A099586
Constant term in (1+x)^n mod (1+x^4).
Original entry on oeis.org
1, 1, 1, 0, -4, -14, -34, -68, -116, -164, -164, 0, 560, 1912, 4616, 9232, 15760, 22288, 22288, 0, -76096, -259808, -627232, -1254464, -2141504, -3028544, -3028544, 0, 10340096, 35303296, 85229696, 170459392, 290992384, 411525376, 411525376, 0, -1405035520
Offset: 1
- A. Erdelyi, Higher Transcendental Functions, McGraw-Hill, 1955, Vol. 3, Chapter XVIII.
- Colin Barker, Table of n, a(n) for n = 1..1000
- John B. Dobson, A matrix variation on Ramus's identity for lacunary sums of binomial coefficients, arXiv preprint arXiv:1610.09361 [math.NT], 2016.
- Vladimir Shevelev, Combinatorial identities generated by difference analogs of hyperbolic and trigonometric functions of order n, arXiv:1706.01454 [math.CO], 2017.
- Vladimir Shevelev, Coefficient of x^k in ((x+1)^n modulo x^N+1), seqfan, Thu Jul 20 2017.
- Gregory Tollisen and Tamás Lengyel, A congruential identity and the 2-adic order of lacunary sums of binomial coefficients, Integers 4 (2004), #A4.
- Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-2).
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seq(eval(rem((1+x)^n, 1+x^4,x),x=0),n=1..40); # Robert Israel, Jul 03 2017
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RecurrenceTable[{a[n] == 4 * a[n - 1] - 6 * a[n - 2] + 4 * a[n - 3] - 2 * a[n - 4], a[1] = 1, a[2] = 1, a[3] = 1, a[4] = 0}, a, {n, 50}] (* G. C. Greubel, Nov 10 2015 *)
a[n_] := HypergeometricPFQ[{(1-n)/4, (2-n)/4, (3-n)/4, -n/4}, {1/4, 1/2, 3/4}, -1]; Array[a, 40] (* Jean-François Alcover, Jul 20 2017, from Vladimir Shevelev's first formula *)
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a(n) = polcoeff(((1+x)^n)%(x^4+1),0)
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Vec(-x*(2*x-1)*(x^2-x+1)/(2*x^4-4*x^3+6*x^2-4*x+1) + O(x^100)) \\ Colin Barker, Nov 08 2015
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a(n) = real(([1,1; I,1])^n)[1,1]; \\ Michel Marcus, Nov 08 2015
A099589
Expansion of x^3 / (1 - 4*x + 6*x^2 - 4*x^3 + 2*x^4).
Original entry on oeis.org
0, 0, 0, 1, 4, 10, 20, 34, 48, 48, 0, -164, -560, -1352, -2704, -4616, -6528, -6528, 0, 22288, 76096, 183712, 367424, 627232, 887040, 887040, 0, -3028544, -10340096, -24963200, -49926400, -85229696, -120532992, -120532992, 0, 411525376, 1405035520, 3392055808
Offset: 0
- A. Erdelyi, Higher Transcendental Functions, McGraw-Hill, 1955, Vol. 3, Chapter XVIII.
- G. C. Greubel, Table of n, a(n) for n = 0..1000
- Vladimir Shevelev, Combinatorial identities generated by difference analogs of hyperbolic and trigonometric functions of order n, arXiv:1706.01454 [math.CO], 2017.
- Vladimir Shevelev, Coefficient of x^k in ((x+1)^n modulo x^N+1), seqfan, Thu Jul 20 2017.
- G. Tollisen and T. Lengyel, A congruential identity and the 2-adic order of lacunary sums of binomial coefficients, Integers 4 (2004), #A4.
- Maran van Heesch, The multiplicative complexity of symmetric functions over a field with characteristic p, Thesis, 2014.
- Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-2).
-
Round@Table[(1/(2*Sqrt[2]))*((2-Sqrt[2])^(n/2)*(Cos[3*Pi*n/8] + Sin[3*Pi*n/8]) + (2+Sqrt[2])^(n/2)*(Sin[Pi*n/8] - Cos[Pi*n/8])), {n, 0, 40}] (* G. C. Greubel, Nov 07 2015 *)
RecurrenceTable[{a[n] == 4*a[n-1] - 6*a[n-2] + 4*a[n-3] - 2*a[n-4], a[0]==0, a[1]==0, a[2]==0, a[3]==1}, a, {n, 0, 40}] (* G. C. Greubel, Nov 10 2015 *)
Table[Sum[(-1)^k*Binomial[n, 4 k + 3], {k, 0, n}], {n, 0, 37}] (* Michael De Vlieger, Jun 30 2017 *)
a[n_] := n*(n-1)*(n-2)/6 HypergeometricPFQ[{(3-n)/4, (4-n)/4, (5-n)/4, (6-n)/4}, {5/4, 3/2, 7/4}, -1]; Array[a, 40, 0] (* Jean-François Alcover, Jul 20 2017, from Vladimir Shevelev's first formula *)
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a(n) = polcoeff(((1+x)^n)%(x^4+1),3)
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concat([0, 0], Vec(x^3/((1-x)^4+x^4) + O(x^50))) \\ Altug Alkan, Nov 08 2015
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a(n) = sum(t=0, (n-3)\4, (-1)^t*binomial(n,4*t+3)); \\ Michel Marcus, Jun 30 2017
A099588
Coefficient of x^2 in (1+x)^n mod 1+x^4.
Original entry on oeis.org
0, 0, 1, 3, 6, 10, 14, 14, 0, -48, -164, -396, -792, -1352, -1912, -1912, 0, 6528, 22288, 53808, 107616, 183712, 259808, 259808, 0, -887040, -3028544, -7311552, -14623104, -24963200, -35303296, -35303296, 0, 120532992, 411525376, 993510144, 1987020288, 3392055808, 4797091328, 4797091328
Offset: 0
- A. Erdelyi, Higher Transcendental Functions, McGraw-Hill, 1955, Vol. 3, Chapter XVIII.
- G. C. Greubel, Table of n, a(n) for n = 0..1000
- Vladimir Shevelev, Combinatorial identities generated by difference analogs of hyperbolic and trigonometric functions of order n, arXiv:1706.01454 [math.CO], 2017.
- Vladimir Shevelev, Coefficient of x^k in ((x+1)^n modulo x^N+1), seqfan, Thu Jul 20 2017.
- G. Tollisen and T. Lengyel, A congruential identity and the 2-adic order of lacunary sums of binomial coefficients, Integers 4 (2004), #A4.
- Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-2).
-
f:= rectoproc({rec, a(0)=0,a(1)=0,a(2)=1,a(3)=3},a(n),remember):
map(f, [$0..100]); # Robert Israel, Jun 30 2017
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RecurrenceTable[{a[n]==4*a[n-1] - 6*a[n-2] + 4*a[n-3] - 2*a[n-4], a[1]=0, a[2]=1, a[3]=3, a[4]=6}, a, {n, 1, 200}] (* G. C. Greubel, Nov 10 2015 *)
Table[Sum[(-1)^k*Binomial[n, 4 k + 2], {k, 0, n}], {n, 0, 36}] (* Michael De Vlieger, Jun 30 2017 *)
a[n_] := n*(n-1)/2 HypergeometricPFQ[{(2-n)/4, (3-n)/4, (4-n)/4, (5-n)/4}, {3/4, 5/4, 3/2}, -1]; Array[a, 40, 0] (* Jean-François Alcover, Jul 20 2017, from Vladimir Shevelev's first formula *)
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x='x+O('x^55); concat([0, 0], Vec(-x^2*(x-1)/(2*x^4-4*x^3+6*x^2-4*x+1))) \\ Altug Alkan, Nov 11 2015
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a(n) = sum(t=0, (n-2)\4, (-1)^t*binomial(n,4*t+2)); \\ Michel Marcus, Jun 30 2017
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a(n)=polcoeff(lift(Mod(1+x,1+x^4)^n),2); \\ Joerg Arndt, Feb 22 2018
A307079
Square array A(n,k), n >= 0, k >= 1, read by antidiagonals, where column k is the expansion of g.f. ((1-x)^(k-2))/((1-x)^k+x^k).
Original entry on oeis.org
1, 1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 2, 3, 0, 1, 1, 2, 3, 3, -4, 1, 1, 2, 3, 4, 0, -8, 1, 1, 2, 3, 4, 4, -9, -8, 1, 1, 2, 3, 4, 5, 0, -27, 0, 1, 1, 2, 3, 4, 5, 5, -14, -54, 16, 1, 1, 2, 3, 4, 5, 6, 0, -48, -81, 32, 1, 1, 2, 3, 4, 5, 6, 6, -20, -116, -81, 32, 1
Offset: 0
Square array begins:
1, 1, 1, 1, 1, 1, 1, 1, 1, ...
1, 2, 2, 2, 2, 2, 2, 2, 2, ...
1, 2, 3, 3, 3, 3, 3, 3, 3, ...
1, 0, 3, 4, 4, 4, 4, 4, 4, ...
1, -4, 0, 4, 5, 5, 5, 5, 5, ...
1, -8, -9, 0, 5, 6, 6, 6, 6, ...
1, -8, -27, -14, 0, 6, 7, 7, 7, ...
1, 0, -54, -48, -20, 0, 7, 8, 8, ...
1, 16, -81, -116, -75, -27, 0, 8, 9, ...
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T[n_, k_] := Sum[(-1)^j * Binomial[n+1, k*j+1], {j, 0, Floor[n/k]}]; Table[T[n-k, k], {n, 0, 12}, {k, n, 1, -1}] // Flatten (* Amiram Eldar, May 20 2021 *)
A290286
Determinant of circulant matrix of order 4 with entries in the first row (-1)^j*Sum_{k>=0}(-1)^k*binomial(n, 4*k+j), j=0,1,2,3.
Original entry on oeis.org
1, 0, 0, 0, -1008, -37120, -473600, 0, 63996160, 702013440, 2893578240, 0, -393379835904, -12971004067840, -160377313820672, 0, 21792325059543040, 239501351489372160, 987061897553510400, 0, -134124249770961666048, -4422152303189489090560
Offset: 0
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seq(LinearAlgebra:-Determinant(Matrix(4,shape=Circulant[seq((-1)^j*
add((-1)^k*binomial(n, 4*k+j),k=0..n/4),j=0..3)])),n=0..50); # Robert Israel, Jul 26 2017
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ro[n_] := Table[Sum[(-1)^(j+k) Binomial[n, 4k+j], {k, 0, n/4}], {j, 0, 3}];
M[n_] := Table[RotateRight[ro[n], m], {m, 0, 3}];
a[n_] := Det[M[n]];
Table[a[n], {n, 0, 21}] (* Jean-François Alcover, Aug 09 2018 *)
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from sympy.matrices import Matrix
from sympy import binomial
def mj(j, n): return (-1)**j*sum((-1)**k*binomial(n, 4*k + j) for k in range(n//4 + 1))
def a(n):
m=Matrix(4, 4, lambda i,j: mj((i-j)%4,n))
return m.det()
print([a(n) for n in range(22)]) # Indranil Ghosh, Jul 31 2017
A307090
Number triangle T(n,k) = Sum_{j=0..n-k} (-1)^j * binomial(k,2*j) * binomial(n-k,2*j).
Original entry on oeis.org
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, -2, -2, 1, 1, 1, 1, -5, -8, -5, 1, 1, 1, 1, -9, -17, -17, -9, 1, 1, 1, 1, -14, -29, -34, -29, -14, 1, 1, 1, 1, -20, -44, -54, -54, -44, -20, 1, 1, 1, 1, -27, -62, -74, -74, -74, -62, -27, 1, 1, 1, 1, -35, -83, -90, -74, -74, -90, -83, -35, 1, 1
Offset: 0
Triangle begins:
n\k | 0 1 2 3 4 5 6 7 8
----+-------------------------------------
0 | 1;
1 | 1, 1;
2 | 1, 1, 1;
3 | 1, 1, 1, 1;
4 | 1, 1, 0, 1, 1;
5 | 1, 1, -2, -2, 1, 1;
6 | 1, 1, -5, -8, -5, 1, 1;
7 | 1, 1, -9, -17, -17, -9, 1, 1;
8 | 1, 1, -14, -29, -34, -29, -14, 1, 1;
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T[n_, k_] := Sum[(-1)^j * Binomial[k, 2*j] * Binomial[n - k, 2*j], {j, 0, n - k}]; Table[T[n, k], {n, 0, 11}, {k, 0, n}] // Flatten (* Amiram Eldar, May 20 2021 *)
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