cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

A000533 a(0)=1; a(n) = 10^n + 1, n >= 1.

Original entry on oeis.org

1, 11, 101, 1001, 10001, 100001, 1000001, 10000001, 100000001, 1000000001, 10000000001, 100000000001, 1000000000001, 10000000000001, 100000000000001, 1000000000000001, 10000000000000001
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

Also, b^n+1 written in base b, for any base b >= 2.
Also, A083318 written in base 2. - Omar E. Pol, Feb 24 2008, Dec 30 2008
Also, palindromes formed from the reflected decimal expansion of the concatenation of 1 and infinite 0's. - Omar E. Pol, Dec 14 2008
It seems that the sequence gives 'all' positive integers m such that m^4 is a palindrome. Note that a(0)^4 = 1 is a palindrome and for n > 0, a(n)^4 = (10^n + 1)^4 = 10^(4n) + 4*10^(3n) + 6*10^(2n) + 4*10^(n) + 1 is a palindrome. - Farideh Firoozbakht, Oct 28 2014
a(n)^2 starts with a(n)+1 for n >= 1. - Dhilan Lahoti, Aug 31 2015
From Peter Bala, Sep 25 2015: (Start)
The simple continued fraction expansion of sqrt(a(2*n)) = [10^n; 2*10^n, 2*10^n, ...] has period 1.
The simple continued fraction expansion of sqrt(a(6*n))/a(2*n) = [10^n - 1; 1, 10^n - 1, 10^n - 1, 1, 2*(10^n - 1), ...] has period 5.
The simple continued fraction expansion of sqrt(a(10*n))/a(2*n) = [10^(3*n) - 10^n; 10^n, 10^n, 2*(10^(3*n) - 10^n), ...] has period 3.
As n increases, these expansions have large partial quotients.
A theorem of Kuzmin in the measure theory of continued fractions says that large partial quotients are the exception in continued fraction expansions.
Empirically, we also see exceptionally large partial quotients in the continued fraction expansions of the m-th root of the numbers a(m*n), for m >= 3. For example, it appears that the continued fraction expansion of a(3*n)^(1/3) begins [10^n; 3*10^(2*n), 10^n, 4.5*10^(2*n), 0.8*10^n, ( 9*10^(2*n + 2) - 144 + 24*(2^mod(n,3) - 1) )/168, ...]. As n increases, the expansion begins with 6 large partial quotients. An example is given below. Cf. A002283, A066138 and A168624.
(End)
a(1) and a(2) are the only prime terms up to n=100000. - Daniel Arribas, Jun 04 2016
Based on factors from A001271, the first abundant number in this sequence should occur in the first M terms, where M is the double factorial M=7607!!. Is any abundant number known in this sequence? - Sergio Pimentel, Oct 04 2019
The (3^5 * 5^2 * 7^2 * 11^2 * 13^2 * 17 * 19 * 23 * 29 * 31 * 37 * 41 * 43 * 47 * 53 * 59 * 61 * 67 * 71 * 73 * 79 * 83 * 89 * 97 * 101 * 103 * 107 * 109 * 113 * 127 * 131 * 137 * 139 * 157 * 163 * 181 * 191 * 241 * 251 * 263)-th term of this sequence is an abundant number. - Jon E. Schoenfield, Nov 19 2019
The only perfect power (i.e., a perfect square, a perfect cube, and so forth) in the present sequence is a(0) by Mihăilescu's theorem (since 10^n is a perfect power not equal to 2^3 - see the links in A001597). - Marco Ripà, Feb 04 2025

Examples

			The continued fraction expansion of a(9)^(1/3) begins [1000; 3000000, 1000, 4500000, 800, 5357142, 1, 6, 14, 6, 1, 5999999, 6, 1, 12, 7, 1, ...] with 5 large partial quotients immediately following the integer part of the number. - _Peter Bala_, Sep 25 2015

Links

Crossrefs

Cf. A002283, A066138, A083318, A138144, A138145, A152756, A168624.

Programs

Formula

a(n) = 10^n + 1 - 0^n. - Reinhard Zumkeller, Jun 10 2003
From Paul Barry, Feb 05 2005: (Start)
G.f.: (1-10*x^2)/((1-x)*(1-10*x));
a(n) = Sum_{k=0..n} binomial(n, k)*0^(k(n-k))*10^k. (End)
a(n) = A178500(n) + 1. - Reinhard Zumkeller, May 28 2010
E.g.f.: exp(x) + exp(10*x) - 1. - Ilya Gutkovskiy, Jun 03 2016

A057934 Number of prime factors of 10^n + 1 (counted with multiplicity).

Original entry on oeis.org

1, 1, 3, 2, 2, 2, 2, 2, 5, 3, 5, 3, 3, 4, 7, 5, 4, 3, 2, 4, 8, 4, 5, 3, 5, 3, 7, 4, 3, 7, 2, 4, 9, 4, 5, 6, 4, 3, 10, 4, 3, 7, 4, 4, 12, 4, 4, 9, 4, 7, 8, 4, 2, 6, 10, 5, 6, 5, 4, 6, 3, 3, 12, 3, 6, 8, 2, 4, 10, 11, 3, 5, 4, 7, 11, 6, 12, 7, 4, 9, 11, 3, 7, 8, 8, 3, 8, 4, 4, 11, 6, 4, 8, 4, 6, 8, 4, 5, 13
Offset: 1

Views

Author

Patrick De Geest, Oct 15 2000

Keywords

Comments

2^(a(2n)-1)-1 predicts the number of pair-solutions of even length L for AB = A^2 + B^2. For instance, with length 18 we have 10^18 + 1 = 101*9901*999999000001 or 3 divisors F which when put into the Mersenne formula 2^(F-1)-1 yields 3 pairs (see reference 'Puzzle 104' for details).

Links

Crossrefs

bigomega(b^n+1): this sequence (b=10), A057935 (b=9), A057936 (b=8), A057937 (b=7), A057938 (b=6), A057939 (b=5), A057940 (b=4), A057941 (b=3), A054992 (b=2).
Cf. A003021, A001271, A046053, A001562, A057951, A119704, A344897.

Programs

Formula

a(n) = A057951(2n) - A057951(n). - T. D. Noe, Jun 19 2003

A087000 Half length of periodic part of decimal expansion of 1/p for those primes having a periodic part of even length.

Original entry on oeis.org

3, 1, 3, 8, 9, 11, 14, 23, 29, 30, 4, 22, 48, 2, 17, 54, 56, 21, 65, 4, 23, 74, 39, 83, 89, 90, 96, 49, 15, 111, 114, 116, 15, 25, 128, 131, 134, 14, 73, 156, 55, 168, 58, 16, 183, 93, 189, 191, 194, 100, 102, 209, 70, 216, 16, 76, 230, 77, 243, 245, 249, 251
Offset: 1

Views

Author

Reinhard Zumkeller, Jul 29 2003

Keywords

Comments

a(n) appears to be the least k such that 10^k+1 is divisible by A028416(n). See A001271. - Michel Marcus, Aug 13 2023

Links

Crossrefs

Cf. A028416, A002371, A086999, A087001, A087002, A001271, A062397 (10^n+1).

Formula

a(n) = A002371(A049084(A028416(n)))/2.
a(n) = A055642(A086999(n))/2.
a(n) = A055642(A087001(n)) = A055642(A087002(n)).
Showing 1-3 of 3 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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