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Revision History for A136616 (Bold, blue-underlined text is an addition; faded, red-underlined text is a deletion.)

Showing entries 1-10 | older changes
a(n) = largest m with H(m) - H(n) <= 1, where H(i) = Sum_{j=1 to i} 1/j, the i-th harmonic number, H(0) = 0.
(history; published version)
#18 by Alois P. Heinz at Mon Apr 06 06:19:59 EDT 2020
STATUS

editing

approved

#17 by Alois P. Heinz at Mon Apr 06 06:19:42 EDT 2020
MAPLE

e:= exp(1):

A136616 := n -> floor( e*n + (e-1)/2 + (e - 1/e)/(24*(n + 1/2))); :

seq(A136616(n), n=0..50);

#16 by Alois P. Heinz at Mon Apr 06 06:16:55 EDT 2020
STATUS

approved

editing

#15 by Susanna Cuyler at Fri Mar 06 08:10:12 EST 2020
STATUS

proposed

approved

#14 by Jinyuan Wang at Fri Mar 06 01:40:59 EST 2020
STATUS

editing

proposed

#13 by Jinyuan Wang at Fri Mar 06 01:40:51 EST 2020
FORMULA

a(n) = A103762(n+1) - 1 = A136617(n+1) + n - 1 for n > 0. - Jinyuan Wang, Mar 06 2020

#12 by Jinyuan Wang at Fri Mar 06 01:37:46 EST 2020
FORMULA

a(n) = floor( e*n + (e-1)/2 + (e - 1/e)/(24*(n + 1/2))), after a suggestion by David Cantrell.

PROG

a(n) = floor(e*n+(e-1)/2+(e-1/e)/(24*n+12)); \\ _Jinyuan Wang_, Mar 06 2020

#11 by Jinyuan Wang at Fri Mar 06 01:37:10 EST 2020
PROG

a(n) = a(n) = floor(e*n+(e-1)/2+(e-1/e)/(24*n+12)); \\

STATUS

proposed

editing

#10 by Jinyuan Wang at Fri Mar 06 01:36:25 EST 2020
STATUS

editing

proposed

#9 by Jinyuan Wang at Fri Mar 06 01:36:09 EST 2020
NAME

a(n) = largest m with H(m) - H(n) <= 1, where H(i) = sumSum_{j=1 to i} 1/j, the i-th harmonic number, H(0) = 0.

FORMULA

a(n) = A103762(n+1) - 1 = A136617(n+1) + n - 1 for n > 0. - Jinyuan Wang, Mar 06 2020

EXAMPLE

a(3) = 9 because H(9) - H(3) = 1/4 + ... + 1/9 < 1 < 1/4 + ... + 1/10 = H(10) - H(3).

PROG

(PARI) default(realprecision, 10^5); e=exp(1);

a(n) = a(n) = floor(e*n+(e-1)/2+(e-1/e)/(24*n+12)); \\

KEYWORD

easy,nonn

nonn,easy

STATUS

approved

editing