A364113 -id:A364113 - cp's OEIS frontend

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

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

Peter Bala, Jul 19 2023

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.

			 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

R. Meštrović, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862--2012), arXiv:1111.3057 [math.NT], 2011.
Wikipedia, Dixon's identity

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).

Cf. A364113, A364298

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);

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.

A364116 a(n) = [x^n] 1/(1 - x) * Legendre_P(n, (1 + x)/(1 - x))^n for n >= 0.

Original entry on oeis.org

1, 3, 73, 5623, 908001, 251831261, 106898093065, 64439674636863, 52344140654486017, 55113399257643294769, 73004404532578627776801, 118810038754810358401521065, 233027150139808176596750408337, 542098915811219991386976197616441

Offset: 0

Peter Bala, Jul 08 2023

Main diagonal of A364113.
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.

Cf. A005258, A005259, A108625, A143007, A364113, A364114, A364115, A364117, A364301.

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

Table[SeriesCoefficient[1/(1 - x) * LegendreP[n, (1 + x)/(1 - x)]^n, {x,0,n}], {n,0,20}] (* Vaclav Kotesovec, Jul 09 2023 *)

Conjectures:
1) a(p) == 2*p + 1 (mod p^4) for all primes p >= 3 (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 >= 5 (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.
From Vaclav Kotesovec, Jul 10 2023: (Start)
a(n) ~ c * d^n * n^(2*n - 1/2), where d = 2.102423770105721036432437141524634595160013830317976222331887376263238499... (the same as for A033935) and c = 1.325068544739430738025458046917491360304162175529817456184402029433873399...
a(n) ~ A033935(n) * exp(2*n + 1) / (2*Pi*n).
a(n) ~ A033935(n) * exp(1) * n^(2*n) / n!^2. (End)

A364298 Square array read by ascending antidiagonals: T(n,k) = [x^k] 1/(1 + x) * Legendre_P(k, (1 - x)/(1 + x))^(-n) for n >= 1, k >= 0.

Original entry on oeis.org

1, 1, 1, 1, 3, 19, 1, 5, 73, 721, 1, 7, 163, 3747, 49251, 1, 9, 289, 10805, 329001, 5370751, 1, 11, 451, 23623, 1179251, 44127003, 859748023, 1, 13, 649, 43929, 3100001, 190464755, 8405999785, 190320431953, 1, 15, 883, 73451, 6751251, 589050007, 42601840975, 2160445363107

Offset: 1

Peter Bala, Jul 18 2023

In the square array A364113, the k-th entry in row n is defined as [x^k] 1/(1 - x) * Legendre_P(k, (1 + x)/(1 - x))^n. Here we essentially extend A364113 to negative values of n.
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 array A364113
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.

			 Square array begins
 n\k|  0    1     2       3          4            5               6
  - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
  1 |  1    1    19     721      49251      5370751       859748023
  2 |  1    3    73    3747     329001     44127003      8405999785
  3 |  1    5   163   10805    1179251    190464755     42601840975
  4 |  1    7   289   23623    3100001    589050007    152184210193
  5 |  1    9   451   43929    6751251   1479318759    434790348679
  6 |  1   11   649   73451   12953001   3219777011   1062573281785

A364299 (row 1), A364300 (row 2), A364301 (main diagonal), A364302 (first subdiagonal). Cf. A005258, A005259, A143007, A364113.

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 = 1..10);
# display as a sequence
seq(seq(T(n-k, k), k = 0..n-1), n = 1..10);

A364114 a(n) = [x^n] 1/(1 - x) * Legendre_P(n, (1 + x)/(1 - x))^3 for n >= 0.

Original entry on oeis.org

1, 7, 163, 5623, 235251, 11009257, 554159719, 29359663991, 1615702377331, 91558286583757, 5310712888211413, 313940484249068761, 18853030977961798359, 1147317139889540758509, 70618205829113737707663, 4389482803713232076789623, 275190242843266217113413491

Offset: 0

Peter Bala, Jul 07 2023

Row 3 of A364113.
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.
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 the present sequence also satisfies the supercongruences 1) and 2).

			Examples of supercongruences:
a(7) - a(1) = 29359663991 - 7 = (2^4)*(7^3)*37*144589 == 0 (mod 7^3).
a(7 - 1) - a(0) = 554159719 - 1 = 2*(3^4)*(7^3)*9973 == 0 (mod 7^3).
a(5^2) - a(5) = 5343160378366596176372561346633696195759257 - 11009257 = (2^4)*(5^6)*21372641513466384705490245386534784739 == 0 (mod 5^6).
a(5^2 - 1) - a(5 - 1) = 81394273032250674032560324508765757297751 - 235251 = (2^2)*(5^6)*7*13*29*6317*78120239161449483411026081851 == 0 (mod 5^6).

Cf. A005258, A005259, A364113, A364115, A364116.

a(n) := coeff(series( 1/(1-x)* LegendreP(n,(1+x)/(1-x))^3, x, 21), x, n):
seq(a(n), n = 0..20);

Table[SeriesCoefficient[1/(1 - x) * LegendreP[n, (1 + x)/(1 - x)]^3, {x,0,n}], {n,0,20}] (* Vaclav Kotesovec, Jul 09 2023 *)

a(n) ~ (1921 + 533*sqrt(13))^(n + 1/2) / (13^(1/4) * Pi^2 * n^2 * 2^(n + 7/2) * 3^(3*n + 1/2)). - Vaclav Kotesovec, Jul 09 2023
Conjectures:
1) 3*a(p) - 11*a(p-1) == 10 (mod p^5) for all primes p >= 7 (checked up to p = 101).
2) a(p)^21 == (7^21)*a(p-1)^11 (mod p^5) for all primes p >= 7 (checked up to p = 101).

A364115 a(n) = [x^n] 1/(1 - x) * Legendre_P(n, (1 + x)/(1 - x))^4 for n >= 0.

Original entry on oeis.org

1, 9, 289, 14409, 908001, 65898009, 5246665201, 445752724041, 39731504675041, 3674479246416009, 349918540195094289, 34125049533650776281, 3394306634561379583281, 343284252364774351717641, 35215197976859176290014289, 3657148830889736882170190409

Offset: 0

Peter Bala, Jul 08 2023

Row 4 of A364113.
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.
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 the present sequence also satisfies the supercongruences 1) and 2).

			Examples of supercongruences:
a(11) - a(1) = 34125049533650776281 - 9  = (2^4)*(3^2)*(11^3)*13*97*11423* 12360541 == 0 (mod 11^3).
a(11 - 1) - a(0) = 349918540195094289 - 1 = (2^4)*(11^3)*103*159526079101 == 0 (mod 11^3).
a(5^2) - a(5) = 823068999686576893970482230168234294266351898009 - 65898009 = (2^7)*(3^2)*(5^6)*11*17*31*311*35978539*2371705409*297232149579326831 == 0 (mod 5^6).
a(5^2 - 1) - a(5 - 1) = 7402345246022867712987394168675984358488158001- 908001 = (2^4)*(5^6)*13*29*911*1459*26046751*925152076787*2452153330349 == 0 (mod 5^6).

Cf. A005258, A005259, A364113, A364114, A364116.

a(n) := coeff(series(1/((1-x))* LegendreP(k,(1+x)/(1-x))^4,x, 21):
seq(a(n), n = 0..20);

Table[SeriesCoefficient[1/(1 - x) * LegendreP[n, (1 + x)/(1 - x)]^4, {x,0,n}], {n,0,20}] (* Vaclav Kotesovec, Jul 09 2023 *)

a(n) = my(x='x+O('x^(n+1))); polcoef((1/(1-x))*pollegendre(n, (1+x)/(1-x))^4, n); \\ Michel Marcus, Jul 12 2023

a(n) ~ phi^(10*n + 5) / (2^(3/2) * 5^(1/4) * Pi^(5/2) * n^(5/2)), where phi = A001622 is the golden ratio. - Vaclav Kotesovec, Jul 09 2023

A364117 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, 5, 163, 14409, 2511251, 730485013, 320259339415, 197591579213969, 163325387776051459, 174310058440646865021, 233402385203650889753429, 383208210107883180333696265, 757120215942256247847040802463, 1772210276849283299764079883683173

Offset: 0

Peter Bala, Jul 08 2023

First subdiagonal of A364113.

Cf. A364113, A364116.

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);

Conjectures:
1) the supercongruences a(p) == 2*p + 3 (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.

cp's OEIS Frontend

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.

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A364116 a(n) = [x^n] 1/(1 - x) * Legendre_P(n, (1 + x)/(1 - x))^n for n >= 0.

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A364298 Square array read by ascending antidiagonals: T(n,k) = [x^k] 1/(1 + x) * Legendre_P(k, (1 - x)/(1 + x))^(-n) for n >= 1, k >= 0.

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A364114 a(n) = [x^n] 1/(1 - x) * Legendre_P(n, (1 + x)/(1 - x))^3 for n >= 0.

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A364115 a(n) = [x^n] 1/(1 - x) * Legendre_P(n, (1 + x)/(1 - x))^4 for n >= 0.

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A364117 a(n) = [x^n] 1/(1 - x) * Legendre_P(n, (1 + x)/(1 - x))^(n+1) for n >= 0.

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