A092425 Decimal expansion of Pi^4.

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

Offset: 2

Author

Mohammad K. Azarian, Mar 22 2004

Keywords

• cons
• easy
• nonn

Examples

			97.40909103400243723644033268870511124972758567268542169146785938997085...
		

Links

• Harry J. Smith, Table of n, a(n) for n = 2..20000
• Mohammad Reza Yegan, On the irrationality of Pi^4 and Pi^6, Journal of Number Theory, Volume 178, September 2017, Pages 5-10.
• Index entries for transcendental numbers

Crossrefs

Cf. A000796 (Pi), A002388 (Pi^2), A091925 (Pi^3), A092731 (Pi^5), A092732 (Pi^6), A092735 (Pi^7), A092736 (Pi^8), A058286 (continued fraction), A013662.

Programs

Magma

R:= RealField(150); (Pi(R))^4; // G. C. Greubel, Mar 09 2018

Magma

R:=RealField(110); SetDefaultRealField(R); n:=Pi(R)^4; Reverse(Intseq(Floor(10^98*n))); // Bruno Berselli, Mar 12 2018

Mathematica

RealDigits[Pi^4, 10, 100][[1]] (* G. C. Greubel, Mar 09 2018 *)

PARI

default(realprecision, 20080); x=Pi^4/10; for (n=2, 20000, d=floor(x); x=(x-d)*10; write("b092425.txt", n, " ", d)); \\ Harry J. Smith, Jun 22 2009

Formula

Equals 120 * Sum_{j>=1} Sum_{i=1..j-1} 1/(i*j)^2. - Enrique Pérez Herrero, Jun 29 2012

Equals Sum_{k>=1} k*(k+1)*(k+2)*zeta(k+3)/2^(k-1). - Amiram Eldar, May 21 2021

From Peter Bala, Oct 21 2023: (Start)

Pi^4 = 90*Sum_{n >= 1} 1/n^4 (Euler).

The following faster converging series representations for the constant Pi^4 may be easily verified using partial fraction expansions of the summands of the series. Presumably, these are the first three cases of an infinite family of similar results.

Let P(n) = n*(n + 1)*(n + 2)/2!. Then Pi^4 = 1575/16 - 15*Sum_{n >= 1} d/dn(P(n))/P(n)^4.

Let Q(n) = n*(n + 1)*(n + 2)*(n + 3)*(n + 4)/4!. Then Pi^4 = 673165/6912 + Sum_{n >= 1} d/dn(Q(n))/Q(n)^4.

Let R(n) = n*(n + 1)*(n + 2)*(n + 3)*(n + 4)*(n + 5)*(n + 6)/6!. Then Pi^4 = 5610787/57600 - (3/56)*Sum_{n >= 1} d/dn(R(n))/R(n)^4.

Taking 10 terms of the last series gives the approximation Pi^4 = 97.4090910340

024372(50...), correct to 16 decimal places. (End)