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.

Showing 1-10 of 106 results. Next

A279785 Number of ways to choose a strict partition of each part of a strict partition of n.

Original entry on oeis.org

1, 1, 1, 3, 4, 7, 11, 18, 28, 47, 71, 108, 166, 252, 382, 587, 869, 1282, 1938, 2832, 4153, 6148, 8962, 12965, 18913, 27301, 39380, 56747, 81226, 115907, 166358, 236000, 334647, 475517, 671806, 947552, 1335679, 1875175, 2630584, 3687589, 5150585, 7183548
Offset: 0

Views

Author

Gus Wiseman, Dec 18 2016

Keywords

Comments

This sequence is obtained from the generalized Euler transform in A266964 by taking f(n) = -1, g(n) = -A000009(n). - Seiichi Manyama, Nov 14 2018

Examples

			The a(6)=11 twice-partitions are:
((6)), ((5)(1)), ((51)), ((4)(2)), ((42)), ((41)(1)),
((3)(2)(1)), ((31)(2)), ((32)(1)), ((321)), ((21)(2)(1)).
		

Crossrefs

Programs

  • Maple
    with(numtheory):
    g:= proc(n) option remember; `if`(n=0, 1, add(add(
          `if`(d::odd, d, 0), d=divisors(j))*g(n-j), j=1..n)/n)
        end:
    b:= proc(n, i) option remember; `if`(n>i*(i+1)/2, 0,
          `if`(n=0, 1, b(n, i-1)+`if`(i>n, 0, g(i)*b(n-i, i-1))))
        end:
    a:= n-> b(n$2):
    seq(a(n), n=0..70);  # Alois P. Heinz, Dec 20 2016
  • Mathematica
    nn=20;CoefficientList[Series[Product[(1+PartitionsQ[k]x^k),{k,nn}],{x,0,nn}],x]
    (* Second program: *)
    g[n_] := g[n] = If[n==0, 1, Sum[Sum[If[OddQ[d], d, 0], {d, Divisors[j]}]* g[n - j], {j, 1, n}]/n]; b[n_, i_] := b[n, i] = If[n > i*(i + 1)/2, 0, If[n==0, 1, b[n, i-1] + If[i>n, 0, g[i]*b[n-i, i-1]]]]; a[n_] := b[n, n]; Table[a[n], {n, 0, 70}] (* Jean-François Alcover, Feb 07 2017, after Alois P. Heinz *)

Formula

G.f.: Product_{k>0} (1 + A000009(k)*x^k). - Seiichi Manyama, Nov 14 2018

A271619 Twice partitioned numbers where the first partition is strict.

Original entry on oeis.org

1, 1, 2, 5, 8, 18, 34, 65, 109, 223, 386, 698, 1241, 2180, 3804, 6788, 11390, 19572, 34063, 56826, 96748, 163511, 272898, 452155, 755928, 1244732, 2054710, 3382147, 5534696, 8992209, 14733292, 23763685, 38430071, 62139578, 99735806, 160183001, 256682598
Offset: 0

Views

Author

Gus Wiseman, Apr 10 2016

Keywords

Comments

Number of sequences of integer partitions of the parts of some strict partition of n.
This sequence is obtained from the generalized Euler transform in A266964 by taking f(n) = -1, g(n) = -A000041(n). - Seiichi Manyama, Nov 15 2018

Examples

			a(6)=34: {(6);(5)(1),(51);(4)(2),(42);(4)(11),(41)(1),(411);(33);(3)(2)(1),(31)(2),(32)(1),(321);(3)(11)(1),(31)(11),(311)(1),(3111);(22)(2),(222);(21)(2)(1),(22)(11),(211)(2),(221)(1),(2211);(21)(11)(1),(111)(2)(1),(211)(11),(1111)(2),(2111)(1),(21111);(111)(11)(1),(1111)(11),(11111)(1),(111111)}
		

Crossrefs

Cf. A000009, A000041, A063834 (twice partitioned numbers), A270995, A279785, A327552, A327607.

Programs

  • Maple
    b:= proc(n, i) option remember; `if`(n>i*(i+1)/2, 0,
          `if`(n=0, 1, b(n, i-1) +`if`(i>n, 0,
           b(n-i, i-1)*combinat[numbpart](i))))
        end:
    a:= n-> b(n$2):
    seq(a(n), n=0..50);  # Alois P. Heinz, Apr 11 2016
  • Mathematica
    With[{n = 50}, CoefficientList[Series[Product[(1 + PartitionsP[i] x^i), {i, 1, n}], {x, 0, n}], x]]

Formula

G.f.: Product_{i>=1} (1 + A000041(i) * x^i).

A075900 Expansion of g.f.: Product_{n>0} 1/(1 - 2^(n-1)*x^n).

Original entry on oeis.org

1, 1, 3, 7, 19, 43, 115, 259, 659, 1523, 3731, 8531, 20883, 47379, 113043, 259219, 609683, 1385363, 3245459, 7344531, 17028499, 38579603, 88585619, 199845267, 457864595, 1028904339, 2339763603, 5256820115, 11896157587, 26626389395
Offset: 0

Views

Author

N. J. A. Sloane, Oct 15 2002

Keywords

Comments

Number of compositions of partitions of n. a(3) = 7: 3, 21, 12, 111, 2|1, 11|1, 1|1|1. - Alois P. Heinz, Sep 16 2019
Also the number of ways to split an integer composition of n into consecutive subsequences with weakly decreasing (or increasing) sums. - Gus Wiseman, Jul 13 2020
This sequence is obtained from the generalized Euler transform in A266964 by taking f(n) = 1, g(n) = 2^(n-1). - Seiichi Manyama, Aug 22 2020

Examples

			From _Gus Wiseman_, Jul 13 2020: (Start)
The a(0) = 1 through a(4) = 19 splittings:
  ()  (1)  (2)      (3)          (4)
           (1,1)    (1,2)        (1,3)
           (1),(1)  (2,1)        (2,2)
                    (1,1,1)      (3,1)
                    (2),(1)      (1,1,2)
                    (1,1),(1)    (1,2,1)
                    (1),(1),(1)  (2,1,1)
                                 (2),(2)
                                 (3),(1)
                                 (1,1,1,1)
                                 (1,1),(2)
                                 (1,2),(1)
                                 (2),(1,1)
                                 (2,1),(1)
                                 (1,1),(1,1)
                                 (1,1,1),(1)
                                 (2),(1),(1)
                                 (1,1),(1),(1)
                                 (1),(1),(1),(1)
(End)
		

Crossrefs

Row sums of A327549.
The strict case is A304961.
Partitions of partitions are A001970.
Splittings with equal sums are A074854.
Splittings of compositions are A133494.
Splittings of partitions are A323583.
Splittings with distinct sums are A336127.
Starting with a reversed partition gives A316245.
Starting with a partition instead of composition gives A336136.

Programs

  • Magma
    m:=80;
    R:=PowerSeriesRing(Integers(), m);
    Coefficients(R!( 1/(&*[1-2^(j-1)*x^j: j in [1..m+2]]) )); // G. C. Greubel, Jan 25 2024
    
  • Maple
    oo := 101; t1 := mul(1/(1-x^n/2),n=1..oo): t2 := series(t1,x,oo-1): t3 := seriestolist(t2): A075900 := n->2^n*t3[n+1];
    with(combinat); A075900 := proc(n) local i,t1,t2,t3; t1 := partition(n); t2 := 0; for i from 1 to nops(t1) do t3 := t1[i]; t2 := t2+2^(n-nops(t3)); od: t2; end;
  • Mathematica
    b[n_]:= b[n]= Sum[d*2^(n - n/d), {d, Divisors[n]}];
    a[0]= 1; a[n_]:= a[n]= 1/n*Sum[b[k]*a[n-k], {k,n}];
    Table[a[n], {n,0,30}] (* Jean-François Alcover, Mar 20 2014, after Vladeta Jovovic, fixed by Vaclav Kotesovec, Mar 08 2018 *)
  • Maxima
    s(m,n):=if nVladimir Kruchinin, Sep 06 2014 */
    
  • PARI
    {a(n)=polcoeff(prod(k=1,n,1/(1-2^(k-1)*x^k+x*O(x^n))),n)} \\ Paul D. Hanna, Jan 13 2013
    
  • PARI
    {a(n)=polcoeff(exp(sum(k=1,n+1,x^k/(k*(1-2^k*x^k)+x*O(x^n)))),n)} \\ Paul D. Hanna, Jan 13 2013
    
  • SageMath
    m=80;
    def A075900_list(prec):
        P. = PowerSeriesRing(QQ, prec)
        return P( 1/product(1-2^(j-1)*x^j for j in range(1,m+1)) ).list()
    A075900_list(m) # G. C. Greubel, Jan 25 2024

Formula

a(n) = Sum_{ partitions n = c_1 + ... + c_k } 2^(n-k). If p(n, m) = number of partitions of n into m parts, a(n) = Sum_{m=1..n} p(n, m)*2^(n-m).
G.f.: Sum_{n>=0} (a(n)/2^n)*x^n = Product_{n>0} 1/(1-x^n/2). - Vladeta Jovovic, Feb 11 2003
a(n) = 1/n*Sum_{k=1..n} A080267(k)*a(n-k). - Vladeta Jovovic, Feb 11 2003
G.f.: exp( Sum_{n>=1} x^n / (n*(1 - 2^n*x^n)) ). - Paul D. Hanna, Jan 13 2013
a(n) = s(1,n), a(0)=1, where s(m,n) = Sum_{k=m..n/2} 2^(k-1)*s(k, n-k) + 2^(n-1), s(n,n) = 2^(n-1), s(m,n)=0, m>. - Vladimir Kruchinin, Sep 06 2014
a(n) ~ 2^(n-2) * (Pi^2 - 6*log(2)^2)^(1/4) * exp(sqrt((Pi^2 - 6*log(2)^2)*n/3)) / (3^(1/4) * sqrt(Pi) * n^(3/4)). - Vaclav Kotesovec, Mar 09 2018

Extensions

More terms from Vladeta Jovovic, Feb 11 2003

A304961 Expansion of Product_{k>=1} (1 + 2^(k-1)*x^k).

Original entry on oeis.org

1, 1, 2, 6, 12, 32, 72, 176, 384, 960, 2112, 4992, 11264, 26112, 58368, 136192, 301056, 688128, 1548288, 3489792, 7766016, 17596416, 38993920, 87293952, 194248704, 432537600, 957349888, 2132803584, 4699717632, 10406068224, 23001563136, 50683969536, 111434268672, 245819768832
Offset: 0

Views

Author

Ilya Gutkovskiy, May 22 2018

Keywords

Comments

Number of compositions of partitions of n into distinct parts. a(3) = 6: 3, 21, 12, 111, 2|1, 11|1. - Alois P. Heinz, Sep 16 2019
Also the number of ways to split a composition of n into contiguous subsequences with strictly decreasing sums. - Gus Wiseman, Jul 13 2020
This sequence is obtained from the generalized Euler transform in A266964 by taking f(n) = -1, g(n) = (-1) * 2^(n-1). - Seiichi Manyama, Aug 22 2020

Examples

			From _Gus Wiseman_, Jul 13 2020: (Start)
The a(0) = 1 through a(4) = 12 splittings:
  ()  (1)  (2)    (3)        (4)
           (1,1)  (1,2)      (1,3)
                  (2,1)      (2,2)
                  (1,1,1)    (3,1)
                  (2),(1)    (1,1,2)
                  (1,1),(1)  (1,2,1)
                             (2,1,1)
                             (3),(1)
                             (1,1,1,1)
                             (1,2),(1)
                             (2,1),(1)
                             (1,1,1),(1)
(End)
		

Crossrefs

The non-strict version is A075900.
Starting with a reversed partition gives A323583.
Starting with a partition gives A336134.
Partitions of partitions are A001970.
Splittings with equal sums are A074854.
Splittings of compositions are A133494.
Splittings with distinct sums are A336127.

Programs

  • Mathematica
    nmax = 33; CoefficientList[Series[Product[(1 + 2^(k - 1) x^k), {k, 1, nmax}], {x, 0, nmax}], x]
  • PARI
    N=40; x='x+O('x^N); Vec(prod(k=1, N, 1+2^(k-1)*x^k)) \\ Seiichi Manyama, Aug 22 2020

Formula

G.f.: Product_{k>=1} (1 + A011782(k)*x^k).
a(n) ~ 2^n * exp(2*sqrt(-polylog(2, -1/2)*n)) * (-polylog(2, -1/2))^(1/4) / (sqrt(6*Pi) * n^(3/4)). - Vaclav Kotesovec, Sep 19 2019

A028377 Expansion of Product_{m>0} (1+q^m)^(m(m+1)/2).

Original entry on oeis.org

1, 1, 3, 9, 19, 46, 100, 218, 460, 965, 1975, 3993, 7975, 15712, 30650, 59150, 113093, 214300, 402812, 751165, 1390714, 2557004, 4670770, 8479232, 15302657, 27462424, 49021252, 87057783, 153850769, 270614429, 473850031, 826125184, 1434286323, 2480145226
Offset: 0

Views

Author

Keywords

Comments

Convolved with aerated A000294: [1, 0, 2, 0, 4, 0, 10, 0, 26, ...] = A000294. - Gary W. Adamson, Jun 13 2009
This sequence is obtained from the generalized Euler transform in A266964 by taking f(n) = -n*(n+1)/2, g(n) = -1. - Seiichi Manyama, Nov 14 2017

Crossrefs

Programs

  • Maple
    b:= proc(n, i) option remember; `if`(n=0, 1, `if`(i<1, 0,
          add(binomial(i*(i+1)/2, j)*b(n-i*j, i-1), j=0..n/i)))
        end:
    a:= n-> b(n$2):
    seq(a(n), n=0..50);  # Alois P. Heinz, Aug 03 2013
  • Mathematica
    b[n_, i_] := b[n, i] = If[n == 0, 1, If[i < 1, 0, Sum[Binomial[i*(i+1)/2, j]*b[n-i*j, i-1], {j, 0, n/i}]]]; a[n_] := b[n, n]; Table[a[n], {n, 0, 50}] (* Jean-François Alcover, Oct 13 2014, after Alois P. Heinz *)

Formula

a(n) ~ 7^(1/8) * exp(2 * 7^(1/4) * Pi * n^(3/4) / (3^(5/4) * 5^(1/4)) + 3^(3/2) * 5^(1/2) * Zeta(3) * n^(1/2) / (2 * 7^(1/2) * Pi^2) - 3^(13/4) * 5^(5/4) * Zeta(3)^2 * n^(1/4) / (4 * 7^(5/4) * Pi^5) + 2025 * Zeta(3)^3 / (98*Pi^8)) / (2^(49/24) * 15^(1/8) * n^(5/8)), where Zeta(3) = A002117. - Vaclav Kotesovec, Mar 11 2015
a(0) = 1 and a(n) = (1/(2*n)) * Sum_{k=1..n} b(k)*a(n-k) where b(n) = Sum_{d|n} d^2*(d+1)*(-1)^(1+n/d). - Seiichi Manyama, Nov 14 2017
G.f.: exp(Sum_{k>=1} (-1)^(k+1)*x^k/(k*(1 - x^k)^3)). - Ilya Gutkovskiy, May 28 2018

A022661 Expansion of Product_{m>=1} (1-m*q^m).

Original entry on oeis.org

1, -1, -2, -1, -1, 5, 1, 13, 4, 0, 2, -8, -61, -31, 13, -156, 21, 11, 223, 92, 91, 426, 972, 165, 141, -1126, 440, 1294, -4684, -2755, -5748, -2414, -6679, 10511, -10048, -19369, 19635, 22629, 14027, 76969, -1990, 40193, -10678, 75795, 215767, -54322, -40882
Offset: 0

Views

Author

Keywords

Comments

Is a(9) the only occurrence of 0 in this sequence? - Robert Israel, Jun 02 2015

Crossrefs

Programs

  • Magma
    Coefficients(&*[(1-m*x^m):m in [1..40]])[1..40] where x is PolynomialRing(Integers()).1; // G. C. Greubel, Feb 18 2018
  • Maple
    P:= mul(1-m*q^m,m=1..100):
    S:= series(P,q,101):
    seq(coeff(S,q,j),j=0..100); # Robert Israel, Jun 02 2015
    # second Maple program:
    b:= proc(n, i) option remember; `if`(i*(i+1)/2n, 0, i*b(n-i, i-1))))
        end:
    a:= n-> b(n$2):
    seq(a(n), n=0..60);  # Sean A. Irvine (after Alois P. Heinz), May 19 2019
  • Mathematica
    nmax = 40; CoefficientList[Series[Product[1 - k*x^k, {k, 1, nmax}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Dec 15 2015 *)
    nmax = 40; CoefficientList[Series[Exp[-Sum[PolyLog[-j, x^j]/j, {j, 1, nmax}]], {x, 0, nmax}], x] (* Vaclav Kotesovec, Dec 15 2015 *)
    (* More efficient program: *) nmax = 50; poly = ConstantArray[0, nmax+1]; poly[[1]] = 1; poly[[2]] = -1; Do[Do[poly[[j+1]] -= k*poly[[j-k+1]], {j, nmax, k, -1}];, {k, 2, nmax}]; poly (* Vaclav Kotesovec, Jan 07 2016 *)
  • PARI
    m=50; q='q+O('q^m); Vec(prod(n=1,m,(1-n*q^n))) \\ G. C. Greubel, Feb 18 2018
    

A266891 Expansion of Product_{k>=1} (1 + k*x^k)^k.

Original entry on oeis.org

1, 1, 4, 13, 29, 81, 188, 456, 1030, 2405, 5295, 11611, 25246, 53552, 113332, 235685, 486011, 990840, 2006567, 4018010, 7992003, 15768511, 30875424, 60060509, 116042548, 222817961, 425200270, 806991037, 1522748592, 2858792520, 5339457208, 9924370365
Offset: 0

Views

Author

Vaclav Kotesovec, Jan 05 2016

Keywords

Comments

This sequence is obtained from the generalized Euler transform in A266964 by taking f(n) = -n, g(n) = -n. - Seiichi Manyama, Nov 18 2017

Crossrefs

Programs

  • Mathematica
    nmax=50; CoefficientList[Series[Product[(1+k*x^k)^k, {k, 1, nmax}], {x, 0, nmax}], x]
    (* More efficient program: *) nmax = 50; s = 1+x; Do[s*=Sum[Binomial[k, j] * k^j * x^(j*k), {j, 0, nmax/k}]; s = Take[Expand[s], Min[nmax + 1, Exponent[s, x] + 1]];, {k, 2, nmax}]; CoefficientList[s, x] (* Vaclav Kotesovec, Jan 07 2016 *)

Formula

a(0) = 1 and a(n) = (1/n) * Sum_{k=1..n} b(k)*a(n-k) where b(n) = Sum_{d|n} d*(-d)^(1+n/d). - Seiichi Manyama, Nov 18 2017
Conjecture: log(a(n)) ~ n^(2/3) * (2*log(3*n) - 3) / (4*3^(1/3)). - Vaclav Kotesovec, May 08 2018

A066815 Number of partitions of n into sums of products.

Original entry on oeis.org

1, 1, 2, 3, 6, 8, 14, 19, 33, 45, 69, 94, 148, 197, 289, 390, 575, 762, 1086, 1439, 2040, 2687, 3712, 4874, 6749, 8792, 11918, 15526, 20998, 27164, 36277, 46820, 62367, 80146, 105569, 135326, 177979, 227139, 296027, 377142, 490554, 622526, 804158
Offset: 0

Views

Author

Vladeta Jovovic, Jan 20 2002

Keywords

Comments

Number of ways to choose a factorization of each part of an integer partition of n. - Gus Wiseman, Sep 05 2018
This sequence is obtained from the generalized Euler transform in A266964 by taking f(n) = 1, g(n) = A001055(n). - Seiichi Manyama, Nov 14 2018

Examples

			From _Gus Wiseman_, Sep 05 2018: (Start)
The a(6) = 14 partitions of 6 into sums of products:
  6, 2*3,
  5+1, 4+2, 2*2+2, 3+3,
  4+1+1, 2*2+1+1, 3+2+1, 2+2+2,
  3+1+1+1, 2+2+1+1,
  2+1+1+1+1,
  1+1+1+1+1+1.
(End)
		

Crossrefs

Programs

  • Mathematica
    facs[n_]:=If[n<=1,{{}},Join@@Table[(Prepend[#1,d]&)/@Select[facs[n/d],Min@@#1>=d&],{d,Rest[Divisors[n]]}]];
    Table[Length[Join@@Table[Tuples[facs/@ptn],{ptn,IntegerPartitions[n]}]],{n,20}] (* Gus Wiseman, Sep 05 2018 *)

Formula

G.f.: Product_{k>=1} 1/(1-A001055(k)*x^k).
a(n) = 1/n*Sum_{k=1..n} a(n-k)*b(k), n > 0, a(0)=1, b(k)=Sum_{d|k} d*(A001055(d))^(k/d).

Extensions

Renamed by T. D. Noe, May 24 2011

A266941 Expansion of Product_{k>=1} 1 / (1 - k*x^k)^k.

Original entry on oeis.org

1, 1, 5, 14, 42, 103, 289, 690, 1771, 4206, 10142, 23449, 54786, 123528, 279480, 619206, 1366405, 2969071, 6425534, 13727775, 29187555, 61439660, 128620370, 267044222, 551527679, 1130806020, 2306746335, 4676096006, 9432394144, 18920266428, 37776372312
Offset: 0

Views

Author

Vaclav Kotesovec, Jan 06 2016

Keywords

Comments

This sequence is obtained from the generalized Euler transform in A266964 by taking f(n) = n, g(n) = n. - Seiichi Manyama, Nov 18 2017

Crossrefs

Programs

  • Mathematica
    nmax = 40; CoefficientList[Series[Product[1/(1-k*x^k)^k, {k, 1, nmax}], {x, 0, nmax}], x]
    nmax = 40; s = 1 - x; Do[s *= Sum[Binomial[k, j]*(-1)^j*k^j*x^(j*k), {j, 0, nmax/k}]; s = Expand[s]; s = Take[s, Min[nmax + 1, Exponent[s, x] + 1, Length[s]]];, {k, 2, nmax}]; CoefficientList[Series[1/s, {x, 0, nmax}], x] (* Vaclav Kotesovec, Aug 27 2018 *)

Formula

From Vaclav Kotesovec, Jan 08 2016: (Start)
a(n) ~ c * n^2 * 3^(n/3), where
c = 3278974684037157122864203.021982619109776972432419491714093... if mod(n,3)=0
c = 3278974684037157122864202.999526122508793149896683112820555... if mod(n,3)=1
c = 3278974684037157122864203.001231135511323719311281438384212... if mod(n,3)=2
(End)
In closed form, a(n) ~ (Product_{k>=4}((1 - k/3^(k/3))^(-k)) / ((1 - 2/3^(2/3))^2 * (1 - 1/3^(1/3))) + Product_{k>=4}((1 - (-1)^(2*k/3)*k/3^(k/3))^(-k)) / ((-1)^(2*n/3) * ((1 + 2*(-1)^(1/3)/3^(2/3))^2 * (1 - (-1)^(2/3)/3^(1/3)))) + Product_{k>=4}((1 - (-1)^(4*k/3)*k/3^(k/3))^(-k)) / ((-1)^(4*n/3) * ((1 + (-1/3)^(1/3)) * (1 - 2*(-1/3)^(2/3))^2))) * 3^(n/3) * n^2 / 54. - Vaclav Kotesovec, Apr 24 2017
a(0) = 1 and a(n) = (1/n) * Sum_{k=1..n} (Sum_{d|k} d^(2+k/d)) * a(n-k) for n > 0. - Seiichi Manyama, Nov 02 2017

A284896 Expansion of Product_{k>=1} 1/(1+x^k)^(k^2) in powers of x.

Original entry on oeis.org

1, -1, -3, -6, 0, 11, 42, 63, 73, -45, -267, -720, -1095, -1239, -66, 2794, 8757, 16017, 22885, 19634, -2359, -61979, -161867, -302190, -421971, -432051, -126712, 690578, 2278273, 4584989, 7269985, 8965464, 7515373, -845659, -19930400, -53474765, -100195759
Offset: 0

Views

Author

Seiichi Manyama, Apr 05 2017

Keywords

Comments

This sequence is obtained from the generalized Euler transform in A266964 by taking f(n) = n^2, g(n) = -1. - Seiichi Manyama, Nov 15 2017

Crossrefs

Product_{k>=1} 1/(1+x^k)^(k^m): A081362 (m=0), A255528 (m=1), this sequence (m=2), A284897 (m=3), A284898 (m=4), A284899 (m=5).

Programs

  • Mathematica
    CoefficientList[Series[Product[1/(1 + x^k)^(k^2) , {k, 40}], {x, 0, 40}], x] (* Indranil Ghosh, Apr 05 2017 *)
  • PARI
    x= 'x + O('x^40); Vec(prod(k=1, 40, 1/(1 + x^k)^(k^2))) \\ Indranil Ghosh, Apr 05 2017

Formula

a(0) = 1, a(n) = -(1/n)*Sum_{k=1..n} A078307(k)*a(n-k) for n > 0. - Seiichi Manyama, Apr 06 2017
G.f.: exp(Sum_{k>=1} (-1)^k*x^k*(1 + x^k)/(k*(1 - x^k)^3)). - Ilya Gutkovskiy, May 30 2018
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