A371809 Number of 666 substrings contained in the decimal expansion of the n-th apocalyptic number, where overlapping substrings are counted as distinct.

1, 1, 1, 4, 3, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 3, 1, 2, 3

Offset: 1

Author

Paolo Xausa, Apr 06 2024

Keywords

• nonn
• easy
• base

Comments

An apocalyptic number is a positive power of 2 containing 666 in its decimal expansion.

See A371807 for a variant counting only nonoverlapping substrings.

Examples

			a(8) = 2 because the 8th apocalyptic number (2^243) contains two (overlapping) 666 substrings in its decimal expansion:
                   ***
14134776518227074636666380005943348126619871175004951664972849610340958208.
                    ***
		

Links

• Paolo Xausa, Table of n, a(n) for n = 1..10000
• Brady Haran and Tony Padilla, Apocalyptic Numbers, YouTube Numberphile video, 2024.
• Eric Weisstein's World of Mathematics, Apocalyptic Number.

Crossrefs

Cf. A007356, A371807, A371808.

Programs

Mathematica

Select[StringCount[IntegerString[2^Range[1000]], "666", Overlaps->True], # > 0 &]

Python

from itertools import islice
def agen(): # generator of terms
    pow2 = 1
    while True:
        s = str(pow2)
        c = sum(1 for i in range(len(s)-2) if s[i:i+3] == "666")
        if c > 0: yield c
        pow2 <<= 1
print(list(islice(agen(), 88))) # Michael S. Branicky, Apr 07 2024

Formula

a(n) >= A371807(n).