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 23 results. Next

A386729 a(n) = [x^n] G(x)^n, where G(x) = Product_{k >= 1} (1 + x^k)^(k^3) is the g.f. of A248882.

Original entry on oeis.org

1, 1, 17, 154, 1377, 13276, 127862, 1249746, 12321121, 122287798, 1220492192, 12235940113, 123133325382, 1243080020352, 12583773308102, 127688996851804, 1298370095026017, 13226355435367992, 134955405683954234, 1379032238329708409, 14110075394718902752, 144544237021110644340
Offset: 0

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Author

Vaclav Kotesovec, Jul 31 2025

Keywords

Crossrefs

Programs

  • Mathematica
    Table[SeriesCoefficient[Product[(1+x^k)^(n*k^3), {k, 1, n}], {x, 0, n}], {n, 0, 25}]
    Table[SeriesCoefficient[Exp[n*Sum[Sum[(-1)^(k/d + 1)*d^4, {d, Divisors[k]}]*x^k/k, {k, 1, n}]], {x, 0, n}], {n, 0, 25}]

Formula

a(n) = [x^n] exp(n*Sum_{k >= 1} s_4(k)*x^k/k), where s_4(n) = Sum_{d divides n} (-1)^(n/d+1)*d^4 = A284900(n).
a(n) ~ c * d^n / sqrt(n), where d = 10.49088673566991578441632677715184699104285539252671173854512548234581416... and c = 0.2449508761900081824436717230940007974244164508939377916825513986093942...

A013663 Decimal expansion of zeta(5).

Original entry on oeis.org

1, 0, 3, 6, 9, 2, 7, 7, 5, 5, 1, 4, 3, 3, 6, 9, 9, 2, 6, 3, 3, 1, 3, 6, 5, 4, 8, 6, 4, 5, 7, 0, 3, 4, 1, 6, 8, 0, 5, 7, 0, 8, 0, 9, 1, 9, 5, 0, 1, 9, 1, 2, 8, 1, 1, 9, 7, 4, 1, 9, 2, 6, 7, 7, 9, 0, 3, 8, 0, 3, 5, 8, 9, 7, 8, 6, 2, 8, 1, 4, 8, 4, 5, 6, 0, 0, 4, 3, 1, 0, 6, 5, 5, 7, 1, 3, 3, 3, 3
Offset: 1

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Keywords

Comments

In a widely distributed May 2011 email, Wadim Zudilin gave a rebuttal to v1 of Kim's 2011 preprint: "The mistake (unfixable) is on p. 6, line after eq. (3.3). 'Without loss of generality' can be shown to work only for a finite set of n_k's; as the n_k are sufficiently large (and N is fixed), the inequality for epsilon is false." In a May 2013 email, Zudilin extended his rebuttal to cover v2, concluding that Kim's argument "implies that at least one of zeta(2), zeta(3), zeta(4) and zeta(5) is irrational, which is trivial." - Jonathan Sondow, May 06 2013
General: zeta(2*s + 1) = (A000364(s)/A331839(s)) * Pi^(2*s + 1) * Product_{k >= 1} (A002145(k)^(2*s + 1) + 1)/(A002145(k)^(2*s + 1) - 1), for s >= 1. - Dimitris Valianatos, Apr 27 2020

Examples

			1/1^5 + 1/2^5 + 1/3^5 + 1/4^5 + 1/5^5 + 1/6^5 + 1/7^5 + ... =
1 + 1/32 + 1/243 + 1/1024 + 1/3125 + 1/7776 + 1/16807 + ... = 1.036927755143369926331365486457...
		

References

  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 262.
  • Milton Abramowitz and Irene A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 811.

Crossrefs

Programs

Formula

From Peter Bala, Dec 04 2013: (Start)
Definition: zeta(5) = Sum_{n >= 1} 1/n^5.
zeta(5) = 2^5/(2^5 - 1)*(Sum_{n even} n^5*p(n)*p(1/n)/(n^2 - 1)^6 ), where p(n) = n^2 + 3. See A013667, A013671 and A013675. (End)
zeta(5) = Sum_{n >= 1} (A010052(n)/n^(5/2)) = Sum_{n >= 1} ((floor(sqrt(n)) - floor(sqrt(n-1)))/n^(5/2)). - Mikael Aaltonen, Feb 22 2015
zeta(5) = Product_{k>=1} 1/(1 - 1/prime(k)^5). - Vaclav Kotesovec, Apr 30 2020
From Artur Jasinski, Jun 27 2020: (Start)
zeta(5) = (-1/30)*Integral_{x=0..1} log(1-x^4)^5/x^5.
zeta(5) = (1/24)*Integral_{x=0..infinity} x^4/(exp(x)-1).
zeta(5) = (2/45)*Integral_{x=0..infinity} x^4/(exp(x)+1).
zeta(5) = (1/(1488*zeta(1/2)^5))*(-5*Pi^5*zeta(1/2)^5 + 96*zeta'(1/2)^5 - 240*zeta(1/2)*zeta'(1/2)^3*zeta''(1/2) + 120*zeta(1/2)^2*zeta'(1/2)*zeta''(1/2)^2 + 80*zeta(1/2)^2*zeta'(1/2)^2*zeta'''(1/2)- 40*zeta(1/2)^3*zeta''(1/2)*zeta'''(1/2) - 20*zeta(1/2)^3*zeta'(1/2)*zeta''''(1/2)+4*zeta(1/2)^4*zeta'''''(1/2)). (End).
From Peter Bala, Oct 29 2023: (Start)
zeta(3) = (8/45)*Integral_{x >= 1} x^3*log(x)^3*(1 + log(x))*log(1 + 1/x^x) dx = (2/45)*Integral_{x >= 1} x^4*log(x)^4*(1 + log(x))/(1 + x^x) dx.
zeta(5) = 131/128 + 26*Sum_{n >= 1} (n^2 + 2*n + 40/39)/(n*(n + 1)*(n + 2))^5.
zeta(5) = 5162893/4976640 - 1323520*Sum_{n >= 1} (n^2 + 4*n + 56288/12925)/(n*(n + 1)*(n + 2)*(n + 3)*(n + 4))^5. Taking 10 terms of the series gives a value for zeta(5) correct to 20 decimal places.
Conjecture: for k >= 1, there exist rational numbers A(k), B(k) and c(k) such that zeta(5) = A(k) + B(k)*Sum_{n >= 1} (n^2 + 2*k*n + c(k))/(n*(n + 1)*...*(n + 2*k))^5. A similar conjecture can be made for the constant zeta(3). (End)
zeta(5) = (694/204813)*Pi^5 - Sum_{n >= 1} (6280/3251)*(1/(n^5*(exp(4*Pi*n)-1))) + Sum_{n >= 1} (296/3251)*(1/(n^5*(exp(5*Pi*n)-1))) - Sum_{n >= 1} (1073/6502)*(1/(n^5*(exp(10*Pi*n)-1))) + Sum_{n >= 1} (37/6502)*(1/(n^5*(exp(20*Pi*n)-1))). - Simon Plouffe, Jan 06 2024
From Peter Bala, Apr 27 2025: (Start)
zeta(5) = 1/5! * Integral_{x >= 0} x^5 * exp(x)/(exp(x) - 1)^2 dx = (16/15) * 1/5! * Integral_{x >= 0} x^5 * exp(x)/(exp(x) + 1)^2 dx.
zeta(5) = 1/6! * Integral_{x >= 0} x^6 * exp(x)*(exp(x) + 1)/(exp(x) - 1)^3 dx = 1/(3^3 * 5^2) * Integral_{x >= 0} x^6 * exp(x)*(exp(x) - 1)/(exp(x) + 1)^3 dx. (End)
zeta(5) = Sum_{i, j >= 1} 1/((i^4)*j*binomial(i+j, i)). More generally, zeta(n+1) = Sum_{i, j >= 1} 1/((i^n)*j*binomial(i+j, i)) for n >= 1. - Peter Bala, Aug 07 2025

A026007 Expansion of Product_{m>=1} (1 + q^m)^m; number of partitions of n into distinct parts, where n different parts of size n are available.

Original entry on oeis.org

1, 1, 2, 5, 8, 16, 28, 49, 83, 142, 235, 385, 627, 1004, 1599, 2521, 3940, 6111, 9421, 14409, 21916, 33134, 49808, 74484, 110837, 164132, 241960, 355169, 519158, 755894, 1096411, 1584519, 2281926, 3275276, 4685731, 6682699, 9501979, 13471239, 19044780, 26850921, 37756561, 52955699
Offset: 0

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Comments

In general, for t > 0, if g.f. = Product_{m>=1} (1 + t*q^m)^m then a(n) ~ c^(1/6) * exp(3^(2/3) * c^(1/3) * n^(2/3) / 2) / (3^(2/3) * (t+1)^(1/12) * sqrt(2*Pi) * n^(2/3)), where c = Pi^2*log(t) + log(t)^3 - 6*polylog(3, -1/t). - Vaclav Kotesovec, Jan 04 2016

Examples

			For n = 4, we have 8 partitions
  01: [4]
  02: [4']
  03: [4'']
  04: [4''']
  05: [3, 1]
  06: [3', 1]
  07: [3'', 1]
  08: [2, 2']
		

Crossrefs

Programs

  • Maple
    with(numtheory):
    b:= proc(n) option remember;
          add((-1)^(n/d+1)*d^2, d=divisors(n))
        end:
    a:= proc(n) option remember;
          `if`(n=0, 1, add(b(k)*a(n-k), k=1..n)/n)
        end:
    seq(a(n), n=0..45);  # Alois P. Heinz, Aug 03 2013
  • Mathematica
    a[n_] := a[n] = 1/n*Sum[Sum[(-1)^(k/d+1)*d^2, {d, Divisors[k]}]*a[n-k], {k, 1, n}]; a[0] = 1; Table[a[n], {n, 0, 41}] (* Jean-François Alcover, Apr 17 2014, after Vladeta Jovovic *)
    nmax=50; CoefficientList[Series[Exp[Sum[(-1)^(k+1)*x^k/(k*(1-x^k)^2),{k,1,nmax}]],{x,0,nmax}],x] (* Vaclav Kotesovec, Feb 28 2015 *)
  • PARI
    N=66; q='q+O('q^N);
    gf= prod(n=1,N, (1+q^n)^n );
    Vec(gf)
    /* Joerg Arndt, Oct 06 2012 */

Formula

a(n) = (1/n)*Sum_{k=1..n} A078306(k)*a(n-k). - Vladeta Jovovic, Nov 22 2002
G.f.: Product_{m>=1} (1+x^m)^m. Weighout transform of natural numbers (A000027). Euler transform of A026741. - Franklin T. Adams-Watters, Mar 16 2006
a(n) ~ zeta(3)^(1/6) * exp((3/2)^(4/3) * zeta(3)^(1/3) * n^(2/3)) / (2^(3/4) * 3^(1/3) * sqrt(Pi) * n^(2/3)), where zeta(3) = A002117. - Vaclav Kotesovec, Mar 05 2015

A261049 Expansion of Product_{k>=1} (1+x^k)^(p(k)), where p(k) is the partition function.

Original entry on oeis.org

1, 1, 2, 5, 9, 19, 37, 71, 133, 252, 464, 851, 1547, 2787, 4985, 8862, 15639, 27446, 47909, 83168, 143691, 247109, 423082, 721360, 1225119, 2072762, 3494359, 5870717, 9830702, 16409939, 27309660, 45316753, 74986921, 123748430, 203686778, 334421510, 547735241
Offset: 0

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Author

Vaclav Kotesovec, Aug 08 2015

Keywords

Comments

Number of strict multiset partitions of integer partitions of n. Weigh transform of A000041. - Gus Wiseman, Oct 11 2018

Examples

			From _Gus Wiseman_, Oct 11 2018: (Start)
The a(1) = 1 through a(5) = 19 strict multiset partitions:
  {{1}}  {{2}}    {{3}}        {{4}}          {{5}}
         {{1,1}}  {{1,2}}      {{1,3}}        {{1,4}}
                  {{1,1,1}}    {{2,2}}        {{2,3}}
                  {{1},{2}}    {{1,1,2}}      {{1,1,3}}
                  {{1},{1,1}}  {{1},{3}}      {{1,2,2}}
                               {{1,1,1,1}}    {{1},{4}}
                               {{1},{1,2}}    {{2},{3}}
                               {{2},{1,1}}    {{1,1,1,2}}
                               {{1},{1,1,1}}  {{1},{1,3}}
                                              {{1},{2,2}}
                                              {{2},{1,2}}
                                              {{3},{1,1}}
                                              {{1,1,1,1,1}}
                                              {{1},{1,1,2}}
                                              {{1,1},{1,2}}
                                              {{2},{1,1,1}}
                                              {{1},{1,1,1,1}}
                                              {{1,1},{1,1,1}}
                                              {{1},{2},{1,1}}
(End)
		

Crossrefs

Programs

  • Maple
    b:= proc(n, i) option remember; `if`(n=0, 1, `if`(i<1, 0, add(
          binomial(combinat[numbpart](i), j)*b(n-i*j, i-1), j=0..n/i)))
        end:
    a:= n-> b(n$2):
    seq(a(n), n=0..40);  # Alois P. Heinz, Aug 08 2015
  • Mathematica
    nmax=40; CoefficientList[Series[Product[(1+x^k)^PartitionsP[k],{k,1,nmax}],{x,0,nmax}],x]

A027998 Expansion of Product_{m>=1} (1+q^m)^(m^2).

Original entry on oeis.org

1, 1, 4, 13, 31, 83, 201, 487, 1141, 2641, 5972, 13309, 29248, 63360, 135688, 287197, 601629, 1247909, 2565037, 5226816, 10565132, 21192569, 42202909, 83466925, 163999684, 320230999, 621579965, 1199659836, 2302765961, 4397132933, 8354234552, 15795913477
Offset: 0

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Comments

In general, if g.f. = Product_{k>=1} (1 + x^k)^(c2*k^2 + c1*k + c0) and c2 > 0, then a(n) ~ exp(2*Pi/3 * (14*c2/15)^(1/4) * n^(3/4) + 3*c1 * Zeta(3) / Pi^2 * sqrt(15*n/(14*c2)) + (Pi * c0 * (5/(14*c2))^(1/4) / (2*3^(3/4)) - 9*c1^2 * Zeta(3)^2 * (15/(14*c2))^(5/4) / Pi^5) * n^(1/4) + 2025 * c1^3 * Zeta(3)^3 / (49 * c2^2 * Pi^8) - 15*c0*c1*Zeta(3) / (28*c2 * Pi^2)) * ((7*c2)/15)^(1/8) / (2^(15/8 + c0/2 + c1/12) * n^(5/8)). - Vaclav Kotesovec, Nov 08 2017

Crossrefs

Programs

  • Magma
    m:=50; R:=PowerSeriesRing(Rationals(), m); Coefficients(R! ( (&*[(1+x^k)^k^2: k in [1..m]]) )); // G. C. Greubel, Oct 31 2018
  • Maple
    with(numtheory):
    b:= proc(n) option remember;
          add((-1)^(n/d+1)*d^3, d=divisors(n))
        end:
    a:= proc(n) option remember;
          `if`(n=0, 1, add(b(k)*a(n-k), k=1..n)/n)
        end:
    seq(a(n), n=0..35);  # Alois P. Heinz, Aug 03 2013
  • Mathematica
    a[0] = 1; a[n_] := a[n] = 1/n*Sum[Sum[(-1)^(k/d+1)*d^3, {d, Divisors[k]}]*a[n-k], {k, 1, n}]; Table[a[n], {n, 0, 31} ] (* Jean-François Alcover, Jan 17 2014, after Vladeta Jovovic *)
    nmax=50; CoefficientList[Series[Product[(1+x^k)^(k^2),{k,1,nmax}],{x,0,nmax}],x] (* Vaclav Kotesovec, Mar 05 2015 *)
  • PARI
    x = 'x + O('x ^ 50); Vec(prod(k=1, 50, (1 + x^k)^(k^2))) \\ Indranil Ghosh, Apr 05 2017
    

Formula

a(n) = 1/n*Sum_{k=1..n} A078307(k)*a(n-k). - Vladeta Jovovic, Nov 22 2002
a(n) ~ 7^(1/8) * exp(2/3 * Pi * (14/15)^(1/4) * n^(3/4)) / (2^(15/8) * 15^(1/8) * n^(5/8)). - Vaclav Kotesovec, Mar 05 2015
G.f.: exp(Sum_{k>=1} (-1)^(k+1)*x^k*(1 + x^k)/(k*(1 - x^k)^3)). - Ilya Gutkovskiy, May 30 2018

A023872 Expansion of Product_{k>=1} (1 - x^k)^(-k^3).

Original entry on oeis.org

1, 1, 9, 36, 136, 477, 1703, 5746, 19099, 61622, 195366, 607069, 1856516, 5586870, 16579850, 48549116, 140438966, 401592524, 1136121837, 3181700219, 8825733603, 24261363403, 66124058839, 178757752892, 479513547399, 1276792213203, 3375707760306, 8864712158225
Offset: 0

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Author

Keywords

Crossrefs

Column k=3 of A144048.
Cf. A248882.

Programs

  • Magma
    m:=30; R:=PowerSeriesRing(Rationals(), m); Coefficients(R! ( (&*[1/(1-x^k)^k^3: k in [1..m]]) )); // G. C. Greubel, Oct 30 2018
    
  • Maple
    with(numtheory):
    a:= proc(n) option remember; `if`(n=0, 1,
          add(add(d*d^3, d=divisors(j)) *a(n-j), j=1..n)/n)
        end:
    seq(a(n), n=0..30);  # Alois P. Heinz, Nov 02 2012
  • Mathematica
    max = 27; Series[ Product[ 1/(1-x^k)^k^3, {k, 1, max}], {x, 0, max}] // CoefficientList[#, x]& (* Jean-François Alcover, Mar 05 2013 *)
  • PARI
    m=30; x='x+O('x^m); Vec(prod(k=1, m, 1/(1-x^k)^k^3)) \\ G. C. Greubel, Oct 30 2018
    
  • SageMath
    # uses[EulerTransform from A166861]
    b = EulerTransform(lambda n: n^3)
    print([b(n) for n in range(30)]) # Peter Luschny, Nov 11 2020

Formula

a(n) ~ (3*Zeta(5))^(59/600) * exp(5 * n^(4/5) * (3*Zeta(5))^(1/5) / 2^(7/5) + Zeta'(-3)) / (2^(41/200) * n^(359/600) * sqrt(5*Pi)), where Zeta(5) = A013663 = 1.036927755143369926..., Zeta'(-3) = ((gamma + log(2*Pi) - 11/6)/30 - 3*Zeta'(4)/Pi^4)/4 = 0.00537857635777430114441697421... . - Vaclav Kotesovec, Feb 27 2015
G.f.: exp( Sum_{n>=1} sigma_4(n)*x^n/n ). - Seiichi Manyama, Mar 04 2017
a(n) = (1/n)*Sum_{k=1..n} sigma_4(k)*a(n-k). - Seiichi Manyama, Mar 04 2017

Extensions

Definition corrected by Franklin T. Adams-Watters and R. J. Mathar, Dec 04 2006

A258343 Expansion of Product_{k>=1} (1+x^k)^(k*(k+1)*(k+2)/6).

Original entry on oeis.org

1, 1, 4, 14, 36, 101, 260, 669, 1669, 4116, 9932, 23636, 55483, 128532, 294422, 667026, 1496232, 3324720, 7323570, 15998749, 34679966, 74622839, 159454379, 338472749, 713956569, 1496950669, 3120663129, 6469901522, 13343153563, 27379250529, 55907749171
Offset: 0

Views

Author

Vaclav Kotesovec, May 27 2015

Keywords

Crossrefs

Programs

  • Maple
    b:= proc(n, i) option remember; `if`(n=0, 1, `if`(i<1, 0, add(
          binomial(binomial(i+2, 3), j)*b(n-i*j, i-1), j=0..n/i)))
        end:
    a:= n-> b(n$2):
    seq(a(n), n=0..30);  # Alois P. Heinz, May 28 2018
  • Mathematica
    nmax=40; CoefficientList[Series[Product[(1+x^k)^(k*(k+1)*(k+2)/6),{k,1,nmax}],{x,0,nmax}],x]

Formula

a(n) ~ (3*Zeta(5))^(1/10) / (2^(523/720) * 5^(2/5) * sqrt(Pi) * n^(3/5)) * exp(-2401 * Pi^16 / (10497600000000 * Zeta(5)^3) + 49*Pi^8 * Zeta(3) / (16200000 * Zeta(5)^2) - Zeta(3)^2 / (150*Zeta(5)) + (343*Pi^12 / (2430000000 * 2^(3/5) * 15^(1/5) * Zeta(5)^(11/5)) - 7*Pi^4 * Zeta(3) / (4500 * 2^(3/5) * 15^(1/5) * Zeta(5)^(6/5))) * n^(1/5) + (-49*Pi^8 / (1080000 * 2^(1/5) * 15^(2/5) * Zeta(5)^(7/5)) + Zeta(3) / (2^(6/5) * (15*Zeta(5))^(2/5))) * n^(2/5) + 7*Pi^4 / (180 * 2^(4/5) * (15*Zeta(5))^(3/5)) * n^(3/5) + 5*(15*Zeta(5))^(1/5) / 2^(12/5) * n^(4/5)), where Zeta(3) = A002117, Zeta(5) = A013663.
G.f.: exp(Sum_{k>=1} (-1)^(k+1)*x^k/(k*(1 - x^k)^4)). - Ilya Gutkovskiy, May 28 2018

A206623 G.f.: Product_{n>0} ( (1+x^n)/(1-x^n) )^(n^3).

Original entry on oeis.org

1, 2, 18, 88, 398, 1768, 7508, 30644, 121310, 467234, 1756080, 6457168, 23274788, 82381584, 286760344, 982874120, 3320800590, 11070619228, 36446345198, 118581503192, 381552358872, 1214868568728, 3829841265428, 11959828895612, 37013411304892, 113570015855642
Offset: 0

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Author

Paul D. Hanna, Feb 12 2012

Keywords

Comments

Convolution of A023872 and A248882. - Vaclav Kotesovec, Aug 19 2015

Examples

			G.f.: A(x) = 1 + 2*x + 18*x^2 + 88*x^3 + 398*x^4 + 1768*x^5 + 7508*x^6 +...
where A(x) = (1+x)/(1-x) * (1+x^2)^8/(1-x^2)^8 * (1+x^3)^27/(1-x^3)^27 *...
Also, A(x) = Euler transform of [2,15,54,120,250,405,686,960,1458,...]:
A(x) = 1/((1-x)^2*(1-x^2)^15*(1-x^3)^54*(1-x^4)^120*(1-x^5)^250*(1-x^6)^405*...).
		

Crossrefs

Cf. A156616, A206622, A206624, A001159 (sigma_4).

Programs

  • Mathematica
    nmax = 40; CoefficientList[Series[Product[((1+x^k)/(1-x^k))^(k^3), {k, 1, nmax}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Aug 19 2015 *)
  • PARI
    {a(n)=polcoeff(prod(m=1,n+1,((1+x^m)/(1-x^m+x*O(x^n)))^(m^3)),n)}
    
  • PARI
    {a(n)=polcoeff(exp(sum(m=1, n, (sigma(2*m, 4)-sigma(m, 4))/8*x^m/m)+x*O(x^n)), n)}
    
  • PARI
    {a(n)=local(InvEulerGF=x*(2+15*x+46*x^2+60*x^3+46*x^4+15*x^5+2*x^6)/(1-x^2+x*O(x^n))^4);polcoeff(1/prod(k=1,n,(1-x^k+x*O(x^n))^polcoeff(InvEulerGF,k)),n)}
    for(n=0,30,print1(a(n),", "))

Formula

G.f.: exp( Sum_{n>=1} (sigma_4(2*n) - sigma_4(n))/8 * x^n/n ), where sigma_4(n) is the sum of 4th powers of divisors of n (A001159).
Inverse Euler transform has g.f.: x*(2 + 15*x + 46*x^2 + 60*x^3 + 46*x^4 + 15*x^5 + 2*x^6)/(1-x^2)^4.
a(n) ~ (93*Zeta(5))^(59/600) * exp(5/4 * (93*Zeta(5)/2)^(1/5) * n^(4/5) + Zeta'(-3)) / (2^(59/100) * sqrt(5*Pi) * n^(359/600)), where Zeta(5) = A013663, Zeta'(-3) = A259068. - Vaclav Kotesovec, Aug 19 2015

A248883 Expansion of Product_{k>=1} (1+x^k)^(k^4).

Original entry on oeis.org

1, 1, 16, 97, 457, 2297, 11113, 52049, 235334, 1039886, 4497930, 19074006, 79418883, 325184763, 1311252535, 5212704708, 20449320159, 79231806015, 303428397505, 1149325838320, 4308477305997, 15993198330782, 58815616643170, 214383601754107, 774837953045873
Offset: 0

Views

Author

Vaclav Kotesovec, Mar 05 2015

Keywords

Crossrefs

Column k=4 of A284992.

Programs

  • Magma
    m:=50; R:=PowerSeriesRing(Rationals(), m); Coefficients(R! ( (&*[(1+x^k)^k^4: k in [1..m]]) )); // G. C. Greubel, Oct 31 2018
  • Maple
    b:= proc(n) option remember; add(
          (-1)^(n/d+1)*d^5, d=numtheory[divisors](n))
        end:
    a:= proc(n) option remember; `if`(n=0, 1,
          add(b(k)*a(n-k), k=1..n)/n)
        end:
    seq(a(n), n=0..35);  # Alois P. Heinz, Oct 16 2017
  • Mathematica
    nmax=50; CoefficientList[Series[Product[(1+x^k)^(k^4),{k,1,nmax}],{x,0,nmax}],x]
  • PARI
    x = 'x + O('x^50); Vec(prod(k=1, 50, (1 + x^k)^(k^4))) \\ Indranil Ghosh, Apr 06 2017
    

Formula

a(n) ~ 31^(1/12) * exp(1/5 * (31/7)^(1/6) * 6^(2/3) * Pi * n^(5/6)) / (2^(7/6) * 3^(2/3) * 7^(1/12) * n^(7/12)).
a(0) = 1, a(n) = (1/n)*Sum_{k=1..n} A284926(k)*a(n-k) for n > 0. - Seiichi Manyama, Apr 06 2017
G.f.: exp(Sum_{k>=1} (-1)^(k+1)*x^k*(1 + 11*x^k + 11*x^(2*k) + x^(3*k))/(k*(1 - x^k)^5)). - Ilya Gutkovskiy, May 30 2018

A248884 Expansion of Product_{k>=1} (1+x^k)^(k^5).

Original entry on oeis.org

1, 1, 32, 275, 1763, 12421, 85808, 561074, 3535678, 21815897, 131733641, 778099521, 4505634324, 25635135074, 143507764032, 791243636305, 4300983535471, 23070300486656, 122213931799869, 639848848696540, 3312824859756453, 16972058378914997, 86082216143323410
Offset: 0

Views

Author

Vaclav Kotesovec, Mar 05 2015

Keywords

Comments

In general, for m > 0, if g.f. = Product_{k>=1} (1+x^k)^(k^m), then a(n) ~ 2^(zeta(-m)) * ((1-2^(-m-1)) * Gamma(m+2) * zeta(m+2))^(1/(2*m+4)) * exp((m+2)/(m+1) * ((1-2^(-m-1)) * Gamma(m+2) * zeta(m+2))^(1/(m+2)) * n^((m+1)/(m+2))) / (sqrt(2*Pi*(m+2)) * n^((m+3)/(2*m+4))).

Crossrefs

Cf. A026007 (m=1), A027998 (m=2), A248882 (m=3), A248883 (m=4).
Column k=5 of A284992.

Programs

  • Magma
    m:=50; R:=PowerSeriesRing(Rationals(), m); Coefficients(R! ( (&*[(1+x^k)^k^5: k in [1..m]]) )); // G. C. Greubel, Oct 31 2018
  • Maple
    b:= proc(n) option remember; add(
          (-1)^(n/d+1)*d^6, d=numtheory[divisors](n))
        end:
    a:= proc(n) option remember; `if`(n=0, 1,
          add(b(k)*a(n-k), k=1..n)/n)
        end:
    seq(a(n), n=0..35);  # Alois P. Heinz, Oct 16 2017
  • Mathematica
    nmax=50; CoefficientList[Series[Product[(1+x^k)^(k^5),{k,1,nmax}],{x,0,nmax}],x]
  • PARI
    m=50; x='x+O('x^m); Vec(prod(k=1, m, (1+x^k)^k^5)) \\ G. C. Greubel, Oct 31 2018
    

Formula

a(n) ~ (5*zeta(7))^(1/14) * 3^(2/7) * exp(zeta(7)^(1/7) * 2^(-9/7) * 3^(-3/7) * 5^(1/7) * 7^(8/7) * n^(6/7)) / (2^(163/252) * 7^(3/7) * sqrt(Pi) * n^(4/7)), where zeta(7) = A013665.
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