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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A270917 Coefficient of x^n in Product_{k>=1} (1 + x^k)^(k^n).

Original entry on oeis.org

1, 1, 4, 35, 457, 12421, 678101, 69540142, 13730026114, 5551573311817, 4379029522335786, 6705866900012021577, 21038900445652125741759, 131183458646068931932668114, 1603688863449847489871671547959, 40294004792352613617780682256221711
Offset: 0

Views

Author

Vaclav Kotesovec, Mar 25 2016

Keywords

Crossrefs

Main diagonal of A284992.

Programs

  • Maple
    b:= proc(n, i, k) option remember; `if`(n=0, 1, `if`(i<1, 0,
          add(b(n-i*j, i-1, k)*binomial(i^k, j), j=0..n/i)))
        end:
    a:= n-> b(n$3):
    seq(a(n), n=0..20);  # Alois P. Heinz, Oct 16 2017
  • Mathematica
    Table[SeriesCoefficient[Product[(1+x^k)^(k^n), {k, 1, n}], {x, 0, n}], {n, 0, 20}]

Formula

Conjecture: limit n->infinity a(n)^(1/n^2) = exp(exp(-1)) = 1.444667861...

A324595 Number of colored integer partitions of 2n such that all colors from an n-set are used and parts differ by size or by color.

Original entry on oeis.org

1, 1, 5, 19, 85, 381, 1751, 8135, 38173, 180415, 857695, 4096830, 19645975, 94523729, 456079769, 2206005414, 10693086637, 51930129399, 252617434619, 1230714593340, 6003931991895, 29325290391416, 143393190367102, 701862880794183, 3438561265961263
Offset: 0

Views

Author

Alois P. Heinz, Sep 03 2019

Keywords

Examples

			a(2) = 5: 2a1a1b, 2b1a1b, 2a2b, 3a1b, 3b1a.
		

Crossrefs

Programs

  • Maple
    b:= proc(n, i, k) option remember; `if`(n=0, 1, `if`(i<1, 0, add((t->
          b(t, min(t, i-1), k)*binomial(k, j))(n-i*j), j=0..min(k, n/i))))
        end:
    a:= n-> add(b(2*n$2, n-i)*(-1)^i*binomial(n, i), i=0..n):
    seq(a(n), n=0..25);
    # second Maple program:
    b:= proc(n) option remember; `if`(n=0, 1, add(b(n-j)*add(
         `if`(d::odd, d, 0), d=numtheory[divisors](j)), j=1..n)/n)
        end:
    g:= proc(n, k) option remember; `if`(k=0, 1, `if`(k=1, b(n+1),
          (q-> add(g(j, q)*g(n-j, k-q), j=0..n))(iquo(k, 2))))
        end:
    a:= n-> g(n$2):
    seq(a(n), n=0..25);  # Alois P. Heinz, Jan 29 2021
  • Mathematica
    b[n_, i_, k_] := b[n, i, k] = If[n == 0, 1, If[i < 1, 0, Sum[Function[t, b[t, Min[t, i - 1], k] Binomial[k, j]][n - i j], {j, 0, Min[k, n/i]}]]];
    a[n_] := Sum[b[2n, 2n, n - i] (-1)^i Binomial[n, i], {i, 0, n}];
    a /@ Range[0, 25] (* Jean-François Alcover, May 06 2020, after Maple *)
    Table[SeriesCoefficient[(-1 + QPochhammer[-1, Sqrt[x]]/2)^n, {x, 0, n}], {n, 0, 25}] (* Vaclav Kotesovec, Jan 15 2024 *)
    (* Calculation of constant d: *) 1/r /. FindRoot[{2 + 2*s == QPochhammer[-1, Sqrt[r*s]], Sqrt[r]*Derivative[0, 1][QPochhammer][-1, Sqrt[r*s]] == 4*Sqrt[s]}, {r, 1/5}, {s, 1}, WorkingPrecision -> 120] (* Vaclav Kotesovec, Jan 15 2024 *)

Formula

a(n) = A308680(2n,n).
a(n) ~ c * d^n / sqrt(n), where d = 5.0032778445310926321307990027... and c = 0.2798596129161126875318997... - Vaclav Kotesovec, Sep 14 2019
a(n) = [x^(2n)] (-1 + Product_{j>=1} (1 + x^j))^n. - Alois P. Heinz, Jan 29 2021

A296163 a(n) = [x^n] Product_{k>=1} ((1 - x^(5*k))/(1 - x^k))^n.

Original entry on oeis.org

1, 1, 5, 22, 105, 501, 2456, 12160, 60801, 306130, 1550255, 7887034, 40281720, 206405967, 1060602800, 5463059772, 28199365873, 145832364580, 755420838614, 3918935839970, 20357605331355, 105878815699042, 551273881133750, 2873161931172668, 14988243880188600
Offset: 0

Views

Author

Ilya Gutkovskiy, Dec 06 2017

Keywords

Crossrefs

Programs

  • Mathematica
    Table[SeriesCoefficient[Product[((1 - x^(5 k))/(1 - x^k))^n, {k, 1, n}], {x, 0, n}], {n, 0, 24}]
    Table[SeriesCoefficient[Product[(1 + x^k + x^(2 k) + x^(3 k) + x^(4 k))^n, {k, 1, n}], {x, 0, n}], {n, 0, 24}]
    (* Calculation of constant d: *) With[{k = 5}, 1/r /. FindRoot[{s == QPochhammer[(r*s)^k] / QPochhammer[r*s], k*(-(s*QPochhammer[r*s]*(Log[1 - (r*s)^k] + QPolyGamma[0, 1, (r*s)^k]) / Log[(r*s)^k]) + (r*s)^k * Derivative[0, 1][QPochhammer][(r*s)^k, (r*s)^k]) == s*QPochhammer[r*s] + s^2*(-(QPochhammer[r*s]*(Log[1 - r*s] + QPolyGamma[0, 1, r*s]) / (s*Log[r*s])) + r*Derivative[0, 1][QPochhammer][r*s, r*s])}, {r, 1/5}, {s, 1}, WorkingPrecision -> 70]] (* Vaclav Kotesovec, Jan 17 2024 *)

Formula

a(n) = [x^n] Product_{k>=1} (1 + x^k + x^(2*k) + x^(3*k) + x^(4*k))^n.
a(n) ~ c * d^n / sqrt(n), where d = 5.3271035802753567624196808294779171420899175782347488197... and c = 0.2712048688090020853684153670711011713396954... - Vaclav Kotesovec, May 13 2018

A304443 Coefficient of x^n in Product_{k>=1} (1+x^k)^(2*n).

Original entry on oeis.org

1, 2, 10, 62, 394, 2562, 16966, 113794, 770442, 5254334, 36042250, 248403586, 1718732998, 11931569028, 83064794746, 579696375972, 4054279504266, 28408328186508, 199390547044342, 1401564307833908, 9865190079554954, 69522550703432476, 490484539061916794
Offset: 0

Views

Author

Vaclav Kotesovec, May 12 2018

Keywords

Crossrefs

Programs

  • Mathematica
    nmax = 25; Table[SeriesCoefficient[Product[(1+x^k)^(2*n), {k, 1, n}], {x, 0, n}], {n, 0, nmax}]
    nmax = 25; Table[SeriesCoefficient[(QPochhammer[-1, x]/2)^(2*n), {x, 0, n}], {n, 0, nmax}]
    (* Calculation of constants {d,c}: *) {1/r, Sqrt[Derivative[0, 1][QPochhammer][-1, r*s] / (Pi*r*(Sqrt[s]*Derivative[0, 1][QPochhammer][-1, r*s]^2 + 2*s*Derivative[0, 2][QPochhammer][-1, r*s]))]} /. FindRoot[{4*s == QPochhammer[-1, r*s]^2, 2*r*Sqrt[s]*Derivative[0, 1][QPochhammer][-1, r*s] == 2}, {r, 1/8}, {s, 2}, WorkingPrecision -> 120] (* Vaclav Kotesovec, Oct 03 2023 *)

Formula

a(n) ~ c * d^n / sqrt(n), where d = 7.21883059750200610514730564495768943165197819880185778427663522275469... and c = 0.300860732623379969554285615234449502629772950943717460278989499...

A304445 Coefficient of x^n in Product_{k>=1} (1+x^k)^(n^2).

Original entry on oeis.org

1, 1, 10, 174, 4132, 126905, 4802046, 216313349, 11313533064, 674172155790, 45102830448510, 3347925678474168, 273085613904116184, 24282144087974698445, 2337736453663291696624, 242272777074973285192935, 26892702305505020726198800, 3183326728470139531614913088
Offset: 0

Views

Author

Vaclav Kotesovec, May 12 2018

Keywords

Crossrefs

Programs

  • Mathematica
    nmax = 20; Table[SeriesCoefficient[Product[(1+x^k)^(n^2), {k, 1, n}], {x, 0, n}], {n, 0, nmax}]
    nmax = 20; Table[SeriesCoefficient[(QPochhammer[-1, x]/2)^(n^2), {x, 0, n}], {n, 0, nmax}]

Formula

a(n) ~ exp(n + 1/2) * n^(n - 1/2) / sqrt(2*Pi).

A380291 a(n) = [x^n] G(x)^n, where G(x) = Product_{k >= 1} (1 + x^k)^(k^2) is the g.f. of A027998.

Original entry on oeis.org

1, 1, 9, 64, 425, 3026, 21672, 157095, 1149289, 8464240, 62683134, 466307865, 3482008904, 26083955002, 195932407939, 1475267031164, 11131100990825, 84140066313620, 637054366975740, 4830417047590165, 36674477204674750, 278779034863684377, 2121418004609211361, 16159262748227985561
Offset: 0

Views

Author

Peter Bala, Jan 19 2025

Keywords

Comments

Given an integer sequence {f(n) : n >= 0} with f(0) = 1, there is a unique power series F(x) with rational coefficients, where F(0) = 1, such that f(n) = [x^n] F(x)^n. F(x) is given by F(x) = series_reversion(x/E(x)), where E(x) = exp(Sum_{n >= 1} f(n)*x^n/n). Furthermore, if the series E(x) has integer coefficients then the series F(x) also has integer coefficients and the sequence {f(n)} satisfies the Gauss congruences: f(n*p^r) == f(n*p^(r-1)) (mod p^r) for all primes p and positive integers n and r (by Stanley, Ch. 5, Ex. 5.2(a), p. 72 and the Lagrange inversion formula).
Thus the present sequence satisfies the Gauss congruences. In fact, stronger congruences appear to hold for the present sequence.
We conjecture that a(p) == 1 (mod p^3) for all primes p >= 7 (checked up to p = 61).
More generally, we conjecture that the supercongruences a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) hold for all primes p >= 7 and positive integers n and r. Some examples are given below.

Examples

			Examples of supercongruences:
a(7) - a(1) = 157095 - 1 = 2*(7^3)*229 == 0 (mod 7^3)
a(11) - a(1) = 466307865 - 1 = (2^3)*(11^3)*43793 == 0 (mod 11^3)
a(3*7) - a(3) = 278779034863684377 - 64 = (7^4)*43*26891*100413601 == 0 (mod 7^4)
		

References

  • R. P. Stanley. Enumerative combinatorics. Vol. 2, volume 62 of Cambridge Studies in Advanced Mathematics. Cambridge University Press, Cambridge, 1999.

Crossrefs

Programs

  • Maple
    with(numtheory):
    s_3 := n-> add((-1)^(n/d+1)*d^3, d in divisors(n)):
    G(x) := series(exp(add(s_3(k)*x^k/k, k = 1..23)), x, 24):
    seq(coeftayl(G(x)^n, x = 0, n), n = 0..23);
  • Mathematica
    Table[SeriesCoefficient[Product[(1 + x^k)^(n*k^2), {k, 1, n}], {x, 0, n}], {n, 0, 25}] (* Vaclav Kotesovec, Jul 30 2025 *)
    (* or *)
    Table[SeriesCoefficient[Exp[n*Sum[Sum[(-1)^(k/d + 1)*d^3, {d, Divisors[k]}]*x^k/k, {k, 1, n}]], {x, 0, n}], {n, 0, 25}] (* Vaclav Kotesovec, Jul 30 2025 *)

Formula

a(n) = [x^n] exp(n*Sum_{k >= 1} s_3(k)*x^k/k), where s_3(n) = Sum_{d divides n} (-1)^(n/d+1)*d^3 = A078307(n).
a(n) ~ c * d^n / sqrt(n), where d = 7.7846790125019502578773343468308844201627754275100035492213697757399421948... and c = 0.2484592487737716543953469621097743519172686743284742545545347906986158... - Vaclav Kotesovec, Jul 30 2025

A303071 a(n) = [x^n] (1/(1 - x))*Product_{k>=1} (1 + x^k)^n.

Original entry on oeis.org

1, 2, 6, 23, 90, 362, 1491, 6225, 26242, 111479, 476466, 2046464, 8825559, 38191467, 165751529, 721177328, 3144703234, 13739010855, 60127642329, 263545670385, 1156732481150, 5083320593976, 22364017244278, 98491038664903, 434160710647831, 1915482295831037, 8457663096970431
Offset: 0

Views

Author

Ilya Gutkovskiy, Apr 18 2018

Keywords

Crossrefs

Programs

  • Mathematica
    Table[SeriesCoefficient[1/(1 - x) Product[(1 + x^k)^n, {k, 1, n}], {x, 0, n}], {n, 0, 26}]
    Table[SeriesCoefficient[1/(1 - x) Exp[n Sum[(-1)^(k + 1) x^k/(k (1 - x^k)), {k, 1, n}]], {x, 0, n}], {n, 0, 26}]

Formula

a(n) = [x^n] (1/(1 - x))*exp(n*Sum_{k>=1} (-1)^(k+1)*x^k/(k*(1 - x^k))).
a(n) = Sum_{j=0..n} A286335(j,n).
a(n) ~ c * d^n / sqrt(n), where d = A270914 = 4.5024767476173544877385939327007... and c = 0.44252758868364961050787771300805... - Vaclav Kotesovec, May 19 2018

A304625 a(n) = [x^n] Product_{k>=1} ((1 - x^(n*k))/(1 - x^k))^n.

Original entry on oeis.org

1, 0, 3, 19, 101, 501, 2486, 12398, 62329, 315436, 1605330, 8207552, 42124368, 216903051, 1119974861, 5796944342, 30068145889, 156250892593, 813310723907, 4239676354631, 22130265931880, 115654632452514, 605081974091853, 3168828466966365, 16610409114771876, 87141919856550506
Offset: 0

Views

Author

Ilya Gutkovskiy, May 15 2018

Keywords

Comments

Number of partitions of n into 2 or more parts of n kinds. - Ilya Gutkovskiy, May 16 2018

Crossrefs

Programs

  • Mathematica
    Table[SeriesCoefficient[Product[((1 - x^(n k))/(1 - x^k))^n, {k, 1, n}], {x, 0, n}], {n, 0, 25}]
    Table[SeriesCoefficient[Product[1/(1 - x^k)^n, {k, 1, n - 1}], {x, 0, n}], {n, 0, 25}]

Formula

a(n) ~ c * d^n / sqrt(n), where d = A270915 = 5.3527013334866426877724... and c = 0.268015212710733315686... - Vaclav Kotesovec, May 16 2018

A304626 a(n) = [x^n] Product_{k>=1} ((1 + x^k)/(1 + x^(n*k)))^n.

Original entry on oeis.org

1, 0, 1, 10, 47, 201, 849, 3578, 15147, 64516, 276268, 1188342, 5130987, 22226036, 96543989, 420368843, 1834203939, 8018057328, 35107961157, 153950675566, 675978772306, 2971700764920, 13078268135661, 57613905606250, 254038914924767, 1121081799217206, 4951199308679965
Offset: 0

Views

Author

Ilya Gutkovskiy, May 15 2018

Keywords

Comments

Number of partitions of n into 2 or more distinct parts, with n types of each part. - Ilya Gutkovskiy, May 16 2018

Crossrefs

Programs

  • Mathematica
    Table[SeriesCoefficient[Product[((1 + x^k)/(1 + x^(n k)))^n, {k, 1, n}], {x, 0, n}], {n, 0, 26}]
    Table[SeriesCoefficient[Product[(1 + x^k)^n, {k, 1, n - 1}], {x, 0, n}], {n, 0, 26}]
    Table[SeriesCoefficient[(QPochhammer[-1, x, 1 + n]/QPochhammer[-1, x^n, 1 + n])^n, {x, 0, n}], {n, 0, 26}]

Formula

a(n) ~ c * d^n / sqrt(n), where d = A270914 = 4.502476747617354487738... and c = 0.2605422331424384694... - Vaclav Kotesovec, May 16 2018

A319671 a(n) = [x^n] Product_{k>=2} (1 + x^k)^n.

Original entry on oeis.org

1, 0, 2, 3, 10, 30, 77, 252, 682, 2145, 6182, 18887, 56317, 170534, 515930, 1563843, 4759338, 14480073, 44203595, 134972504, 412984510, 1264601502, 3877302717, 11898761051, 36548512477, 112358685555, 345673541514, 1064250223230, 3278695047218, 10107173174013, 31174889414807
Offset: 0

Views

Author

Ilya Gutkovskiy, Sep 25 2018

Keywords

Comments

Number of partitions of n into distinct parts > 1, with n types of each part.

Crossrefs

Programs

  • Mathematica
    Table[SeriesCoefficient[Product[(1 + x^k)^n, {k, 2, n}], {x, 0, n}], {n, 0, 30}]
    Table[SeriesCoefficient[1/((1 + x) QPochhammer[x, x^2])^n, {x, 0, n}], {n, 0, 30}]
    Table[SeriesCoefficient[Exp[n Sum[(Sum[Mod[d, 2] d, {d, Divisors[k]}] + (-1)^k) x^k/k, {k, 1, n}]], {x, 0, n}], {n, 0, 30}]

Formula

a(n) = [x^n] exp(n*Sum_{k>=1} (A000593(k) + (-1)^k)*x^k/k).
a(n) ~ c * d^n / sqrt(n), where d = 3.136240011804974455586379053639831470878466... and c = 0.220695581251514154138820799337758703024... - Vaclav Kotesovec, Oct 06 2018
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