cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

A377442 Square array read by rising antidiagonals: T(n, k) = A377441(-n, k), an extension of A377441 into the domain of negative n.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 1, 2, 5, 1, 1, 2, 4, 14, 1, 1, 2, 3, 9, 42, 1, 1, 2, 2, 6, 22, 132, 1, 1, 2, 1, 5, 12, 57, 429, 1, 1, 2, 0, 6, 6, 26, 154, 1430, 1, 1, 2, -1, 9, -2, 15, 59, 429, 4862, 1, 1, 2, -2, 14, -18, 24, 24, 138, 1223, 16796, 1, 1, 2, -3, 21, -48, 77, -23, 53, 332, 3550, 58786, 1, 1, 2, -4, 30, -98, 222, -226, 102, 107, 814, 10455, 208012, 1, 1, 2
Offset: 0

Views

Author

Thomas Scheuerle, Nov 04 2024

Keywords

Comments

The main entry for this array is A377441.

Examples

			The array begins:
  [ 0] 1, 1, 2,  5, 14,  42,  132,   429,   1430, ... = A000108
  [-1] 1, 1, 2,  4,  9,  22,   57,   154,    429, ... = A105633
  [-2] 1, 1, 2,  3,  6,  12,   26,    59,    138, ... = A152172
  [-3] 1, 1, 2,  2,  5,   6,   15,    24,     53, ...
  [-4] 1, 1, 2,  1,  6,  -2,   24,   -23,    102, ...
  [-5] 1, 1, 2,  0,  9, -18,   77,  -226,    765, ...
  [-6] 1, 1, 2, -1, 14, -48,  222,  -921,   3914, ...
  [-7] 1, 1, 2, -2, 21, -98,  531, -2756,  14373, ...
Row index written as [m] is corresponding to A377441(m, k).
		

Crossrefs

Cf. A377441 (The main entry for this sequence).
Cf. A105633 (row -1), A152172 (row -2).
Cf. A000108 (row 0), A254316 (row 1).
Cf. A000012 (Hankel transform of row 0), A006720 (Hankel transform of row 1).
Cf. A330025 (Hankel transform of row -1), A328380 (Hankel transform of row -2).

A377443 Triangular array T(n,k) read by rows, satisfies A377441(n, k+2) = Sum_{m=0..k} T(k, m)*n^m.

Original entry on oeis.org

2, 5, 1, 14, 6, 1, 42, 27, 8, 1, 132, 111, 45, 10, 1, 429, 441, 222, 67, 12, 1, 1430, 1728, 1029, 382, 93, 14, 1, 4862, 6733, 4608, 2005, 599, 123, 16, 1, 16796, 26181, 20199, 10018, 3495, 881, 157, 18, 1, 58786, 101763, 87270, 48445, 19188, 5641, 1236, 195, 20, 1
Offset: 0

Views

Author

Thomas Scheuerle, Nov 04 2024

Keywords

Examples

			Triangle T(n, k) starts:
[0]     2
[1]     5,      1
[2]    14,      6,     1
[3]    42,     27,     8,     1
[4]   132,    111,    45,    10,     1
[5]   429,    441,   222,    67,    12,    1
[6]  1430,   1728,  1029,   382,    93,   14,    1
[7]  4862,   6733,  4608,  2005,   599,  123,   16,   1
[8] 16796,  26181, 20199, 10018,  3495,  881,  157,  18,  1
[9] 58786, 101763, 87270, 48445, 19188, 5641, 1236, 195, 20, 1
		

Crossrefs

Cf. A254316 (row sums).

Programs

  • PARI
    A377441(n, max_k) = Vec(-2*((n+1)*x-1)/((x-1)*(n*x-1)+((n*x^2-(n+1)*x+1)^2-4*x*(x-1)*((n+1)*x-1)+O(x^max_k))^(1/2)))
    T(n, k) = Vec(A377441(y, n+5)[n+3])[n-k+1]
    
  • PARI
    A091894(n, k) = 2^(n-2*k-1)*binomial(n-1, 2*k)*(binomial(2*k, k)/(k + 1))
    A175136(n, k) = sum(m=0,(n - k)/2,A091894(n-k, m)*binomial(n-m-1, n-k))
    T(n, k) = sum(m=1, n+1-k, A175136(n+2-k, n-m+2-k)*binomial(m+k-1, m-1))+(k==0)

Formula

G.f.: (-(y*x^3-(y+1)*x^2+2*x+1) + sqrt((y*x^3-(y+1)*x^2+x)^2 - 4*(x^3-x^2)*((y+1)*x^2-x)))/(2*(x^3-x^2))/x^2.
T(n, 0) = A000108(n+2).
T(n, 1) = A371965(n+2).
T(n, 2) G.f.: x^2*1/( (x - 1)^2*(1 - 4*x)^(3/2) ).
T(n, 3) G.f.: x^3*(3*x - 1)/( (x - 1)^3*(1 - 4*x)^(5/2) ).
T(n, 4) G.f.: x^4*(x^3 + (3*x - 1)^2)/( (x - 1)^4*(1 - 4*x)^(7/2) ).
T(n, 5) G.f.: x^5*(3*x^3*(3*y - 1) + (3*x - 1)^3)/( (x - 1)^5*(1 - 4*x)^(9/2) ).
T(n, 6) G.f.: x^6*(2*x^6 + 6*x^3*(3*x - 1)^2 + (3*x - 1)^4)/( (x - 1)^6*(1 - 4*x)^(11/2) ).
T(n, 7) G.f.: x^7*(10*x^6*(3*x - 1) + 10*x^3*(3*x - 1)^3 + (3*x - 1)^5)/( (x - 1)^7*(1 - 4*x)^(13/2) ).
0 = Sum_{n=0..k} T(n+k, n)*(-1)^n*binomial(k, n).
The diagonal k terms below main diagonal has G.f.: 1 + Sum_{m=1..k+1} A175136(k+2, k-m+2)*(1 - x)^k.
T(n+k, n) = Sum_{m=1..k+1} A175136(k+2, k-m+2)*binomial(m+n-1, m-1), for k > 0.

A378125 Triangle T(n, k) read by rows. Let m be a nonzero rational number then T(n, m mod (n+1)) is the n-th coefficient in the Hasse-Weil L-series (q^(n+1) in the q-expansion) associated to the elliptic equation -4*x^3 + ((m+1)^2 + 8)*x^2 - 2*(m+3)*x + 1 - y^2 = 0.

Original entry on oeis.org

1, -1, -2, 0, -3, -1, 1, 2, 1, 2, -1, -2, -1, -3, 1, 0, 6, 1, 0, 3, 2, 1, -1, -3, 1, -2, -2, -2, -1, 0, -1, 0, -1, 0, -1, 0, 0, 6, -2, 0, 6, -2, 0, 6, -2, 1, 4, 1, 6, -1, 2, 2, 2, 3, -2, -1, -5, 4, 3, 1, -2, -4, -5, -3, -1, 1, 0, -6, -1, 0, -3, -2, 0, -6, -1, 0, -3, -2, 1, -2, -7, 0, 2, -2, -1, 0, -5, -2, -5, 3, 4, -1, 2, 3, -2, 2, 4, 2, -2, 1, 6, -1, 4, 2, 4
Offset: 0

Views

Author

Thomas Scheuerle, Nov 17 2024

Keywords

Comments

Unfortunately, if m is a fraction m = b/c, this triangle can only be used for those coefficients where c and (n+1) are coprime. This is not only because the modulo operation is undefined otherwise, but also because rows of the triangle where (n+1) divides c contain these coefficients with the wrong sign.
The parametrization model for elliptic equations defined by -4*x^3 + ((m+1)^2 + 8)*x^2 - 2*(m+3)*x + 1 - y^2 is also used in A377441. From its relation to Somos-4 sequences, it is known that there is at least one generator point of infinite order if m is an integer > 0 or < -1. If we assume the Birch and Swinnerton-Dyer conjecture to be true, then we expect the associated L-function L(E, s) to be zero at s = 1 for such m.
The relation of m to the J-invariant is given by J(m) = (m^12 + 12*m^11 + 114*m^10 + 628*m^9 + 2823*m^8 + 8184*m^7 + 19036*m^6 + 24552*m^5 + 25407*m^4 + 16956*m^3 + 9234*m^2 + 2916*m + 729)/(m^5 + 4*m^4 + 23*m^3 + 9*m^2) for rational m.
The row sums of the triangle show some connection to the Dedekind psi function (A001615), but will deviate for at least many nonsquarefree n+1.
A short table which shows the Cremona label which corresponds to the L-series obtained for some rational m:
.
m | label
-------------
-5 655a1
-4 166a1
-3 153a1
-2 58a1
-1 11a3
-1/2 26b1
-1/3 141a1
1 37a1
2 158b1
3 423g1
4 458a1
5 1745b1
.

Examples

			The triangle T(n, k) begins:
  q^(n+1) 0, 1  2  3  4  5  6  7  8  9 10 11 12 13 14 15     sum  A001615
  --------------------------------------------------------------------
  [q^1]   1                                                   1     1
  [q^2]  -1,-2                                               -3     3
  [q^3]   0,-3,-1                                            -4     4
  [q^4]   1, 2, 1, 2                                          6     6
  [q^5]  -1,-2,-1,-3, 1                                       6     6
  [q^6]   0, 6, 1, 0, 3, 2                                   12    12
  [q^7]   1,-1,-3, 1,-2,-2,-2                                -8     8
  [q^8]  -1, 0,-1, 0,-1, 0,-1, 0                             -4    12 <- not equal
  [q^9]   0, 6,-2, 0, 6,-2, 0, 6,-2                          12    12
  [q^10]  1, 4, 1, 6,-1, 2, 2, 2, 3,-2                       18    18
  [q^11] -1,-5, 4, 3, 1,-2,-4,-5,-3,-1, 1                    12    12
  [q^12]  0,-6,-1, 0,-3,-2, 0,-6,-1, 0,-3,-2                -24    24
  [q^13]  1,-2,-7, 0, 2,-2,-1, 0,-5,-2,-5, 3, 4             -14    14
  [q^14] -1, 2, 3,-2, 2, 4, 2,-2, 1, 6,-1, 4, 2, 4           24    24
  [q^15]  0, 6, 1, 0,-3, 1, 0, 3, 3, 0, 3, 2, 0, 9,-1        24    24
  [q^16]  1,-4, 1,-4, 1,-4, 1,-4, 1,-4, 1,-4, 1,-4, 1,-4     24    24
		

Crossrefs

Cf. A006571 (main diagonal), A007653 (column 1).

Programs

  • PARI
    T(n, k) = ellak(ellinit(ellfromeqn(-4*x^3 + ((k+n+2)^2 + 8)*x^2 - 2*(k+n+4)*x + 1 - y^2)),n+1);

Formula

T(n, n) = A006571(n), case m =-1. Also the expansion of (eta(q) * eta(q^11))^2 in powers of q.
T(n, 1) = A007653(n), case m = 1.
T(2*n, n) = A251913(2*n+1), case m = -1/2. See first comment.
Let p be an odd prime with good reduction, then T(p-1, k) is odd iff -4*x^3 + ((k+1)^2 + 8)*x^2 - 2*(k+3)*x + 1 == 0 (mod p) has no solution.
Showing 1-3 of 3 results.