A212334 -id:A212334 - cp's OEIS frontend

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

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.

A208673 Number of words A(n,k), either empty or beginning with the first letter of the k-ary alphabet, where each letter of the alphabet occurs n times and letters of neighboring word positions are equal or neighbors in the alphabet; square array A(n,k), n>=0, k>=0, read by antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 5, 10, 1, 1, 1, 1, 9, 37, 35, 1, 1, 1, 1, 15, 163, 309, 126, 1, 1, 1, 1, 25, 640, 3593, 2751, 462, 1, 1, 1, 1, 41, 2503, 36095, 87501, 25493, 1716, 1, 1, 1, 1, 67, 9559, 362617, 2336376, 2266155, 242845, 6435, 1, 1

Offset: 0

Alois P. Heinz, Feb 29 2012

Also the number of (n*k-1)-step walks on k-dimensional cubic lattice from (1,0,...,0) to (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.

All rows are linear recurrences with constant coefficients and for n > 0 the order of the recurrence is bounded by 2*n-1. For n up to at least 20 this upper bound is exact. - Andrew Howroyd, Feb 22 2022

			A(0,0) = A(n,0) = A(0,k) = 1: the empty word.
A(2,3) = 5:
  +------+   +------+   +------+   +------+   +------+
  |aabbcc|   |aabcbc|   |aabccb|   |ababcc|   |abccba|
  +------+   +------+   +------+   +------+   +------+
  |122222|   |122222|   |122222|   |112222|   |111112|
  |001222|   |001122|   |001112|   |011222|   |011122|
  |000012|   |000112|   |000122|   |000012|   |001222|
  +------+   +------+   +------+   +------+   +------+
  |xx    |   |xx    |   |xx    |   |x x   |   |x    x|
  |  xx  |   |  x x |   |  x  x|   | x x  |   | x  x |
  |    xx|   |   x x|   |   xx |   |    xx|   |  xx  |
  +------+   +------+   +------+   +------+   +------+
Square array A(n,k) begins:
  1,  1,    1,     1,       1,         1,           1, ..
  1,  1,    1,     1,       1,         1,           1, ..
  1,  1,    3,     5,       9,        15,          25, ..
  1,  1,   10,    37,     163,       640,        2503, ..
  1,  1,   35,   309,    3593,     36095,      362617, ..
  1,  1,  126,  2751,   87501,   2336376,    62748001, ..
  1,  1,  462, 25493, 2266155, 164478048, 12085125703, ..

Andrew Howroyd, Table of n, a(n) for n = 0..1325

Columns k=0+1, 2-4 give: A000012, A088218, A208675, A212334.
Rows n=0+1, 2-3 give: A000012, A001595, A208674.
Main diagonal gives A351759.
Cf. A208879 (cyclic alphabet), A331562.

b:= proc(t, l) option remember; local n; n:= nops(l);
     `if`(n<2 or {0}={l[]}, 1,
     `if`(l[t]>0, b(t, [seq(l[i]-`if`(i=t, 1, 0), i=1..n)]), 0)+
     `if`(t0,
                  b(t+1, [seq(l[i]-`if`(i=t+1, 1, 0), i=1..n)]), 0)+
     `if`(t>1 and l[t-1]>0,
                  b(t-1, [seq(l[i]-`if`(i=t-1, 1, 0), i=1..n)]), 0))
    end:
A:= (n, k)-> `if`(n=0 or k=0, 1, b(1, [n-1, n$(k-1)])):
seq(seq(A(n, d-n), n=0..d), d=0..10);

b[t_, l_List] := b[t, l] = Module[{n = Length[l]}, If[n < 2 || {0} == Union[l], 1, If[l[[t]] > 0, b[t, Table[l[[i]] - If[i == t, 1, 0], {i, 1, n}]], 0] + If[t < n && l[[t + 1]] > 0, b[t + 1, Table[l[[i]] - If[i == t + 1, 1, 0], {i, 1, n}]], 0] + If[t > 1 && l[[t - 1]] > 0, b[t - 1, Table[l[[i]] - If[i == t - 1, 1, 0], {i, 1, n}]], 0]]]; A[n_, k_] := If[n == 0 || k == 0, 1, b[1, Join[{n - 1}, Array[n&, k - 1]]]]; Table[Table[A[n, d - n], {n, 0, d}], {d, 0, 10}] // Flatten (* Jean-François Alcover, Dec 27 2013, translated from Maple *)

F(u)={my(n=#u); sum(k=1, n,u[k]*binomial(n-1,k-1))}
step(u, c)={my(n=#u); vector(n, k, sum(i=max(0, 2*k-c-n), k-1, sum(j=0, n-2*k+i+c, u[k-i+j]*binomial(n-1, 2*k-1-c-i+j)*binomial(k-1, k-i-1)*binomial(k-i+j-c, j) ))) }
R(n,k)={my(r=vector(n+1), u=vector(k), v=vector(k)); u[1]=v[1]=r[1]=r[2]=1; for(n=3, #r, u=step(u,1); v=step(v,0)+u; r[n]=F(v)); r}
T(n,k)={if(n==0||k==0, 1, R(k,n)[1+k])} \\ Andrew Howroyd, Feb 22 2022

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.

A361712 a(n) = Sum_{k = 0..n-1} binomial(n,k)^2binomial(n+k,k)binomial(n+k-1,k).

Original entry on oeis.org

0, 1, 25, 649, 16921, 448751, 12160177, 336745053, 9513822745, 273585035755, 7988828082775, 236367018090017, 7072779699975601, 213701611408357567, 6511338458568750853, 199850727914988936149, 6173376842290368719385, 191776434791965521115235, 5987554996434696230487955

Offset: 0

Peter Bala, Mar 21 2023

Conjecture 1: the supercongruence a(p) == a(1) (mod p^5) holds for all primes p >= 7 (checked up to p = 199).
Conjecture 2: for r >= 2, the supercongruence a(p^r) == a(p^(r-1)) (mod p^(3*r+3)) holds for all primes p >= 5.
Compare with the Apéry numbers A005259(n) = Sum_{k = 0..n} binomial(n,k)^2 * binomial(n+k,k)^2, which satisfy the weaker supercongruences A005259(p^r) == A005259(p^(r-1)) (mod p^(3*r)) for all primes p >= 5.

			a(7) - a(1) = (2^2)*(7^5)*5009 == 0 (mod 7^5)
a(11) - a(1) = (2^5)*(11^5)*45864163 == 0 (mod 11^5)
a(7^2) - a(7) = (2*3)*(7^9)*377052719*240136524699189343838527* 17965610580703155723668147409587 == 0 (mod 7^9)

Paolo Xausa, Table of n, a(n) for n = 0..650
Peter Bala, Recurrence equation for A361712

Cf. A005259, A212334, A361713, A361714, A361715, A361717.

Cf. A361877, A361878.

seq(add(binomial(n,k)^2*binomial(n+k,k)*binomial(n+k-1,k), k = 0..n-1), n = 0..25);
# Alternative:
A361712 := n -> hypergeom([-n, -n, n, n + 1], [1, 1, 1], 1) - binomial(2*n, n)*binomial(2*n-1, n): seq(simplify(A361712(n)), n = 0..18); # Peter Luschny, Mar 27 2023

A361712[n_] := HypergeometricPFQ[{-n, -n, n, n+1}, {1, 1, 1}, 1] - Binomial[2*n, n]*Binomial[2*n-1, n]; Array[A361712, 20, 0] (* Paolo Xausa, Jul 10 2024 *)

a(n) = (1/12)*(7*A005259(n) + A005259(n-1)) - (1/2)*binomial(2*n,n)^2.
a(n) ~ 2^(1/4)*(1 + sqrt(2))^(4*n+1)/(4*Pi^(3/2)*n^(3/2)).
a(n) = hypergeom([-n, -n, n, n + 1], [1, 1, 1], 1) - binomial(2*n, n)*binomial(2*n - 1, n) = A361878(n) - A361877(n). - Peter Luschny, Mar 27 2023

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

cp's OEIS Frontend

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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A208673 Number of words A(n,k), either empty or beginning with the first letter of the k-ary alphabet, where each letter of the alphabet occurs n times and letters of neighboring word positions are equal or neighbors in the alphabet; square array A(n,k), n>=0, k>=0, read by antidiagonals.

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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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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).

A361712 a(n) = Sum_{k = 0..n-1} binomial(n,k)^2binomial(n+k,k)binomial(n+k-1,k).