A352655 -id:A352655 - cp's OEIS frontend

A212334 Number of words, either empty or beginning with the first letter of the 4-ary alphabet, where each letter of the alphabet occurs n times and letters of neighboring word positions are equal or neighbors in the alphabet.

1, 1, 9, 163, 3593, 87501, 2266155, 61211095, 1704838665, 48605519665, 1411522695509, 41606511550803, 1241591466423467, 37435593955828069, 1138713916992923679, 34901292375152457663, 1076813644170756916745, 33416749492077957930105, 1042376218505671236116985

Offset: 0

Alois P. Heinz, Aug 07 2012

nonn

Also the number of (4*n-1)-step walks on 4-dimensional cubic lattice from (1,0,0,0) to (n,n,n,n) with positive unit steps in all dimensions such that the absolute difference of the dimension indices used in consecutive steps is <= 1.
It appears that for primes p >= 5, a(p) == 1 (mod p^5). Cf. A352655. - Peter Bala, Dec 12 2021
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

Alois P. Heinz, Table of n, a(n) for n = 0..656

Column k = 4 of A208673.

Cf. A005259, A352655.

a:= proc(n) option remember; `if`(n<3, [1, 1, 9][n+1],
      ((26682*n^4 -102687*n^3 +149385*n^2 -109413*n +31101) *a(n-1)
      +(-161058*n^4 +1392915*n^3 -4418826*n^2 +6030348*n -2931516) *a(n-2)
      +(4718*n^4 -47957*n^3 +176841*n^2 -275751*n +148365) *a(n-3)) /
      (n^3 *(646*n -1057)))
    end:
seq(a(n), n=0..30);

a[n_] := a[n] = If[n < 3, {1, 1, 9}[[n + 1]], ((26682 n^4 - 102687 n^3 + 149385 n^2 - 109413 n + 31101) a[n-1] + (-161058 n^4 + 1392915 n^3 - 4418826 n^2 + 6030348 n - 2931516)a[n-2] + (4718 n^4 - 47957 n^3 + 176841 n^2 - 275751 n + 148365)a[n-3])/(n^3 (646 n - 1057))];
a /@ Range[0, 30] (* Jean-François Alcover, May 14 2020, after Maple *)

a(n) ~ (1 + sqrt(2))^(4*n-1) / (2^(7/4) * (Pi*n)^(3/2)). - Vaclav Kotesovec, Aug 13 2013, simplified Apr 06 2022
From Peter Bala, Apr 17 2022: (Start)
a(n) = (1/12)*(A005259(n) + 7*A005259(n-1)) for n >= 1.
The supercongruences a(n*p^k) == a(n*p^(k-1)) (mod p^(3*k)) hold for all primes p >= 5 and positive integers n and k.
a(n) = (1/3)*Sum_{k = 0..n} binomial(n,k)^2*binomial(n + k,k)^2*(2*n^2 - 3*k*n + 2*k^2)/(n + k)^2.
(24*n^3 - 102*n^2 + 148*n - 73)*n^3*a(n) = 4*(204*n^6 - 1173*n^5 + 2668*n^4 - 3065*n^3 + 1905*n^2 - 634*n + 86)*a(n-1) - (24*n^3 - 30*n^2 + 16*n-3)*(n - 2)^3*a(n-2) with a(0) = a(1) = 1. (End)
a(n) = Sum_{k=0..n-1} binomial(n,k)*binomial(n-1,k)*binomial(n+k-1,k)^2 for n>=1. - Peter Bala, Mar 22 2023

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);

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);

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

A212334 Number of words, either empty or beginning with the first letter of the 4-ary alphabet, where each letter of the alphabet occurs n times and letters of neighboring word positions are equal or neighbors in the alphabet.

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

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Maple

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

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Maple

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

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Maple

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

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Maple

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

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Maple

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

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Maple

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

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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).