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 1-4 of 4 results.

A364119 a(n) = 7*A364115(n) - 17*A364115(n-1).

Original entry on oeis.org

46, 1870, 95950, 6111054, 445850046, 35606390254, 3031075759870, 270542736416590, 25045919145436366, 2386963634176587870, 232926731552238831054, 23180020599857593886190, 2345286553765877009107710, 240670553547813070050900126, 25001383450621552178261089950
Offset: 1

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Author

Peter Bala, Jul 12 2023

Keywords

Comments

It is conjectured that the sequence {A364115(n)} and the shifted sequence {A364115(n-1)} both satisfy the supercongruences A364115(p^r) == A364115(p^(r-1)) (mod p^(3*r)) for all primes p >= 5 and positive integers r. Stronger supercongruences may hold for the present sequence, a linear combination of A364115(n) and A364115(n-1).
Conjectures: 1) the supercongruences a(p) == a(1) (mod p^5) hold for all primes p >= 7 (checked up to p = 101).
2) for r >= 2, the supercongruences a(p^r) == a(p^(r-1)) (mod p^(3*r+3)) hold for all primes p >= 7. Cf. A212334.
There is also a multiplicative version of this sequence. Define a sequence of rational numbers {b(n) : n >= 1} by b(n) = A364115(n)^63 / A364115(n-1)^17. Then we conjecture that the above pair of supercongruences also hold for the sequence {b(n)}.

Crossrefs

Programs

  • Maple
    A364115 := n -> coeff(series( 1/(1-x)* LegendreP(n,(1+x)/(1-x))^4, x, 21), x, n):
    seq(7*A364115(n) - 17*A364115(n-1), n = 1..20);

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

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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);

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

Views

Author

Peter Bala, Jul 08 2023

Keywords

Comments

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.

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

Formula

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)

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

Views

Author

Peter Bala, Jul 07 2023

Keywords

Comments

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

			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).
		

Crossrefs

Programs

  • Maple
    a(n) := coeff(series( 1/(1-x)* LegendreP(n,(1+x)/(1-x))^3, x, 21), x, n):
    seq(a(n), n = 0..20);
  • Mathematica
    Table[SeriesCoefficient[1/(1 - x) * LegendreP[n, (1 + x)/(1 - x)]^3, {x,0,n}], {n,0,20}] (* Vaclav Kotesovec, Jul 09 2023 *)

Formula

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).
Showing 1-4 of 4 results.