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A358464
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a(n) is the greatest m such that Sum_{k = 1..m} 1/(1 + n*k) <= 1.
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1
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2, 6, 16, 42, 110, 288, 761, 2020, 5388, 14417, 38681, 103994, 280032, 755031, 2037848, 5504884, 14880978, 40250609, 108926101, 294902398, 798703663, 2163878141
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OFFSET
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1,1
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COMMENTS
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This sequence coincidences with 2*Fibonacci(2*n) (A025169) for the first 6 terms.
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LINKS
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FORMULA
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ceiling(digamma(a(n)+(1/n)+1) - digamma((1/n)+1)) = n.
Integral_{x=0..oo} Product_{k=0..m} sinc(x/(n*k+1)) dx = Pi for 0 <= m <= a(n). See links Schmid and Borwein.
ceiling(Sum_{m = 0..oo} ( 1/(m+1) * Sum_{k = 0..m} (-1)^k*binomial(m, k)*log( (a(n)+(1/n)+1+k) / ((1/n)+1+k) ) )) = n.
a(n) ~ floor(exp(n + digamma(1+(1/n))) - (1/2) - (1/n)). This appears to be accurate for at least n < 22.
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EXAMPLE
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a(2) = 6 because Sum_{m = 1..a(2)} 1/(1+2*m) = 43024/45045 < 1, but Sum_{m = 1..a(2)+1} 1/(1+2*m) = 46027/45045 > 1.
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PROG
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(PARI) a(n) = {my(k = 2*fibonacci(2*n)-1); my(b = (psi(k+(1/n))-psi(1+(1/n)))/n); until(b > 1, b = b+(1/(1+n*k)); k=k+1 ); k-2}
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CROSSREFS
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KEYWORD
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nonn,more
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AUTHOR
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STATUS
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approved
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