A357567 -id:A357567 - cp's OEIS frontend

A357568 a(n) = 9binomial(2n,n)^2 - 8binomial(3n,n).

1, 12, 204, 2928, 40140, 547512, 7535472, 105077376, 1484848332, 21237645000, 306972655704, 4477160465856, 65802123629424, 973487343836448, 14483651478207360, 216550246159148928, 3251660678391659724, 49011343741651501800, 741221951008966181160, 11243583961952559386400

Offset: 0

Peter Bala, Oct 21 2022

Conjectures:
1) a(p^r) == a(p^(r-1)) ( mod p^(3*r+3) ) for r >= 2 and all primes p >= 3.
These are stronger supercongruences than those satisfied separately by the sequences {binomial(2*n,n)} = A000984 and {binomial(3*n,n)} = A005809.
2) More generally, for k >= 1, the sequence {9*binomial(2*n,n)^k - k*(2^k)*binomial(3*n,n): n >= 0} may satisfy the same supercongruences. This is the case k = 2. See A357509 for the case k = 1.

			Examples of supercongruences:
a(11) - a(1) = 4477160465856 - 12 = (2^2)*3*(11^5)*101*22937 == 0 (mod 11^5).
a(5^2) - a(5) = 143816772358933669354266172512 - 547512 = (2^3)*3*(5^9)*167191* 194659*94271599039 == 0 (mod 5^9).

Paolo Xausa, Table of n, a(n) for n = 0..800
C. Helou and G. Terjanian, On Wolstenholme’s theorem and its converse, J. Number Theory 128 (2008), 475-499.
Romeo Meštrović, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2011), arXiv preprint arXiv:1111.3057 [math.NT], 2011.

Cf. A000984, A002894, A005809, A357509, A357567, A357569, A357955.

seq(9*binomial(2*n,n)^2 - 8*binomial(3*n,n), n = 0..20);

A357568[n_] := 9*Binomial[2*n, n]^2 - 8*Binomial[3*n, n];
Array[A357568, 25, 0] (* Paolo Xausa, Jul 17 2024 *)

a(n) = 9*A002894(n) - 8*A005809(n) = 9*A000984(n)^2 - 8*A005809(n). .
a(k*p^r) == a(k*p^(r-1)) ( mod p^(3*r) ) for positive integers k and r and for all primes p >= 5 (see Meštrović, Section 6, equation 39).
a(p) == a(1) (mod p^5) for all primes p >= 7 (apply Helou and Terjanian, Section 3, Proposition 2).
From Stefano Spezia, Jul 17 2024: (Start)
G.f.: 16*cos(arccos(1-27*x/2)/6)/sqrt(4-27*x) + 18*EllipticK(16*x)/Pi.
E.g.f.: 9*hypergeom([1/2, 1/2], [1, 1], 16*x) - 8*hypergeom([1/3, 2/3], [1/2, 1], 27*x/4).
a(n) ~ 9*2^(4*n)/(n*Pi). (End)

A357956 a(n) = 5A005259(n) - 2A005258(n).

Original entry on oeis.org

3, 19, 327, 6931, 162503, 4072519, 107094207, 2919528211, 81819974343, 2343260407519, 68285241342827, 2018360803903111, 60366625228511423, 1823565812734012639, 55557838850469305327, 1705172303553678726931, 52672608711829111519943, 1636296668756812403477839, 51088496012515356589705107

Offset: 0

Peter Bala, Oct 24 2022

Conjectures:
1) a(p - 1) == 3 (mod p^5) for all primes p >= 5.
2) a(p^r - 1) == a(p^(r-1) - 1) ( mod p^(3*r+3) ) for r >= 2 and for all primes p >= 5.
These are stronger supercongruences than those satisfied separately by the two types of Apéry numbers A005258 and A005259. Cf. A357567.

Peter Bala, Some conjectural supercongruences for the Apéry numbers.

Cf. A005258, A005259, A212334, A352655, A357567, A357568, A357569, A357957, A357958, A357959, A357960.

seq(add(5*binomial(n,k)^2*binomial(n+k,k)^2 - 2*binomial(n,k)^2* binomial(n+k,k), k = 0..n), n = 0..20);
# Alternatively:
a := n -> 5*hypergeom([-n, -n, 1 + n, 1 + n], [1, 1, 1], 1) - 2*hypergeom([1 + n, -n, -n], [1, 1], 1): seq(simplify(a(n)), n = 0..18); # Peter Luschny, Nov 01 2022

a(n) = 5*Sum_{k = 0..n} binomial(n,k)^2*binomial(n+k,k)^2 - 2*Sum_{k = 0..n} binomial(n,k)^2*binomial(n+k,k).
a(n*p^r - 1) == a(n*p^(r-1) - 1) ( mod p^(3*r) ) for positive integers n and r and for all primes p >= 5.
a(n) = 5*hypergeom([-n, -n, 1 + n, 1 + n], [1, 1, 1], 1) - 2*hypergeom([1 + n, -n, -n], [1, 1], 1). - Peter Luschny, Nov 01 2022

A357959 a(n) = 5A005259(n-1) + 2A005258(n).

Original entry on oeis.org

11, 63, 659, 9727, 187511, 4304943, 109312739, 2941124607, 82033399631, 2345394917563, 68306797052879, 2018580243252847, 60368874298729631, 1823588997226603663, 55558079041172790659, 1705174802761490321407, 52672634815976274443711, 1636296942340074307669443

Offset: 1

Peter Bala, Oct 25 2022

Conjectures:
1) a(p) == a(1) (mod p^5) for all primes p >= 5 (checked up to p = 271).
2) a(p^r) == a(p^(r-1)) ( mod p^(3*r+3) ) for r >= 2, and for all primes p >= 5.
These are stronger supercongruences than those satisfied separately by the two types of Apéry numbers A005258 and A005259. Cf. A357959.
There is also a product version of these conjectures:
3) the sequence {u(n): n >= 1} defined by u(n) = A005259(n-1)^5 * A005258(n)^6 also satisfies the congruences in 1) and 2) above. See A357960.

			Examples of supercongruences:
a(13) - a(1) = 60368874298729631 - 11 = (2^2)*3*5*(13^5)*131*20685869 == 0 (mod 13^5).
a(5^2) - a(5) = 51292638914356604042099497031437511 - 187511 = (2^4)*3*(5^10)* 37*72974432287*40526706713533 == 0 (mod 5^10).

Cf. A005258, A005259, A212334, A352655, A357567, A357956, A356957, A357958, A357960.

seq( add( 5*binomial(n-1,k)^2*binomial(n+k-1,k)^2 + 2*binomial(n,k)^2* binomial(n+k,k), k = 0..n ), n = 1..20);

a(n) = 5*Sum_{k = 0..n-1} binomial(n-1,k)^2*binomial(n+k-1,k)^2 + 2*Sum_{k = 0..n} binomial(n,k)^2*binomial(n+k,k).

a(n*p^r) == a(n*p^(r-1)) ( mod p^(3*r) ) for positive integers n and r and for all primes p >= 5.

A357507 a(n) = A005259(n)^5 * (A005259(n-1))^7.

Original entry on oeis.org

3125, 161958718203125, 69598400094777710760545478125, 514885225734532980507136994998009584838203125, 15708056924221066705174364772957342407662356116035885781253125, 1125221282019374727979322420623179115437017599670596496532725068048858642578125

Offset: 1

Peter Bala, Oct 01 2022

The Apéry numbers A(n) = A005259(n) satisfy the supercongruences A(p) == 5 (mod p^3) and A(p-1) == 1 (mod p^3) for all primes p >= 5 (see, for example, Straub, Introduction). It follows that a(p) == 3125 (mod p^3) for all primes p >= 5. We conjecture that, in fact, the stronger congruence a(p) == 3125 (mod p^5) holds for all primes p >= 3 (checked up to p = 251). Compare with the congruence A(p) + 7*A(p-1) == 12 (mod p^5) conjectured to hold for all primes p >= 5. See A212334.

Conjecture: a(p^r) == a(p^(r-1)) ( mod p^(3*r+3) ) for r >= 2 and all primes p >= 5. - Peter Bala, Oct 26 2022

A. Straub, Multivariate Apéry numbers and supercongruences of rational functions, arXiv:1401.0854 [math.NT] (2014).

Cf. A005259, A212334, A339946, A352655, A357506, A357508, A357509, A357567, A357568, A357569, A357956, A357957, A357958, A357959.

A005259 := n -> add(binomial(n,k)^2*binomial(n+k,k)^2, k = 0..n):
seq(A005259(n)^5 * A005259(n-1)^7, n = 1..10);

A357569 a(n) = binomial(3n,n)^2 - 27binomial(2*n,n).

Original entry on oeis.org

-26, -45, 63, 6516, 243135, 9011205, 344597148, 13520945736, 540917244351, 21966327267885, 902702921361813, 37456461969311736, 1566697064604277788, 65973795093057780936, 2794203818388994498200, 118933541228931589568016, 5084343623375039833670079, 218184481964802862563857685

Offset: 0

Peter Bala, Oct 21 2022

Conjectures:
1) a(p^r) == a(p^(r-1)) ( mod p^(3*r+3) ) for r >= 2 and all primes p >= 3.
These are stronger supercongruences than those satisfied separately by the sequences {binomial(2*n,n)} = A000984 and {binomial(3*n,n)} = A005809.
2) More generally, for k >= 1, the sequence {2*binomial(3*n,n)^k - k*(3^(k+1))*binomial(2*n,n): n >= 0} may satisfy the same supercongruences. This is the case k = 2. See A357509 for the case k = 1.

			Examples of supercongruences:
a(13) - a(1) = 65973795093057780936 + 45 = (3^2)*(13^5)*163*121122434651 == 0 (mod 13^5).
a(5^2) - a(5) = 2765555290416839473031163791322085183080 - 9011205 = (3^2)*(5^9)* 229*2333*6840413*74974087*574203805501 == 0 (mod 5^9).

C. Helou and G. Terjanian, On Wolstenholme’s theorem and its converse, J. Number Theory 128 (2008), 475-499.
Romeo Meštrović, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2011), arXiv preprint arXiv:1111.3057 [math.NT], 2011.

Cf. A000984, A005809, A188662, A357509, A357567, A357568, A357955.

seq(binomial(3*n,n)^2 - 27*binomial(2*n,n), n = 0..20);

Table[Binomial[3n,n]^2-27*Binomial[2n,n],{n,0,30}] (* Harvey P. Dale, Jun 12 2023 *)

a(n) = 3*A188662(n) - 27*A000984(n) = 3*A005809(n)^2 - 27*A000984(n).
a(k*p^r) == a(k*p^(r-1)) ( mod p^(3*r) ) for positive integers k and r and for all primes p >= 5 (see Meštrović, Section 6, equation 39).
a(p) == a(1) (mod p^5) for all primes p >= 7 (apply Helou and Terjanian, Section 3, Proposition 2).

A357958 a(n) = 5A005259(n) + 14A005258(n-1).

Original entry on oeis.org

39, 407, 7491, 167063, 4112539, 107461667, 2923006251, 81853622423, 2343591359499, 68288538877907, 2018394003648391, 60366962358086243, 1823569260750104179, 55557874330437332267, 1705172670555862322491, 52672612525369663916183

Offset: 1

Peter Bala, Oct 25 2022

Conjectures:
1) a(p) == a(1) (mod p^5) for all primes p >= 5 (checked up to p = 271).
2) a(p^r) == a(p^(r-1)) ( mod p^(3*r+3) ) for r >= 2 and for all primes p >= 3.
These are stronger supercongruences than those satisfied separately by the two types of Apéry numbers A005258 and A005259. Cf. A357959.
There is also a product version of these conjectures:
3) the sequence {u(n): n>= 1} defined by u(n) = A005259(n)^25 * A005258(n-1)^14 conjecturally satisfies the congruences in 1) and 2) above.

			Examples of supercongruences:
a(13) - a(1) = 1823569260750104179 - 39 = (2^2)*5*7*(13^5)*35081444357 == 0 (mod 13^5).
a(7^2) - a(7) = (2^3)*(7^9)* 10412078726049425470554760052126170543547100055154203726400782433 == 0 (mod 7^9).

Cf. A005258, A005259, A212334, A352655, A357567, A357956, A357957, A357959, A357960.

seq( add( 5*binomial(n,k)^2*binomial(n+k,k)^2 + 14*binomial(n-1,k)^2* binomial(n+k-1,k), k = 0..n ), n = 1..20);

a(n) = 5*Sum_{k = 0..n} binomial(n,k)^2*binomial(n+k,k)^2 + 14*Sum_{k = 0..n-1} binomial(n-1,k)^2*binomial(n+k-1,k).

a(p^r) == a(p^(r-1)) ( mod p^(3*r) ) for positive integer r and for all primes p >= 5.

A357957 a(n) = A005259(n)^5 - A005258(n)^2.

Original entry on oeis.org

0, 3116, 2073071232, 6299980938881516, 39141322964380888600000, 368495989505416178203682748116, 4552312485541626792249211584618373944, 68109360474242016374599574592870648425552876, 1174806832391451114413440151405736019461523615095744

Offset: 0

Peter Bala, Oct 24 2022

Conjectures:
1) a(p - 1) == 0 (mod p^5) for all primes p >= 5 (checked up to p = 271).
2) a(p^r - 1) == a(p^(r-1) - 1) ( mod p^(3*r+3) ) for r >= 2 and all primes p >= 5.
These are stronger supercongruences than those satisfied separately by the two types of Apéry numbers A005258 and A005259.
3) Put u(n) = A005259(n)^5 / A005258(n)^2. Then u(p^r - 1) == u(p^(r-1) - 1) ( mod p^(3*r+3) ) for r >= 2 and all primes p >= 5.

			a(7) = 4552312485541626792249211584618373944 = (2^3)*(3^3)*(7^5)*29*107* 404116272977592231282158029 == 0 (mod 7^5).

Cf. A005258, A005259, A212334, A352655, A357567, A357568, A357569, A357956, A357958, A357959, A357960.

seq(add(binomial(n,k)^2*binomial(n+k,k)^2, k = 0..n)^5 - add(binomial(n,k)^2*binomial(n+k,k), k = 0..n)^2, n = 0..20);
# Alternatively:
a := n -> hypergeom([-n, -n, 1 + n, 1 + n], [1, 1, 1], 1)^5 - hypergeom([1 + n, -n, -n], [1, 1], 1)^2: seq(simplify(a(n)), n=0..8); # Peter Luschny, Nov 01 2022

a(n) = ( Sum_{k = 0..n} binomial(n,k)^2*binomial(n+k,k)^2 )^5 - ( Sum_{k = 0..n} binomial(n,k)^2*binomial(n+k,k) )^2.
a(n*p^r - 1) == a(n*p^(r-1) - 1) ( mod p^(3*r) ) for positive integers n and r and for all primes p >= 5.
a(n) = hypergeom([-n, -n, 1 + n, 1 + n], [1, 1, 1], 1)^5 - hypergeom([1 + n, -n, -n], [1, 1], 1)^2. - Peter Luschny, Nov 01 2022

A357960 a(n) = A005259(n-1)^5 * A005258(n)^6.

Original entry on oeis.org

729, 147018378125, 20917910914764786689697, 24148107115850058575342740485778125, 79477722547796770983047586179643766765851375729, 492664048531500749211923278756418311980637289373757041378125, 4671227340507161302417161873394448514470099313382652883508175438056640625

Offset: 1

Peter Bala, Oct 25 2022

Conjectures:
1) a(p) == a(1) (mod p^5) for all primes p >= 3 (checked up to p = 271).
2) a(p^r) == a(p^(r-1)) ( mod p^(3*r+3) ) for r >= 2 and for all primes p >= 3. These are stronger supercongruences than those satisfied separately by the two types of Apéry numbers A005258 and A005259.

Cf. A005258, A005259, A212334, A352655, A357567, A357956, A357957, A357958, A357959.

seq( add(binomial(n-1,k)^2*binomial(n+k-1,k)^2, k = 0..n-1)^5 * add(binomial(n, k)^2*binomial(n+k,k), k = 0..n)^6, n = 1..20);

a(n) = ( Sum_{k = 0..n-1} binomial(n-1,k)^2*binomial(n+k-1,k)^2 )^5 * ( Sum_{k = 0..n} binomial(n,k)^2*binomial(n+k,k) )^6.

a(n*p^r) == a(n*p^(r-1)) ( mod p^(3*r) ) for positive integers n and r and for all primes p >= 5.

A357955 a(n) = 3binomial(4n,n) - 20binomial(3n,n) + 54binomial(2n,n).

Original entry on oeis.org

37, 60, 108, 60, -660, 60, 82404, 1411848, 17540460, 191318820, 1952058108, 19175376324, 184118073828, 1743153802320, 16359157606200, 152693295412560, 1420516291306860, 13190159377278324, 122358232382484420, 1134645084249344400, 10522118980232969340

Offset: 0

Peter Bala, Oct 22 2022

Conjectures:
1) a(p) == a(1) (mod p^7) for all primes p >= 3 except p = 7.
2) a(p^r) == a(p^(r-1)) ( mod p^(3*r+5) ) for r >= 2 and all primes p >= 3.
These are stronger supercongruences than those satisfied separately by the sequences {binomial(2*n,n)} = A000984, {binomial(3*n,n)} = A005809 and {binomial(4*n,n)} = A005810.
Conjecture 1) was proved by Aidagulov and Alekseyev; see the remarks following Corollary 2. - Peter Bala, Oct 29 2022

			Examples of supercongruences:
a(11) - a(1) = 19175376324 - 60 = (2^3)*3*(11^7)*41 == 0 (mod 11^7).
a(5^2) - a(5) = 726506045044361132812560 - 60 = (2^2)*3*(5^11)*41*30241552444123 == 0 (mod 5^11).

Paolo Xausa, Table of n, a(n) for n = 0..1000
R. R. Aidagulov and M. A. Alekseyev, On p-adic approximation of sums of binomial coefficients, Journal of Mathematical Sciences 233:5 (2018), 626-634; arXiv:1602.02632 [math.NT], 2018.
C. Helou and G. Terjanian, On Wolstenholme’s theorem and its converse, J. Number Theory 128 (2008), 475-499.
Romeo Meštrović, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2011), arXiv preprint arXiv:1111.3057 [math.NT], 2011.

Cf. A000984, A005809, A005810, A268590, A357509, A357567, A357568, A357569.

seq(3*binomial(4*n,n) - 20*binomial(3*n,n) + 54*binomial(2*n,n), n = 0..20);

A357955[n_] := 3*Binomial[4*n, n] - 20*Binomial[3*n, n] + 54*Binomial[2*n, n];
Array[A357955, 25, 0] (* Paolo Xausa, Jul 17 2024 *)

from math import comb
def A357955(n): return 54*comb(m:=n<<1,n)+3*comb(m<<1,n)-20*comb(m+n,n) # Chai Wah Wu, Oct 24 2022

a(n) = 3*A005810(n) - 20*A005809(n) + 54*A000984(n).
a(k*p^r) == a(k*p^(r-1)) ( mod p^(3*r) ) for positive integers k and r and for all primes p >= 5 (see Meštrović, Section 6, equation 39).
a(p) == a(1) (mod p^6) for all primes p >= 7 (apply Helou and Terjanian, Section 3, Proposition 2).
From Stefano Spezia, Jul 17 2024: (Start)
G.f.: 54/sqrt(1-4*x) - 40*cos(arccos(1-27*x/2)/6)/sqrt(4-27*x) + 3*hypergeom([1/4, 1/2, 3/4], [1/3, 2/3], 4^4*x/3^3).
E.g.f.: 54*exp(2*x)*BesselI(0, 2*x) - 20*hypergeom([1/3, 2/3], [1/2, 1], 27*x/4) + 3*hypergeom([1/4, 1/2, 3/4], [1/3, 2/3, 1], 4^4*x/3^3).
a(n) ~ exp(2*n*arctanh(229/283))*sqrt(6/(n*Pi)). (End)

cp's OEIS Frontend

A357568 a(n) = 9*binomial(2*n,n)^2 - 8*binomial(3*n,n).

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A357956 a(n) = 5*A005259(n) - 2*A005258(n).

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A357959 a(n) = 5*A005259(n-1) + 2*A005258(n).

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A357507 a(n) = A005259(n)^5 * (A005259(n-1))^7.

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A357569 a(n) = binomial(3*n,n)^2 - 27*binomial(2*n,n).

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A357958 a(n) = 5*A005259(n) + 14*A005258(n-1).

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A357957 a(n) = A005259(n)^5 - A005258(n)^2.

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A357960 a(n) = A005259(n-1)^5 * A005258(n)^6.

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A357568 a(n) = 9binomial(2n,n)^2 - 8binomial(3n,n).

A357956 a(n) = 5A005259(n) - 2A005258(n).

A357959 a(n) = 5A005259(n-1) + 2A005258(n).

A357569 a(n) = binomial(3n,n)^2 - 27binomial(2*n,n).

A357958 a(n) = 5A005259(n) + 14A005258(n-1).

A357955 a(n) = 3binomial(4n,n) - 20binomial(3n,n) + 54binomial(2n,n).