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# Calendar for Sequence of the Day in October

Template:Sequence of the Day for October 1
 { 0, 1, –1, 2, –2, 3, –3, 4, –4, 5, –5, 6, –6, 7, –7, ... }

Questions of infinity can too easily lead to confusing, paradoxical thoughts. Consider a finite version of this sequence by picking a maximum positive integer, say ${\displaystyle m=2^{p}-1}$ with ${\displaystyle p=127}$. Our subset ${\displaystyle Z}$ of ${\displaystyle \mathbb {Z} }$ thus runs from ${\displaystyle -m}$ to ${\displaystyle m}$, and our finite subsequence starts 0, 1, –1, 2, –2, and ends ${\displaystyle 2^{127}-1}$, ${\displaystyle (-2)^{127}+1}$ (and incidentally, each number in this subsequence can be represented in a signed 256-bit word). What is the length ${\displaystyle {\mathcal {L}}}$ of ${\displaystyle Z}$? Easy: ${\displaystyle 2m+1=2^{256}-1}$. But we can't really apply that to the length of A001057, as it turns out that ${\displaystyle 2\infty +1}$ is the "wrong" answer. The "right" answer is just ${\displaystyle \infty }$?

Template:Sequence of the Day for October 2

A039661: Gelfond's constant: ${\displaystyle e^{\pi }}$.

 23.1406926327792690...

M. F. Hasler noticed the interesting coincidence that ${\displaystyle e^{\pi }-\pi }$ is almost an integer.

Template:Sequence of the Day for October 3

A084468: Odd numbers with exactly 3 ones in binary expansion.

 { 7, 11, 13, 19, 21, 25, 35, 37, ... }

How can this sequence be computed most efficiently? We could try taking three 1s and then listing the various ways of inserting a varying number of 0s between the most significant bit and the least significant bit. The middle 1 would also have to be made to wander. Numbers like 10001010 don't enter our consideration here because they're even.

And that should lead the train of thought to an easier way: since these numbers are odd, the one power of 2 we always need to have for this sequence is 1. Thus, we can take the numbers that are the sum of two distinct powers of 2 (see A018900), multiply them by 2 and add 1, thus obtaining the requisite binary weight of 3.

Template:Sequence of the Day for October 4
 { 1, 2, 10, 56, 346, 2252, 15184, ... }

Matthijs Coster realized that this is the Taylor expansion of a special point on a curve described by Beauville.

Template:Sequence of the Day for October 5

A004001: Hofstadter-Conway $10000 sequence: ${\displaystyle a(n)=a(a(n-1))+a(n-a(n-1))}$, with ${\displaystyle a(1)=a(2)=1}$.  { 1, 1, 2, 2, 3, 4, 4, 4, 5, 6, 7, 7, 8, 8, 8, 8, ... } This sequence acquired its name because John Horton Conway offered a prize of$10,000 to anyone who could demonstrate a particular result about its asymptotic behavior. The prize, since reduced to $1,000, was claimed by Collin Mallows. (Comment from Hofstadter–Conway$10,000 sequence on Wikipedia.)

Template:Sequence of the Day for October 6

A1098765: Sequence name.

 { 1, 3, 2, 6, 7, 4, 5, ... }

Paragraph or two of info.

Template:Sequence of the Day for October 7

A082020: Decimal expansion of ${\displaystyle \scriptstyle {\frac {15}{\pi ^{2}}}\,}$.

 1.519817754635...

This is one of those numbers which can be expressed both as an infinite sum and as an infinite product:

${\displaystyle \sum _{n=1}^{\infty }{\frac {|\mu (n)|}{n^{2}}}={\frac {15}{\pi ^{2}}},\,}$
${\displaystyle \prod _{p \atop {p{\rm {~prime}}}}\left(1+{\frac {1}{p^{2}}}\right)={\frac {15}{\pi ^{2}}},\,}$

where ${\displaystyle \mu (n)}$ is the Möbius function and the infinite product is over all primes.

Yet another expression for this number is ${\displaystyle \scriptstyle {\frac {\zeta (2)}{\zeta (4)}}={\frac {\pi ^{2}/6}{\pi ^{4}/90}}\,}$, where ${\displaystyle \zeta (s)}$ is the Riemann zeta function.

Template:Sequence of the Day for October 8

A061248: Primes at which sum of base 10 digits strictly increases.

 { ..., 17, 19, 29, 59, 79, 89, 199, ... }

Most of those primes have all their digits (except the first digit) as some 8's and mostly 9's (with 9 as final digit) and terms with 7 as a final digit 7 become rarer. (Is there any [albeit extremely slim] chance that a prime ending in 1 or 3 appears?)

Template:Sequence of the Day for October 9

A010888: Base 10 (additive) digital root of ${\displaystyle \scriptstyle n,\,n\,\geq \,0,\,}$

 { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, ... }

Repeatedly add digits of ${\displaystyle \scriptstyle n\,}$ until reaching a single digit. For the most part, in this procedure, 0s and 9s can be ignored (unless of course the number consists entirely of a 0 or several 9s; in the latter case the digital root is 9).

The (additive) digital root has the property

${\displaystyle {\rm {dr}}(n)\,\equiv \,n{\pmod {9}},\,}$

which allows us to find whether ${\displaystyle \scriptstyle n\,}$ is divisible by 3 or 32 without doing any division (using a table).

Template:Sequence of the Day for October 10

A002113: Palindromes in base 10.

 { ..., 101, 111, 121, 131, 141, 151, 161, 171, 181, 191, 202, 212, ... }

The sequence of course also includes 1-digit (trivially) and 2-digits (repdigits) terms. Obviously, all repdigits, which include repunits, are palindromic.

Template:Sequence of the Day for October 11

A001020: ${\displaystyle 11^{n}}$.

 { 1, 11, 121, 1331, 14641, 161051, ... }

The decimal digits of the powers of 11 give binomial coefficients (which appear in Pascal's triangle), until some binomial coefficients have more than one digit, where overlapping occurs.

 ${\displaystyle n=0}$ 1 ${\displaystyle 1}$ 1 1 ${\displaystyle 2}$ 1 2 1 ${\displaystyle 3}$ 1 3 3 1 ${\displaystyle 4}$ 1 4 6 4 1

After ${\displaystyle 11^{4}}$ it doesn't quite work out, since some binomial coefficients in {1, 5, 10, 10, 5, 1} have more than one decimal digits.

 1, 5, 1 0, 1 0, 5, 1 ---------------- 1 6 1 0 5 1 

In fact, the powers of 11 correspond to a Pascal's triangle with only one digit per entry, and carry over to the next cell to the left in case of a value > 9.

In general, if ${\displaystyle b}$ appears in Pascal's triangle in the middle of row ${\displaystyle k}$, then ${\displaystyle (b+1)^{n}}$ can be found in Pascal's triangle written in base ${\displaystyle b}$ up to ${\displaystyle (b+1)^{k-1}}$.

Template:Sequence of the Day for October 12

 { ... 10, 12, 18, 20, 21, 24, 27, ... }

These numbers are divisible by the sum of their base 10 digits.

Template:Sequence of the Day for October 13

A003462: ${\displaystyle \scriptstyle {\frac {3^{n}-1}{2}}\,=\,\sum _{i=0}^{n-1}3^{i},\,n\,\geq \,1.\,}$

 { 1, 4, 13, 40, 121, 364, 1093, ... }

This sequence gives the repunits in base 3, i.e. the sum of the first ${\displaystyle n}$ powers of 3, with A000244 being the sequence of powers of 3.

Template:Sequence of the Day for October 14

A008619: ${\displaystyle \scriptstyle 1+\lfloor {\frac {n}{2}}\rfloor ,\,n\,\geq \,0.\,}$ (Positive integers repeated.)

 { 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7, ... }

Smallest positive integer whose harmonic mean with another positive integer is ${\displaystyle n}$. For example, ${\displaystyle a(6)=4}$ is already given (as 4 is the smallest positive integer such that the harmonic mean of 4 (with 12) is 6)—but the harmonic mean of 2 (with –6) is also 6 and 2 < 4, so the two positive integer restrictions need to be imposed to rule out both 2 and –6.

Note also that this sequence is the second outermost diagonal of Lozanić's triangle.

Template:Sequence of the Day for October 15

A011371: ${\displaystyle n}$ minus its binary weight, ${\displaystyle n}$ ≥ 0.

 { 0, 0, 1, 1, 3, 3, 4, 4, 7, 7, ... }

This is also the exponent of the highest power of 2 dividing ${\displaystyle n!}$

${\displaystyle \sum _{i=1}^{\lfloor \log _{2}(n)\rfloor }\left\lfloor {\frac {n}{2^{i}}}\right\rfloor .\,}$

This sequence shows why, in binary, 0 and 1 are the only two numbers ${\displaystyle n}$ such that ${\displaystyle n}$ equals the sum of its digits raised to the consecutive powers. 1 raised to any consecutive power is still 1 and thus any sum of digits raised to consecutive powers for any ${\displaystyle n>1}$ falls short of equaling the value of ${\displaystyle n}$ by the ${\displaystyle n}$-th number of this sequence.

Template:Sequence of the Day for October 16

A117972: Numerator of ${\displaystyle \scriptstyle \zeta '(-2n),\,n\,\geq \,0\,}$.

 { 1, –1, 3, –45, 315, –14175, 467775, –42567525, 638512875, ... }

This sequence is related to the correlation function in Montgomery's pair correlation conjecture for the nontrivial zeros of the Riemann zeta function

${\displaystyle \scriptstyle R_{2}(u)\,=\,1-\mathrm {sinc} _{\pi }^{2}\,u+\delta (u)\,=\,1-\left({\frac {\sin \pi u}{\pi u}}\right)^{2}+\delta (u),\,}$

where ${\displaystyle \scriptstyle \mathrm {sinc} _{\pi }\,u\,}$ is the normalized sinc function.

Maclaurin series for ${\displaystyle \scriptstyle 1-\left({\frac {\sin x}{x}}\right)^{2}\,=\,{\frac {x^{2}}{3}}-{\frac {2x^{4}}{45}}+{\frac {x^{6}}{315}}-{\frac {2x^{8}}{14175}}+{\frac {2x^{10}}{467775}}-\cdots \,}$

• Numerators of Maclaurin series for ${\displaystyle \scriptstyle 1-\left({\frac {\sin x}{x}}\right)^{2}\,}$: A048896 2A000120(${\displaystyle \scriptstyle n\,}$+1) − 1, ${\displaystyle \scriptstyle n\,\geq \,1.\,}$ Also, maximal power of 2 dividing ${\displaystyle \scriptstyle n\,}$th Catalan number.
{1, 2, 1, 2, 2, 4, 1, 2, 2, 4, 2, 4, 4, 8, 1, 2, 2, 4, 2, 4, 4, 8, 2, 4, 4, ...}
• Denominators of Maclaurin series for ${\displaystyle \scriptstyle 1-\left({\frac {\sin x}{x}}\right)^{2}\,}$: A117972 Numerator of ${\displaystyle \scriptstyle \zeta '(-2n),\,n\,\geq \,2\,}$.
{ 3, –45, 315, –14175, 467775, –42567525, 638512875, –97692469875, ... }

where A000120 is number of 1's in binary expansion of ${\displaystyle \scriptstyle n\,}$ (or the binary weight of ${\displaystyle \scriptstyle n,\,n\,\geq \,2\,}$)

{1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, ...}

Why are the denominators of Maclaurin series for ${\displaystyle \scriptstyle 1-\left({\frac {\sin x}{x}}\right)^{2}\,}$ corresponding to the numerators of ${\displaystyle \scriptstyle \zeta '(-2n),\,n\,\geq \,2\,}$?

Why are the numerators of Maclaurin series for ${\displaystyle \scriptstyle 1-\left({\frac {\sin x}{x}}\right)^{2}\,}$ corresponding to maximal power of 2 dividing ${\displaystyle \scriptstyle n\,}$th, ${\displaystyle \scriptstyle n\,\geq \,1\,}$, Catalan number?

Template:Sequence of the Day for October 17

A019973: Decimal expansion of tangent of 75 degrees (${\displaystyle \scriptstyle {\frac {5\pi }{12}}\,}$ radians), or of cotangent of 15 degrees (${\displaystyle \scriptstyle {\frac {\pi }{12}}\,}$ radians).

 3.732050807568877...

This number ${\displaystyle \scriptstyle c\,=\,2+{\sqrt {3}}\,}$, one of the two roots of ${\displaystyle \scriptstyle x^{2}-4x+1\,=\,0\,}$, is connected to the Lucas-Lehmer sequence via the formula

${\displaystyle a(n)=c^{(2^{n})}+c^{-(2^{n})}.}$

Template:Sequence of the Day for October 18

A092290: Decimal expansion of the [positive] solution to ${\displaystyle \scriptstyle {\frac {n}{x}}\,=\,x-n\,}$ for ${\displaystyle \scriptstyle n\,=\,7\,}$.

 7.887482193696061...

Since ${\displaystyle \scriptstyle {\frac {n}{x}}\,=\,x-n\,}$ gives the quadratic equation ${\displaystyle \scriptstyle x^{2}-nx-n\,=\,0\,}$ with roots

${\displaystyle x={\frac {n\pm {\sqrt {n^{2}+4n}}}{2}},\,}$

for ${\displaystyle n=7}$, the positive root is

${\displaystyle x={\frac {7+{\sqrt {77}}}{2}}.\,}$

Note that for ${\displaystyle n=1}$, the positive root is

${\displaystyle x={\frac {1+{\sqrt {5}}}{2}},\,}$

which is the golden ratio.

Template:Sequence of the Day for October 19

A065918: Logarithm of ${\displaystyle \scriptstyle 2+{\sqrt {3}}\,}$.

 1.3169578969248...

This number figures in the search for Mersenne primes thus: the Mersenne number ${\displaystyle M=2^{n}-1}$ is prime if and only if ${\displaystyle M}$ divides

${\displaystyle \cosh(2^{n-2}\log(2+{\sqrt {3}}))=\,}$
${\displaystyle {\frac {e^{\{2^{n-2}\log(2+{\sqrt {3}})\}}+e^{-\{2^{n-2}\log(2+{\sqrt {3}})\}}}{2}}=\,}$
${\displaystyle {\frac {(2+{\sqrt {3}})^{\{2^{n-2}\}}+(2+{\sqrt {3}})^{-\{2^{n-2}\}}}{2}},\,}$

where ${\displaystyle \scriptstyle \cosh(x)\,}$ is the hyperbolic cosine function. David Wilson has proposed calling the number ${\displaystyle \scriptstyle \log(2+{\sqrt {3}})\,}$ the Helms constant.

The number ${\displaystyle \scriptstyle 2+{\sqrt {3}}\,=\,\tan {\frac {5\pi }{12}}\,=\,\cot {\frac {\pi }{12}}\,}$ is sometimes called the Kasner constant (see A019973 for decimal expansion).

Template:Sequence of the Day for October 20

A023194: Numbers ${\displaystyle n}$ such that ${\displaystyle \sigma (n)}$ is prime.

 { 2, 4, 9, 16, 25, 64, 289, 729, 1681, ... }

In 2005, Zak Seidov wondered why all terms except the first are squares. Gabe Cunningham provided the answer:

"From the fact that (...) the sum-of-divisors function is multiplicative, we can derive that ${\displaystyle \sigma (n)}$ is even except when ${\displaystyle n}$ is a square or twice a square."

"If ${\displaystyle n=2(2k+1)^{2}}$, that is, ${\displaystyle n}$ is twice an odd square, then ${\displaystyle \sigma (n)=3\sigma ((2k+1)^{2})}$. If ${\displaystyle n=2(2k)^{2}}$, that is, ${\displaystyle n}$ is twice an even square, then ${\displaystyle \sigma (n)}$ is only prime if ${\displaystyle n}$ is a power of 2; otherwise we have ${\displaystyle \scriptstyle \sigma (n)\,=\,\sigma (8\times 2^{m})\sigma ({\frac {k}{2^{m}}})}$ for some positive integer ${\displaystyle m}$."

"So the only possible candidates for values of ${\displaystyle n}$ other than squares such that ${\displaystyle \sigma (n)}$ is prime are odd powers of 2. But ${\displaystyle \sigma (2^{(2m+1)})}$ = ${\displaystyle 2^{(2m+2)}-1}$ = ${\displaystyle (2^{(m+1)}+1)(2^{(m+1)}-1)}$, which is only prime when ${\displaystyle m=0}$, that is, when ${\displaystyle n=2}$. So 2 is the only nonsquare ${\displaystyle n}$ such that ${\displaystyle \sigma (n)}$ is prime."

Template:Sequence of the Day for October 21

A081119: Number of integral solutions to Mordell's equation ${\displaystyle y^{2}=x^{3}+n}$.

 { 5, 2, 2, 2, 2, 0, 0, 7, 10, 2, ... }
Some Mordell curves in Grapher for Mac OS X.

Mordell's equation has a finite number of integral solutions for all nonzero ${\displaystyle n}$. For example, for ${\displaystyle n=1}$, there are five solutions: (2, –3), (0, –1), (–1, 0), (0, 1) and (2, 3). For ${\displaystyle n=2}$ to 5, there are just two solutions each and none for ${\displaystyle n=6}$ or 7.

Template:Sequence of the Day for October 22

A192189: Polynomial-like numbers: numbers whose sequence of number-derivatives is monotonically decreasing

 { 1, 2, 3, 5, 6, 7, 9, 10, 11, 13, 14, 17, 19, 21, ... }

IN THIS SPACE, VLADIMIR SHEVELEV WILL MAKE SOME BRIEF REMARKS ABOUT THIS SEQUENCE. Ipse lore factum. Ipse lore factum. Ipse lore factum. Ipse lore factum. Ipse lore factum. Ipse lore factum. Ipse lore factum. Ipse lore factum. Ipse lore factum. Ipse lore factum. Ipse lore factum. Ipse lore factum. Ipse lore factum. Ipse lore factum. Ipse lore factum. Ipse lore factum. Ipse lore factum attibutum non contendare.

Template:Sequence of the Day for October 23

A105027 Write numbers in binary under each other; to get the next block of ${\displaystyle 2^{k}}$ ${\displaystyle (k>=0)}$ terms of the sequence, start at ${\displaystyle 2^{k}}$, read diagonals in upward direction and convert to decimal.

 { 0, 1, 3, 2, 6, 5, 4, 7, ... }

This is a permutation of the nonnegative integers (and for ${\displaystyle n>0}$, a permutation of the positive integers). Furthermore, for each block of numbers with ${\displaystyle k}$ binary digits, we have a permutation of this block.

........0
........1
......1 0    <- Starting here, the upward diagonals
......1 1    read 11, 10 giving the block 3, 2.
....1 0 0    <- Starting here, the upward diagonals
....1 0 1    read 110, 101, 100, 111,
....1 1 0    giving the block 6, 5, 4, 7.
....1 1 1
..1 0 0 0
..1 0 0 1
..1 0 1 0
..1 0 1 1
..1 1 0 0
..1 1 0 1
..1 1 1 0
..1 1 1 1


Template:Sequence of the Day for October 24

A1098765: Sequence name.

 { 1, 3, 2, 6, 5, 4, 7, ... }

Paragraph or two of info.

Template:Sequence of the Day for October 25

A006508: ${\displaystyle a(n)=a(n-1)}$-th composite number, with ${\displaystyle a(0)=1}$.

 { 1, 4, 9, 16, 26, 39, 56, 78, ... }

This sequence is generated by a sieve: start with the natural numbers, remove those terms which occupy positions which are prime, leaving {1, 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, ...}; remove those terms whose positions are primes plus one; leaving {1, 4, 9, 12, 15, 16, 18, ...}; remove those whose positions are primes plus two; and so on...

What is the asymptotic behavior of this sequence? The Bojarincev asymptotic formula for the composite numbers allows a formula for ${\displaystyle a(n+k)}$ for any fixed ${\displaystyle k}$ in terms of ${\displaystyle a(n)}$. For example,

${\displaystyle a(n+10)=a(n)\left(1+{\frac {10}{\log a(n)}}+{\frac {65}{\log ^{2}a(n)}}+O\left({\frac {1}{\log ^{3}a(n)}}\right)\right).\,}$

But is there a reasonable asymptotic for ${\displaystyle a(n)}$ without using earlier values?

Template:Sequence of the Day for October 26

A1098765: Sequence name.

 { 1, 3, 2, 6, 5, 4, 7, ... }

Paragraph or two of info.

Template:Sequence of the Day for October 27
0
1
10
11
22100
101
110
111
1000
1001
...

A105027: Write numbers in binary under each other. Starting with the first digit of each line, read diagonals upward and convert to decimal.

 { 0, 1, 3, 2, 6, 5, 4, 7, ... }

## Contents

### Example

After the first 4 terms resulting from the 1- and 2-binary-digit numbers, start with the initial (binary) digit of 22 = 100[2] and read diagonal-wise upwards: 110[2]= 6 = a(4), then starting with the first digit of 101, one reads 101[2]= 5 = a(5), then 100[2]= 4 = a(6) and finally 111[2]= 7 = a(7).

This is a permutation of the nonnegative integers.

The structure of the sequence is as follows: blocks of size ${\displaystyle 2^{k}-1}$ taken from A102370, interspersed with terms of A102371.— Philippe DELEHAM, Nov 17 2007

### References

Template:Sequence of the Day for October 28

A026150${\displaystyle (n+1)}$/A002605${\displaystyle (n)}$: Convergents to ${\displaystyle \scriptstyle {\sqrt {3}}\,}$.

 ${\displaystyle \left\{{\frac {1}{1}},{\frac {4}{2}},{\frac {10}{6}},{\frac {28}{16}},{\frac {76}{44}},{\frac {208}{120}},{\frac {568}{328}},{\frac {1552}{2448}},\cdots \right\}\,}$

Obviously these fractions can be expressed in lower terms. But by leaving them as is, I wish to highlight that both the numerators ${\displaystyle a(n)}$ and the denominators ${\displaystyle b(n)}$ are obtained by recurrence relations of order 2 (actually, the recurrences are the same, only the initial conditions differ):

${\displaystyle a(0)=1,a(1)=1;\,}$
${\displaystyle a(n)=2(a(n-1)+a(n-2)),\quad n\geq 2.\,}$
${\displaystyle b(0)=0,b(1)=1;\,}$
${\displaystyle b(n)=2(b(n-1)+b(n-2)),\quad n\geq 2.\,}$

(The 0th term of the sequence would be ${\displaystyle \scriptstyle {\frac {1}{0}}\,:=\,\infty \,}$.)

Note also the recurrence (involving only the previous numerator/denominator)

${\displaystyle a(1)=1,b(1)=1;\,}$
${\displaystyle a(n)=a(n-1)+3b(n-1),\quad n\geq 2;\,}$
${\displaystyle b(n)=a(n-1)+b(n-1),\quad n\geq 2.\,}$

Template:Sequence of the Day for October 29

A072449: Decimal expansion of the limit of the nested radical ${\displaystyle \scriptstyle {\sqrt {1+{\sqrt {2+{\sqrt {3+{\sqrt {4+\cdots }}}}}}}}\,}$.

 1.757932756618...

This limit is referred to as the nested radical constant. This is not your typical infinitely nested radical, in that the number nested at each step is different. But much more importantly, no closed form is known for this nested radical, further setting it apart from the typical nested radical.

Template:Sequence of the Day for October 30

A985647: Sequence name.

 { 1, 3, 2, 6, 5, 4, 7, ... }

Paragraph or two of info.

Template:Sequence of the Day for October 31

A008595: Multiples of 13.

 {13, 26, 39, 52, 65, 78, 91, ... }

In this day and age, there are still people afraid of the number 13. That's why I've chosen this sequence for Sequence of the Day on Halloween.

A decimal rule for divisibility by 13:

To find out wether a number is a multiple of 13 (and/or a multiple of 7 and/or 11 for that matter), consider

1001 = 1111 − 110 = 11 × 101 − 11 × 10 = 11 × 91 = 7 × 11 × 13,

which means that if you take the decimal digits by groups of three and alternatively add and subtract the obtained three digits numbers, iterating the procedure will lead to a three digit number that is divisible by 13 (and/or by 7 and/or by 11).

For example, with 183440731043453, we have

+ 183 − 440 + 731 − 043 + 453 = 884 = 68 × 13 = 2^2 × 13 × 17,

so 183440731043453 is a multiple of 13, but neither a multiple of 7 nor a multiple of 11.

With 14124936290345881, we have

− 14 + 124 − 936 + 290 − 345 + 881 = 0,

so 14124936290345881 is a multiple of 7, 11 and 13.