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A245330
Decimal expansion of phi(0), an auxiliary constant associated with Shapiro's cyclic sum constant lambda.
10
9, 8, 9, 1, 3, 3, 6, 3, 4, 4, 4, 6, 9, 9, 3, 0, 5, 2, 2, 4, 3, 4, 9, 0, 3, 0, 8, 2, 6, 6, 3, 3, 7, 9, 8, 1, 3, 8, 0, 3, 4, 8, 0, 9, 8, 0, 4, 4, 1, 8, 2, 2, 1, 9, 0, 3, 9, 3, 5, 7, 8, 7, 8, 0, 8, 7, 3, 8, 2, 8, 9, 5, 4, 2, 6, 7, 5, 7, 9, 5, 8, 1, 5, 3, 8, 0, 3, 7, 6, 7, 5, 0, 8, 8, 0, 8, 0, 3, 8, 9, 4, 2, 4
OFFSET
0,1
COMMENTS
From Petros Hadjicostas, Jun 01 2020: (Start)
The calculations in sequences A319568 and A319569 are needed for the estimation of the constant phi(0) = 2*lambda = 2*A086277. This was done in Drinfel'd (1971).
Similar calculations were done by Elbert (1973) for the Shapiro cyclic sum constant mu = psi(0) = A086278.
For more information, see my comments in sequence A319568. Based on those comments, we may create the PARI program below that may be used to calculate phi(0). (End)
REFERENCES
Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, Section 3.1, Shapiro-Drinfeld Constant, p. 209.
LINKS
V. G. Drinfel'd, A cyclic inequality, Mathematical Notes of the Academy of Sciences of the USSR, 9 (1971), 68-71.
Á. Elbert, On a cyclic inequality, Periodica Mathematica Hungarica, 4 (1973), 163-168.
Á. Elbert, On a cyclic inequality, Periodica Mathematica Hungarica, 4 (1973), 163-168.
R. A. Rankin, 2743. An inequality, Mathematical Gazette, 42(339) (1958), 39-40.
H. S. Shapiro, Proposed problem for solution 4603, American Mathematical Monthly, 61(8) (1954), 571.
H. S. Shapiro, Solution to Problem 4603: An invalid inequality, American Mathematical Monthly, 63(3) (1956), 191-192; counterexample provided by M. J. Lighthill.
B. A. Troesch, The validity of Shapiro's cyclic inequality, Mathematics of Computation, 53 (1989), 657-664.
Eric Weisstein's MathWorld, Shapiro's Cyclic Sum Constant.
FORMULA
lambda = A086277 = phi(0)/2.
phi(0) = exp(-c)*(c+1), where c = A319569. - Petros Hadjicostas, Jun 01 2020
EXAMPLE
0.989133634446993052243490308266337981380348098044182219039357878...
MATHEMATICA
digits = 103; phi[0] = (2 + x + 2*E^(x/2)*(1+x))/(E^(x/2)*(1+E^(x/2))^2) /. FindRoot[E^(x/2)*(1+E^(x/2))^2/(1+2*E^(x/2)) == E^(1/(1+2*E^(x/2)) + x), {x, 0}, WorkingPrecision -> digits+10]; RealDigits[phi[0], 10, digits] // First
PROG
(PARI) c(b) = b + exp(b/2)/(2*exp(b)+exp(b/2));
a=c(solve(b=-2, 2, exp(-c(b))*(1-b+c(b))-2/(exp(b)+exp(b/2))));
exp(-a)*(a+1) \\ Petros Hadjicostas, Jun 02 2020
CROSSREFS
KEYWORD
nonn,cons
AUTHOR
STATUS
approved