A005259 -id:A005259 - cp's OEIS frontend

A357960 a(n) = A005259(n-1)^5 * A005258(n)^6.

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.

A267220 Expansion of exp( Sum_{n >= 1} A005259(n)*x^n/n ).

Original entry on oeis.org

1, 5, 49, 685, 11807, 232771, 5031415, 116222699, 2822056474, 71230971002, 1854334597966, 49503739725470, 1349449053997654, 37438177610268014, 1054371288632733022, 30081609844254013942, 867990837575171240525, 25295504900322451251793

Offset: 0

Peter Bala, Jan 12 2016

The expansion of exp( Sum_{n >= 1} A005259(n-1)*x^n/n ) may also have only integer coefficients. See the Example section.

This is proved in Beukers, p. 143. - Peter Bala, Mar 06 2020

			exp( Sum_{n >= 1} A005259(n-1)*x^n/n ) = 1 + 3*x + 27*x^2 + 390*x^3 + .... The coefficient list begins [1, 3, 27, 390, 7038, 144550, 3232294, 76768575, 1907248655, 49067872253, 1298071849821, 35131589549434, 969031280176162, 27161049933788970, 771837331889465586, 22196147941000214583, 644991902540171273463, ...].
The Apéry number sequence A005259 begins [1, 5, 73, 1445, 33001, 819005, ...]. Let F(x) = x/( series reversion of x*A(x) ) = 1 + 5*x + 24*x^2 + 200*x^3 + 2430*x^4 + ....
Truncation of F(x)^n:
  F(x)^0: 1
  F(x)^1: 1 + 5*x
  F(x)^2: 1 + 10*x + 73*x^2
  F(x)^3: 1 + 15*x + 147*x^2 + 1445*x^3
  F(x)^4: 1 + 20*x + 246*x^2 + 2740*x^3 + 33001*x^4
The coefficient array
      1
      5    1
     73   10   1
   1445  147  15  1
  33001 2740 246 20 1
  ...
has the Apéry numbers as the first column.
It is a Riordan array belonging to the hitting-time subgroup of the Riordan group - see Peart and Woan.

F. Beukers, Some congruences for the Apery numbers, Journal of Number Theory, Vol. 21, Issue 2, Oct. 1985, pp. 141-155.
P. Peart and W.-J. Woan, A divisibility property for a subgroup of Riordan matrices, Discrete Applied Mathematics, Vol. 98, Issue 3, Jan 2000, 255-263.

Cf. A005259, A267219.

#A267220
#define the Apéry numbers
A005259 := proc (n) option remember; if n = 0 then 1 elif n = 1 then 5 else ((34*(n-1)^3+51*(n-1)^2+27*n-22)*A005259(n-1)-(n-1)^3*A005259(n-2))/n^3 end if; end proc:
exp(add(A005259(n)*x^n/n, n = 1 .. 17)):
seq(coeftayl(%, x = 0, n), n = 0 .. 17);

n*a(n) = Sum_{k = 0..n-1} A005259(n-k)*a(k).
O.g.f.: A(x) = exp( Sum_{n >= 1} A005259(n)*x^n/n ) = 1 + 5*x + 49*x^2 + 685*x^3 + 11807*x^4 + 232771*x^5 + 5031415*x^6 + ....
The o.g.f. A(x) satisfies 1 + x*d/dx(log(A(x))) = Sum_{n >= 0} A005259(n)*x^n.
Let F(x) = x/( series reversion of x*A(x) ) = 1 + 5*x + 24*x^2 + 200*x^3 + 2430*x^4 + 36096*x^5 + 605620*x^6 + 11024496*x^7 + 212758245*x^8 + .... Then A005259(n) = [x^n]( F(x)^n ). See the example section.
From Peter Bala, Oct 17 2024: (Start)
For integer m, define a sequence {u_m(n) : n >= 0} by u_m(n) = [x^n] A(x)^(m*n). Conjecture: the supercongruences u_m(n*p^r) == u_m(n*p^(r-1)) (mod p^(2*r)) hold for all primes p >= 5 and positive integers n and r.
For integer m, define a sequence {v_m(n) : n >= 0} by v_m(n) = [x^n] F(x)^(m*n). Conjecture: the supercongruences v_m(n*p^r) == v_m(n*p^(r-1)) (mod p^(2*r)) hold for all primes p >= 5 and positive integers n and r. (End)

A379153 The binary weights of the Apéry numbers (A005259).

Original entry on oeis.org

1, 2, 3, 6, 6, 14, 15, 15, 20, 19, 23, 23, 27, 34, 35, 44, 40, 36, 40, 44, 41, 48, 52, 62, 64, 66, 57, 66, 72, 79, 71, 75, 77, 78, 79, 78, 88, 86, 92, 100, 103, 103, 92, 116, 96, 116, 117, 113, 129, 117, 123, 128, 123, 126, 130, 133, 129, 142, 147, 134, 135, 148

Offset: 0

Amiram Eldar, Dec 17 2024

Amiram Eldar, Table of n, a(n) for n = 0..10000
Arnold Knopfmacher and Florian Luca, Digit sums of binomial sums, Journal of Number Theory, Vol. 132, No. 2 (2012), pp. 324-331.
Florian Luca and Igor E. Shparlinski, On the g-ary expansions of Apéry, Motzkin, Schröder and other combinatorial numbers, Annals of Combinatorics, Vol. 14 (2010), pp. 507-524.

Cf. A000120, A005259.

Similar sequences: A011373, A079584, A082481, A379151, A379152.

a[n_] := DigitCount[Sum[(Binomial[n, k] * Binomial[n+k, k])^2, {k, 0, n}], 2, 1]; Array[a, 100, 0]

a(n) = hammingweight(sum(k=0, n, (binomial(n, k)*binomial(n+k, k))^2));

a(n) = A000120(A005259(n)).
a(n) > c * (log(n)/log(log(n)))^(1/4) holds on a set of n of asymptotic density 1, where c > 0 is a constant (Luca and Shparlinski, 2010).
a(n) > c * log(n)/log(log(n)) holds on a set of n of asymptotic density 1, where c > 0 is a constant (Knopfmacher and Luca, 2012).
Conjecture: Limit_{m->oo} (1/m^2) * Sum_{k=1..m} a(k) = log(sqrt(2) + 1)/log(2) = 1.2715533... (Knopfmacher and Luca, 2012).

A276323 a(n) = (binomial(2 * prime(n + 3), prime(n + 3)) * A005259(prime(n + 3) - 1) - 2)/prime(n + 3)^5 for n >= 1.

Original entry on oeis.org

4382314, 59821998476834, 338197165389273486, 17314015796594772560245514, 145853326344012138627669357202, 12936469013977571458378002436843685186, 15931675838688077485749893663903436780403973163302

Offset: 1

Seiichi Manyama, Aug 30 2016

nonn

Let p be a prime > 5. Binomial(2 * p, p) * A005259(p - 1) == 2 (mod p^5). So a(n) is an integer.

			a(1) = (binomial(14, 7) * A005259(6) - 2)/7^5 = (3432 * 21460825 - 2)/7^5 = 4382314.

Seiichi Manyama, Table of n, a(n) for n = 1..88
Julian Rosen, Periods, supercongruences, and their motivic lifts, arXiv:1608.06864 [math.NT], 2016.

Cf. A000984, A005259.

Table[(Binomial[2 Prime[n + 3], Prime[n + 3]] Sum[(Binomial[#, k] Binomial[# + k, k])^2, {k, 0, #}] &[Prime[n + 3] - 1] - 2)/Prime[n + 3]^5, {n, 7}] (* Michael De Vlieger, Aug 30 2016 *)

require 'prime'
def C(n, r)
  r = [r, n - r].min
  return 1 if r == 0
  return n if r == 1
  numerator = (n - r + 1..n).to_a
  denominator = (1..r).to_a
  (2..r).each{|p|
    pivot = denominator[p - 1]
    if pivot > 1
      offset = (n - r) % p
      (p - 1).step(r - 1, p){|k|
        numerator[k - offset] /= pivot
        denominator[k] /= pivot
      }
    end
  }
  result = 1
  (0..r - 1).each{|k|
    result *= numerator[k] if numerator[k] > 1
  }
  return result
end
def A005259(n)
  i = 0
  a, b = 1, 5
  ary = [1]
  while i < n
    i += 1
    a, b = b, ((((34 * i + 51) * i + 27) * i + 5) * b - i ** 3 * a) / (i + 1) ** 3
    ary << a
  end
  ary
end
def A276323(n)
  p_ary = Prime.take(n + 3)[3..-1]
  a = A005259(p_ary[-1] - 1)
  ary = []
  p_ary.each{|i|
    j = C(2 * i, i) * a[i - 1] - 2
    break if j % i ** 5 > 0
    ary << j / i ** 5
  }
  ary
end

A289277 a(n) = A005259(n) mod 2*n+1.

Original entry on oeis.org

0, 2, 3, 3, 7, 0, 9, 5, 16, 6, 1, 13, 4, 26, 24, 26, 22, 30, 23, 32, 7, 9, 43, 11, 37, 29, 23, 0, 49, 40, 1, 44, 20, 54, 19, 18, 8, 20, 22, 55, 4, 70, 80, 62, 2, 31, 37, 20, 7, 44, 51, 62, 64, 76, 77, 41, 75, 75, 115, 68, 0, 35, 42, 11, 88, 59, 101, 35, 119, 11

Offset: 0

Seiichi Manyama, Jul 01 2017

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..10000
F. Beukers, Another congruence for the Apéry numbers, J. Number Theory 25 (1987), no. 2, 201-210.
Eric Weisstein's World of Mathematics, Apéry Number

Cf. A005259, A030211, A289278.

Table[Mod[Sum[(Binomial[n, k] Binomial[n + k, k])^2, {k, 0, n}], 2n + 1], {n, 0, 100}] (* Indranil Ghosh, Jul 01 2017 *)

a(n) = sum(k=0, n, (binomial(n, k)*binomial(n+k, k))^2) % (2*n+1); \\ Michel Marcus, Jul 01 2017

If m = 2*n + 1 is a prime, a(n) = A030211(n) mod m.

A289278 a(n) = A005259(n) mod (2*n+1)^2.

Original entry on oeis.org

0, 5, 23, 24, 34, 77, 22, 140, 50, 44, 169, 473, 354, 539, 198, 801, 385, 135, 1207, 617, 1483, 52, 2023, 528, 723, 2273, 2567, 1265, 1303, 2813, 550, 233, 1775, 188, 2365, 728, 154, 1520, 4180, 5585, 571, 236, 3650, 2672, 714, 4581, 4966, 2490, 8931, 4796, 1566

Offset: 0

Seiichi Manyama, Jul 01 2017

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..10000
Eric Weisstein's World of Mathematics, Apéry Number

Cf. A005259, A030211, A289277.

Table[Mod[Sum[(Binomial[n, k] Binomial[n + k, k])^2, {k, 0, n}], (2n + 1)^2], {n, 0, 100}] (* Indranil Ghosh, Jul 01 2017 *)

a(n) = sum(k=0, n, (binomial(n, k)*binomial(n+k, k))^2) % (2*n+1)^2; \\ Michel Marcus, Jul 01 2017

If m = 2*n + 1 is a prime, a(n) = A030211(n) mod m^2.

A362723 a(n) = [x^n] ( E(x)/E(-x) )^n where E(x)= exp( Sum_{k >= 1} A005259(k)*x^k/k ).

Original entry on oeis.org

1, 10, 200, 7390, 260800, 10263010, 407520920, 16758685030, 697767370240, 29525605934410, 1261570539980200, 54419751094210270, 2364396136291654720, 103393259758470870770, 4545671563318715532280, 200804420082143353690390, 8907295723280072012247040, 396570344897237949249382010

Offset: 0

Peter Bala, May 01 2023

It is known that the sequence of Apéry numbers A005259 satisfies the Gauss congruences A005259(n*p^r) == A005259(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} A005259(k)*x^k/k ) = 1 + 5*x + 49*x^2 + 685*x^3 + 11807*x^4 + ... has integer coefficients. See A267220. For a proof 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: A005259(n*p^r) == A005259(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.

Frits 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. A005259, A267220, A362722 - A362733.

A005259 := proc(n) add(binomial(n, k)^2*binomial(n+k,k)^2, k = 0..n) end;
E(n,x) := series(exp(n*add(2*A005259(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*A005259(2*k+1)*x^(2*k+1) )/(2*k+1) ).
Conjectures:
1) the supercongruence a(p) == a(1) (mod p^3) holds for all primes p >= 5 (checked up to p = 101).
2) for n >= 2, a(n*p) == a(n) (mod p^2) holds for all primes p >= 5.
3) for n >= 1, r >= 2, the supercongruence a(n*p^r) == a(n*p^(r-1)) (mod p^(2*r)) holds for all primes p >= 5.

A228143 Determinant of the (n+1) X (n+1) Hankel-type matrix with (i,j)-entry equal to A005259(i+j) for all i,j = 0,...,n.

Original entry on oeis.org

1, 48, 161856, 39002646528, 674708032182398976, 839431510934341028210638848, 75178263784150214825106859877233852416, 484905075185415831301477770434885768003422223597568, 225327830550164300895512117291590826401931052058453494726924435456, 7544971365077550026405694467600069733983243666195122776655161969325034606646263808

Offset: 0

Zhi-Wei Sun, Aug 14 2013

Conjecture: a(n)/24^n is always a positive integer. Similarly, if b(n) denotes the (n+1) X (n+1) Hankel-type determinant with (i,j)-entry equal to A005258(i+j) for all i,j = 0,...,n, then b(n)/10^n is always a positive integer; also, if p is a prime with floor(p/10) odd and p is not congruent to 31 or 39 modulo 40, then p divides b((p-1)/2).

Conjecture: if A(x) = 1 + 48*x + 161856*x^2 + ... denotes the o.g.f. then A(x/3)^(1/8) has integer coefficients (checked up to x^30). - Peter Bala, Apr 22 2018

			a(0) = 1 since A005259(0+0) = 1.
A(x/3)^(1/8) = 1 + 2*x + 2234*x^2 + 180536476*x^3 + 1041213553880806*x^4 + 431806318205326490858140*x^5 + 12890648790962619413782473229673892*x^6 + 27715196341006992690056202634389754569453086008*x^7 + 4292939920556011562306504817069205738464230629574745210785030*x^8 + 47915532217380103151430239883031701095737468980424637791531495548671526291244*x^9 + .... - _Peter Bala_, Apr 22 2018

Cf. A005258, A005259, A225776.

A[n_]:=Sum[Binomial[n,k]^2*Binomial[n+k,k]^2,{k,0,n}]; a[n_]:=Det[Table[A[i+j],{i,0,n},{j,0,n}]]; Table[a[n],{n,0,10}]

A289289 a(n) = A005259(n) mod (n+1)^3.

Original entry on oeis.org

0, 5, 19, 37, 1, 149, 1, 165, 559, 5, 1, 1373, 1, 5, 2698, 2725, 1, 581, 1, 2445, 3160, 5, 1, 8285, 1751, 5, 15139, 9677, 1, 18005, 1, 31397, 16045, 5, 13450, 2669, 1, 5, 52801, 20365, 1, 20501, 1, 65333, 59425, 5, 1, 86621, 48707, 99005, 59029, 54173, 1, 99725

Offset: 0

Seiichi Manyama, Jul 01 2017

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..10000
Eric Weisstein's World of Mathematics, Apéry Number

Cf. A005259, A276323.

a(n) = sum(k=0, n, (binomial(n, k)*binomial(n+k, k))^2) % (n+1)^3; \\ Michel Marcus, Jul 02 2017

If m = n + 1 is a prime and m >= 5, a(n) = 1.

A322518 Binomial transform of the Apéry numbers (A005259).

Original entry on oeis.org

1, 6, 84, 1680, 39240, 999216, 26899896, 752939424, 21691531800, 638947312080, 19155738105504, 582589712312064, 17930566188602136, 557417298916695600, 17477836958370383280, 552090876791399769600, 17552554240486710112920, 561230779055361080132880

Offset: 0

Sarah Arpin, Dec 13 2018

Starting with the a(3) term, each term is divisible by 8. (Empirical observation.)

The above is true and follows easily from the pair of known congruences for the Apéry numbers A(n): A(2*n) == 1 (mod 8) and A(2n+1) == 5 (mod 8). - Peter Bala, Jan 06 2020

			a(2) = binomial(2,0)*A(0) + binomial(2,1)*A(1) + binomial(2,2)*A(2), where A(k) denotes the k-th Apéry number. Using this definition:
a(2) = binomial(2,0)*(binomial(0,0)*binomial(0,0))^2 + binomial(2,1)*((binomial(1,0)*binomial(1,0))^2 + (binomial(1,1)*binomial(2,1))^2) + binomial(2,2)*((binomial(2,0)*binomial(2,0))^2 + (binomial(2,1)*binomial(3,1))^2 + (binomial(2,2)*binomial(4,2))^2) = 84.

Jackson Earles, Justin Ford, Poramate Nakkirt, Marlo Terr, Dr. Ilia Mishev, Sarah Arpin, Binomial Transforms of Sequences, Fall 2018.
N. J. A. Sloane, Transforms

Cf. A005259, A322519.

function BinomialTransform(seq)
    N = length(seq)
    bt = Array{BigInt,1}(undef,N)
    bt[1] = seq[1]
    for k in 1:N-1
        next = BigInt(0)
        for j in 0:k next += binomial(k, j)*seq[j+1] end
        bt[k+1] = next
    end
bt end
BinomialTransform([A005259(n) for n in 0:18]) |> println # Peter Luschny, Jan 06 2020

a[n_] := Sum[Binomial[n, k] * Sum[(Binomial[k, j] * Binomial[k+j, j])^2, {j, 0, k}], {k, 0, n}]; Array[a, 20, 0] (* Amiram Eldar, Dec 13 2018 *)

def OEISbinomial_transform(N, seq):
    BT = [seq[0]]
    k = 1
    while k< N:
        next = 0
        j = 0
        while j <=k:
            next = next + ((binomial(k,j))*seq[j])
            j = j+1
        BT.append(next)
        k = k+1
    return BT
Apery = oeis('A005259')
OEISBinom = OEISbinomial_transform(18,Apery.first_terms(20))

a(n) ~ 2^(n - 3/4) * 3^(n + 3/2) * (1 + sqrt(2))^(2*n - 1) / (Pi*n)^(3/2). - Vaclav Kotesovec, Dec 17 2018

The Gauss congruences hold: a(n*p^k) == a(n*p^(k-1)) (mod p^k) for all primes p and n a positive integer. - Peter Bala, Jan 06 2020

cp's OEIS Frontend

A357960 a(n) = A005259(n-1)^5 * A005258(n)^6.

Views

Author

Keywords

Comments

Crossrefs

Programs

Maple

Formula

A267220 Expansion of exp( Sum_{n >= 1} A005259(n)*x^n/n ).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Formula

A379153 The binary weights of the Apéry numbers (A005259).

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A276323 a(n) = (binomial(2 * prime(n + 3), prime(n + 3)) * A005259(prime(n + 3) - 1) - 2)/prime(n + 3)^5 for n >= 1.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Ruby

A289277 a(n) = A005259(n) mod 2*n+1.

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A289278 a(n) = A005259(n) mod (2*n+1)^2.

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A362723 a(n) = [x^n] ( E(x)/E(-x) )^n where E(x)= exp( Sum_{k >= 1} A005259(k)*x^k/k ).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Formula

A228143 Determinant of the (n+1) X (n+1) Hankel-type matrix with (i,j)-entry equal to A005259(i+j) for all i,j = 0,...,n.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs