A364116 -id:A364116 - cp's OEIS frontend

A033935 Sum of squares of coefficients in full expansion of (z1+z2+...+zn)^n.

1, 1, 6, 93, 2716, 127905, 8848236, 844691407, 106391894904, 17091486402849, 3410496772665940, 827540233598615691, 239946160014513220896, 81932406267721802925925, 32541656017173091541743368, 14874686717916861528415671285, 7753005946480818323895940923376

Offset: 0

Warren D. Smith, Dec 11 1999

Two samples of size n are taken from an urn containing infinitely many marbles of n distinct colors. a(n)/n^(2*n) is the probability that the two samples match. That is, they contain the same number of each color of marbles without regard to order. - Geoffrey Critzer, Apr 19 2014

Alois P. Heinz, Table of n, a(n) for n = 0..237

Column k=2 of A245397.
Main diagonal of A287316.
Cf. A364116.

b:= proc(n, i) option remember; `if`(n=0 or i=1, 1,
      add(b(n-j, i-1)*binomial(n, j)^2, j=0..n))
    end:
a:= n-> b(n$2):
seq(a(n), n=0..20);  # Alois P. Heinz, Jul 21 2014
A033935:= proc(n) series(hypergeom([],[1],z)^n, z=0, n+1): n!^2*coeff(%,z,n) end: seq(A033935(n), n=0..16); # Peter Luschny, May 31 2017

Table[nn=n;n!^2 Coefficient[Series[(Sum[x^k/k!^2,{k,0,nn}])^n,{x,0,nn}],x^n],{n,1,20}] (* Geoffrey Critzer, Apr 19 2014 *)
Flatten[{1,Table[n!^2*Coefficient[Series[BesselI[0,2*Sqrt[x]]^n,{x,0,n}],x^n],{n,1,20}]}] (* Vaclav Kotesovec, Jul 29 2014 *)
Table[SeriesCoefficient[HypergeometricPFQ[{},{1},x]^n, {x,0,n}] n!^2, {n,0,16}] (* Peter Luschny, May 31 2017 *)

a(n) is coefficient of x^n in expansion of n!^2*(1 + x/1!^2 + x^2/2!^2 + x^3/3!^2 + ... + x^n/n!^2)^n. - Vladeta Jovovic, Jun 09 2000
a(n) ~ c * d^n * (n!)^2 / sqrt(n), where d = 2.1024237701057210364324371415246345951600138303179762223318873762632384990..., c = 0.487465475752598098146353111500372156824276600165331887960705498284416... - Vaclav Kotesovec, Jul 29 2014, updated Jul 10 2023
a(n) = n!^2 * [z^n] hypergeom([], [1], z)^n. - Peter Luschny, May 31 2017

More terms from James Sellers, Jun 01 2000 and Vladeta Jovovic, Jun 05 2000

a(0)=1 inserted by Alois P. Heinz, Jul 21 2014

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

Peter Bala, Jul 07 2023

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.

			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

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

Cf. A108625, A143007, A364298.

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

Peter Bala, Jul 18 2023

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.

Cf. A005258, A005259, A108625, A143007, A364116, A364298, A364302.

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

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.

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

A033935 Sum of squares of coefficients in full expansion of (z1+z2+...+zn)^n.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Mathematica

Formula

Extensions

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.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Maple

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

Views

Author

Keywords

Comments

Crossrefs

Programs

Maple

Formula

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

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Maple

Mathematica

Formula

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

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

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

Views

Author

Keywords

Comments

Crossrefs

Programs

Maple

Formula