A256968 Let b(n) = Product_{i=1..n} p_i/(p_i - 1), p_i = i-th prime; a(n) = minimum k such that b(k) >= n.

1, 1, 1, 2, 4, 6, 9, 14, 22, 35, 55, 89, 142, 230, 373, 609, 996, 1637, 2698, 4461, 7398, 12301, 20503, 34253, 57348, 96198, 161659, 272124, 458789, 774616, 1309627, 2216968, 3757384, 6375166

Offset

0,4

Comments

A001611 is similar but strictly different.

Equal to A256969 except for n = 2 and n = 3. The following argument shows that they are equal for n > 3. First note that b(k+1) > b(k). Next, Product_{i=1..k} p_i is 2 times an odd number, i.e., it is not divisible by 4. Similarly since p_i - 1 is even for i > 1, Product_{i=1..k} (p_i - 1) is divisible by 2^(k-1), i.e., it is divisible by 4 for k >= 3. Thus b(k) is not an integer for k >= 3. Since b(3) = 15/4 > 3, this means that A256969(n) = A256968(n) for n > 3 - Chai Wah Wu, Apr 17 2015

Links

Table of n, a(n) for n=0..33.

Popular Computing (Calabasas, CA), Problem 182 (Suggested by Victor Meally), Annotated and scanned copy of page 10 of Vol. 5 (No. 53, Aug 1977).

Example

The b(n) sequence for n >= 0 begins 1, 2, 3, 15/4, 35/8, 77/16, 1001/192, 17017/3072, 323323/55296, 676039/110592, 2800733/442368, 86822723/13271040, 3212440751/477757440, 131710070791/19110297600, 5663533044013/802632499200, ... = A060753/A038110. So a(3) = 2.

Prog

(Python)
from sympy import prime
A256968_list, count, bn, bd = [], 0, 1, 1
for k in range(1, 10**4):
....p = prime(k)
....bn *= p
....bd *= p-1
....while bn >= count*bd:
........A256968_list.append(k)
........count += 1 # Chai Wah Wu, Apr 17 2015

Crossrefs

Cf. A060753, A038110, A256969, A001611.

Sequence in context: A119737 A038718 A042942 * A005687 A164139 A218605

Adjacent sequences: A256965 A256966 A256967 * A256969 A256970 A256971

Keyword

nonn,more

Author

N. J. A. Sloane, Apr 17 2015

Extensions

More terms from Chai Wah Wu, Apr 17 2015

a(32)-a(33) from Chai Wah Wu, Apr 19 2015

Status

approved