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.

0, 0, 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, 10828012, 18409028, 31326514, 53354259, 90945529, 155142139

Offset: 0

Author

N. J. A. Sloane, Apr 17 2015

Keywords

• nonn

Comments

A001611 is similar but different.

Equal to A005579 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 a(n) = A005579(n) for n > 3 - Chai Wah Wu, Apr 17 2015

Examples

			The sequence b(n) 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.
		

Links

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

Crossrefs

Cf. A005579, A060753, A038110, A001611.

Programs

Python

from sympy import prime
A256968_list, count, bn, bd = [0,0], 2, 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; corrected by Max Alekseyev, Jan 26 2025

Extensions

More terms from Chai Wah Wu, Apr 17 2015

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

a(0)-a(1) corrected and a(34)-a(39) copied over from A005579 by Max Alekseyev, Jan 26 2025