cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A242207 Least prime divisor of the n-th Domb number D(n) which does not divide any D(k) with k < n, or 1 if such a primitive prime divisor of D(n) does not exist.

Original entry on oeis.org

2, 7, 1, 97, 11, 23, 19, 643, 659, 1753, 4922329, 613, 341447, 1193, 2213, 2040452101603, 491, 82461839, 733, 113, 1108394340978316050481, 1034497328556150923437, 593, 73, 17117, 804943, 422291, 1559, 858631, 337655751557
Offset: 1

Views

Author

Zhi-Wei Sun, May 07 2014

Keywords

Comments

Conjecture: a(n) is prime except for n = 3.

Examples

			 a(4) = 97 since D(4) = 2^2*7*97 with 97 dividing none of D(1) = 2^2, D(2) = 2^2*7 and D(3) = 2^8.
		

Crossrefs

Programs

  • Mathematica
    d[n_]:=Sum[Binomial[n,k]^2*Binomial[2k,k]*Binomial[2(n-k),n-k],{k,0,n}]
    f[n_]:=FactorInteger[d[n]]
    p[n_]:=Table[Part[Part[f[n],k],1],{k,1,Length[f[n]]}]
    Do[If[d[n]<2,Goto[cc]];Do[Do[If[Mod[d[i],Part[p[n],k]]==0,Goto[aa]],{i,1,n-1}];Print[n," ",Part[p[n],k]];Goto[bb];Label[aa];Continue,{k,1,Length[p[n]]}];Label[cc];Print[n," ",1];Label[bb];Continue,{n,1,30}]

A336639 Sum_{n>=0} a(n) * x^n / (n!)^2 = 1 / BesselJ(0,2*sqrt(x))^4.

Original entry on oeis.org

1, 4, 36, 544, 12196, 377904, 15438816, 803602944, 51908768676, 4074743122384, 382079412133936, 42184889139337344, 5417567866536188896, 800808722921088352384, 135006904500993157933056, 25751088570167886107910144, 5517695042810314282550432676
Offset: 0

Views

Author

Ilya Gutkovskiy, Jul 28 2020

Keywords

Comments

In general, if k>=1 and Sum_{n>=0} a(n) * x^n / n!^2 = 1 / BesselJ(0, 2*sqrt(x))^k, then a(n) ~ n!^2 * n^(k-1) / ((k-1)! * r^(n + k/2) * BesselJ(1, 2*sqrt(r))^k), where r = BesselJZero(0,1)^2 / 4 = A115368^2/4. - Vaclav Kotesovec, Jul 11 2025

Crossrefs

Column k=4 of A340986.

Programs

  • Mathematica
    nmax = 16; CoefficientList[Series[1/BesselJ[0, 2 Sqrt[x]]^4, {x, 0, nmax}], x] Range[0, nmax]!^2
    a[0] = 1; a[n_] := a[n] = Sum[(-1)^(k + 1) Binomial[n, k]^2 Binomial[2 k, k] HypergeometricPFQ[{1/2, -k, -k, -k}, {1, 1, 1/2 - k}, 1] a[n - k], {k, 1, n}]; Table[a[n], {n, 0, 16}]

Formula

a(0) = 1; a(n) = Sum_{k=1..n} (-1)^(k+1) * binomial(n,k)^2 * A002895(k) * a(n-k).
a(n) ~ n!^2 * n^3 / (6 * r^(n+2) * BesselJ(1, 2*sqrt(r))^4), where r = BesselJZero(0,1)^2 / 4 = A115368^2/4 = 1.4457964907366961302939989396139517587... - Vaclav Kotesovec, Jul 11 2025

A336637 Sum_{n>=0} a(n) * x^n / (n!)^2 = exp(BesselI(0,2*sqrt(x))^4 - 1).

Original entry on oeis.org

1, 4, 60, 1648, 71612, 4448384, 370135632, 39480942848, 5227020747708, 837878205997216, 159457868003008640, 35459969754432262208, 9093585253916177728592, 2659611377508767798535488, 878768601771275773332660736, 325350340926291926090183214848
Offset: 0

Views

Author

Ilya Gutkovskiy, Jul 28 2020

Keywords

Crossrefs

Programs

  • Mathematica
    nmax = 15; CoefficientList[Series[Exp[BesselI[0, 2 Sqrt[x]]^4 - 1], {x, 0, nmax}], x] Range[0, nmax]!^2
    a[0] = 1; a[n_] := a[n] = (1/n) Sum[Binomial[n, k]^2 Binomial[2 k, k] HypergeometricPFQ[{1/2, -k, -k, -k}, {1, 1, 1/2 - k}, 1] k a[n - k], {k, 1, n}]; Table[a[n], {n, 0, 15}]

Formula

a(0) = 1; a(n) = (1/n) * Sum_{k=1..n} binomial(n,k)^2 * A002895(k) * k * a(n-k).

A364111 a(n) = Sum_{k = 0..n} binomial(n+k-1,k)^2 * binomial(2*n-2*k,n-k) * binomial(2*k,k).

Original entry on oeis.org

1, 4, 76, 2560, 106060, 4864504, 237354880, 12079462560, 633885607500, 34050190896040, 1863047125801576, 103465470769890112, 5817117095161011328, 330450303019252600240, 18937657945720403830240, 1093557503049551583194560, 63566414131528881235953228, 3716526456851323626808570632
Offset: 0

Views

Author

Peter Bala, Jul 07 2023

Keywords

Comments

Compare with the Domb numbers A002895, which are defined by A002895(n) = Sum_{k = 0..n} binomial(n,k)^2 * binomial(2*n-2*k,n-k) * binomial(2*k,k).
The supercongruences A002895(n*p^r) == A002895(n*p^(r-1)) (mod p^(3*r)) hold for all primes p >= 5 and positive integers n and r (see Osburn and Sahu).
We conjecture that the present sequence satisfies the same supercongruences.
More generally, let A >= 2, B and C be positive integers. Then we conjecture that the sequence whose terms are given by Sum_{k = 0..n} binomial(n+k-1,k)^A * binomial(2*n-2*k,n-k)^B * binomial(2*k,k)^C also satisfies the same supercongruences.

Crossrefs

Programs

  • Maple
    seq(add(binomial(n+k-1,k)^2 * binomial(2*n-2*k,n-k) * binomial(2*k,k)), n = 0..20);
    # faster program for large n
    seq(simplify(binomial(2*n,n)*hypergeom([-n, n, n, 1/2], [1, 1, 1/2 - n], 1)), n = 0..20);
  • Mathematica
    Table[Binomial[2*n,n] * HypergeometricPFQ[{-n, n, n, 1/2}, {1, 1, 1/2 - n}, 1], {n, 0, 20}] (* Vaclav Kotesovec, Jul 09 2023 *)

Formula

a(n) = Sum_{k = 0..n} binomial(-n,k)^2 * binomial(2*n-2*k,n-k) * binomial(2*k,k).
a(n) = binomial(2*n,n)*hypergeom([-n, n, n, 1/2], [1, 1, 1/2 - n], 1).
a(n) ~ 2^(6*n-1) / (sqrt(3) * Pi^(3/2) * n^(3/2)). - Vaclav Kotesovec, Jul 09 2023

A174123 Partial sums of A002893.

Original entry on oeis.org

1, 4, 19, 112, 751, 5404, 40573, 313408, 2471167, 19791004, 160459069, 1313922064, 10847561089, 90174127684, 754009158019, 6336733626112, 53489159252671, 453258909448636, 3854034482891725, 32871004555812112, 281127047928811201, 2410285909684370788
Offset: 0

Views

Author

Jonathan Vos Post, Mar 08 2010

Keywords

Crossrefs

Programs

  • Mathematica
    Accumulate[Table[Sum[Binomial[n,k]^2 Binomial[2k,k],{k,0,n}],{n,0,20}]] (* Harvey P. Dale, May 05 2013 *)
  • PARI
    a(n)=sum(m=0,n,sum(k=0,m,binomial(m,k)^2*binomial(2*k,k)))

Formula

a(n) = Sum_{i=0..n} A002893(i).
From Sergey Perepechko, Feb 16 2011: (Start)
O.g.f.: 2*sqrt(2)/Pi/(1-z)/sqrt(1-6*z-3*z^2+sqrt((1-z)^3*(1-9*z)))* EllipticK(8*z^(3/2)/(1-6*z-3*z^2+sqrt((1-z)^3*(1-9*z)))).
9*(n+2)^2*a(n) - (99+86*n+19*n^2)*a(n+1) + (72+56*n+11*n^2)*a(n+2) - (n+3)^2*a(n+3)=0. (End)
a(n) ~ 3^(2*n + 7/2) / (32*Pi*n). - Vaclav Kotesovec, Jul 11 2016

A329475 a(n) = Sum_{k=0..n} C(n,k)^2*T(k)*T(n-k), where T(k) = A002426(k) is the coefficient of x^k in the expansion of (x^2+x+1)^k.

Original entry on oeis.org

1, 2, 10, 68, 586, 5252, 49204, 475400, 4723786, 47937812, 494786260, 5177188040, 54794164660, 585565913480, 6309889976680, 68484312535568, 747985368753226, 8214968193003860, 90669516557975524, 1005156080857529768, 11187435500257898836, 124964856185950621832
Offset: 0

Views

Author

Zhi-Wei Sun, Nov 13 2019

Keywords

Comments

The author introduced this sequence in arXiv:1911.05456 and made the following conjecture.
Conjecture: Let p be an odd prime and let S = Sum_{k=0..p-1}a(k)/(-4)^k. If p == 1 (mod 12) and p = x^2 + 9*y^2 with x and y integers, then S == 4*x^2-2*p (mod p^2). If p == 5 (mod 12) and p = x^2 + y^2 with x == y (mod 3), then S == 4*x*y (mod p^2). If p == 3 (mod 4), then S == 0 (mod p^2).
Note that if p > 3 is a prime, then a(p-1) == Sum_{k=0..p-1} T(k)*T(p-1-k) == Legendre(p/3)*Sum_{k=0..p-1}T(k)^2/(-3)^k == 1 (mod p) by (1.7) and (2.3) of the author's 2014 paper in Sci. China Math.

Examples

			a(1) = 2 since Sum_{k=0,1} C(1,k)^2*T(k)*T(1-k) = C(1,0)^2*T(0)*T(1) + C(1,1)^2*T(1)*T(0) = 2*T(0)*T(1) = 2*1*1 = 2.
		

Crossrefs

Programs

  • Mathematica
    T[0]=1; T[1]=1; T[n_]:=T[n]=((2n-1)T[n-1]+3*(n-1)*T[n-2])/n;
    a[n_]:=a[n]=Sum[Binomial[n,k]^2*T[k]*T[n-k],{k,0,n}];
    Table[a[n],{n,0,21}]

Formula

a(n) ~ (3/2)*12^n/(n*Pi)^(3/2) as n tends to the infinity.

A336622 a(n) = Sum_{k=0..n} Sum_{i=0..k} Sum_{j=0..i} (binomial(n,k) * binomial(k,i) * binomial(i,j))^n.

Original entry on oeis.org

1, 4, 28, 1192, 591460, 3441637504, 219272057247376, 185528149944660881488, 2405748000504972140803769860, 349789137657321307953339196885516144, 652520795984468974632890750361094911319873648
Offset: 0

Views

Author

Ilya Gutkovskiy, Jul 29 2020

Keywords

Crossrefs

Programs

  • Magma
    B:=Binomial; [(&+[(&+[(&+[(B(n,j)*B(n-j,k-j)*B(k-j,k-i))^n: j in [0..i]]): i in [0..k]]): k in [0..n]]): n in [0..20]]; // G. C. Greubel, Aug 31 2022
    
  • Mathematica
    Table[Sum[Sum[Sum[(Binomial[n, k] Binomial[k, i] Binomial[i, j])^n, {j, 0, i}], {i, 0, k}], {k, 0, n}], {n, 0, 10}]
    Table[(n!)^n SeriesCoefficient[Sum[x^k/(k!)^n, {k, 0, n}]^4, {x, 0, n}], {n, 0, 10}]
  • SageMath
    b=binomial
    def A336622(n): return sum(sum(sum( (b(n,j)*b(n-j,k-j)*b(k-j,k-i))^n for j in (0..i)) for i in (0..k)) for k in (0..n))
    [A336622(n) for n in (0..20)] # G. C. Greubel, Aug 31 2022

Formula

a(n) = (n!)^n * [x^n] (Sum_{k>=0} x^k / (k!)^n)^4.

A372941 Numbers k that divide the k-th Domb number.

Original entry on oeis.org

1, 2, 4, 14, 28, 112, 133, 176, 224, 368, 388, 448, 616, 704, 784, 896, 1216, 1568, 1792, 3563, 4256, 5144, 6272, 8624, 8924, 9856, 11264, 11776, 13927, 16702, 23408, 32936, 38509, 42238, 43456, 43652, 43904, 46424, 67328, 73784, 76912, 78848, 81466, 110614, 118256
Offset: 1

Views

Author

Amiram Eldar, May 17 2024

Keywords

Comments

Numbers k such that k | A002895(k).

Examples

			2 is a term since A002895(2) = 28 = 2 * 14 is divisible by 2.
4 is a term since A002895(4) = 2716 = 4 * 679 is divisible by 4.
		

Crossrefs

Cf. A002895.
Similar sequences: A014847 (Catalan), A016089 (Lucas), A023172 (Fibonacci), A051177 (partition), A232570 (tribonacci), A246692 (Pell), A266969 (Motzkin).

Programs

  • Mathematica
    seq[kmax_] := Module[{d0 = 1, d1 = 4, d2, s = {1}}, Do[d2 = ((20*k^3 - 30*k^2 + 18*k - 4)*d1 - 64*(k-1)^3*d0)/k^3; If[Divisible[d2, k], AppendTo[s, k]]; d0 = d1; d1 = d2, {k, 2, kmax}]; s]; seq[5000]
  • PARI
    lista(kmax) = {my(d0 = 1, d1 = 4, d2); print1("1, "); for(k = 2, kmax, d2 = ((20*k^3 - 30*k^2 + 18*k - 4)*d1 - 64*(k-1)^3*d0)/k^3; if(!(d2 % k), print1(k, ", ")); d0 = d1; d1 = d2);}

A385087 2-adic valuation of A039699.

Original entry on oeis.org

3, 3, 10, 3, 6, 8, 12, 3, 6, 6, 11, 8, 11, 12, 16, 3, 6, 6, 12, 6, 9, 13, 17, 8, 11, 11, 16, 12, 15, 16, 20, 3, 6, 6, 14, 6, 9, 11, 15, 6, 9, 9, 14, 13, 16, 17, 21, 8, 11, 11, 17, 11, 14, 17, 21, 12, 15, 15, 20, 16, 19, 20, 24, 3, 6, 6, 13, 6, 9, 11, 15, 6, 9, 9, 14, 11, 14, 15, 19, 6
Offset: 1

Views

Author

Michel Marcus, Jun 17 2025

Keywords

Crossrefs

Programs

  • PARI
    C=binomial;
    A002895(n) = sum(k=0, n, C(n, k)^2 * C(2*n-2*k, n-k) * C(2*k, k) );
    a(n) = hammingweight(n) + valuation(A002895(n), 2);
    
  • Python
    from math import comb
    def A385087(n): return (~(a:=(sum(comb(n,k)**2*comb(n-k<<1,n-k)*comb(m:=k<<1,k) for k in range(n+1>>1))<<1) + (0 if n&1 else comb(n,n>>1)**4)) & a-1).bit_length() + n.bit_count() # Chai Wah Wu, Jun 17 2025

Formula

a(n) = A007814(A039699(n)) = A000120(n) + A007814(A002895(n)).

A385286 a(n) = (n!)^2 [x^n] hypergeom([], [1], x)^8.

Original entry on oeis.org

1, 8, 120, 2528, 66424, 2039808, 70283424, 2643158400, 106391894904, 4518833256512, 200396211454720, 9205443151733760, 435368682010660000, 21100379936684418560, 1044115187294444772480, 52597451834668445910528, 2691037806733052553149304, 139567074682665782246950080
Offset: 0

Views

Author

Peter Luschny, Jun 24 2025

Keywords

Comments

We regard this sequence in the list of sequences n -> A287316(n, 2^k) for k = 3.

Crossrefs

Cf. A000012 (k=0), A000984 (k=1), A002895 (k=2), this sequence (k=3), A287316.

Programs

  • Maple
    A385286_list := proc(len) local n; series(hypergeom([], [1], x)^8, x, len);
    seq((n!)^2*coeff(%, x, n), n = 0..len-1) end: A385286_list(18);
  • Mathematica
    nmax = 20; CoefficientList[Series[BesselI[0, 2*Sqrt[x]]^8, {x, 0, nmax}], x] * Range[0, nmax]!^2 (* Vaclav Kotesovec, Jun 24 2025 *)
  • PARI
    a(n) = my(x='x+O('x^(n+1))); n!^2*polcoeff(hypergeom([], [1], x)^8, n); \\ Michel Marcus, Jun 24 2025

Formula

a(n) = (n!)^2 [x^n] BesselI(0, 2*sqrt(x))^8.
a(n) = A287316(n, 2^3).
a(n) ~ 2^(6*n+5) / (Pi^(7/2) * n^(7/2)). - Vaclav Kotesovec, Jun 24 2025
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