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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 1-8 of 8 results.

A376458 a(n) = Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)*A108625(n-1, n-k).

Original entry on oeis.org

1, 1, -7, 1, 569, -3749, -45151, 806737, 1052729, -130060889, 740060243, 16076432923, -238772815711, -1050791121197, 49401000432497, -171944622257999, -7658491447803847, 87632552103603679, 768037618172427023, -22023427875902878553, 19183786570616924819, 4030690809877385503081, -33792039667279104716677, -520860578851790657166869
Offset: 0

Views

Author

Peter Bala, Sep 23 2024

Keywords

Comments

Compare with the following identity relating the sequence of Apéry numbers to the table of crystal ball sequences for the A_n lattices: A005259(n) = Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)*A108625(n, k), which can be verified by using the MulZeil procedure in Zeilberger's MultiZeilberger Maple package to find a recurrence for the double sum on the right-hand side of the identity.
The sequence of Apéry numbers A005259 satisfies the supercongruences A005259(n*p^r) == A005259(n*p^(r-1)) (mod p^(3*r)) for all primes p >= 5 and all positive integers n and r.
We conjecture that the present sequence satisfies same supercongruences and in addition satisfies the stronger congruences a(p) == a(1) (mod p^5) for all primes p >= 7 (checked up to p = 199) and a(p^r) == a(p^(r-1)) (mod p^(3*r+3)) for all primes p >= 5 and integers r >= 2.

Examples

			Examples of supercongruences:
a(7) - a(1) = 806737 - 1 = (2^4)*3*(7^5) == 0 (mod 7^5).
a(11) - a(1) = 16076432923 - 1 =  2*3*(11^5)*127*131 == 0 (mod 11^5).
a(5^2) - a(5) = 22511570786292886382808751 - (-3749) = (2^2)*(3^2)*(5^9)*67*97*
7741*49223*129289 == 0 (mod 5^9).

Crossrefs

Cf. A005259, A108625, A183204, A352653, A376459 - A376466.

Programs

  • Maple
    A108625(n, k) := add(binomial(n, i)^2 * binomial(n+k-i, k-i), i = 0..k):
a(n) := add((-1)^(n+k)*binomial(n, k)*binomial(n+k, k)*A108625(n-1, n-k), k = 0..n):
seq(a(n), n = 0..25);

Formula

a(n) = Sum_{0 <= i <= k <= n} (-1)^k * binomial(n, k) * binomial(2*n-k, n-k) * binomial(n-1, i)^2 * binomial(n+k-i-1, k-i).
P-recursive: (2*n - 3)*n^3*(n - 1)^2*(473*n^5 - 4988*n^4 + 20888*n^3 - 43462*n^2 + 45019*n - 18634)*a(n) = - 2*(n - 1)^2*(3784*n^9 - 51256*n^8 + 303801*n^7 - 1037327*n^6 + 2252744*n^5 - 3220636*n^4 + 3006247*n^3 - 1739455*n^2 + 555714*n - 75024)*a(n-1) - 2*(n - 2)*(2*n - 1)*(52030*n^9 - 756800*n^8 + 4787337*n^7 - 17271387*n^6 + 39143817*n^5 - 57806236*n^4 + 55708921*n^3 - 33926177*n^2 + 11955879*n - 1890360)*a(n-2) - 2*(n - 2)*(n - 3)^3*(2*n - 1)*(2*n - 3)*(473*n^5 - 2623*n^4 + 5666*n^3 - 5996*n^2 + 3172*n - 704)*a(n-3) with a(0) = 1, a(1) = -1 and a(2) = 7.

A376459 a(n) = Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)*A108625(n, n-k).

Original entry on oeis.org

1, -1, -17, 143, 751, -20251, 30871, 2584847, -21586193, -251907751, 5176221733, 5498864117, -913327142441, 5540080670669, 120825094592983, -1860921180719857, -8346832617144593, 401702184476719649, -1403893237226212151, -64680833271083055607, 743195619082337134501, 6754996433001423371159, -192371016736634220839987, 139058974519768723621493, 36163089652079749214625751, -298797649039016749340832751
Offset: 0

Views

Author

Peter Bala, Sep 24 2024

Keywords

Comments

Compare with the following identity relating the sequence of Apéry numbers to the table of crystal ball sequences for the A_n lattices: A005259(n) = Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)*A108625(n, k), which can be verified by using the MulZeil procedure in Zeilberger's MultiZeilberger Maple package to find a recurrence for the double sum on the right-hand side of the identity.
The sequence of Apéry numbers A005259 satisfies the pair of supercongruences
1) A005259(n*p^r) == A005259(n*p^(r-1)) (mod p^(3*r)) for all primes p >= 5 and all positive integers n and r
and
2) A005259(n*p^r - 1) == A005259(n*p^(r-1) - 1) (mod p^(3*r)) for all primes p >= 5 and all positive integers n and r.
We conjecture that the present sequence satisfies the same pair of supercongruences. Some examples are given below.

Examples

			Examples of supercongruences:
(1a) a(11) - a(1) = 5498864117 - (-1) = 2*(3^3)*(11^3)*76507 == 0 (mod 11^3);
(1b) a(10) - a(0) = 5176221733 - 1 = (2^2)*(3^5)*(11^3)*4001 == 0 (mod 11^3).
(2a) a(5^2) - a(5) = -298797649039016749340832751 - (-20251) = -(2^2)*3*(5^6)*(11^2)*47*89*1683049*1870707593 == 0 (mod 5^6);
(2b) a(5^2 - 1) - a(5 - 1) = 36163089652079749214625751 - 751 = (2^3)*3*(5^6)*7*11*17*101*729412564491671 == 0 (mod 5^6).

Crossrefs

Cf. A005259, A108625, A376458, A376460 - A376466.

Programs

  • Maple
    A108625(n, k) := add(binomial(n, i)^2 * binomial(n+k-i, k-i), i = 0..k):
a(n) := add((-1)^(n+k)*binomial(n, k)*binomial(n+k, k)*A108625(n, n-k), k = 0..n):
seq(a(n), n = 0..25);

Formula

P-recursive: n^3*(n - 1)*(946*n^2 - 3053*n + 2475)*a(n) = -2*(n - 1)*(3784*n^5 - 17888*n^4 + 31787*n^3 - 26726*n^2 + 11051*n - 1824)*a(n-1) - 2*(104060*n^6 - 752070*n^5 + 2212238*n^4 - 3374927*n^3 + 2802671*n^2 - 1196821*n + 205920)*a(n-2) - 2*(n - 2)^3*(2*n - 3)*(946*n^2 - 1161*n + 368)*a(n-3) with a(0) = 1, a(1) = -1 and a(2) = -17.

A376460 a(n) = Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)^2*A108625(n, k).

Original entry on oeis.org

1, 11, 559, 42923, 3996751, 416148761, 46607742091, 5494953548459, 673006663951183, 84887862132737261, 10960384987436995309, 1442356991928315362009, 192832695330486199076011, 26126080107002322789528197, 3580244300364122205875141059, 495479786625880934937914634923, 69162655828624756613259127413583
Offset: 0

Views

Author

Peter Bala, Sep 24 2024

Keywords

Comments

Cf. A005258(n) = Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)^2.
The sequence of Apéry numbers A005258 satisfies the pair of supercongruences
1) A005258(n*p^r) == A005258(n*p^(r-1)) (mod p^(3*r)) for all primes p >= 5 and all positive integers n and r
and
2) A005258(n*p^r - 1) == A005258(n*p^(r-1) - 1) (mod p^(3*r)) for all primes p >= 5 and all positive integers n and r.
We conjecture that the present sequence satisfies the same pair of supercongruences. Some examples are given below.

Examples

			Examples of supercongruences:
(1a) a(13) - a(1) = 26126080107002322789528197 - 11 = 2*3*(7^3)*11*(13^3)*101* 5200975019533651  == 0 (mod 13^3);
(1b) a(13 - 1) - a(1 - 1) = 192832695330486199076011 - 1 = 2*3*5*(13^3)*107*457* 1366649*43779661 == 0 (mod 13^3).
(2a) a(5^2) - a(5) = 1806832316206138774124773283611717478292644188023761 - 416148761 = (2^3)*(3^3)*(5^8)*953*22470418607359387809254496551076506667197 = 0 (mod 5^8);
(2b) a(5^2 - 1) - a(5 - 1) = 12360782465873293068286987426696670402529985934251 - 3996751 = (2^2)*3*(5^6)*11779*1268093*9320371*634752259*746015256807592939 == 0 (mod 5^6).

Crossrefs

Cf. A005258, A376458 - A376466.

Programs

  • Maple
    A108625(n, k) := add(binomial(n, i)^2 * binomial(n+k-i, k-i), i = 0..k):
a(n) := add((-1)^(n+k)*binomial(n, k)*binomial(n+k, k)^2*A108625(n, k), k = 0..n):
seq(a(n), n = 0..25);

A376465 a(n) = Sum_{k = 0..n} binomial(n, k)^2*binomial(n+k, k)^2*A108625(n-1 k).

Original entry on oeis.org

1, 5, 289, 32909, 4846001, 824723005, 154077084505, 30725590395197, 6429986799134257, 1396511202452212733, 312375607924873231289, 71567605248444374973205, 16725718218774077760354953, 3974882968204513021199112653, 958241214962583413382405975337, 233879553339714596653197104362909
Offset: 0

Views

Author

Peter Bala, Sep 25 2024

Keywords

Comments

The sequence of Apéry numbers A005259 defined by A005259(n) = Sum_{k = 0..n} binomial(n, k)^2*binomial(n+k, k)^2 satisfies the pair of supercongruences
1) A005259(n*p^r) == A005259(n*p^(r-1)) (mod p^(3*r)) for all primes p >= 5 and all positive integers n and r
2) A005259(n*p^r - 1) == A005259(n*p^(r-1) - 1) (mod p^(3*r)) for all primes p >= 5 and all positive integers n and r.
We conjecture that the present sequence satisfies the same pair of supercongruences. Some examples are given below.

Examples

			Examples of supercongruences:
a(11) - a(1) = 71567605248444374973205 - 5 = (2^4)*(5^2)*(11^3)*19*56040893*126246629 == 0 (mod 11^3).
a(10) - a(0) = 312375607924873231289 - 1 = (2^3)*(11^3)*17*1725679541724893 == 0 (mod 11^3).

Crossrefs

Cf. A005259, A376458 - A376466.

Programs

  • Maple
    A108625(n, k) := add(binomial(n, i)^2 * binomial(n+k-i, k-i), i = 0..k):
a(n) := add(binomial(n, k)^2*binomial(n+k, k)^2*A108625(n-1, k), k = 0..n):
seq(a(n), n = 0..25);

A376461 a(n) = Sum_{k = 0..n} binomial(n, k)^2*binomial(n+k, k)*A108625(n, k).

Original entry on oeis.org

1, 7, 199, 8359, 423751, 23973757, 1457477659, 93222362023, 6192061490503, 423467225669257, 29640623864436949, 2114316676856525533, 153205735737013589371, 11249571292274407086001, 835446050985954811273699, 62655191693751473885672359, 4739286937335653870416516423, 361196527809084536022005145121
Offset: 0

Views

Author

Peter Bala, Sep 24 2024

Keywords

Comments

The sequence of Apéry numbers A005258 defined by A005258(n) = Sum_{k = 0..n} binomial(n, k)^2*binomial(n+k, k) satisfies the pair of supercongruences
1) A005258(n*p^r) == A005258(n*p^(r-1)) (mod p^(3*r)) for all primes p >= 5 and all positive integers n and r
and
2) A005258(n*p^r - 1) == A005258(n*p^(r-1) - 1) (mod p^(3*r)) for all primes p >= 5 and all positive integers n and r.
We conjecture that the present sequence satisfies the same pair of supercongruences. Some examples are given below.

Crossrefs

Cf. A005258, A376458 - A376466.

Programs

  • Maple
    A108625(n, k) := add(binomial(n, i)^2 * binomial(n+k-i, k-i), i = 0..k):
a(n) := add(binomial(n, k)^2*binomial(n+k, k)*A108625(n, k), k = 0..n):
seq(a(n), n = 0..25);

Formula

Examples of supercongruences:
a(11) - a(1)= 2114316676856525533 - 7 = 2*(3^2)*7*(11^3)*12607281056471 == 0 (mod 11^3).
a(10) - a(0) = 29640623864436949 - 1 = (2^2)*3*(11^3)*504001*3682109 == 0 (mod 11^3).
From Vaclav Kotesovec, Sep 29 2024: (Start)
Recurrence: n^4*(171*n^2 - 551*n + 446)*a(n) = 7*(2052*n^6 - 10716*n^5 + 21806*n^4 - 22301*n^3 + 12415*n^2 - 3630*n + 440)*a(n-1) + (27189*n^6 - 196365*n^5 + 577739*n^4 - 882118*n^3 + 733309*n^2 - 313390*n + 53944)*a(n-2) + (n-2)^4*(171*n^2 - 209*n + 66)*a(n-3).
a(n) ~ (5/12 + sqrt(37)*cos(arccos(433/(74*sqrt(37)))/3)/6) * (28 + 6*sqrt(93)*cos(arccos((199*sqrt(3/31))/62)/3))^n / (Pi^2*n^2). (End)

A376462 a(n) = Sum_{k = 0..n} binomial(n, k)^2*binomial(n+k, k)*A108625(n, n-k).

Original entry on oeis.org

1, 5, 109, 3317, 121501, 4954505, 216867925, 9981053045, 476860000285, 23451310381505, 1180189308268609, 60519806861966105, 3152285573768063461, 166371462775232899553, 8880340127444426907109, 478649327347386225075317, 26019989011889817463755805, 1425143757811438999747555313, 78578956793385528989609594089
Offset: 0

Views

Author

Peter Bala, Sep 24 2024

Keywords

Comments

The sequence of Apéry numbers A005258 defined by A005258(n) = Sum_{k = 0..n} binomial(n, k)^2*binomial(n+k, k) satisfies the pair of supercongruences
1) A005258(n*p^r) == A005258(n*p^(r-1)) (mod p^(3*r)) for all primes p >= 5 and all positive integers n and r
and
2) A005258(n*p^r - 1) == A005258(n*p^(r-1) - 1) (mod p^(3*r)) for all primes p >= 5 and all positive integers n and r.
We conjecture that the present sequence satisfies the same pair of supercongruences. Some examples are given below.

Examples

			Examples of supercongruences:
a(11) - a(1) = 60519806861966105 - 5 = (2^2)*(3^2)*(5^2)*(11^3)*197*256454747 == 0 (mod 11^3).
a(10) - a(0) = 1180189308268609 - 1 = (2^6)*3*(11^3)*37*2789*44753 == 0 (mod 11^3).

Crossrefs

Cf. A005258, A376458 - A376466.

Programs

  • Maple
    A108625(n, k) := add(binomial(n, i)^2 * binomial(n+k-i, k-i), i = 0..k):
a(n) := add(binomial(n, k)^2*binomial(n+k, k)*A108625(n, n-k), k = 0..n):
seq(a(n), n = 0..25);

A376463 a(n) = Sum_{k = 0..n} binomial(n, k)^2*binomial(n+k, k)^2*A108625(n, k).

Original entry on oeis.org

1, 13, 937, 110173, 16431001, 2815533013, 528281347609, 105661979187421, 22160058768609049, 4820836639111911013, 1079739020625352737937, 247635383880853678809541, 57923551410778898112945769, 13775523966484086307239139141, 3322958149149086403877851762937, 811467759428066412526078761086173
Offset: 0

Views

Author

Peter Bala, Sep 24 2024

Keywords

Comments

The sequence of Apéry numbers A005259 defined by A005259(n) = Sum_{k = 0..n} binomial(n, k)^2*binomial(n+k, k)^2 satisfies the pair of supercongruences
1) A005259(n*p^r) == A005259(n*p^(r-1)) (mod p^(3*r)) for all primes p >= 5 and all positive integers n and r
and
2) A005259(n*p^r - 1) == A005259(n*p^(r-1) - 1) (mod p^(3*r)) for all primes p >= 5 and all positive integers n and r.
We conjecture that the present sequence satisfies the same pair of supercongruences. Some examples are given below.

Examples

			Examples of supercongruences:
a(11) - a(1) = 247635383880853678809541 - 13 = (2^3)*(3^2)*(11^3)*13*178697* 389023*2859343 == 0 (mod 11^3).
a(10) - a(0) = 1079739020625352737937 - 1 = (2^4)*3*(11^3)*13*1459*601831*1480561 == 0 (mod 11^3).

Crossrefs

Cf. A005259, A376458 - A376466.

Programs

  • Maple
    A108625(n, k) := add(binomial(n, i)^2 * binomial(n+k-i, k-i), i = 0..k):
a(n) := add(binomial(n, k)^2*binomial(n+k, k)^2*A108625(n, k), k = 0..n):
seq(a(n), n = 0..25);

A376464 a(n) = Sum_{k = 0..n} binomial(n, k)^2*binomial(n+k, k)^2*A108625(n, n-k).

Original entry on oeis.org

1, 7, 307, 20167, 1667251, 157682257, 16284029959, 1790285284039, 206214335367859, 24625438825569757, 3026463150129458557, 380777119628689277809, 48850700118326352063559, 6371139828304049908181989, 842708419059535649574061807, 112830751689834347753141396167, 15268616117602266095803172887219
Offset: 0

Views

Author

Peter Bala, Sep 24 2024

Keywords

Comments

The sequence of Apéry numbers A005259 defined by A005259(n) = Sum_{k = 0..n} binomial(n, k)^2*binomial(n+k, k)^2 satisfies the pair of supercongruences
1) A005259(n*p^r) == A005259(n*p^(r-1)) (mod p^(3*r)) for all primes p >= 5 and all positive integers n and r
and
2) A005259(n*p^r - 1) == A005259(n*p^(r-1) - 1) (mod p^(3*r)) for all primes p >= 5 and all positive integers n and r.
We conjecture that the present sequence satisfies the same pair of supercongruences. Some examples are given below.

Crossrefs

Cf. A005259, A376458 - A376466.

Programs

  • Maple
    A108625(n, k) := add(binomial(n, i)^2 * binomial(n+k-i, k-i), i = 0..k):
a(n) := add(binomial(n, k)^2*binomial(n+k, k)^2*A108625(n, n-k), k = 0..n):
seq(a(n), n = 0..25);

Formula

Examples of supercongruences:
a(11) - a(1) = 380777119628689277809 - 7 = 2*(3^3)*7*(11^3)*(19^2)*83*103*587*417773 == 0 (mod 11^3).
a(10) - a(0) = 3026463150129458557 - 1 = (2^2)*3*(11^3)*17*19*191*251*12236761 == 0 (mod 11^3).
Showing 1-8 of 8 results.

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