A005258 -id:A005258 - cp's OEIS frontend

A291275 Primes p such that p does not divide any term of the Apéry-like sequence A005258.

2, 5, 13, 17, 29, 37, 41, 61, 73, 89, 101, 109, 137, 149, 173, 181, 197, 229, 233, 269, 277, 313, 337, 349, 353, 373, 397, 401, 409, 433, 457, 461, 541, 557, 601, 613, 641, 661, 673, 677, 701, 709, 733, 761, 769, 797, 821, 829, 853, 857, 877, 929, 941, 977

Offset: 1

N. J. A. Sloane, Aug 21 2017

nonn

Amita Malik and Armin Straub, Mathematica notebook for generating A133370 and A260793, A291275-A291284
Amita Malik and Armin Straub, Lists of all primes up to 10000 in A133370 and A260793, A291275-A291284, together with Mathematica code.
Amita Malik and Armin Straub, Divisibility properties of sporadic Apéry-like numbers, Research in Number Theory, 2016, 2:5.

For primes that do not divide the terms of the sequences A000172, A005258, A002893, A081085, A006077, A093388, A125143, A229111, A002895, A290575, A290576, A005259 see A260793, A291275-A291284 and A133370 respectively.

maxPrime = 977;
maxPi = PrimePi @ maxPrime;
okQ[p_] := AllTrue[Range[3 maxPi (* coeff 3 is empirical *)], GCD[HypergeometricPFQ[{# + 1, -#, -#}, {1, 1}, 1], p] == 1&];
Select[Prime[Range[maxPi]], okQ] (* Jean-François Alcover, Jan 13 2020 *)

A362722 a(n) = [x^n] ( E(x)/E(-x) )^n where E(x) = exp( Sum_{k >= 1} A005258(k)*x^k/k ).

Original entry on oeis.org

1, 6, 72, 1266, 23232, 445506, 8740728, 174366114, 3519799296, 71696570010, 1470795168072, 30344633110710, 628994746308288, 13089254107521234, 273292588355096760, 5722454505166750266, 120119862431845048320, 2526922404360157374738, 53260275108329790626952

Offset: 0

Peter Bala, May 01 2023

It is known that the sequence of Apéry numbers A005258 satisfies the Gauss congruences A005258(n*p^r) == A005258(n*p^(r-1)) (mod p^r) for all primes p and positive integers n and r.
One consequence is that the power series expansion of E(x) = exp( Sum_{k
>= 1} A005258(k)*x^k/k ) = 1 + 3*x + 14*x^2 + 82*x^3 + 551*x^4 + ... has integer coefficients (see, for example, Beukers, Proposition, p. 143). Therefore, the power series expansion of E(x)/E(-x) also has integer coefficients and so a(n) = [x^n] ( E(x)/E(-x) )^n is an integer.
In fact, the Apéry numbers satisfy stronger congruences than the Gauss congruences known as supercongruences: A005258(n*p^r) == A005258(n*p^(r-1)) (mod p^(3*r)) for all primes p >= 5 and positive integers n and r (see Straub, Section 1).
We conjecture below that {a(n)} satisfies supercongruences similar to (but weaker than) the above supercongruences satisfied by the Apéry numbers.

F. Beukers, Some congruences for the Apery numbers, Journal of Number Theory, Vol. 21, Issue 2, Oct. 1985, pp. 141-155. local copy
Armin Straub, Multivariate Apéry numbers and supercongruences of rational functions, Algebra & Number Theory, Vol. 8, No. 8 (2014), pp. 1985-2008; arXiv preprint, arXiv:1401.0854 [math.NT], 2014.

Cf: A005258, A362723 - A362733.

A005258 := proc(n) add(binomial(n, k)^2*binomial(n+k,k), k = 0..n) end proc:
E(n,x) := series(exp(n*add(2*A005258(2*k+1)*x^(2*k+1)/(2*k+1), k = 0..10)), x, 21):
seq(coeftayl(E(n,x), x = 0, n), n = 0..20);

a(n) = [x^n] exp( Sum_{k >= 1} n*( 2*A005258(2*k+1)*x^(2*k+1) )/(2*k+1) ).
Conjectures:
1) the supercongruence a(p^r) == a(p^(r-1)) (mod p^(2*r+1)) holds for all primes p >= 5.
2) for n >= 2, a(n*p) == a(n) (mod p^2) holds for all primes p >= 3.
3) for r >= 2, the supercongruence a(n*p^r) == a(n*p^(r-1)) (mod p^(2*r)) holds for all primes p >= 3 and n >= 1.

A352655 a(n) = (1/2)*(A005258(n) + A005258(n-1)).

Original entry on oeis.org

2, 11, 83, 699, 6252, 58106, 554633, 5399099, 53356322, 533627511, 5388927513, 54859837434, 562267554552, 5796123147756, 60047675871333, 624801952898619, 6526036790730942, 68395815476047901, 718992874207884953, 7578808590187108199

Offset: 1

Peter Bala, Apr 17 2022

The Apéry numbers A005258 satisfy the supercongruences A005258(p) == 3 (mod p^3) and A005258(p-1) == 1 (mod p^3) for primes p >= 5. It easily follows that a(p) == 2 (mod p^3) for primes p >= 3. We conjecture that the stronger supercongruences a(p) == 2 (mod p^5) hold for primes p >= 5. See A212334 for the corresponding conjecture for the Apéry numbers A005259.

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

			Examples of superconguences:
a(5) - 2 = 6252 - 2 = 2*(5^5) == 0 (mod 5^5).
a(7) - 2 = 554633 - 2 = 3*(7^5)*11 == 0 (mod 7^5).
a(11) - 2 = 5388927513 - 2 = (11^5)*33461 == 0 (mod 11^5).

Cf. A005258, A103882, A108628, A208675, A212334, A352654.

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

f(n) = sum(k=0, n, binomial(n, k)^2 * binomial(n+k, k)); \\ A005258
a(n) = (f(n) + f(n-1))/2; \\ Michel Marcus, Apr 20 2022

a(n) = (1/2)*Sum_{k = 0..n} (2*n^2 - k*n + k^2)/(n*(n + k)) * binomial(n,k)^2 * binomial(n + k,k).
a(n) = (1/2)*Sum_{k = 0..n-1} (4*n + k)*(n - k)/(n*(n + k)) * binomial(n,k)^2* binomial(n + k,k) for n >= 1.
a(n) = (1/2)*(A108628(n-1) + 3*A208675(n)) for n >= 1.
a(n) = (1/2)*(2*A103882(n) - A352654(n)).
a(n) ~ 5^(3/4)*(13 + 5*sqrt(5))/(20*sqrt(22 + 10*sqrt(5))*Pi*n) * ((11 + 5*sqrt(5))/2)^n.
(11*n^2 - 31*n + 22)*n^2*a(n) = (121*n^4 - 462*n^3 + 607*n^2 - 322*n + 64)*a(n-1) + (11*n^2 - 9*n + 2)*(n - 2)^2*a(n-2) with a(1) = 2 and a(2) = 11.
The g.f. A(x) = 2*x + 11*x^2 + 83*x^3 + ... satisfies the differential equation
(x^5 + 13*x^4 + 22*x^3 + 9*x^2 - x)*A''(x) + (x^4 + 4*x^3 + 26*x^2 + 22*x - 1)*A'(x) + (2*x^2 - 16*x + 4)*A(x) + x^2 - 8*x + 2 = 0, with A(0) = 2 and A'(0) = 11.

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

Original entry on oeis.org

-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.

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.

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);

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.

cp's OEIS Frontend

A291275 Primes p such that p does not divide any term of the Apéry-like sequence A005258.

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A362722 a(n) = [x^n] ( E(x)/E(-x) )^n where E(x) = exp( Sum_{k >= 1} A005258(k)*x^k/k ).

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A352655 a(n) = (1/2)*(A005258(n) + A005258(n-1)).

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A357567 a(n) = 5*A005259(n) - 14*A005258(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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A357506 a(n) = A005258(n)^3 * A005258(n-1).

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Maple

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

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

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

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

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