A357509 -id:A357509 - 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)

A357506 a(n) = A005258(n)^3 * A005258(n-1).

Original entry on oeis.org

27, 20577, 60353937, 287798988897, 1782634331587527, 13011500170881726987, 106321024671550496694837, 943479109706472533832704097, 8916177779855571182824077866307, 88547154924474394601268826256953077, 915376390434997094066775480671975209017

Offset: 1

Peter Bala, Oct 01 2022

The Apéry numbers B(n) = A005258(n) satisfy the supercongruences B(p) == 3 (mod p^3) and B(p-1) == 1 (mod p^3) for all primes p >= 5 (see, for example, Straub, Example 3.4). It follows that a(p) == 27 (mod p^3) for all primes p >= 5. We conjecture that, in fact, the stronger congruence a(p) == 27 (mod p^5) holds for all primes p >= 3 (checked up to p = 251). Compare with the congruence B(p) + B(p-1) == 4 (mod p^5) conjectured to hold for all primes p >= 5. See A352655.

Conjecture: for r >= 2, a(p^r) == a(p^(r-1)) ( mod p^(3*r+3) ) for all primes p >= 5. - Peter Bala, Oct 13 2022

			Example of a supercongruence:
a(7) - a(1) = 106321024671550496694837 - 27 = 2*(3^3)*5*(7^5)* 11*18143* 117398731273 == 0 (mod 7^5)

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

Cf. A005258, A212334, A339946, A352655, A357507, A357508, A357509.

A005258 := n -> add(binomial(n,k)^2*binomial(n+k,k), k = 0..n):
seq(A005258(n)^3*A005258(n-1), n = 1..20);

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).

A357671 a(n) = Sum_{k = 0..n} ( binomial(n+k-1,k) + binomial(n+k-1,k)^2 ).

Original entry on oeis.org

2, 4, 20, 166, 1812, 22504, 297362, 4067298, 56897300, 809019580, 11649254520, 169444978124, 2485270719570, 36707044807996, 545386321069862, 8144809732228666, 122177690210103060, 1839933274439787940, 27804610626798500372, 421476329345312885304, 6406685025104178888312

Offset: 0

Peter Bala, Oct 10 2022

Conjectures:
1) a(p) == 4 (mod p^5) for all primes p >= 7 (checked up to p = 499). Note that A000984(p) == 2 (mod p^3) and A333592(p) == 2 (mod p^3) for all primes p >= 5.
2) For r >= 2, and all primes p >= 5, a(p^r) == a(p^(r-1)) ( mod p^(3*r+3) ).
3) More generally, let m be a positive integer and set u(n) = 2*m*Sum_{k = 0..m*n} binomial(n+k-1,k) + (m + 1)*Sum_{k = 0..m*n} binomial(n+k-1,k)^2. Then the sequence {u(n)} satisfies the supercongruence u(p) == u(1) (mod p^5) for all primes p >= 7. This is the case m = 1. See A357673 for the case m = 2.
4) For r >= 2, and all primes p >= 5, u(p^r) == u(p^(r-1)) ( mod p^(3*r+3) ).

			Examples of supercongruences:
a(19) - a(1) = 421476329345312885304 - 4 = (2^2)*(5^2)*(19^5)*1913*2383*373393 == 0 (mod 19^5).
a(25) - a(5) = 5375188503768910125546897504 - 22504 = (2^3)*(5^10)*1858537* 37019662696111 == 0 (mod 5^10).

Cf. A000984, A333592, A357509, A357565, A357566, A357672, A357673, A357674.

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

a(n) = sum(k = 0, n, binomial(n+k-1,k) + binomial(n+k-1,k)^2); \\ Michel Marcus, Oct 24 2022

from math import comb
def A357671(n): return comb(n<<1,n)+sum(comb(n+k-1,k)**2 for k in range(n+1)) if n else 2 # Chai Wah Wu, Oct 28 2022

a(n) = A000984(n) + A333592(n).
a(p) == 4 (mod p^3) for all primes p >= 5.
a(n) ~ 2^(4*n) / (3*Pi*n).

A357673 a(n) = 4Sum_{k = 0..2n} binomial(n+k-1,k) + 3Sum_{k = 0..2n} binomial(n+k-1,k)^2.

Original entry on oeis.org

7, 21, 225, 5124, 162657, 5812521, 219004812, 8516056500, 338508840801, 13679415485805, 559978704877725, 23162632151271480, 966309241173439500, 40602415885424806824, 1716435895297948558812, 72941388509291664563124, 3113826813351114598588257, 133458673478315967012049245

Offset: 0

Peter Bala, Oct 11 2022

Conjectures:
1) a(p) == a(1) (mod p^5) for all primes p >= 3 (checked up to p = 499).
2) For r >= 2, and all primes p >= 3, a(p^r) == a(p^(r-1)) ( mod p^(3*r+3) ).
3) More generally, let m be a positive integer and set u(n) = 2*m*Sum_{k = 0..m*n} binomial(n+k-1,k) + (m + 1)*Sum_{k = 0..m*n} binomial(n+k-1,k)^2. Then the sequence {u(n)} satisfies the supercongruence u(p) == u(1) (mod p^5) for all primes p >= 7. This is the case m = 2. See A357673 for the case m = 1.
4) For r >= 2, and all primes p >= 5, u(p^r) == u(p^(r-1)) ( mod p^(3*r+3) ).

			Examples of supercongruences:
a(17) - a(1) = 133458673478315967012049245 - 21 = (2^3)*3*7*(17^5)*61*109*4441*86491*219071 == 0 (mod 17^5).
a(25) - a(5) = 1681058690656849873108154414589433546896 - 5812521 = 3*(5^9)*17*124471*39410141*65963867*52155532801 == 0 (mod 5^9).

Cf. A005809, A357509, A357565, A357566, A357671, A357672, A357674.

seq(add( 4*binomial(n+k-1,k) + 3*binomial(n+k-1,k)^2, k = 0..2*n ), n = 0..20);

Table[4 Sum[Binomial[n+k-1,k],{k,0,2n}]+3*Sum[Binomial[n+k-1,k]^2,{k,0,2n}],{n,0,20}] (* Harvey P. Dale, Oct 29 2022 *)

a(n) = 4*sum(k = 0, 2*n, binomial(n+k-1,k)) + 3*sum(k = 0, 2*n, binomial(n+k-1,k)^2); \\ Michel Marcus, Oct 24 2022

a(n) = 4*A005809(n) + 3*Sum_{k = 0..2*n} binomial(n+k-1,k)^2.

A357508 a(n) = binomial(4n,2n) - 2binomial(4n,n).

Original entry on oeis.org

-1, -2, 14, 484, 9230, 153748, 2434964, 37748520, 580043790, 8886848740, 136151207764, 2088760285456, 32108266614164, 494648505828904, 7637081136832840, 118158193386475984, 1831647087068431374, 28444051172077725444, 442429676097305612324

Offset: 0

Peter Bala, Oct 01 2022

Sun and Wan's supercongruence stated below apparently generalizes as follows:
Let m be an integer and k a positive integer. Define u(n) = binomial((m+2)*n,(k+1)*n) - binomial(m,k)*binomial((m+2)*n,n). We conjecture that u(n) == u(1) (mod p^5) for all primes p >= 7. [added 22 Oct 2022: the conjecture is true: apply Helou and Terjanian, Section 3, Proposition 2.]
Conjecture: for r >= 2, u(p^r) == u(p^(r-1)) ( mod p^(3*r+3) ) for all primes p >= 5. - Peter Bala, Oct 13 2022

C. Helou and G. Terjanian, On Wolstenholme’s theorem and its converse, J. Number Theory 128 (2008), 475-499.
Z.-W. Sun and D. Wan, On Fleck quotients, arXiv:math/0603462 [math.NT], 2006-2007.

Cf. A001448, A005810, A357509.

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

a(n) = A001448(n) - 2*A005810(n).

a(p) == -2 (mod p^5) for all primes p >= 7. (Sun and Wan, Corollary 1.5.)

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

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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A357506 a(n) = A005258(n)^3 * A005258(n-1).

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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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A357671 a(n) = Sum_{k = 0..n} ( binomial(n+k-1,k) + binomial(n+k-1,k)^2 ).

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A357673 a(n) = 4*Sum_{k = 0..2*n} binomial(n+k-1,k) + 3*Sum_{k = 0..2*n} binomial(n+k-1,k)^2.

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A357508 a(n) = binomial(4*n,2*n) - 2*binomial(4*n,n).

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

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

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

A357673 a(n) = 4Sum_{k = 0..2n} binomial(n+k-1,k) + 3Sum_{k = 0..2n} binomial(n+k-1,k)^2.

A357508 a(n) = binomial(4n,2n) - 2binomial(4n,n).

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