A126933 Quotients arising from sequence A053312.

1, 3, 14, 132, 691, 1908, 16579, 47352, 414301, 1183713, 5474669, 27151397, 135646011, 678174568, 6442602909, 18480090517, 85533990571, 424236721848, 4026815626549, 11550150977337, 53458791308981, 265147974756053, 1324666882885839, 6622797918981982, 62916043734881616, 329481245744393933

Offset

1,2

Comments

Take the decimal number formed by the first n digits of A023396 in reverse order and divide by 2^n.

The sequence A053312 gives n-digit numbers consisting entirely of 1s and 2s which are divisible by 2^n. The quotients upon division form the present sequence. The parity of the n-th term here determines the next term in A023396; if odd, it is a 1 and if even, a 2.

This was set as a problem in the All Union Mathematical Olympiad of 1971 and can be found in the reference cited here.

References

J. B. Tabov and P. J. Taylor, Methods of Problem Solving, Book 1, Australian Mathematics Trust, 1996.

Links

David A. Corneth, Table of n, a(n) for n = 1..1431 (terms <= 10^1000)

Index to sequences related to Olympiads.

Formula

a(n) < 0.3 * 5^n. - David A. Corneth, Jun 11 2020

Example

a(4) = A053312(4) / 2^4 = 2112 / 16 = 132. - David A. Corneth, Jun 11 2020

Prog

(Python)
from itertools import count, islice
def A126933_gen(): # generator of terms
    a, b = 2, 10
    for n in count(1):
        a+=b if (c:=a>>n)&1 else b<<1
        b *= 10
        yield c
A126933_list = list(islice(A126933_gen(), 20)) # Chai Wah Wu, Mar 16 2023

Crossrefs

Cf. A023396, A053312.

Sequence in context: A061029 A096657 A365997 * A073550 A319361 A367422

Adjacent sequences: A126930 A126931 A126932 * A126934 A126935 A126936

Keyword

nonn,easy,base

Author

Gerry Leversha, Mar 18 2007

Extensions

Name changed and other minor edits by Ray Chandler, Jun 17 2020

Status

approved