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A013671 Decimal expansion of zeta(13).

%I A013671 #29 Jul 08 2025 02:50:27
%S A013671 1,0,0,0,1,2,2,7,1,3,3,4,7,5,7,8,4,8,9,1,4,6,7,5,1,8,3,6,5,2,6,3,5,7,
%T A013671 3,9,5,7,1,4,2,7,5,1,0,5,8,9,5,5,0,9,8,4,5,1,3,6,7,0,2,6,7,1,6,2,0,8,
%U A013671 9,6,7,2,6,8,2,9,8,4,4,2,0,9,8,1,2,8,9,2,7,1,3,9,5,3,2,6,8,1,3
%N A013671 Decimal expansion of zeta(13).
%D A013671 M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 811.
%H A013671 M. Abramowitz and I. A. Stegun, eds., <a href="http://www.convertit.com/Go/ConvertIt/Reference/AMS55.ASP">Handbook of Mathematical Functions</a>, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
%F A013671 From _Peter Bala_, Dec 04 2013: (Start)
%F A013671 Definition: zeta(13) = sum {n >= 1} 1/n^13.
%F A013671 zeta(13) = 2^13/(2^13 - 1)*( sum {n even} n^9*p(n)*p(1/n)/(n^2 - 1)^14 ), where p(n) = n^6 + 21*n^4 + 35*n^2 + 7. (End)
%F A013671 zeta(13) = Sum_{n >= 1} (A010052(n)/n^(13/2)) = Sum_{n >= 1} ( (floor(sqrt(n))-floor(sqrt(n-1)))/n^(13/2) ). - _Mikael Aaltonen_, Feb 22 2015
%F A013671 zeta(13) = Product_{k>=1} 1/(1 - 1/prime(k)^13). - _Vaclav Kotesovec_, May 02 2020
%e A013671 1.0001227133475784891467518365263573957142751058955098451367026716208967...
%t A013671 RealDigits[Zeta[13],10,120][[1]] (* _Harvey P. Dale_, Dec 24 2016 *)
%o A013671 (PARI) zeta(13) \\ _Charles R Greathouse IV_, Apr 25 2016
%Y A013671 Cf. A013663, A013667, A013669, A013671, A013675, A013677.
%K A013671 cons,nonn
%O A013671 1,6
%A A013671 _N. J. A. Sloane_