A357569 -id:A357569 - cp's OEIS frontend

A357567 a(n) = 5A005259(n) - 14A005258(n).

-9, -17, 99, 5167, 147491, 3937483, 105834699, 2907476527, 81702447651, 2342097382483, 68273597307599, 2018243113678027, 60365426282638091, 1823553517258576723, 55557712038989195099, 1705170989220937925167, 52672595030914982754851, 1636296525812843554700323

Offset: 0

Peter Bala, Oct 19 2022

Conjectures:
1) a(p) == a(1) (mod p^5) for all primes p >= 5.
2) For r >= 2, a(p^r) == a(p^(r-1)) ( mod p^(3*r+3) ) for all primes p >= 5.
These are stronger supercongruences than those satisfied separately by the two types of Apéry numbers A005258 and A005259.
From Peter Bala, Oct 25 2022: (Start)
Additional conjectures:
3) the sequence {u(n): n>= 1} defined by u(n) = (3^42)*A005259(n)^25 - (5^25)* A005258(n)^42 also satisfies the congruences in 1) and 2) above.
4) u(n) == 0 (mod n^5) for integer n of the form 3^i*5^j (see A003593). (End)

			a(11) - a(1) = 2018243113678027 + 17 = (2^2)*(3^2)*(11^5)*17*20476637 == 0 (mod 11^5).

Cf. A005258, A005259, A212334, A352655, A357506, A357507, A357508, A357509, A357568, A357569, A357956, A357957, A357958, A357959, A357960.

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

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

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

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

Original entry on oeis.org

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

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);

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

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

A357567 a(n) = 5*A005259(n) - 14*A005258(n).

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

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

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

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A357567 a(n) = 5A005259(n) - 14A005258(n).

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

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

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