Displaying 1-10 of 81 results found.
Numerator of the coefficient of z^(2n) in the Stirling-like asymptotic expansion of the hyperfactorial function A002109.
+20
10
1, 1, -1433, 1550887, -365236274341, 31170363588856607, -2626723351027654662151, 127061942835077684151157039, -5696145248370283185291966600124423, 254326794362835881966596504823903633657, -33203124408022060010631772664020406983485604379
REFERENCES
Mohammad K. Azarian, On the Hyperfactorial Function, Hypertriangular Function, and the Discriminants of Certain Polynomials, International Journal of Pure and Applied Mathematics, Vol. 36, No. 2, 2007, pp. 251-257. Mathematical Reviews, MR2312537. Zentralblatt MATH, Zbl 1133.11012.
FORMULA
Let B_n be the Bernoulli number, and define the sequence {c_n} by the recurrence
c_0 = 1, c_n = (-1/(2*n)) * Sum_{k=0..n-1} B_{2*n-2*k+2}*c_k/((2*n-2*k+1)*(2*n-2*k+2)) for n > 0.
a(n) is the numerator of c_n. (End)
EXAMPLE
(Glaisher*(1 - 1433/(7257600*z^4) + 1/(720*z^2))*z^(1/12 + (z*(1 + z))/2))/e^(z^2/4).
c_1 = -1/2 * (B_4*c_0/(3*4)) = 1/720, so a(1) = 1.
c_2 = -1/4 * (B_6*c_0/(5*6) + B_4*c_1/(3*4)) = -1433/7257600, so a(2) = -1433. (End)
Denominator of the coefficient of z^(2n) in the Stirling-like asymptotic expansion of the hyperfactorial function A002109.
+20
10
1, 720, 7257600, 15676416000, 3476402012160000, 162695614169088000000, 4919915372473221120000000, 60219764159072226508800000000, 507464726196802564122476544000000000, 3288371425755280615513648005120000000000
COMMENTS
In Glaisher (1878) equation (2) is "1^1.2^2.3^3 ... n^n = A n^(n^2/2 + n/2 + 1/12) e^(-n^4/4) (1 + 1/(720n^2) - 1433/(7257600n^4) + &c.)" - Michael Somos, Jun 24 2012
REFERENCES
Mohammad K. Azarian, On the Hyperfactorial Function, Hypertriangular Function, and the Discriminants of Certain Polynomials, International Journal of Pure and Applied Mathematics, Vol. 36, No. 2, 2007, pp. 251-257. Mathematical Reviews, MR2312537. Zentralblatt MATH, Zbl 1133.11012.
J. W. L. Glaisher, On The Product 1^1.2^2.3^3 ... n^n, Messenger of Mathematics, 7 (1878), pp. 43-47, see p. 43 eq. (2)
FORMULA
Let B_n be the Bernoulli number, and define the sequence {c_n} by the recurrence
c_0 = 1, c_n = (-1/(2*n)) * Sum_{k=0..n-1} B_{2*n-2*k+2}*c_k/((2*n-2*k+1)*(2*n-2*k+2)) for n > 0.
a(n) is the denominator of c_n. (End)
EXAMPLE
(Glaisher*(1 - 1433/(7257600*z^4) + 1/(720*z^2))*z^(1/12 + (z*(1 + z))/2))/e^(z^2/4).
c_1 = -1/2 * (B_4*c_0/(3*4)) = 1/720, so a(1) = 720.
c_2 = -1/4 * (B_6*c_0/(5*6) + B_4*c_1/(3*4)) = -1433/7257600, so a(2) = 7257600. (End)
0, 0, 2, 2, 10, 10, 16, 16, 40, 40, 50, 50, 74, 74, 88, 88, 152, 152, 170, 170, 210, 210, 232, 232, 304, 304, 330, 330, 386, 386, 416, 416, 576, 576, 610, 610, 682, 682, 720, 720, 840, 840, 882, 882, 970, 970, 1016, 1016, 1208, 1208, 1258, 1258, 1362, 1362, 1416, 1416, 1584, 1584, 1642, 1642
COMMENTS
This is the function ord_2(D*_n) listed in the leftmost column of Table 7.1 in Lagarias & Mehta 2014 paper (on page 19).
FORMULA
a(n) = A174605(n) + A187059(n). [Lagarias and Mehta theorem 4.1 for p=2] a(n) = Sum_{i=1..n} i*v_2(i), where v_2(i) = A007814(i) is the exponent of the highest power of 2 dividing i. - Ridouane Oudra, Oct 17 2019 a(n) ~ (n^2+2n)/2 as n -> infinity. - Luca Onnis, Oct 17 2021 a(2n) = a(2n+1) = 2*a(n) + n*(n+1).
a(n) = ( n^2 + Sum_{j=1..k} (e[j]-2*j+1) * 2^e[j] )/2, where binary expansion n = 2^e[1] + ... + 2^e[k] with ascending exponents e[1] < e[2] < ... < e[k] ( A133457). (End)
a(n) = Sum_{j=1..floor(log_2(n))} j*2^j*round(n/2^(j+1))^2, for n>=1. - Ridouane Oudra, Oct 01 2022
MAPLE
with(padic): seq(add(i*ordp(i, 2), i=1..n), n=0..60); # Ridouane Oudra, Oct 17 2019
MATHEMATICA
Table[i=0; Hyperfactorial@n//.x_/; EvenQ@x:>(i++; x/2); i, {n, 0, 60}] (* Giorgos Kalogeropoulos, Oct 28 2021 *)
PROG
(Scheme, two alternative implementations)
(Magma) [0] cat [&+[i*Valuation(i, 2):i in [1..n]]:n in [1..60]]; // Marius A. Burtea, Oct 18 2019 (PARI) a(n) = sum(i=1, n, i*valuation(i, 2)); \\ Michel Marcus, Sep 14 2021 (PARI) a(n) = my(v=binary(n), t=0); forstep(j=#v, 1, -1, if(v[j], v[j]=t--, t++)); (n^2 + fromdigits(v, 2))>>1; \\ Kevin Ryde, Nov 03 2021 (Python)
def A249152(n): return sum(i*(~i&i-1).bit_length() for i in range(2, n+1, 2)) # Chai Wah Wu, Jul 11 2022
A000142 (n+1) * A002109(n), a product of factorials and hyperfactorials.
+20
5
1, 2, 24, 2592, 3317760, 62208000000, 20316635136000000, 133852981198454784000000, 20211123400293732996612096000000, 78302033109811407811828935756349440000000, 8613223642079254859301182933198438400000000000000000
COMMENTS
row products of the triangle A245334.
FORMULA
a(n) ~ A * sqrt(2*Pi) * n^(n^2/2+3*n/2+19/12) / exp(n*(n+4)/4), where A = 1.2824271291... is the Glaisher-Kinkelin constant (see A074962). - Vaclav Kotesovec, Nov 14 2014
MATHEMATICA
Table[(n+1)*(n!)^(n+1)/BarnesG[n+1], {n, 0, 10}] (* Vaclav Kotesovec, Nov 14 2014 *)
PROG
(Haskell)
a240993 n = a000142 (n + 1) * a002109 n
Exponent of 2 in the hyperfactorial of 2n: a(n) = A007814( A002109(2n)).
+20
4
0, 2, 10, 16, 40, 50, 74, 88, 152, 170, 210, 232, 304, 330, 386, 416, 576, 610, 682, 720, 840, 882, 970, 1016, 1208, 1258, 1362, 1416, 1584, 1642, 1762, 1824, 2208, 2274, 2410, 2480, 2696, 2770, 2922, 3000, 3320, 3402, 3570, 3656, 3920, 4010, 4194, 4288, 4768, 4866, 5066, 5168, 5480, 5586, 5802, 5912
MATHEMATICA
Table[IntegerExponent[Hyperfactorial[2*n], 2], {n, 0, 55}] (* Amiram Eldar, Sep 10 2024 *)
PROG
(Scheme, two alternative versions)
(Python)
from sympy import multiplicity
for i in range(2, 20002, 2):
n += multiplicity(2, i)*i
Number of trailing zeros in A002109(n).
+20
3
0, 0, 0, 0, 0, 5, 5, 5, 5, 5, 15, 15, 15, 15, 15, 30, 30, 30, 30, 30, 50, 50, 50, 50, 50, 100, 100, 100, 100, 100, 130, 130, 130, 130, 130, 165, 165, 165, 165, 165, 205, 205, 205, 205, 205, 250, 250, 250, 250, 250, 350, 350, 350, 350, 350, 405, 405, 405, 405
FORMULA
a(n) = Sum_{i=1..n} i*v_5(i), where v_5(i) = A112765(i) is the exponent of the highest power of 5 dividing i. After a similar formula in A249152. (End)
MATHEMATICA
(n=#; k=0; While[Mod[n, 10]==0, n=n/10; k++]; k)&/@Hyperfactorial@Range[0, 60] (* Giorgos Kalogeropoulos, Sep 14 2021 *)
PROG
(Python)
def a(n):
..s = 1
..for k in range(n+1):
....s *= k**k
..i = 1
..while not s % 10**i:
....i += 1
..return i-1
n = 1
while n < 100:
..print(a(n), end=', ')
(Python)
from sympy import multiplicity
for n in range(5, 10**3, 5):
....p5 += multiplicity(5, n)*n
(PARI) a(n) = sum(i=1, n, i*valuation(i, 5)); \\ Michel Marcus, Sep 14 2021
a(n) = Product_{k=1..n} A002109(k).
+20
2
1, 1, 4, 432, 11943936, 1031956070400000, 4159895825138319360000000000, 13809882382682787973867537170432000000000000000, 769161257109634779902443718589603914508004789479014400000000000000000000, 16596916396875768196482032091931000424134701157007816971266990744831779993781534720000000000000000000000000
FORMULA
a(n) = BarnesG(n+2)^n / Product_{k=1..n+1} BarnesG(k)^2.
a(n) ~ A^(n+1) * n^(n^3/6 + n^2/2 + 5*n/12 + 1/12) / exp(5*n^3/36 + n^2/4 + n/12 + zeta(3)/(4*Pi^2)), where A is the Glaisher-Kinkelin constant A074962. (End)
MAPLE
seq(mul(mul(mul(k, j=1..k), k=1..m), m=1..n), n=0..9); # Zerinvary Lajos, Jun 01 2007
MATHEMATICA
Table[Product[Gamma[1 + k]^k/BarnesG[1 + k], {k, 1, n}], {n, 0, 10}] (* Vaclav Kotesovec, Nov 19 2023 *) Table[BarnesG[n + 2]^n/Product[BarnesG[k]^2, {k, 1, n + 1}], {n, 0, 10}] (* Vaclav Kotesovec, Nov 19 2023 *)
Possible number of trailing zeros in hyperfactorials ( A002109).
+20
2
0, 5, 15, 30, 50, 100, 130, 165, 205, 250, 350, 405, 465, 530, 600, 750, 830, 915, 1005, 1100, 1300, 1405, 1515, 1630, 1750, 2125, 2255, 2390, 2530, 2675, 2975, 3130, 3290, 3455, 3625, 3975, 4155, 4340, 4530, 4725, 5125, 5330, 5540, 5755, 5975, 6425, 6655
COMMENTS
The number of trailing zeros in A002109 increases every 5 terms since the exponent of the factor 5 increases every 5 terms and the exponent of the factor 2 increases every 2 terms.
PROG
(Python)
from sympy import multiplicity
for n in range(5, 5*10**3, 5):
....p5 += multiplicity(5, n)*n
Logarithmic derivative of the hyperfactorials ( A002109).
+20
1
1, 7, 313, 110143, 431860201, 24185951471887, 23238336572015738041, 445571476975584446962639039, 194201470505208674769594891331807753, 2157794122078406207016487628429579826176795887, 677208230450612019931822374477208301572175793625037599321
COMMENTS
Hyperfactorial A002109(n) = Product_{k=0..n} k^k.
REFERENCES
Mohammad K. Azarian, On the Hyperfactorial Function, Hypertriangular Function, and the Discriminants of Certain Polynomials, International Journal of Pure and Applied Mathematics, Vol. 36, No. 2, 2007, pp. 251-257. Mathematical Reviews, MR2312537. Zentralblatt MATH, Zbl 1133.11012.
FORMULA
a(n) ~ A * n^(n*(n+1)/2 + 13/12) / exp(n^2/4), where A = A074962 = 1.2824271291... is the Glaisher-Kinkelin constant. - Vaclav Kotesovec, Jul 10 2015
EXAMPLE
L.g.f.: L(x) = x + 7*x^2/2 + 313*x^3/3 + 110143*x^4/4 + 431860201*x^5/5 +...
where
exp(L(x)) = 1 + x + 4*x^2 + 108*x^3 + 27648*x^4 + 86400000*x^5 + 4031078400000*x^6 +...+ n^n*(n-1)^(n-1)*(n-2)^(n-2)*...*3^3*2^2*1^1*0^0**x^n +...
MATHEMATICA
nmax=15; Rest[CoefficientList[Series[Log[Sum[Product[j^j, {j, 1, k}]*x^k, {k, 0, nmax}]], {x, 0, nmax}], x] * Range[0, nmax]] (* Vaclav Kotesovec, Jul 10 2015 *)
PROG
(PARI) {a(n)=n*polcoeff(log(sum(k=0, n+1, prod(j=0, k, j^j)*x^k)+x*O(x^n)), n)}
for(n=1, 21, print1(a(n), ", "))
Partial sums of hyperfactorials ( A002109).
+20
1
1, 2, 6, 114, 27762, 86427762, 4031164827762, 3319770429936027762, 55696441261496986915227762, 21577941278638297470665013744027762, 215779412250996503370318565758665013744027762, 61564384586850833363801728392684283449726665013744027762
FORMULA
a(n) = Sum_{k = 0..n} (k!)^k/Barnes G-Function(k + 1).
EXAMPLE
a(0) = 1;
a(1) = 1 + 1^1 = 2;
a(2) = 1 + 1^1 + 1^1*2^2 = 6;
a(3) = 1 + 1^1 + 1^1*2^2 + 1^1*2^2*3^3 = 114;
a(4) = 1 + 1^1 + 1^1*2^2 + 1^1*2^2*3^3 + 1^1*2^2*3^3*4^4 = 27762, etc.
MATHEMATICA
Table[Sum[Hyperfactorial[k], {k, 0, n}], {n, 0, 11}]
Accumulate[Hyperfactorial[Range[0, 15]]] (* Harvey P. Dale, Sep 22 2021 *)
PROG
(PARI) a(n) = sum(k=0, n, prod(j=2, k, j^j)); \\ Altug Alkan, Nov 27 2015
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