Sarah Arpin - cp's OEIS frontend

A307800 Binomial transform of least common multiple sequence (A003418), starting with a(1).

1, 3, 11, 37, 153, 551, 2023, 7701, 29417, 107083, 384771, 1408133, 5457961, 22466367, 92977823, 365613181, 1342359393, 4677908531, 16159185307, 58676063493, 231520762361, 967464685783, 4052593703511, 16354948948517, 62709285045913, 229276436653851

Offset: 0

Sarah Arpin, Apr 29 2019

nonn

			For n = 3, a(3) = binomial(3,0)*1 + binomial(3,1)*2 + binomial(3,2)*6 + binomial(3,3)*12 = 37.

Jackson Earles, Aaron Li, Adam Nelson, Marlo Terr, Sarah Arpin, and Ilia Mishev Binomial Transforms of Sequences, CU Boulder Experimental Math Lab, Spring 2019.

Binomial transform of A003418 (shifted).

b:= proc(n) option remember; `if`(n=0, 1, ilcm(n, b(n-1))) end:
a:= n-> add(b(i+1)*binomial(n, i), i=0..n):
seq(a(n), n=0..30);  # Alois P. Heinz, Apr 29 2019

Table[Sum[Binomial[n, k]*Apply[LCM, Range[k+1]], {k, 0, n}], {n, 0, 30}] (* Vaclav Kotesovec, Jun 06 2019 *)

a(n) = sum(k=0, n, binomial(n, k)*lcm(vector(k+1, i, i))); \\ Michel Marcus, Apr 30 2019

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
LCMSeq = []
for k in range(1,26):
    LCMSeq.append(lcm(range(1,k+1)))
OEISbinomial_transform(25, LCMSeq)

a(n) = Sum_{k=0..n} binomial(n,k)*A003418(k+1).
Formula for values modulo 10: (Proof by considering the formula modulo 10)
    a(n) (mod 10) = 1, if n = 0, 2 (mod 5),
    a(n) (mod 10) = 3, if n = 1, 4 (mod 5),
    a(n) (mod 10) = 7, if n = 3 (mod 5).

A307803 Inverse binomial transform of least common multiple sequence.

Original entry on oeis.org

1, -1, 3, 1, 41, 171, 799, 2633, 7881, 24391, 99611, 461649, 2252953, 10773491, 46602711, 176413201, 596116769, 1899975183, 6302881171, 24136694081, 105765310281, 476455493179, 2033813426063, 8019234229401, 29410337173561, 102444237073751, 347418130583499

Offset: 0

Sarah Arpin, Apr 29 2019

sign

			For n = 3, a(3) = binomial(3,0)*1 - binomial(3,1)*2 + binomial(3,2)*6 - binomial(3,3)*12 = 1.

Alois P. Heinz, Table of n, a(n) for n = 0..1000
Jackson Earles, Aaron Li, Adam Nelson, Marlo Terr, Sarah Arpin, and Ilia Mishev Binomial Transforms of Sequences, CU Boulder Experimental Math Lab, Spring 2019.

Inverse binomial transform of A003418 (shifted).

b:= proc(n) option remember; `if`(n=0, 1, ilcm(n, b(n-1))) end:
a:= n-> add(b(i+1)*binomial(n, i)*(-1)^i, i=0..n):
seq(a(n), n=0..30);  # Alois P. Heinz, Apr 29 2019

b[n_] := b[n] = If[n == 0, 1, LCM[n, b[n - 1]]];
a[n_] := Sum[b[i + 1] Binomial[n, i] (-1)^i, {i, 0, n}];
a /@ Range[0, 30] (* Jean-François Alcover, Nov 27 2020, after Alois P. Heinz *)

a(n) = sum(k=0, n, (-1)^k*binomial(n, k)*lcm(vector(k+1, i, i))); \\ Michel Marcus, Apr 30 2019

def SIbinomial_transform(N, seq):
    BT = [seq[0]]
    k = 1
    while k< N:
        next = 0
        j = 0
        while j <=k:
            next = next + (((-1)^j)*(binomial(k,j))*seq[j])
            j = j+1
        BT.append(next)
        k = k+1
    return BT
LCMSeq = []
for k in range(1,26):
    LCMSeq.append(lcm(range(1,k+1)))
SIbinomial_transform(25, LCMSeq)

a(n) = Sum_{k=0..n} (-1)^k*binomial(n,k)*A003418(k+1).
Formula for values modulo 10: (Proof by considering the formula modulo 10)
    a(n) (mod 10) = 1, if n = 0, 3, 4 (mod 5),
    a(n) (mod 10) = 9, if n = 1 (mod 5),
    a(n) (mod 10) = 3, if n = 2 (mod 5).

A306810 Inverse binomial transform of the continued fraction expansion of e.

Original entry on oeis.org

2, -1, 2, -4, 7, -8, -2, 41, -134, 296, -485, 512, 82, -2107, 6562, -13852, 21871, -22600, -2186, 83105, -255878, 531440, -826685, 846368, 59050, -2952451, 9034498, -18600436, 28697815, -29229256, -1594322, 98848025, -301327046, 617003000, -947027861, 961376768, 43046722

Offset: 0

Sarah Arpin, Mar 11 2019

			For n = 3, a(3) = -binomial(3,0)*2 + binomial(3,1)*1 - binomial(3,2)*2 + binomial(3,3)*1 = -4.

Jackson Earles, Aaron Li, Adam Nelson, Marlo Terr, Sarah Arpin, and Ilia Mishev Binomial Transforms of Sequences, CU Boulder Experimental Math Lab, Spring 2019.

Continued fraction of e: A003417.

Binomial transform of continued fraction of e: A306809.

nmax = 50; A003417 = ContinuedFraction[E, nmax+1]; Table[Sum[(-1)^(n + k)*Binomial[n, k]*A003417[[k + 1]], {k, 0, n}], {n, 0, nmax}] (* Vaclav Kotesovec, Apr 23 2020 *)

def OEISInverse(N, seq):
    BT = [seq[0]]
    k = 1
    while k< N:
        next = 0
        j = 0
        while j <=k:
            next = next + (((-1)^(j+k))*(binomial(k,j))*seq[j])
            j = j+1
        BT.append(next)
        k = k+1
    return BT
econt = oeis('A003417')
OEISInverse(50,econt)

a(n) = Sum{k=0...n}(-1)^(n+k)*binomial(n,k)*b(k), where b(k) is the k-th term of the continued fraction expansion of e.
Conjectures from Colin Barker, Mar 12 2019: (Start)
G.f.: (2 + 13*x + 37*x^2 + 55*x^3 + 42*x^4 + 14*x^5 + 2*x^6) / ((1 + x)*(1 + 3*x + 3*x^2)^2).
a(n) = - 7*a(n-1) - 21*a(n-2) - 33*a(n-3) - 27*a(n-4) - 9*a(n-5) for n>6.
(End)

A306809 Binomial transform of the continued fraction expansion of e.

Original entry on oeis.org

2, 3, 6, 12, 23, 46, 98, 213, 458, 972, 2051, 4322, 9098, 19113, 40054, 83748, 174767, 364086, 757298, 1572861, 3262242, 6757500, 13981019, 28894090, 59652314, 123032913, 253522382, 521957844, 1073741831, 2207135966, 4533576578, 9305762469, 19088743546, 39131924268

Offset: 0

Sarah Arpin, Mar 11 2019

Taking this sequence as a continued fraction it seems to converge to 2.31601650488979...

			For n = 3, the a(3) = binomial(3,0)*2 + binomial(3,1)*1 + binomial(3,2)*2 + binomial(3,3)*1 = 12.

Jackson Earles, Aaron Li, Adam Nelson, Marlo Terr, Sarah Arpin, and Ilia Mishev Binomial Transforms of Sequences, CU Boulder Experimental Math Lab, Spring 2019.

Cf. A003417 (continued fraction for e).

nmax = 50; A003417 = ContinuedFraction[E, nmax+1]; Table[Sum[Binomial[n, k]*A003417[[k + 1]], {k, 0, n}], {n, 0, nmax}] (* Vaclav Kotesovec, Apr 23 2020 *)

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
econt = oeis('A003417')
OEISbinomial_transform(50,econt)

a(n) = Sum_{k=0..n} binomial(n,k)*b(k), where b(k) is the k-th term of the continued fraction expansion of e.
Conjectures from Colin Barker, Mar 12 2019: (Start)
G.f.: (2 - 11*x + 27*x^2 - 41*x^3 + 40*x^4 - 22*x^5 + 6*x^6) / ((1 - x)*(1 - 2*x)^2*(1 - x + x^2)^2).
a(n) = 7*a(n-1) - 21*a(n-2) + 37*a(n-3) - 43*a(n-4) + 33*a(n-5) - 16*a(n-6) + 4*a(n-7) for n>6.
(End)

A322519 Inverse binomial transform of the Apéry numbers (A005259).

Original entry on oeis.org

1, 4, 64, 1240, 27640, 667744, 17013976, 450174736, 12250723480, 340711148320, 9641274232384, 276704848753216, 8035189363318936, 235655550312118720, 6970100090159566480, 207674717284507191520, 6227433643414033714840, 187795334412416019255520

Offset: 0

Sarah Arpin, Dec 13 2018

nonn

Starting with the a(2) term, each term is divisible by 8. (Empirical observation.)

			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) = 64.

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

Cf. A005259, A322518.

[(&+[(-1)^(n-k)*Binomial(n,k)*(&+[(Binomial(k,j)*Binomial(k+j,j))^2: j in [0..k]]): k in [0..n]]): n in [0..20]]; // G. C. Greubel, Dec 13 2018

a:=n->add(binomial(n,i)*(-1)^i*add((binomial(n-i,k)*binomial(n-i+k,k))^2,k=0..n-i),i=0..n): seq(a(n),n=0..20); # Muniru A Asiru, Dec 22 2018

a[n_] := Sum[(-1)^(n-k) * 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 *)

{a(n) = sum(k=0,n, (-1)^(n-k)*binomial(n,k)*sum(j=0,k, (binomial(k,j)*binomial(k+j,j))^2))};
for(n=0, 20, print1(a(n), ", ")) \\ G. C. Greubel, Dec 13 2018

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

[sum((-1)^(n-k)*binomial(n,k)*sum((binomial(k,j)* binomial(k+j,j))^2 for j in (0..k)) for k in (0..n)) for n in (0..20)] # G. C. Greubel, Dec 13 2018

a(n) = Sum_{i=0..n} C(n,i) * (-1)^i * A005259(n-i).

a(n) ~ 2^((5*n + 3)/2) * (1 + sqrt(2))^(2*n - 1) / (Pi*n)^(3/2). - Vaclav Kotesovec, Dec 17 2018

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

User: Sarah Arpin

A307800 Binomial transform of least common multiple sequence (A003418), starting with a(1).

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Maple

Mathematica

PARI

Sage

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A307803 Inverse binomial transform of least common multiple sequence.

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Maple

Mathematica

PARI

Sage

Formula

A306810 Inverse binomial transform of the continued fraction expansion of e.

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Mathematica

Sage

Formula

A306809 Binomial transform of the continued fraction expansion of e.

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Examples

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Programs

Mathematica

Sage

Formula

A322519 Inverse binomial transform of the Apéry numbers (A005259).

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Magma

Maple

Mathematica

PARI

Sage

Sage

Formula

A322518 Binomial transform of the Apéry numbers (A005259).

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Julia

Mathematica

Sage

Formula