Coefficients of $(1+x+x^2)^2018$ The Next CEO of Stack OverflowHow to compute $I(n)=int_0^1x_1^aint_x_1^1x_2^acdotsint_x_n-1^1x_n^adx_ncdots dx_2dx_1$?Number of possible results in election with one of candidates getting more then 50% votesLinear recursion with coefficients depending on nCoefficients of exponential generating functionsHow do you solve a two variable 2nd degree polynomialDo all polynomials of degree n with indeterminate coefficients have Galois groups that are isomorphic to Sn?Closed Form Expression of sum with binomial coefficientSolutions of $lfloor 4xrfloor+lfloor 3xrfloor=1$Trigonometric/polynomial equations and the algebraic nature of trig functionscoefficients of $x^k$ and $x^k+1$Divisiblity of binomial coefficients
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Coefficients of $(1+x+x^2)^2018$
The Next CEO of Stack OverflowHow to compute $I(n)=int_0^1x_1^aint_x_1^1x_2^acdotsint_x_n-1^1x_n^adx_ncdots dx_2dx_1$?Number of possible results in election with one of candidates getting more then 50% votesLinear recursion with coefficients depending on nCoefficients of exponential generating functionsHow do you solve a two variable 2nd degree polynomialDo all polynomials of degree n with indeterminate coefficients have Galois groups that are isomorphic to Sn?Closed Form Expression of sum with binomial coefficientSolutions of $lfloor 4xrfloor+lfloor 3xrfloor=1$Trigonometric/polynomial equations and the algebraic nature of trig functionscoefficients of $x^k$ and $x^k+1$Divisiblity of binomial coefficients
$begingroup$
The question is
How many of the coefficients of $(1+x+x^2)^2018$ are not divisible by 3?
Somebody asked me the question, and I have no idea how to solve it. I am not sure if the coefficients are known to have a closed form either.
I guess that the question might be solvable using a combinatorics way, but not sure either. Thanks.
sequences-and-series combinatorics algebra-precalculus polynomials binomial-coefficients
asked Mar 19 at 21:24
Kay K.Kay K.
6,9401337
$endgroup$
add a comment |
$begingroup$
The question is
How many of the coefficients of $(1+x+x^2)^2018$ are not divisible by 3?
Somebody asked me the question, and I have no idea how to solve it. I am not sure if the coefficients are known to have a closed form either.
I guess that the question might be solvable using a combinatorics way, but not sure either. Thanks.
sequences-and-series combinatorics algebra-precalculus polynomials binomial-coefficients
asked Mar 19 at 21:24
Kay K.Kay K.
6,9401337
$endgroup$
$begingroup$
Mathematica says it's 576:Length[Select[(Mod[#, 3] & /@ CoefficientList[(1 + x + x^2)^2018, x]), # != 0 &]].
$endgroup$
– JimB
Mar 19 at 21:50
add a comment |
$begingroup$
The question is
How many of the coefficients of $(1+x+x^2)^2018$ are not divisible by 3?
Somebody asked me the question, and I have no idea how to solve it. I am not sure if the coefficients are known to have a closed form either.
I guess that the question might be solvable using a combinatorics way, but not sure either. Thanks.
sequences-and-series combinatorics algebra-precalculus polynomials binomial-coefficients
asked Mar 19 at 21:24
Kay K.Kay K.
6,9401337
$endgroup$
The question is
How many of the coefficients of $(1+x+x^2)^2018$ are not divisible by 3?
Somebody asked me the question, and I have no idea how to solve it. I am not sure if the coefficients are known to have a closed form either.
I guess that the question might be solvable using a combinatorics way, but not sure either. Thanks.
sequences-and-series combinatorics algebra-precalculus polynomials binomial-coefficients
sequences-and-series combinatorics algebra-precalculus polynomials binomial-coefficients
asked Mar 19 at 21:24
Kay K.Kay K.
6,9401337
asked Mar 19 at 21:24
Kay K.Kay K.
6,9401337
asked Mar 19 at 21:24
Kay K.Kay K.
6,9401337
asked Mar 19 at 21:24
Kay K.Kay K.
6,9401337
asked Mar 19 at 21:24
Kay K.Kay K.
6,9401337
6,9401337
$begingroup$
Mathematica says it's 576:Length[Select[(Mod[#, 3] & /@ CoefficientList[(1 + x + x^2)^2018, x]), # != 0 &]].
$endgroup$
– JimB
Mar 19 at 21:50
add a comment |
$begingroup$
Mathematica says it's 576:Length[Select[(Mod[#, 3] & /@ CoefficientList[(1 + x + x^2)^2018, x]), # != 0 &]].
$endgroup$
– JimB
Mar 19 at 21:50
$begingroup$
Mathematica says it's 576:
Length[Select[(Mod[#, 3] & /@ CoefficientList[(1 + x + x^2)^2018, x]), # != 0 &]].$endgroup$
– JimB
Mar 19 at 21:50
$begingroup$
Mathematica says it's 576:
Length[Select[(Mod[#, 3] & /@ CoefficientList[(1 + x + x^2)^2018, x]), # != 0 &]].$endgroup$
– JimB
Mar 19 at 21:50
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If $p$ is a prime and $m geq 1$, then by the Freshman's dream,
$$ (X + Y)^p^m equiv X^p^m + Y^p^m pmodp. $$
So, expanding $2018 = sum_kgeq 0 a_k 3^k$, then
beginalign*
(1+x+x^2)^2018
= prod_k geq 0 left( (1 + x + x^2)^3^k right)^a_k
equiv prod_k geq 0 left( 1 + x^3^k + x^2cdot 3^k right)^a_k pmod3
endalign*
Now, from $ 2018 = 2 + 2cdot3^2 + 2cdot3^3 + 2cdot3^5 + 2cdot3^6 $, we have $a_k = 2$ when $k = 0, 2, 3, 5, 6$ and $a_k = 0$ otherwise. Moreover,
$$ left( 1 + x^3^k + x^2cdot 3^k right)^2
equiv left( 1 + 2x^3^k right)left(1 + 2 x^3^k+1right) pmod3 $$
Plugging this back,
beginalign*
(1 + x + x^2)^2018
&equiv prod_k in 0, 2, 3, 5, 6 left( 1 + 2x^3^k right)left(1 + 2 x^3^k+1right) pmod3 \
&= (1 + 2x) (1 + 2x^3) (1 + 2x^3^2) (1 + 2x^3^3)^2 \
&quad times (1 + 2x^3^4) (1 + 2x^3^5) (1 + 2x^3^6)^2 (1 + 2x^3^7) \
&equiv (1 + 2x^3^0) (1 + 2x^3^1) (1 + 2x^3^2) (1 + x^3^3 + x^2cdot 3^3) \
&quad times (1 + 2x^3^4) (1 + 2x^3^5) (1 + x^3^6 + x^2cdot 3^6) (1 + 2x^3^7) pmod3
endalign*
If we expand the last product, no two terms have the same exponent, and so, there are total $2^6 times 3^2 = 576$ terms with coefficients not divisible by $3$.
answered Mar 19 at 22:47
Sangchul LeeSangchul Lee
96.3k12171282
$endgroup$
$begingroup$
Wow. This is amazing. Thanks a lot!
$endgroup$
– Kay K.
Mar 19 at 22:51
$begingroup$
@KayK. No problem :)
$endgroup$
– Sangchul Lee
Mar 19 at 22:59
1
$begingroup$
@SangchulLee: Nice presentation. (+1)
$endgroup$
– Markus Scheuer
Mar 20 at 9:27
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $p$ is a prime and $m geq 1$, then by the Freshman's dream,
$$ (X + Y)^p^m equiv X^p^m + Y^p^m pmodp. $$
So, expanding $2018 = sum_kgeq 0 a_k 3^k$, then
beginalign*
(1+x+x^2)^2018
= prod_k geq 0 left( (1 + x + x^2)^3^k right)^a_k
equiv prod_k geq 0 left( 1 + x^3^k + x^2cdot 3^k right)^a_k pmod3
endalign*
Now, from $ 2018 = 2 + 2cdot3^2 + 2cdot3^3 + 2cdot3^5 + 2cdot3^6 $, we have $a_k = 2$ when $k = 0, 2, 3, 5, 6$ and $a_k = 0$ otherwise. Moreover,
$$ left( 1 + x^3^k + x^2cdot 3^k right)^2
equiv left( 1 + 2x^3^k right)left(1 + 2 x^3^k+1right) pmod3 $$
Plugging this back,
beginalign*
(1 + x + x^2)^2018
&equiv prod_k in 0, 2, 3, 5, 6 left( 1 + 2x^3^k right)left(1 + 2 x^3^k+1right) pmod3 \
&= (1 + 2x) (1 + 2x^3) (1 + 2x^3^2) (1 + 2x^3^3)^2 \
&quad times (1 + 2x^3^4) (1 + 2x^3^5) (1 + 2x^3^6)^2 (1 + 2x^3^7) \
&equiv (1 + 2x^3^0) (1 + 2x^3^1) (1 + 2x^3^2) (1 + x^3^3 + x^2cdot 3^3) \
&quad times (1 + 2x^3^4) (1 + 2x^3^5) (1 + x^3^6 + x^2cdot 3^6) (1 + 2x^3^7) pmod3
endalign*
If we expand the last product, no two terms have the same exponent, and so, there are total $2^6 times 3^2 = 576$ terms with coefficients not divisible by $3$.
answered Mar 19 at 22:47
Sangchul LeeSangchul Lee
96.3k12171282
$endgroup$
$begingroup$
Wow. This is amazing. Thanks a lot!
$endgroup$
– Kay K.
Mar 19 at 22:51
$begingroup$
@KayK. No problem :)
$endgroup$
– Sangchul Lee
Mar 19 at 22:59
1
$begingroup$
@SangchulLee: Nice presentation. (+1)
$endgroup$
– Markus Scheuer
Mar 20 at 9:27
add a comment |
$begingroup$
If $p$ is a prime and $m geq 1$, then by the Freshman's dream,
$$ (X + Y)^p^m equiv X^p^m + Y^p^m pmodp. $$
So, expanding $2018 = sum_kgeq 0 a_k 3^k$, then
beginalign*
(1+x+x^2)^2018
= prod_k geq 0 left( (1 + x + x^2)^3^k right)^a_k
equiv prod_k geq 0 left( 1 + x^3^k + x^2cdot 3^k right)^a_k pmod3
endalign*
Now, from $ 2018 = 2 + 2cdot3^2 + 2cdot3^3 + 2cdot3^5 + 2cdot3^6 $, we have $a_k = 2$ when $k = 0, 2, 3, 5, 6$ and $a_k = 0$ otherwise. Moreover,
$$ left( 1 + x^3^k + x^2cdot 3^k right)^2
equiv left( 1 + 2x^3^k right)left(1 + 2 x^3^k+1right) pmod3 $$
Plugging this back,
beginalign*
(1 + x + x^2)^2018
&equiv prod_k in 0, 2, 3, 5, 6 left( 1 + 2x^3^k right)left(1 + 2 x^3^k+1right) pmod3 \
&= (1 + 2x) (1 + 2x^3) (1 + 2x^3^2) (1 + 2x^3^3)^2 \
&quad times (1 + 2x^3^4) (1 + 2x^3^5) (1 + 2x^3^6)^2 (1 + 2x^3^7) \
&equiv (1 + 2x^3^0) (1 + 2x^3^1) (1 + 2x^3^2) (1 + x^3^3 + x^2cdot 3^3) \
&quad times (1 + 2x^3^4) (1 + 2x^3^5) (1 + x^3^6 + x^2cdot 3^6) (1 + 2x^3^7) pmod3
endalign*
If we expand the last product, no two terms have the same exponent, and so, there are total $2^6 times 3^2 = 576$ terms with coefficients not divisible by $3$.
answered Mar 19 at 22:47
Sangchul LeeSangchul Lee
96.3k12171282
$endgroup$
$begingroup$
Wow. This is amazing. Thanks a lot!
$endgroup$
– Kay K.
Mar 19 at 22:51
$begingroup$
@KayK. No problem :)
$endgroup$
– Sangchul Lee
Mar 19 at 22:59
1
$begingroup$
@SangchulLee: Nice presentation. (+1)
$endgroup$
– Markus Scheuer
Mar 20 at 9:27
add a comment |
$begingroup$
If $p$ is a prime and $m geq 1$, then by the Freshman's dream,
$$ (X + Y)^p^m equiv X^p^m + Y^p^m pmodp. $$
So, expanding $2018 = sum_kgeq 0 a_k 3^k$, then
beginalign*
(1+x+x^2)^2018
= prod_k geq 0 left( (1 + x + x^2)^3^k right)^a_k
equiv prod_k geq 0 left( 1 + x^3^k + x^2cdot 3^k right)^a_k pmod3
endalign*
Now, from $ 2018 = 2 + 2cdot3^2 + 2cdot3^3 + 2cdot3^5 + 2cdot3^6 $, we have $a_k = 2$ when $k = 0, 2, 3, 5, 6$ and $a_k = 0$ otherwise. Moreover,
$$ left( 1 + x^3^k + x^2cdot 3^k right)^2
equiv left( 1 + 2x^3^k right)left(1 + 2 x^3^k+1right) pmod3 $$
Plugging this back,
beginalign*
(1 + x + x^2)^2018
&equiv prod_k in 0, 2, 3, 5, 6 left( 1 + 2x^3^k right)left(1 + 2 x^3^k+1right) pmod3 \
&= (1 + 2x) (1 + 2x^3) (1 + 2x^3^2) (1 + 2x^3^3)^2 \
&quad times (1 + 2x^3^4) (1 + 2x^3^5) (1 + 2x^3^6)^2 (1 + 2x^3^7) \
&equiv (1 + 2x^3^0) (1 + 2x^3^1) (1 + 2x^3^2) (1 + x^3^3 + x^2cdot 3^3) \
&quad times (1 + 2x^3^4) (1 + 2x^3^5) (1 + x^3^6 + x^2cdot 3^6) (1 + 2x^3^7) pmod3
endalign*
If we expand the last product, no two terms have the same exponent, and so, there are total $2^6 times 3^2 = 576$ terms with coefficients not divisible by $3$.
answered Mar 19 at 22:47
Sangchul LeeSangchul Lee
96.3k12171282
$endgroup$
If $p$ is a prime and $m geq 1$, then by the Freshman's dream,
$$ (X + Y)^p^m equiv X^p^m + Y^p^m pmodp. $$
So, expanding $2018 = sum_kgeq 0 a_k 3^k$, then
beginalign*
(1+x+x^2)^2018
= prod_k geq 0 left( (1 + x + x^2)^3^k right)^a_k
equiv prod_k geq 0 left( 1 + x^3^k + x^2cdot 3^k right)^a_k pmod3
endalign*
Now, from $ 2018 = 2 + 2cdot3^2 + 2cdot3^3 + 2cdot3^5 + 2cdot3^6 $, we have $a_k = 2$ when $k = 0, 2, 3, 5, 6$ and $a_k = 0$ otherwise. Moreover,
$$ left( 1 + x^3^k + x^2cdot 3^k right)^2
equiv left( 1 + 2x^3^k right)left(1 + 2 x^3^k+1right) pmod3 $$
Plugging this back,
beginalign*
(1 + x + x^2)^2018
&equiv prod_k in 0, 2, 3, 5, 6 left( 1 + 2x^3^k right)left(1 + 2 x^3^k+1right) pmod3 \
&= (1 + 2x) (1 + 2x^3) (1 + 2x^3^2) (1 + 2x^3^3)^2 \
&quad times (1 + 2x^3^4) (1 + 2x^3^5) (1 + 2x^3^6)^2 (1 + 2x^3^7) \
&equiv (1 + 2x^3^0) (1 + 2x^3^1) (1 + 2x^3^2) (1 + x^3^3 + x^2cdot 3^3) \
&quad times (1 + 2x^3^4) (1 + 2x^3^5) (1 + x^3^6 + x^2cdot 3^6) (1 + 2x^3^7) pmod3
endalign*
If we expand the last product, no two terms have the same exponent, and so, there are total $2^6 times 3^2 = 576$ terms with coefficients not divisible by $3$.
answered Mar 19 at 22:47
Sangchul LeeSangchul Lee
96.3k12171282
answered Mar 19 at 22:47
Sangchul LeeSangchul Lee
96.3k12171282
answered Mar 19 at 22:47
Sangchul LeeSangchul Lee
96.3k12171282
answered Mar 19 at 22:47
Sangchul LeeSangchul Lee
96.3k12171282
96.3k12171282
$begingroup$
Wow. This is amazing. Thanks a lot!
$endgroup$
– Kay K.
Mar 19 at 22:51
$begingroup$
@KayK. No problem :)
$endgroup$
– Sangchul Lee
Mar 19 at 22:59
1
$begingroup$
@SangchulLee: Nice presentation. (+1)
$endgroup$
– Markus Scheuer
Mar 20 at 9:27
add a comment |
$begingroup$
Wow. This is amazing. Thanks a lot!
$endgroup$
– Kay K.
Mar 19 at 22:51
$begingroup$
@KayK. No problem :)
$endgroup$
– Sangchul Lee
Mar 19 at 22:59
1
$begingroup$
@SangchulLee: Nice presentation. (+1)
$endgroup$
– Markus Scheuer
Mar 20 at 9:27
$begingroup$
Wow. This is amazing. Thanks a lot!
$endgroup$
– Kay K.
Mar 19 at 22:51
$begingroup$
Wow. This is amazing. Thanks a lot!
$endgroup$
– Kay K.
Mar 19 at 22:51
$begingroup$
@KayK. No problem :)
$endgroup$
– Sangchul Lee
Mar 19 at 22:59
$begingroup$
@KayK. No problem :)
$endgroup$
– Sangchul Lee
Mar 19 at 22:59
1
1
$begingroup$
@SangchulLee: Nice presentation. (+1)
$endgroup$
– Markus Scheuer
Mar 20 at 9:27
$begingroup$
@SangchulLee: Nice presentation. (+1)
$endgroup$
– Markus Scheuer
Mar 20 at 9:27
add a comment |
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$begingroup$
Mathematica says it's 576:
Length[Select[(Mod[#, 3] & /@ CoefficientList[(1 + x + x^2)^2018, x]), # != 0 &]].$endgroup$
– JimB
Mar 19 at 21:50