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A328655
Least k such that Sum_{m=1..k} 1/m^2 > Product_{i=1..n} 1/(1 - 1/prime(i)^2).
2
3, 7, 12, 20, 27, 37, 47, 59, 72, 84, 98, 112, 125, 141, 157, 173, 188, 205, 222, 239, 258, 277, 297, 316, 335, 354, 374, 395, 418, 442, 463, 484, 506, 528, 550, 573, 595, 618, 641, 665, 688, 713, 737, 761, 787, 813, 838, 862, 887, 912, 938, 964, 991, 1017, 1044, 1070, 1097, 1125, 1152, 1181
OFFSET
1,1
COMMENTS
Euler proved that for every s > 1, Sum_{m>=1} 1/m^s = Product_{n>=1} 1/(1 - 1/prime(n)^s) = zeta(s).
This sequence compares partial sums with partial products for the case s = 2.
For the case s = 1 and partial sums of harmonic series and Euler’s partial products see A328684.
LINKS
Leonhard Euler, Variae observationes circa series infinitas, (Various Observations about Infinite Series), published by St Petersburg Academy, 1737.
EXAMPLE
a(2)=7 because Product_{n=1..2} 1/(1 - 1/prime(n)^2) < Sum_{m=1..7} 1/m^2, since 3/2 = 1.5 < 266681/176400 = 1.5118...
MAPLE
N:= 100: # for a(1)..a(N)
p:= 1: P:= 1:
for n from 1 to N do
p:= nextprime(p);
P:= P * 1/(1-1/p^2);
PP[n]:= P;
od:
k:= 1: S:= 0: notdone:= true:
for n from 1 while notdone do
S:= S + 1/n^2;
while S > PP[k] do
A[k]:= n;
k:= k+1;
if k > N then notdone:= false; break fi
od
od:
seq(A[i], i=1..N); # Robert Israel, Dec 10 2019
MATHEMATICA
dd = {}; h = 1; hh = 1; k = 1; m = 1; Do[k = k (1/(1 - Prime[n]^-2));
kk = N[k, 30];
While[kk > hh, h = h + 1/(m + 1)^2; hh = N[h, 30]; m++];
AppendTo[dd, m], {n, 1, 68}]; dd
PROG
(PARI) a(n) = my(k=1, pp = prod(i=1, n, 1/(1 - 1/prime(i)^2)), s = 1); while (s <= pp, k++; s += 1/k^2); k; \\ Michel Marcus, Oct 29 2019
(PARI) apply( {A328655(n, s=2, p=1/prod(k=1, n, 1-prime(k)^-s))=for(k=1, oo, (p-=k^-s)<0&&return(k))}, [1..60]) \\ optional 2nd arg allows to compute analog for other powers s (float avoids exact calculation using fractions, use with care). - M. F. Hasler, Oct 31 2019
CROSSREFS
Cf. A328684.
Sequence in context: A011899 A002498 A172115 * A091369 A036698 A279169
KEYWORD
nonn
AUTHOR
Artur Jasinski, Oct 24 2019
STATUS
approved