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 11 results. Next

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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Author

Keywords

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

A078306 a(n) = Sum_{d divides n} (-1)^(n/d+1)*d^2.

Original entry on oeis.org

1, 3, 10, 11, 26, 30, 50, 43, 91, 78, 122, 110, 170, 150, 260, 171, 290, 273, 362, 286, 500, 366, 530, 430, 651, 510, 820, 550, 842, 780, 962, 683, 1220, 870, 1300, 1001, 1370, 1086, 1700, 1118, 1682, 1500, 1850, 1342, 2366, 1590, 2210, 1710, 2451, 1953
Offset: 1

Views

Author

Vladeta Jovovic, Nov 22 2002

Keywords

Crossrefs

Programs

  • Mathematica
    a[n_] := Sum[(-1)^(n/d+1)*d^2, {d, Divisors[n]}]; Array[a, 50] (* Jean-François Alcover, Apr 17 2014 *)
    Table[CoefficientList[Series[-Log[Product[1/(x^k + 1)^k, {k, 1, 90}]], {x, 0, 80}], x][[n + 1]] n, {n, 1, 80}] (* Benedict W. J. Irwin, Jul 05 2016 *)
  • PARI
    a(n) = sumdiv(n, d, (-1)^(n/d+1)*d^2); \\ Michel Marcus, Jul 06 2016
    
  • Python
    from sympy import divisors
    print([sum((-1)**(n//d + 1)*d**2 for d in divisors(n)) for n in range(1, 51)]) # Indranil Ghosh, Apr 05 2017

Formula

G.f.: Sum_{n >= 1} n^2*x^n/(1+x^n).
Multiplicative with a(2^e) = (2*4^e+1)/3, a(p^e) = (p^(2*e+2)-1)/(p^2-1), p > 2.
L.g.f.: -log(Product_{ k>0 } 1/(x^k+1)^k) = Sum_{ n>0 } (a(n)/n)*x^n. - Benedict W. J. Irwin, Jul 05 2016
G.f.: Sum_{n >= 1} (-1)^(n+1) * x^n*(1 + x^n)/(1 - x^n)^3. - Peter Bala, Jan 14 2021
From Vaclav Kotesovec, Aug 07 2022: (Start)
Dirichlet g.f.: zeta(s) * zeta(s-2) * (1 - 2^(1-s)).
Sum_{k=1..n} a(k) ~ zeta(3) * n^3 / 4. (End)

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

A284900 a(n) = Sum_{d|n} (-1)^(n/d+1)*d^4.

Original entry on oeis.org

1, 15, 82, 239, 626, 1230, 2402, 3823, 6643, 9390, 14642, 19598, 28562, 36030, 51332, 61167, 83522, 99645, 130322, 149614, 196964, 219630, 279842, 313486, 391251, 428430, 538084, 574078, 707282, 769980, 923522, 978671, 1200644, 1252830, 1503652, 1587677
Offset: 1

Views

Author

Seiichi Manyama, Apr 05 2017

Keywords

Comments

Multiplicative because this sequence is the Dirichlet convolution of A000583 and A062157 which are both multiplicative. - Andrew Howroyd, Jul 20 2018

Crossrefs

Sum_{d|n} (-1)^(n/d+1)*d^k: A000593 (k=1), A078306 (k=2), A078307 (k=3), this sequence (k=4), A284926 (k=5), A284927 (k=6), A321552 (k=7), A321553 (k=8), A321554 (k=9), A321555 (k=10), A321556 (k=11), A321557 (k=12).

Programs

  • Mathematica
    Table[Sum[(-1)^(n/d + 1)*d^4, {d, Divisors[n]}], {n, 50}] (* Indranil Ghosh, Apr 05 2017 *)
    f[p_, e_] := (p^(4*e + 4) - 1)/(p^4 - 1); f[2, e_] := (7*2^(4*e + 1) + 1)/15; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 50] (* Amiram Eldar, Nov 11 2022 *)
  • PARI
    a(n) = sumdiv(n, d, (-1)^(n/d + 1)*d^4); \\ Indranil Ghosh, Apr 05 2017
    
  • Python
    from sympy import divisors
    print([sum([(-1)**(n//d + 1)*d**4 for d in divisors(n)]) for n in range(1, 51)]) # Indranil Ghosh, Apr 05 2017

Formula

G.f.: Sum_{k>=1} k^4*x^k/(1 + x^k). - Ilya Gutkovskiy, Apr 07 2017
From Amiram Eldar, Nov 11 2022: (Start)
Multiplicative with a(2^e) = (7*2^(4*e+1)+1)/15, and a(p^e) = (p^(4*e+4) - 1)/(p^4 - 1) if p > 2.
Sum_{k=1..n} a(k) ~ c * n^5, where c = 3*zeta(5)/16 = 0.194423... . (End)

A284926 a(n) = Sum_{d|n} (-1)^(n/d+1)*d^5.

Original entry on oeis.org

1, 31, 244, 991, 3126, 7564, 16808, 31711, 59293, 96906, 161052, 241804, 371294, 521048, 762744, 1014751, 1419858, 1838083, 2476100, 3097866, 4101152, 4992612, 6436344, 7737484, 9768751, 11510114, 14408200, 16656728, 20511150, 23645064, 28629152, 32472031, 39296688
Offset: 1

Views

Author

Seiichi Manyama, Apr 06 2017

Keywords

Comments

Multiplicative because this sequence is the Dirichlet convolution of A000584 and A062157 which are both multiplicative. - Andrew Howroyd, Jul 20 2018

Crossrefs

Sum_{d|n} (-1)^(n/d+1)*d^k: A000593 (k=1), A078306 (k=2), A078307 (k=3), A284900 (k=4), this sequence (k=5), A284927 (k=6), A321552 (k=7), A321553 (k=8), A321554 (k=9), A321555 (k=10), A321556 (k=11), A321557 (k=12).

Programs

  • Mathematica
    Table[Sum[(-1)^(n/d + 1)*d^5, {d, Divisors[n]}], {n, 50}] (* Indranil Ghosh, Apr 06 2017 *)
    f[p_, e_] := (p^(5*e + 5) - 1)/(p^5 - 1); f[2, e_] := (15*2^(5*e + 1) + 1)/31; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 50] (* Amiram Eldar, Nov 11 2022 *)
  • PARI
    a(n) = sumdiv(n, d, (-1)^(n/d + 1)*d^5); \\ Indranil Ghosh, Apr 06 2017
    
  • Python
    from sympy import divisors
    print([sum((-1)**(n//d + 1)*d**5 for d in divisors(n)) for n in range(1, 51)]) # Indranil Ghosh, Apr 06 2017

Formula

G.f.: Sum_{k>=1} k^5*x^k/(1 + x^k). - Ilya Gutkovskiy, Apr 07 2017
From Amiram Eldar, Nov 11 2022: (Start)
Multiplicative with a(2^e) = (15*2^(5*e+1)+1)/31, and a(p^e) = (p^(5*e+5) - 1)/(p^5 - 1) if p > 2.
Sum_{k=1..n} a(k) ~ c * n^6, where c = 31*zeta(6)/192 = 0.164258... . (End)

Extensions

Keyword:mult added by Andrew Howroyd, Jul 23 2018

A284927 a(n) = Sum_{d|n} (-1)^(n/d+1)*d^6.

Original entry on oeis.org

1, 63, 730, 4031, 15626, 45990, 117650, 257983, 532171, 984438, 1771562, 2942630, 4826810, 7411950, 11406980, 16510911, 24137570, 33526773, 47045882, 62988406, 85884500, 111608406, 148035890, 188327590, 244156251, 304089030, 387952660, 474247150, 594823322
Offset: 1

Views

Author

Seiichi Manyama, Apr 06 2017

Keywords

Comments

Multiplicative because this sequence is the Dirichlet convolution of A001014 and A062157 which are both multiplicative. - Andrew Howroyd, Jul 20 2018

Crossrefs

Sum_{d|n} (-1)^(n/d+1)*d^k: A000593 (k=1), A078306 (k=2), A078307 (k=3), A284900 (k=4), A284926 (k=5), this sequence (k=6), A321552 (k=7), A321553 (k=8), A321554 (k=9), A321555 (k=10), A321556 (k=11), A321557 (k=12).

Programs

  • Mathematica
    Table[Sum[(-1)^(n/d + 1)*d^6, {d, Divisors[n]}], {n, 50}] (* Indranil Ghosh, Apr 06 2017 *)
    f[p_, e_] := (p^(6*e + 6) - 1)/(p^6 - 1); f[2, e_] := (31*2^(6*e + 1) + 1)/63; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 50] (* Amiram Eldar, Nov 11 2022 *)
  • PARI
    a(n) = sumdiv(n, d, (-1)^(n/d + 1)*d^6); \\ Indranil Ghosh, Apr 06 2017
    
  • Python
    from sympy import divisors
    print([sum([(-1)**(n//d + 1)*d**6 for d in divisors(n)]) for n in range(1, 51)]) # Indranil Ghosh, Apr 06 2017

Formula

G.f.: Sum_{k>=1} k^6*x^k/(1 + x^k). - Ilya Gutkovskiy, Apr 07 2017
From Amiram Eldar, Nov 11 2022: (Start)
Multiplicative with a(2^e) = (31*2^(6*e+1)+1)/63, and a(p^e) = (p^(6*e+6) - 1)/(p^6 - 1) if p > 2.
Sum_{k=1..n} a(k) ~ c * n^7, where c = 9*zeta(7)/64 = 0.141799... . (End)

Extensions

Keyword:mult added by Andrew Howroyd, Jul 23 2018

A321552 a(n) = Sum_{d|n} (-1)^(n/d+1)*d^7.

Original entry on oeis.org

1, 127, 2188, 16255, 78126, 277876, 823544, 2080639, 4785157, 9922002, 19487172, 35565940, 62748518, 104590088, 170939688, 266321791, 410338674, 607714939, 893871740, 1269938130, 1801914272, 2474870844, 3404825448, 4552438132, 6103593751, 7969061786, 10465138360, 13386707720, 17249876310
Offset: 1

Views

Author

N. J. A. Sloane, Nov 23 2018

Keywords

Crossrefs

Sum_{k>=1} k^b*x^k/(1 + x^k): A000593 (b=1), A078306 (b=2), A078307 (b=3), A284900 (b=4), A284926 (b=5), A284927 (b=6), this sequence (b=7), A321553 (b=8), A321554 (b=9), A321555 (b=10), A321556 (b=11), A321557 (b=12).
Cf. A321543 - A321565, A321807 - A321836 for similar sequences.
Cf. A013666.

Programs

  • Mathematica
    f[p_, e_] := (p^(7*e + 7) - 1)/(p^7 - 1); f[2, e_] := (63*2^(7*e + 1) + 1)/127; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 50] (* Amiram Eldar, Nov 11 2022 *)
  • PARI
    apply( A321552(n)=sumdiv(n, d, (-1)^(n\d-1)*d^7), [1..30]) \\ M. F. Hasler, Nov 26 2018

Formula

G.f.: Sum_{k>=1} k^7*x^k/(1 + x^k). - Seiichi Manyama, Nov 23 2018
From Amiram Eldar, Nov 11 2022: (Start)
Multiplicative with a(2^e) = (63*2^(7*e+1)+1)/127, and a(p^e) = (p^(7*e+7) - 1)/(p^7 - 1) if p > 2.
Sum_{k=1..n} a(k) ~ c * n^8, where c = 127*zeta(8)/1024 = 0.124529... . (End)

A309337 a(n) = n^3 if n odd, 3*n^3/4 if n even.

Original entry on oeis.org

0, 1, 6, 27, 48, 125, 162, 343, 384, 729, 750, 1331, 1296, 2197, 2058, 3375, 3072, 4913, 4374, 6859, 6000, 9261, 7986, 12167, 10368, 15625, 13182, 19683, 16464, 24389, 20250, 29791, 24576, 35937, 29478, 42875, 34992, 50653, 41154, 59319, 48000, 68921, 55566, 79507, 63888, 91125
Offset: 0

Views

Author

Ilya Gutkovskiy, Jul 24 2019

Keywords

Comments

Moebius transform of A078307.

Crossrefs

Programs

  • Mathematica
    a[n_] := If[OddQ[n], n^3, 3 n^3/4]; Table[a[n], {n, 0, 45}]
    nmax = 45; CoefficientList[Series[x (1 + 6 x + 23 x^2 + 24 x^3 + 23 x^4 + 6 x^5 + x^6)/(1 - x^2)^4, {x, 0, nmax}], x]
    LinearRecurrence[{0, 4, 0, -6, 0, 4, 0, -1}, {0, 1, 6, 27, 48, 125, 162, 343}, 46]
    Table[n^3 (7 - (-1)^n)/8, {n, 0, 45}]

Formula

G.f.: x * (1 + 6*x + 23*x^2 + 24*x^3 + 23*x^4 + 6*x^5 + x^6)/(1 - x^2)^4.
G.f.: Sum_{k>=1} J_3(k) * x^k/(1 + x^k), where J_3() is the Jordan function (A059376).
Dirichlet g.f.: zeta(s-3) * (1 - 2^(1-s)).
a(n) = n^3 * (7 - (-1)^n)/8.
a(n) = Sum_{d|n} (-1)^(n/d + 1) * J_3(d).
Sum_{n>=1} 1/a(n) = 25*zeta(3)/24 = 1.252142607457910713958...
Multiplicative with a(2^e) = 3*2^(3*e-2), and a(p^e) = p^(3*e) for odd primes p. - Amiram Eldar, Oct 26 2020
a(n) = Sum_{1 <= i, j, k <= n} (-1)^(1 + gcd(i,j,k,n)) = Sum_{d | n} (-1)^(d+1) * J_3(n/d). Cf. A129194. - Peter Bala, Jan 16 2024

A322081 Square array A(n,k), n >= 1, k >= 0, read by antidiagonals: A(n,k) = Sum_{d|n} (-1)^(n/d+1)*d^k.

Original entry on oeis.org

1, 1, 0, 1, 1, 2, 1, 3, 4, -1, 1, 7, 10, 1, 2, 1, 15, 28, 11, 6, 0, 1, 31, 82, 55, 26, 4, 2, 1, 63, 244, 239, 126, 30, 8, -2, 1, 127, 730, 991, 626, 196, 50, 1, 3, 1, 255, 2188, 4031, 3126, 1230, 344, 43, 13, 0, 1, 511, 6562, 16255, 15626, 7564, 2402, 439, 91, 6, 2, 1, 1023, 19684, 65279, 78126, 45990, 16808, 3823, 757, 78, 12, -2
Offset: 1

Views

Author

Ilya Gutkovskiy, Nov 26 2018

Keywords

Examples

			Square array begins:
   1,  1,   1,    1,     1,     1,  ...
   0,  1,   3,    7,    15,    31,  ...
   2,  4,  10,   28,    82,   244,  ...
  -1,  1,  11,   55,   239,   991,  ...
   2,  6,  26,  126,   626,  3126,  ...
   0,  4,  30,  196,  1230,  7564,  ...
		

Crossrefs

Programs

  • Mathematica
    Table[Function[k, Sum[(-1)^(n/d + 1) d^k, {d, Divisors[n]}]][i - n], {i, 0, 12}, {n, 1, i}] // Flatten
    Table[Function[k, SeriesCoefficient[Sum[j^k x^j/(1 + x^j), {j, 1, n}], {x, 0, n}]][i - n], {i, 0, 12}, {n, 1, i}] // Flatten
  • PARI
    T(n,k)={sumdiv(n, d, (-1)^(n/d+1)*d^k)}
    for(n=1, 10, for(k=0, 8, print1(T(n, k), ", ")); print); \\ Andrew Howroyd, Nov 26 2018

Formula

G.f. of column k: Sum_{j>=1} j^k*x^j/(1 + x^j).

A321438 a(n) = Sum_{d|n} (-1)^(n/d+1)*d^n.

Original entry on oeis.org

1, 3, 28, 239, 3126, 45990, 823544, 16711423, 387440173, 9990235398, 285311670612, 8913939907598, 302875106592254, 11111328602501550, 437893920912786408, 18446462594437808127, 827240261886336764178, 39346258082220810086373, 1978419655660313589123980, 104857499999905732078938574
Offset: 1

Views

Author

Ilya Gutkovskiy, Nov 09 2018

Keywords

Crossrefs

Programs

  • Magma
    m:=20; R:=PowerSeriesRing(Integers(), m); Coefficients(R!( (&+[(k*x)^k/(1+(k*x)^k): k in [1..m]]) ));  // G. C. Greubel, Nov 11 2018
  • Mathematica
    Table[Sum[(-1)^(n/d + 1) d^n, {d, Divisors[n]}], {n, 20}]
    nmax = 20; Rest[CoefficientList[Series[Sum[(k x)^k/(1 + (k x)^k), {k, 1, nmax}], {x, 0, nmax}], x]]
    nmax = 20; Rest[CoefficientList[Series[Log[Product[(1 + k^k x^k)^(1/k), {k, 1, nmax}]], {x, 0, nmax}], x] Range[0, nmax]]
  • PARI
    a(n) = sumdiv(n, d, (-1)^(n/d+1)*d^n); \\ Michel Marcus, Nov 09 2018
    

Formula

G.f.: Sum_{k>=1} (k*x)^k/(1 + (k*x)^k).
L.g.f.: log(Product_{k>=1} (1 + k^k*x^k)^(1/k)) = Sum_{n>=1} a(n)*x^n/n.
a(n) ~ n^n. - Vaclav Kotesovec, Nov 10 2018
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