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 1-6 of 6 results.

A364303 Square array read by ascending antidiagonals: T(n,k) = [x^k] (1 - x)^(2*k) * Legendre_P(n*k, (1 + x)/(1 - x)) for n, k >= 0.

Original entry on oeis.org

1, 1, -2, 1, 0, 6, 1, 4, -6, -20, 1, 10, 36, 0, 70, 1, 18, 300, 400, 90, -252, 1, 28, 1050, 11440, 4900, 0, 924, 1, 40, 2646, 77616, 485100, 63504, -1680, -3432, 1, 54, 5544, 316540, 6370650, 21841260, 853776, 0, 12870, 1, 70, 10296, 972400, 42031990, 554822268, 1022041020, 11778624, 34650, -48620
Offset: 0

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Author

Peter Bala, Jul 19 2023

Keywords

Comments

The first row of the table is a signed version of the central binomial coefficients A000984. The central binomial coefficients satisfy the supercongruences u(n*p^r) == u(n*p^(r-1)) (mod p ^(3*r)) for all primes p >= 5 and all positive integers n and r (see Meštrović, equation 39). We conjecture that each row sequence of the table satisfies the same supercongruences.

Examples

			 Square array begins:
 n\k|  0   1      2        3           4             5
  - + - - - - - - - - - - - - - - - - - - - - - - - - -
  0 |  1  -2      6      -20          70          -252    ...  (-1)^k*A000984(k)
  1 |  1   0     -6        0          90             0    ...  A245086
  2 |  1   4     36      400        4900         63504    ...  A002894
  3 |  1  10    300    11440      485100      21841260    ...  A275652
  4 |  1  18   1050    77616     6370650     554822268    ...  A275653
  5 |  1  28   2646   316540    42031990    5921058528    ...  A275654
  6 |  1  40   5544   972400   189290920   39089615040    ...  A275655
  7 |  1  54  10296  2484000   665091000  188907932304    ...  A364304
  8 |  1  70  17550  5567380  1960044750  732012601320    ...  A364305
		

Crossrefs

Cf. A000984 (row 0 unsigned), A245086 (row 1), A002894 (row 2), A275652 (row 3), A275653 (row 4), A275654 (row 5), A275655 (row 6), A364304 (row 7), A364305 (row 8).

Programs

  • Maple
    T(n,k) := coeff(series( (1 - x)^(2*k) * LegendreP(n*k, (1 + x)/(1 - x)), x, 11), x, k):
    # display as a square array
    seq(print(seq(T(n, k), k = 0..10)), n = 0..10);
    # display as a sequence
    seq(seq(T(n-k, k), k = 0..n), n = 0..10);

Formula

T(n,k) = Sum_{i = 0..k} binomial(n*k, k-i)^2 * binomial((n-2)*k+i-1, i).
For n >= 2, T(n,k) = binomial((n-1)*k, k)^2 * hypergeom([a, b, b], [1 + a - b, 1 + a - b], 1), where a = (n - 3)*k and b = -k.
For n >= 3, T(n,k) = ((n - 1)*k)! * ((n + 1)*k/2)! * ((n - 3)*k/2)! / ( ((n - 1)*k/2)!^2 * k!^2 * ((n - 3)*k)! ) by Dixon's 3F2 summation theorem, where fractional factorials are defined in terms of the gamma function.

A364113 Square array read by ascending antidiagonals: T(n,k) = [x^k] 1/(1 - x) * Legendre_P(k, (1 + x)/(1 - x))^n for n, k >= 0.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 5, 19, 1, 1, 7, 73, 147, 1, 1, 9, 163, 1445, 1251, 1, 1, 11, 289, 5623, 33001, 11253, 1, 1, 13, 451, 14409, 235251, 819005, 104959, 1, 1, 15, 649, 29531, 908001, 11009257, 21460825, 1004307, 1, 1, 17, 883, 52717, 2511251, 65898009, 554159719, 584307365, 9793891, 1
Offset: 0

Views

Author

Peter Bala, Jul 07 2023

Keywords

Comments

The two types of Apéry numbers A005258 and A005259 are related to the Legendre polynomials by A005258(k) = [x^k] 1/(1 - x) * Legendre_P(k, (1 + x)/(1 - x)) and A005259(k) = [x^k] 1/(1 - x) * Legendre_P(k, (1 + x)/(1 - x))^2 and thus form rows 1 and 2 of the present array.
Both types of Apéry numbers satisfy the supercongruences
1) u(n*p^r) == u(n*p^(r-1)) (mod p^(3*r))
and the shifted supercongruences
2) u(n*p^r - 1) == u(n*p^(r-1) - 1) (mod p^(3*r))
for all primes p >= 5 and positive integers n and r.
We conjecture that each row sequence of the present table satisfies the same pair of supercongruences.

Examples

			Square array begins
 n\k|  0   1    2      3        4          5             6               7
  - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
  0 |  1   1    1      1        1          1             1               1
  1 |  1   3   19    147     1251      11253        104959         1004307
  2 |  1   5   73   1445    33001     819005      21460825       584307365
  3 |  1   7  163   5623   235251   11009257     554159719     29359663991
  4 |  1   9  289  14409   908001   65898009    5246665201    445752724041
  5 |  1  11  451  29531  2511251  251831261   28224521263   3423024241627
  6 |  1  13  649  52717  5665001  730485013  106898093065  17144295476461
		

Crossrefs

Cf. A005258 (row 1), A005259 (row 2), A364114 (row 3), A364115 (row 4), A364116 (main diagonal), A364117 (first subdiagonal).

Programs

  • Maple
    T(n,k) := coeff(series(1/(1-x)* LegendreP(k,(1+x)/(1-x))^n, x, 11), x, k):
    # display as a square array
    seq(print(seq(T(n, k), k = 0..10)), n = 0..10);
    # display as a sequence
    seq(seq(T(n-k, k), k = 0..n), n = 0..10);

A364301 a(n) = [x^n] 1/(1 + x) * Legendre_P(n, (1 - x)/(1 + x))^(-n) for n >= 0.

Original entry on oeis.org

1, 1, 73, 10805, 3100001, 1479318759, 1062573281785, 1073267499046525, 1451614640844881665, 2534009926232394596267, 5548110762587726241026801, 14890865228866506199602545427, 48084585660733078332263158771313, 183923731031112887024255817209295155, 822427361894711201025101782425695273529
Offset: 0

Views

Author

Peter Bala, Jul 18 2023

Keywords

Comments

Main diagonal of A364298 (with extra initial term 1). Compare with A364116.
Compare with the two types of Apéry numbers A005258 and A005259, which are related to the Legendre polynomials by A005258(n) = [x^n] 1/(1 - x) * Legendre_P(n, (1 + x)/(1 - x)) and A005259(n) = [x^k] 1/(1 - x) * Legendre_P(n, (1 + x)/(1 - x))^2.
A005258 is the main diagonal of A108625 and A005259 is the main diagonal of A143007.

Crossrefs

Programs

  • Maple
    a(n) := coeff(series( 1/(1 + x) * LegendreP(n, (1 - x)/(1 + x))^(-n), x, 21), x, n):
    seq(a(n), n = 0..20);

Formula

Conjectures:
1) a(p) == 2*p - 1 (mod p^4) for all primes p >= 5 (checked up to p = 101).
More generally, the supercongruence a(p^k) == 2*p^k - 1 (mod p^(3+k)) may hold for all primes p >= 5 and all k >= 1.
2) a(p-1) == 1 (mod p^3) for all primes p except p = 3 (checked up to p = 101).
More generally, the supercongruence a(p^k - p^(k-1)) == 1 (mod p^(2+k)) may hold for all primes p >= 5 and all k >= 1.

A364302 a(n) = [x^n] 1/(1 + x) * Legendre_P(n, (1 - x)/(1 + x))^(-n-1) for n >= 0.

Original entry on oeis.org

1, 3, 163, 23623, 6751251, 3219777011, 2313306332191, 2337707082109071, 3163417897474821763, 5524913023443862515019, 12101947272421487464092429, 32493996621780038121738419591, 104964758754905547830609842389527, 401618040258524641485654323795309235
Offset: 0

Views

Author

Peter Bala, Jul 18 2023

Keywords

Comments

First subdiagonal of A364298.

Crossrefs

Programs

  • Maple
    a(n) := coeff(series( 1/(1 + x) * LegendreP(n, (1 - x)/(1 + x))^(-n-1), x, 21), x, n):
    seq(a(n), n = 0..20);

Formula

Conjectures:
1) the supercongruences a(p) == 2*p + 1 (mod p^3) hold for all primes p >= 5 (checked up to p = 101).
2) the supercongruences a(p - 1) == 1 (mod p^4) hold for all primes p >= 3 (checked up to p = 101).
3) more generally, the supercongruences a(p^k - 1) == 1 (mod p^(3+k)) may hold for all primes p >= 3 and all k >= 1.

A364299 a(n) = [x^n] 1/(1 + x) * Legendre_P(n, (1 - x)/(1 + x))^(-1) for n >= 0.

Original entry on oeis.org

1, 1, 19, 721, 49251, 5370751, 859748023, 190320431953, 55743765411043, 20884452115700251, 9745388924112505269, 5543574376457462884111, 3776677001062829977964007, 3036161801705682492174749691, 2844274879825369072829081331519
Offset: 0

Views

Author

Peter Bala, Jul 18 2023

Keywords

Comments

Row 1 of A364298.
Compare with the Apéry numbers A005258, which are related to the Legendre polynomials by A005258(n) = [x^n] 1/(1 - x) * Legendre_P(n, (1 + x)/(1 - x)).
A005258 satisfies the supercongruences
1) u (n*p^r) == u(n*p^(r-1)) (mod p^(3*r))
and the shifted supercongruences
2) u (n*p^r - 1) == u(n*p^(r-1) - 1) (mod p^(3*r))
for all primes p >= 5 and positive integers n and r.
We conjecture that the present sequence also satisfies the supercongruences 1) and 2).

Crossrefs

Programs

  • Maple
    a(n) := coeff(series( 1/(1 + x) * LegendreP(n, (1 - x)/(1 + x))^(-1), x, 21), x, n):
    seq(a(n), n = 0..20);

Formula

Conjectures:
1) 13*a(p) - 7*a(p-1) == 6 (mod p^5) for all primes p >= 3 (checked up to p = 101).
2) for r >= 2, 13*a(p^r) - 7*a(p^r - 1) == 13*a(p^(r-1)) - 7*a(p^(r-1) - 1) (mod p^(3*r+3)) for all primes p >= 5.
3) a(p)^13 == a(p-1)^7 (mod p^5) for all primes p >= 3 (checked up to p = 101).
4) for r >= 2, a(p^r)^13 * a(p^(r-1) - 1)^7 == a(p^(r-1))^13 * a(p^r - 1)^7 (mod p^(3*r+3)) for all primes p >= 5.

A364300 a(n) = [x^n] 1/(1 + x) * Legendre_P(n, (1 - x)/(1 + x))^(-2) for n >= 0.

Original entry on oeis.org

1, 3, 73, 3747, 329001, 44127003, 8405999785, 2160445363107, 720972846685225, 303256387595475003, 157007652309393485073, 98141188253799911132091, 72882030213423405890701449, 63436168183711463443127520699, 63968150042375034921379294100073, 73985402858435691329113991048739747
Offset: 0

Views

Author

Peter Bala, Jul 18 2023

Keywords

Comments

Row 2 of A364298.
Compare with the Apéry numbers A005259, which are related to the Legendre polynomials by A005259(n) = [x^n] 1/(1 - x) * Legendre_P(n, (1 + x)/(1 - x))^2.
A005259 satisfies the supercongruences
1) u (n*p^r) == u(n*p^(r-1)) (mod p^(3*r))
and the shifted supercongruences
2) u (n*p^r - 1) == u(n*p^(r-1) - 1) (mod p^(3*r))
for all primes p >= 5 and positive integers n and r.
We conjecture that the present sequence also satisfies the supercongruences 1) and 2).

Crossrefs

Programs

  • Maple
    a(n) := coeff(series( 1/(1 + x) * LegendreP(n, (1 - x)/(1 + x))^(-2), x, 21), x, n):
    seq(a(n), n = 0..20);

Formula

Conjectures:
1) 17*a(p) - 11*a(p-1) == 40 (mod p^5) for all primes p >= 7 (checked up to p = 101).
2) for r >= 2, 17*a(p^r) - 11*a(p^r - 1) == 17*a(p^(r-1)) - 11*a(p^(r-1) - 1) (mod p^(3*r+3)) for all primes p >= 5.
3) a(p)^(3*17) == a(1)^(3*17) * a(p-1)^11 (mod p^5) for all primes p except p = 5 (checked up to p = 101).
4) for r >= 2, a(p^r)^(3*17) * a(p^(r-1) - 1)^11 == a(p^(r-1))^(3*17) * a(p^r - 1)^11 (mod p^(3*r+3)) for all primes p >= 5.
Showing 1-6 of 6 results.