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 1-8 of 8 results.

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)

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)

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

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)

A284898 Expansion of Product_{k>=1} 1/(1+x^k)^(k^4) in powers of x.

Original entry on oeis.org

1, -1, -15, -66, -54, 725, 4580, 12739, 3346, -149076, -791226, -2182124, -1656973, 16553206, 100646954, 318795473, 506196578, -818806580, -9148048880, -36415709566, -87180585636, -70923559814, 484810027389, 2992082912770, 9866919438716, 19936695359140
Offset: 0

Views

Author

Seiichi Manyama, Apr 05 2017

Keywords

Crossrefs

Cf. A248883.
Product_{k>=1} 1/(1+x^k)^(k^m): A081362 (m=0), A255528 (m=1), A284896 (m=2), A284897 (m=3), this sequence (m=4), A284899 (m=5).

Programs

  • Mathematica
    CoefficientList[Series[Product[1/(1 + x^k)^(k^4) , {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^4))) \\ Indranil Ghosh, Apr 05 2017

Formula

a(0) = 1, a(n) = -(1/n)*Sum_{k=1..n} A284926(k)*a(n-k) for n > 0. - Seiichi Manyama, Apr 06 2017

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
Showing 1-8 of 8 results.