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For questions about sequences of integers. References are often made to the online resource oeis.org.

1 vote
0 answers
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On the parity of $(2^{\varphi(n)}-1) \bmod{n^2}$

For odd integer $n$ define the function $$ J(n)=(2^{\varphi(n)}-1) \bmod{n^2}$$ $J(n)$ is integer in $[0,n^2-1]$ and it is divisible by $n$. Integer $n$ is Wieferich number iff $J(n)=0$ and if $n$ is …
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5 votes
1 answer
163 views

On vanishing of $p$-adic logarithms

Might be related to Wieferich primes. Let $p$ be odd prime and define the Fermat quotient $$F(n)=\frac{(2^{n-1} -1)}{n} \mod n=\frac{(2^{n-1} \bmod n^2 )-1}{n}$$ For integer $b$ let $L_p(b)$ be the $p …
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0 votes

Conjectured Somos-like closed form of recurrences with polynomial coefficients

I am investigating Groebner basis free approach, using only linear algebra. This might be more human friendly, since linear algebra is more intuitive than the heavy machinery of Groebner basis. Set $f …
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2 votes
0 answers
28 views

On doubling or addition formulas for the sequence $a(n)=(b_1 n +b_2)a(n-1)+(b_3 n + b_4)a(n-2)$

We are interested which integer sequences are efficiently computable possibly over finite rings. Define the integer sequence $a(n)=(b_1 n +b_2)a(n-1)+(b_3 n + b_4)a(n-2)$ with initial terms $a(0),a(1) …
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6 votes
2 answers
357 views

Conjectured Somos-like closed form of recurrences with polynomial coefficients

From Our short paper For polynomial $F$ with integer coefficients, define the recurrence $f(n)=F(n,f(n-1),f(n-2),...,f(n-d))$. We conjecture that $f(n)$ satisfy Somos like sequence $f(n)=\frac{G(f(n-1 …
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0 votes
0 answers
28 views

Short periods modulo primes of linear recurrences with polynomial coefficients

Let $f_i(x)$ be polynomials with integer coefficients. Define the integer linear recurrence with polynomial coefficients: $$ a(n)=f_1(n) a(n-1)+f_2(n)a(n-2)+\cdots +f_d(n) a(n-d) $$ and the initial te …
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6 votes

Nonexistence of short integer program sequence which generates squares

If you can do this efficiently, you can factor integers efficiently. Set $X-Y=N$. If $X=u^2,Y=v^2$ then $(u^2-v^2)=(u-v)(u+v)=N$ will give the factorization of $N$. To avoid the trivial solution, add …
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2 votes
0 answers
151 views

Conjecture: $x^4+1$ is never Wieferich prime

Related to this question and Alexander Kalmynin's answer. For natural $n$ define $J(n)=(2^{n-1}-1) \bmod n^2$ and if $n$ is power of two define $J(2^n)=1$ (this is artificial, just to avoid triviality …
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2 votes
1 answer
255 views

Small solutions of $x^2-a^3 y^2=\pm 1$

We are interested in small integer solutions to the Pell equation: $$x^2-a^3 y^2=\pm 1 \qquad (1)$$ Where in $\pm 1$ you can chose either sign. $(x^2,a^3 y^2)$ are consecutive powerful numbers. $abc$ …
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0 votes
0 answers
62 views

Linear recurrences in coefficients of powers of quotients of polynomial rings

It is known that linear recurrences with constant coefficients can be computed via powers in $\mathbb{Z}[x]/f(x)$. We believe that this generalizes to quotients of multivariate polynomial rings. Let $ …
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2 votes
2 answers
196 views

On the primality of $j(n)=\varphi(p_n+1-n)+1$ when $j(n) \equiv 19 \pmod {100}$

Related to Power of primes. Let $p_n$ denote n-th prime and $\varphi$ the totient function. For natural $n$, define $j(n)=\varphi(p_n+1-n)+1$. For $n$ up to $10^9$ if $j(n) \equiv 19 \pmod {100}$ then …
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0 votes
1 answer
104 views

Non-Wieferich primes with Euler quotient modulo $p$ two and alternating harmonic numbers

Let $b(n)$ denote the Euler quotient modulo $n$. In OEIS we have A128465 Numbers k such that k divides the numerator of alternating Harmonic number H'((k+1)/2) For $n>1$ we have $b(A128465(n))=2$. Als …
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0 votes
0 answers
106 views

What will be the set of non-Wieferich numbers if the set of non-Wieferich primes is finite?

Integer $n$ is Wieferich number if $2^{\phi(n)}-1 \equiv 0 \pmod {n^2}$. Wieferich prime is Wieferich number with $n$ prime. It is an open problem if there are infinitely many Wieferich primes and inf …
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3 votes

How many non-isomorphic, simple, connected graphs with 6 vertices are there?

The answer to the question is $112$. This is available at OEIS: Number of connected graphs with n nodes You can enumerate small graphs with Nauty: https://www.mankier.com/1/nauty-geng Or try the follo …
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1 vote
1 answer
179 views

On the sequence $a(n)=\gcd(2^n-1,\phi(2^n-1))$

For natural $n$, define the sequence $$ a(n)=\gcd(2^n-1,\phi(2^n-1)) $$ It doesn't appear to be in OEIS and starts $1,1,1,1,9,1,1,1,3,1,9,1,3,1,1,1,27,1,75,49$ Q1 Can we unconditionally prove $a(n)=1 …
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