This site is supported by donations to The OEIS Foundation.

# Calendar for Sequence of the Day in November

Template:Sequence of the Day for November 1

A042972: Decimal expansion of ${\displaystyle \scriptstyle {\sqrt[{i}]{i}}\,=\,i^{\frac {1}{i}}\,=\,i^{-i}\,=\,(i^{i})^{-1}\,=\,{\frac {1}{i^{i}}}\,}$.

 4.81047738096535165547...

Since

${\displaystyle i^{i}={\{e^{\frac {i\pi }{2}}\}}^{i}=e^{-{\frac {\pi }{2}}},\,}$

this works out to ${\displaystyle \scriptstyle e^{\frac {\pi }{2}}\,}$ which is a real number!

We all know about the square root of 2, and perhaps some of us have given some thought to the cubic root of 3, the fourth root of 4, the fifth root of 5, etc. Now, here's the ${\displaystyle \scriptstyle i\,}$th root of the imaginary unit ${\displaystyle \scriptstyle i={\sqrt {-1}}\,}$!

By now, you probably want to find out the ${\displaystyle \scriptstyle e\,}$th root of ${\displaystyle \scriptstyle e\,}$ or the ${\displaystyle \scriptstyle \pi \,}$th root of ${\displaystyle \scriptstyle \pi \,}$... or are you looking for the ${\displaystyle \scriptstyle \phi \,}$th root of ${\displaystyle \scriptstyle \phi \,}$ or the ${\displaystyle \scriptstyle {\sqrt {2}}\,}$th root of ${\displaystyle \scriptstyle {\sqrt {2}}\,}$?

Template:Sequence of the Day for November 2

A1102345: Sequence name.

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

Template:Sequence of the Day for November 3

A119258: Triangle read by rows ${\displaystyle n}$ ≥ 0: ${\displaystyle T(n,0)=T(n,n)=1}$ and for ${\displaystyle 0: ${\displaystyle T(n,k)=2T(n-1,k-1)+T(n-1,k)}$.

 1 1 1 1 3 1 1 5 7 1 1 7 17 15 1 1 9 31 49 31 1 1 11 49 111 129 63 1

This sequence is a Pascal-like triangle (see Pascal triangle). It appears naturally in combinatorics, topology, representation theory, computer science and numerical analysis. The natural formula for the terms seems to depend heavily on the context in which they appear, but it is not hard to show by hand that all these formulae satisfy the defining recursive formula.

Template:Sequence of the Day for November 4

A008292: The triangle of Eulerian numbers

                        1
1     1
1     4      1
1    11     11     1
1    26     66     26    1
1    57    302    302    57    1
1    120  1191   2416   1191  120    1
...               ...                ...


is given by the coefficients of the Eulerian polynomials

${\displaystyle E_{n}=\sum _{m=1}^{n}E(n,m)\ X^{n-m},\,}$

which appear as the numerator in an expression for the generating function of the sequence ${\displaystyle \scriptstyle \{1^{n},\,2^{n},\,3^{n},\,\ldots \}\,}$.

The Eulerian number ${\displaystyle \scriptstyle E(n,m)\,=\,\left\langle {n \atop m}\right\rangle \,}$ is the number of permutations of the numbers 1 to ${\displaystyle n}$ in which exactly ${\displaystyle m}$ elements are greater than the previous element.

The subsequence of Eulerian numbers > 1, which are those not lying on the border of the triangle, i.e., with ${\displaystyle 1, is A014449 = {4, 11, 11, ...}.

### Example

For ${\displaystyle n=4}$, the sequence ${\displaystyle (k^{n})_{k=1,2,3,...}}$ {1, 16, 81, 256, 625, 1296, 2401, 4096, 6561, ...} has the generating function

${\displaystyle {\frac {x^{3}+11\,x^{2}+11\,x+1}{1-5\,x+10\,x^{2}-10\,x^{3}+5\,x^{4}-x^{5}}}~.\,}$

Template:Sequence of the Day for November 5

A066342: Number of triangulations of the cyclic polytope ${\displaystyle \scriptstyle C(n,\,n-4)\,}$.

 { 1, 2, 4, 8, 14, 25, 40, 67, 102, 165, 244, 387, 562, 881, ... }

Formerly only generated by brute force in TOPCOM or the like, a surprisingly easy formula was conjectured by Santos, and finally proved in 2002 by M. Azaola and F. Santos. And closed formulas for the numbers of triangulations of other infinite classes of polytopes are rare: The Catalan numbers count the number of triangulations of n-gons—but that's about it.

Template:Sequence of the Day for November 6

A1102345: Title.

 {11, 6, 2013, 4, 5, 6, 7, ... }

Paragraph or two of info.

Template:Sequence of the Day for November 7

A002620: Quarter-squares: ${\displaystyle \scriptstyle \left\lfloor {\frac {n}{2}}\right\rfloor \left\lceil {\frac {n}{2}}\right\rceil \,}$. Equivalently, ${\displaystyle \scriptstyle \left\lfloor {\frac {n^{2}}{4}}\right\rfloor ,\,n\,\geq \,0\,}$.

 { 0, 0, 1, 2, 4, 6, 9, 12, 16, 20, 25, 30, 36, 42, 49, 56, 64, 72, 81, 90, 100... }

This sequence is fairly simple, just the interleaving of two already simple sequences, namely the squares (A000290) and the pronic numbers (A002378), i.e. the [even index/odd index] bisections of this sequence are

${\displaystyle a(2n)=n^{2},\quad n\,\geq \,0,\,}$
${\displaystyle a(2n+1)=n(n+1)=2t_{n},\quad n\,\geq \,0,\,}$

where ${\displaystyle \scriptstyle t_{n}\,}$ is the ${\displaystyle \scriptstyle n\,}$th triangular number.

The recurrences for the [even index/odd index] bisections are

A000290(0) = 0; A000290(${\displaystyle n}$) = A000290(${\displaystyle n}$–1) + 2${\displaystyle n}$ – 1, ${\displaystyle n}$ ≥ 1, and
A002378(0) = 0; A002378(${\displaystyle n}$) = A002378(${\displaystyle n}$–1) + 2${\displaystyle n}$, ${\displaystyle n}$ ≥ 1,

which leads to the simple recurrence

A002620(0) = 0; A002620(${\displaystyle n}$) = A002620(${\displaystyle n}$–1) + floor(${\displaystyle n}$/2), ${\displaystyle n}$ ≥ 1,

where floor(${\displaystyle n}$/2) is given by A004526.

Suppose you try to fill an ${\displaystyle \scriptstyle n\times n}$ discrete grid, node by node, keeping what is covered as compact and simple as possible. If you start with a square covering ${\displaystyle \scriptstyle n\times n}$ nodes, you'll add one column (of ${\displaystyle \scriptstyle n}$ nodes) covering ${\displaystyle \scriptstyle n\times (n+1)}$ nodes, then add another row (of ${\displaystyle \scriptstyle n+1}$ nodes), to now cover ${\displaystyle \scriptstyle (n+1)\times (n+1)}$ nodes. Covering nodes in this 'herring-bone' pattern generates the quarter squares.

Template:Sequence of the Day for November 8

A1102345: Title.

 {11, 8, 2013, 4, 5, 6, 7, ... }

Paragraph or two of info.

Template:Sequence of the Day for November 9

A193018: The largest integer that cannot be written as the sum of squares of integers larger than ${\displaystyle n}$, ${\displaystyle n}$ ≥ 2.

 { 23, 87, 119, 201, 312, 376, 455, 616, 760, 840, 1055, 1136, 1248, 1472, 1719, 1959, ... }

What is the true order of this sequence? Obviously, the smallest integer that can be written thusly is ${\displaystyle (n+1)^{2}}$ (or ${\displaystyle 2(n+1)^{2}}$ if we want two terms). The upper bound ${\displaystyle a(n) can be obtained via Sylvester's theorem, so ${\displaystyle \scriptstyle n^{2}\,\ll \,a(n)\,\ll \,n^{4}\,}$ (the lower bound being trivial).

Template:Sequence of the Day for November 10

A184997: Number of distinct remainders that are possible when a safe prime ${\displaystyle p}$ is divided by ${\displaystyle n}$ (for ${\displaystyle p>2n+1}$).

 { 1, 1, 1, 1, 3, 1, 5, 2, 3, 3, 9, 1, 11, 5, 3, ... }

Per the Chinese remainder theorem, this sequence is multiplicative.

Today marks the 236th anniversary of the founding of the U. S. Marine Corps (back then the Continental Marines) at Tun Tavern.

Template:Sequence of the Day for November 11

A053664: Smallest number ${\displaystyle \scriptstyle m\,}$ such that ${\displaystyle \scriptstyle m\,\equiv \,i{\pmod {p_{i}}}\,}$ for ${\displaystyle \scriptstyle 1\,\leq \,i\,\leq \,n.\,}$

{ 1, 5, 23, 53, 1523, 29243, 299513, ... }

This is a sequence that was suggested to Joe Crump by the Chinese remainder theorem. I've heard theories that the Chinese proved that theorem because it had a practical application as a means to quickly and efficiently count soldiers.

On this Veterans' Day, we salute all those who have served in the Army, Air Force, Navy and Marine Corps.

Template:Sequence of the Day for November 12

A158624: Upper limit of backward value (base 10) of ${\displaystyle \scriptstyle 5^{n}\,}$.

 0.52656795787...

The "backward value (base 10)" of 25 is 0.52; of 125, 0.521; of 625, 0.526; of 3125, 0.5213; etc. Since ${\displaystyle \scriptstyle 5^{n}\,\equiv \,5{\pmod {10}},\,n\,\geq \,1,\,}$ it follows that the backward value of ${\displaystyle \scriptstyle 5^{n},\,n\,\geq \,1,\,}$ starts out 0.5. Furthermore, since ${\displaystyle \scriptstyle 5^{n}\,\equiv \,25{\pmod {100}},\,n\,\geq \,2,\,}$ the next to least significant digit is 2 for all ${\displaystyle \scriptstyle n\,\geq \,2,\,}$ and the backward value therefore in those cases always starts out 0.52. It is a little harder to prove that any subsequent digit of this constant is either 5, 6, 7, 8 or 9 but not 0, 1, 2, 3 or 4.

See also: A158625 Lower limit of backward value (base 10) of ${\displaystyle \scriptstyle 5^{n}\,}$.

Template:Sequence of the Day for November 13

A031347: Multiplicative digital root of ${\displaystyle n}$ (keep multiplying [base 10] digits of ${\displaystyle n}$ until reaching a single digit).

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

This is one of the original base-dependent integer sequences in the OEIS, but not much seems to be known about it (at least, judging from its entry). What is the asymptotic behavior of ${\displaystyle a_{i}}$? (Of course it is ${\displaystyle \Theta (1)}$, but more specifically?) How often does ${\displaystyle a_{n}=a_{n+1}>0}$? Etc.

Since among the ${\displaystyle 10^{k}}$ nonnegative integers in [0, ${\displaystyle 10^{k}-1}$] with ${\displaystyle k}$, ${\displaystyle k}$ ≥ 2, digits [base 10], there are (the first digit being nonzero)

${\displaystyle \left({\frac {9}{10}}\right)^{k-1}\,10^{k}=10\cdot 9^{k-1},\quad k\geq 2,\,}$

integers not containing the digit 0, and

${\displaystyle \lim _{k\to \infty }\left({\frac {9}{10}}\right)^{k-1}=0,\,}$

this implies that, asymptotically, 100% of the multiplicative digital roots are 0, i.e. the asymptotic density of nonzero multiplicative digital roots is 0.

${\displaystyle A_{n}:=\sum _{i=1}^{n}a_{i}=\ ?,\,}$

which will then grow by a nonzero finite amount (1 to 9) asymptotically 0% of the time.

Template:Sequence of the Day for November 14

A1102345: Sequence name.

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

Template:Sequence of the Day for November 15

A049310: Triangle of coefficients of Chebyshev's S(n, x) := U(n, x/2) polynomials.

 { 1, 0, 1, –1, 0, 1, 0, –2, 0, 1, ... }

This sequence appears in linear atomic chains with N uniformly harmonic interacting atoms of the same mass. The eigenmodes have scaled frequency squares x given by the zeros of S(N, 2(1-x)). The recurrence for the oscillations with frequency omega and displacement q_n from the equilibrium position at site no. n, q_n(t) = q_n * exp(i omega t) (i being the complex unit) is

q_{n+1} - 2 (1-x) q_n + q_{ n-1} = 0.

Here x := omega^2 / (2 (omega_0)^2), the normalized frequency squared, with omega_0 := k/m, where k is the uniform spring constant and m is the atom's mass. This leads to a so called 2x2 transfer matrix

R(x) := [[2(1-x), -1], [1, 0]], with [q_{n+1}, q_{n}]^T = R(x) [q_{n}, q_{n-1}]^T (T for matrix transpose).

Iteration yields

[q_{n+1}, q_{n}]^T = M_n(x) [q_{1}, q_{0}]^T,

with M_n(x) = R(x)^n, and the two arbitrary inputs q_{1} and q_{0}. It follows that

M_n(x) = [[S(n, 2*(1-x)), S(n-1, 2*(1-x))], [-S(n-1, 2*(1-x)), -S(n-2, 2(1-x))]]

due to the recurrence for the S-polynomials:

S(n,x) = x * S(n-1, x) - S(n-2, x), S(-1, x) = 0, S(0, x) = 1, n >= 1.

Thus one obtains the general solution for the displacements

q_{n+1}(x) = S(n, 2(1-x)) * q_1 - S(n-1, 2(1-x)) * q_0.

For finite N-chains, with fixed boundary conditions q_0 = 0 = q_{N+1}, one therefore has to solve S(N, 2(1-x)) = 0, and thus obtains the N normalized eigenfrequency squares for the N-chain:

x^{(N)}_{k} = 2 (sin(Pi*k / (2*(N+1)))^2, k = 1, ..., N.

A side remark: because Det R(x) = 1, also Det M_n(x) = 1, identically, therefore on has the so called Cassini-Simson identity

(S(n-1, y))^2 - S(n, y) * S(n-2, y) = 1, for each n >= 0.

For this and another nine applications of this sequence and the row polynomials S(n, x) see the a link under A049310.

Template:Sequence of the Day for November 16

A047679: Denominator in full Stern-Brocot tree.

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

This sequence includes the terms 16, 11, 11 and 11, 16. As such, it can represent both 16 November '11 and November 16. Beyond that, the Stern-Brocot tree is a fascinating way to approximate real numbers using fractions. This image (http://www.kerrymitchellart.com/gallery24/inspired.html) shows a representation of approximating phi, the Golden Ratio. In the image, the widths of the spires are related to the denominators of the fractions encountered. Smaller denominators lead to larger structures.

Template:Sequence of the Day for November 17

A1102345: Sequence name.

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

Template:Sequence of the Day for November 18

A060003: Odd numbers not of the form ${\displaystyle p+2b^{2}}$ for ${\displaystyle p}$ prime and ${\displaystyle b>0}$.

 { 1, 3, 17, 137, 227, 977, 1187, 1493, 5777, 5993, ¿...? }

On November 18, 1752, Christian Goldbach wrote a letter to Leonhard Euler in which he conjectured that every odd integer can be expressed as the sum of a prime and twice a square. Euler verified this lesser known conjecture of Goldbach's up to 2500 and found no counterexamples. Goldbach did allow the square to be zero and considered 1 a prime number, and thus 3, 5, 7 are taken care of with 02, 9 = 1 + 2 × 22 or 7 + 2 × 12, etc.

When Moritz Stern read the Goldbach-Euler correspondence, he became interested in this problem and checked up to 9000, finding the composite numbers 5777 and 5993. With 5777, we can now quickly verify that not only is each ${\displaystyle 5777-2b^{2}}$ composite, quite a few of them are not squarefree. And if we considered 1 prime like they did back then, it would not help here, since 5776 = 24 × 192.

By requiring ${\displaystyle b>0}$, a prime needs a smaller prime for its representation as ${\displaystyle p+2b^{2}}$. For the primes in this sequence, now known as the Stern primes, there is no smaller prime such that the difference is twice a square.

With modern computers, M. F. Hasler and Benjamin Chaffin have verified there are no more terms up to 2 × 1013.

Template:Sequence of the Day for November 19

A065444: Decimal expansion of ${\displaystyle \scriptstyle 9\sum _{k=1}^{\infty }{\frac {1}{10^{k}-1}}.\,}$

 1.1009181908362...

This is the sum of the reciprocals of the base 10 repunits.

Template:Sequence of the Day for November 20
 { 1, 1, 0, 4, 42, 9050, 6965359 ... }

The sequence counts the distinct closed paths that visit every cell of an ${\displaystyle n}$-by-${\displaystyle n}$ square lattice at least once, that never cross any edge between adjacent squares more than once, and that do not self-intersect. Paths related by rotation and/or reflection of the square lattice are not considered distinct.

This sequence is the official selection to commemorate the milestone of the OEIS containing two hundred thousand sequences, which was reached exactly a year ago today.

Template:Sequence of the Day for November 21

A1102345: Sequence name.

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

Template:Sequence of the Day for November 22

A1134567: Sequence title.

 { 11, 22, 2013, 4, 5, 6, 7, ... }

Template:Sequence of the Day for November 23

A001302: Number of ways of making change for ${\displaystyle n}$ cents using coins of 1, 2, 5, 10, 25, 50 cents.

 { ... 407, 428, 457, 478, 507, 540, 569, 602, 631, ... }

Here we are showing this sequence starting at ${\displaystyle a(50)}$ because before that point it is exactly the same as A001301. Having some half dollar pieces in the cash register gives slightly more options for making change when making change for more than 49 cents.

Template:Sequence of the Day for November 24

A002324: Number of divisors of ${\displaystyle \scriptstyle n\,}$ congruent to 1 modulo 3 minus number of divisors of ${\displaystyle \scriptstyle n\,}$ congruent to 2 modulo 3.

 { 1, 0, 1, 1, 0, 0, 2, 0, 1, 0, 0, 1, 2, ... }

These are the coefficients ${\displaystyle \scriptstyle a(n)\,=\,\chi (n),\,n\,\geq \,1,\,}$ in expansion of Dirichlet series

${\displaystyle L(s,\chi )\equiv \sum _{n=1}^{\infty }{\frac {\chi (n)}{n^{s}}}}$
${\displaystyle =\prod _{i=1}^{\infty }{\frac {1}{(1-(\delta _{m}^{p_{i}}+1)~{p_{i}}^{-s}+\delta _{m}^{p_{i}}~{p_{i}}^{-2s})}}}$
${\displaystyle =\left(\prod _{i=1}^{\infty }{\frac {1}{1-\delta _{m}^{p_{i}}~{p_{i}}^{-s}}}\right)\cdot \left(\prod _{i=1}^{\infty }{\frac {1}{1-{p_{i}}^{-s}}}\right)}$
${\displaystyle =\left({\frac {1}{1-3^{-s}}}\right)\cdot \prod _{i=1}^{\infty }{\frac {1}{1-{p_{i}}^{-s}}},\quad m=3,\,}$

where ${\displaystyle \scriptstyle p_{i}\,}$ is the ${\displaystyle \scriptstyle i\,}$th prime.

Note: m was -3 in the formula, it makes more sense for it to be 3, would someone please confirm wether the formula is right.

Template:Sequence of the Day for November 25

A001788: ${\displaystyle \scriptstyle n\,(n+1)\,2^{n-2}\,}$.

 { 1, 6, 24, 80, 240, 672, 1792, 4608, ... }

This is the sum, over all non-empty subsets ${\displaystyle E}$ of {1, 2, ..., ${\displaystyle n}$}, of all elements of ${\displaystyle E}$. For example, ${\displaystyle a(3)=24}$: the non-empty subsets are {1, 2, 3}, {1, 2}, {1, 3}, {2, 3}, {1}, {2}, {3} and 1 + 2 + 3 + 1 + 2 + 1 + 3 + 2 + 3 + 1 + 2 + 3 = 24. Equivalently, this is the sum of all nodes (except the last one, equal to ${\displaystyle n+1}$) of all integer compositions of ${\displaystyle n+1}$.

Template:Sequence of the Day for November 26

A1202345: Sequence name.

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

Template:Sequence of the Day for November 27

A164003: Decimal expansion of ${\displaystyle \scriptstyle e^{\frac {-\pi ^{2}}{2}}\,}$.

 0.00719188335582...

One has to be careful about branches of multivalued complex functions. By definition ${\displaystyle \scriptstyle (ei)^{\pi i}\,}$ is ${\displaystyle \scriptstyle e^{\pi i\log(ei)}\,}$ (using any of the branches of the logarithm function) ${\displaystyle \scriptstyle =\,e^{\pi i(1+\pi {\frac {i}{2}}+2n\pi i)}\,}$ (for any integer ${\displaystyle n}$) ${\displaystyle \scriptstyle =\,-(e^{{\frac {-\pi ^{2}}{2}}-2n\pi ^{2}})\,}$. There is no imaginary part in any of its branches. If ${\displaystyle n=0}$ we get (–1) times the present constant.

Template:Sequence of the Day for November 28

A000055: Number of trees with ${\displaystyle \scriptstyle n,\,n\,\geq \,0,\,}$ unlabeled nodes.

 { 1, 2, 3, 6, 11, 23, 47, 106, 235, ... }

This is the sequence for the example search on the front page of the OEIS. This Thanksgiving we are thankful for, among other things, the OEIS, which is an invaluable resource in many mathematical and scientific endeavors.

Template:Sequence of the Day for November 29

A054551: Prime number spiral (clockwise, North spoke).

 { 2, 3, 31, 107, 241, 443, 709, ... }

I WILL HAVE TO PUT IN THE RIGHT NUMBERS BELOW ANOTHER DAY

 60 59 58 57 33 32 31 56 14 13 30 55 3 12 29 54 2 11 28 53

Of course the spiral may be rotated, but the sequences though pointing in different directions, will remain the same.

Template:Sequence of the Day for November 30

A051682: 11-gonal numbers, ${\displaystyle \scriptstyle {\frac {9n^{2}-7n}{2}}\,}$.

 { 1, 11, 30, 58, 95, 141, 196, ... }

T. D. Noe has proved that there are no 11-gonal numbers greater than 1 that are also triangular numbers.