny, Is the Gamma-function misdefined? Or: Hadamard vs Euler - Who found the better Gamma function?.
Hywel Normington, Python Code, 2023.
Hywel Normington, Reciprocal Gamma Integrals, 2023.

Cf. A360375, A058655.

L := proc(x) local G, S, y; if x = 0 then return 0.5 fi; y := x * 0.5; if is(x < 0) then y := -y fi; G := y * (Psi(y + 0.5) - Psi(y)) - 0.5; if is(x < 0) then return G/(-x)! fi; y := Pi * x; S := sin(y) / y; (1 - S * G) * x! end: RL := x -> 1 / L(x):
IntRL := n -> evalf[n](Int(RL, 0..n, method = Gquad)): IntRL(40); # _Peter Luschny, Feb 22 2023

RealDigits[NIntegrate[1 / (Gamma[x+1] * (1 - (x/2 * (PolyGamma[0, (x+1)/2] - PolyGamma[0, x/2]) - 1/2) * Sin[Pi*x]/(Pi*x))), {x, 0, Infinity}, WorkingPrecision -> 110, MaxRecursion -> Infinity]][[1]] (* Vaclav Kotesovec, Feb 22 2023 *)

default(realprecision, 500); intnum(x=0, [[1], 1], 1 / (gamma(x+1) * (1 - (x/2 * (psi((x+1)/2) - psi(x/2)) - 1/2) * sin(Pi*x)/(Pi*x)))) \\ (default(realprecision, 200) is enough for 59 valid digits, \p 500 for 102 valid digits, \p 1000 for 148 valid digits). - Vaclav Kotesovec, Feb 22 2023

L(x) = Gamma(x+1)P(x), where P(x) = 1 - g(x)*sin(Pi*x)/(Pi*x) and g(x) = (x/2)*(Psi((x+1)/2) - Psi(x/2)) - 1/2.

Equals Integral_{0..oo} 1/L(x) dx.

More digits from Vaclav Kotesovec, Feb 22 2023

A360375 Decimal expansion of the area under the curve of the reciprocal of the Hadamard gamma function from zero to infinity.

Original entry on oeis.org

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

Offset: 1

Hywel Normington, Feb 04 2023

Close to 3 + (1/e) = 3.367879...

Sum_{n>=0} 1/H(n) = 1/log(2) + e - 1 = 3.1609768... This integral may have a similar representation to the Fransen-Robinson constant.

			3.368202929607022792162205962209362548476...

J. Hadamard, (1894), Oeuvre de Jacques Hadamard, Centre National de la Recherche Scientifiques, Paris, 1968.

Philip J. Davis, Leonhard Euler's Integral: A Historical Profile Of The Gamma Function, Am. Math. Monthly 66, 849-869 (1959).
J. Hadamard, Sur l’Expression du Produit 1*2*3...*(n-1) par une Fonction Entière (in French, obtainable from Peter Luschny's Website).
Hywel Normington, Reciprocal Gamma Integrals, 2023.
Hywel Normington, Julia code, 2023.
Hywel Normington, Python code, 2023.
Wikipedia, Hadamard's gamma function.

Cf. A058655.

H := x -> 1/((sin(x*Pi)*(Psi(x/2) - Psi(1/2 + x/2)) + 2*Pi) * GAMMA(x)):
evalf[80](2*Pi*Int(H, 0..60, method = Gquad)); # _Peter Luschny, Feb 20 2023

RealDigits[NIntegrate[2*Gamma[1-x]/(PolyGamma[0, 1 - x/2] - PolyGamma[0, 1/2 - x/2]), {x, 0, Infinity}, WorkingPrecision -> 105, MaxRecursion -> Infinity]][[1]] (* Vaclav Kotesovec, Feb 19 2023 *)

default(realprecision, 200); intnum(x=0,[[1], 1], 2*gamma(1-x) / (psi(1-x/2) - psi(1/2-x/2))) \\ (default(realprecision, 200) is enough for 40 valid digits, \p 500 for 71 valid digits, \p 1000 for 110 valid digits, \p 2000 for 171 valid digits). - Vaclav Kotesovec, Feb 19 2023

Hadamard function definitions:
  H(x) = (1/Gamma(1-x)) * (d/dx) log(Gamma(1/2 - x/2)/Gamma(1-x/2)).
  H(x) = Gamma(x)*(1 + (sin(Pi*x)/(2*Pi)) * (Psi(x/2) - Psi((x+1)/2))).
Equals Integral_{0..oo} 1/H(x) dx.

More digits from Vaclav Kotesovec, Feb 19 2023

A337559 Number of length three 1..n vectors that contain their harmonic mean.

Original entry on oeis.org

1, 2, 3, 4, 5, 18, 19, 20, 21, 22, 23, 36, 37, 38, 51, 52, 53, 66, 67, 80, 81, 82, 83, 96, 97, 98, 99, 112, 113, 138, 139, 140, 141, 142, 155, 168, 169, 170, 171, 184, 185, 210, 211, 212, 237, 238, 239, 252, 253, 254, 255, 256, 257, 270, 271, 284, 285, 286, 287, 324, 325, 326, 339

Offset: 1

Hywel Normington, Aug 31 2020

nonn

			For n = 1, the only solution is (1,1,1).
For n = 6, the a(6) = 18 solutions are (k,k,k) for k=1,..,6, the 6 permutations of (2,3,6) and the 6 permutations of (3,4,6).
For n = 40, the a(40)-a(39) = 13 new solutions are (40,40,40), the 6 permutations of (10,16,40) and the 6 permutations of (24,30,40).

Hywel Normington, Python code, 2020.

Cf. A248434, A337110, A337111, A337560.

Empirical: If A174903(n) = 0, a(n) = a(n-1) + 1.

a(n)-a(n-1) = 1 (mod 6).

A337560 Number of length four 1..n vectors that contain their harmonic mean.

Original entry on oeis.org

1, 2, 3, 28, 29, 78, 79, 104, 105, 154, 155, 252, 253, 254, 351, 376, 377, 426, 427, 620, 717, 718, 719, 912, 913, 914, 915, 1084, 1085, 1374, 1375, 1400, 1425, 1426, 1619, 1812, 1813, 1814, 1839, 2128, 2129, 2442, 2443, 2564, 2829, 2830, 2831, 3192, 3193

Offset: 1

Hywel Normington, Aug 31 2020

nonn

			For n = 1, the a(1) = 1 solution is (1,1,1,1).
For n = 4, the a(4) = 28 solutions are (k,k,k,k) for k=1,2,3,4, the 12 permutations of (1,2,4,4) and the 12 permutations of (2,3,4,4).
For n = 40, the a(40) - a(39) = 289 new solutions are:
  the 24 permutations of (4, 8, 10, 40),
  the 12 permutations of (4, 10, 40, 40),
  the 12 permutations of (5, 12, 40, 40),
  the 24 permutations of (8, 10, 12, 40),
  the 24 permutations of (8, 15, 20, 40),
  the 12 permutations of (10, 16, 16, 40),
  the 24 permutations of (10, 18, 24, 40),
  the 12 permutations of (10, 20, 40, 40),
  the 24 permutations of (12, 20, 24, 40),
  the 24 permutations of (14, 24, 35, 40),
  the 24 permutations of (15, 24, 30, 40),
  the 12 permutations of (16, 16, 20, 40),
  the 12 permutations of (20, 20, 24, 40),
  the 12 permutations of (20, 30, 40, 40),
  the 12 permutations of (24, 30, 30, 40),
  the 12 permutations of (28, 35, 40, 40),
  the 12 permutations of (30, 36, 40, 40),
  and (40, 40, 40, 40).

Hywel Normington, Python code, 2020.

Cf. A248434, A337110, A337111, A337559.

a(n) - a(n-1) = 1 (mod 12).

a(43)-a(49) from Alois P. Heinz, Feb 04 2023

A337110 Number of length three 1..n vectors that contain their geometric mean.

Original entry on oeis.org

1, 2, 3, 10, 11, 12, 13, 20, 33, 34, 35, 42, 43, 44, 45, 64, 65, 78, 79, 86, 87, 88, 89, 96, 121, 122, 135, 142, 143, 144, 145, 164, 165, 166, 167, 198, 199, 200, 201, 208, 209, 210, 211, 218, 231, 232, 233, 252, 289, 314, 315, 322, 323, 336, 337, 344, 345, 346

Offset: 1

Hywel Normington, Aug 16 2020

From David A. Corneth, Aug 26 2020: (Start)
If x^2 == 0 (mod n) has only 1 solution then a(n) = a(n-1) + 1. Proof:
Let (a, b, n) be such a tuple. Let without loss of generality b be the geometric mean of the tuple. Then a*b*n = b^3 and as b is not 0 we have b^2 = a*n. So then b^2 == 0 (mod n). If b^2 == 0 (mod n) has only 1 solution then b = n. This gives the tuple (n, n, n) which has 1 permutation. So giving a(n) = a(n-1) + 1. (End)

			For n = 2, the a(2) = 2 solutions are: (1,1,1) and (2,2,2).
For n = 4, the a(4) = 10 solutions are: (1,1,1),(2,2,2),(3,3,3),(4,4,4) and the 6 permutations of (1,2,4).

Hywel Normington, Python code, 2020.
Hywel Normington, Julia code, 2023.

Cf. A120486, A248434, A337111, A013929, A057918, A000188.

first(n) = {my(s = 0, res = vector(n)); for(i = 1, n, s+=b(i); res[i] = s ); res }
b(n) = { my(s = factorback(factor(n)[, 1]), res = 1); for(i = 1, n \ s - 1, c = (s*i)^2/n; if(denominator(c) == 1 && c <= n, res+=6; ) ); res } \\ David A. Corneth, Aug 26 2020

a(n) = a(n-1) + 1 + 6*A057918(n).

A337111 Number of length four 1..n vectors that contain their geometric mean.

Original entry on oeis.org

1, 2, 3, 16, 17, 18, 19, 56, 105, 106, 107, 144, 145, 146, 147, 208, 209, 306, 307, 320, 321, 322, 323, 432, 529, 530, 723, 736, 737, 738, 739, 968, 969, 970, 971, 1176, 1177, 1178, 1179, 1288, 1289, 1290, 1291, 1304, 1401, 1402, 1403, 1608, 1777, 2018, 2019, 2032

Offset: 1

Hywel Normington, Aug 16 2020

From David A. Corneth, Aug 27 2020: (Start)
Let (a, b, g, n) be a tuple where g is the geometric mean of a*b*g*n and a, b, g <= n. Then there are two cases: g < n and g = n.
If g = n then (a, b, g, n) = (n, n, n, n) adding 1 to the number of permutations.
If g < n then a*b = g^4 / (g*n) = g^3 / n. Furthermore let s be the squarefree part of n (Cf. A007947). Then s | g and so candidates for g are (s*i) where 1 <= i < floor(n/s), depending on whether g^3 / n = (s*i)^3 / n is an integer.
It follows that for suitable values of i, (a, b) are a pair of divisors of (s*i)^3 / n where a*b = (s*i)^3 / n and max(a, b) <= n. (End)
Bounds: n + 12*(floor(n/4) + 2*floor(n/8) + 4*floor(n/9) + 2*floor(n/12) + 2*floor(n/16) + 4*floor(n/18)) <= a(n) <= n^4 - 14*(n-1). - Hywel Normington, Jan 25 2021

			For n = 2 the a(2) = 2 solutions are: (1,1,1,1) and (2,2,2,2).
For n = 4 the a(4) = 16 solutions are:
  (1, 1, 1, 1), (2, 2, 2, 2), (3, 3, 3, 3), (4, 4, 4, 4),
  and the 12 permutations of (1, 2, 2, 4).
For n = 40, the a(40)-a(39) = 109 new solutions are:
  (40,40,40,40),
  the 24 permutations of (1, 10, 25, 40),
  the 12 permutations of (5, 5, 10, 40),
  the 12 permutations of (5, 20, 40, 40),
  the 24 permutations of (8, 20, 25, 40),
  the 12 permutations of (10, 20, 20, 40),
  and the 24 permutations of (25, 27, 30, 40).

David A. Corneth, Table of n, a(n) for n = 1..10000
Hywel Normington, Python code, 2020.
Hywel Normington, Julia code, 2023.

Cf. A007947, A000189, A248435, A337110.

first(n) = { my(res = vector(n)); s = 0; for(i = 1, n, s += b(i);  res[i] = s; ); res }
b(n) = {my(resa = 1); my(s = factorback(factor(n)[, 1])); for(i = 1, n \ s - 1, s4 = (s*i)^3; if(s4 % n == 0, c = tuples((s*i)^3/n, s*i, n); for(i = 1, #c, resa+=qperms(c[i]) ) ) ); resa }
qperms(v) = {my(r=1,t); v = vecsort(v); for(i=1,#v-1,if(v[i]==v[i+1],t++,r*=binomial(i,t+1);t=0));r*=binomial(#v,t+1)}
tuples(n, s, u) = {my(res = List(), u4n, d, i); d = divisors(n); i = (#d + 1) \ 2; while(i > 0 && d[#d - i + 1] <= u, c = vecsort([d[i], d[#d - i + 1], s, u]); listput(res, c); i--); res} \\ David A. Corneth, Aug 28 2020

Empirical: if A000189(n) = 1 then a(n) = a(n-1) + 1.
From David A. Corneth, Aug 25 2020: (Start)
The above holds. That is: if x^3 == 0 (mod n) has only one solution then a(n) = a(n-1) + 1. Proof:
Let (a, b, c, n) be such a tuple. Let without loss of generality c be the geometric mean of the tuple. Then a*b*c*n = c^4 and as c is not 0 we have c^3 = a*b*n. So then c^3 == 0 (mod n). If c^3 == 0 (mod n) has only 1 solution then c = n. This gives the tuple (n, n, n, n) which has 1 permutation. So giving a(n) = a(n-1) + 1. (End)
a(n) - a(n-1) == 1 (mod 12). - Hywel Normington, Sep 28 2020

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User: Hywel Normington

A360809 Decimal expansion of the area under the curve of the reciprocal of the Luschny factorial function from zero to infinity.

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A360375 Decimal expansion of the area under the curve of the reciprocal of the Hadamard gamma function from zero to infinity.

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A337559 Number of length three 1..n vectors that contain their harmonic mean.

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A337560 Number of length four 1..n vectors that contain their harmonic mean.

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A337110 Number of length three 1..n vectors that contain their geometric mean.

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A337111 Number of length four 1..n vectors that contain their geometric mean.

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