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 1-4 of 4 results.

A002348 Degree of rational Poncelet porism of n-gon.

Original entry on oeis.org

1, 2, 3, 4, 6, 8, 9, 12, 15, 16, 21, 24, 24, 32, 36, 36, 45, 48, 48, 60, 66, 64, 75, 84, 81, 96, 105, 96, 120, 128, 120, 144, 144, 144, 171, 180, 168, 192, 210, 192, 231, 240, 216, 264, 276, 256, 294, 300, 288, 336, 351, 324, 360, 384, 360, 420, 435, 384, 465
Offset: 3

Views

Author

N. J. A. Sloane

Keywords

Examples

			For a triangle the degree is 1, thus a(3) = 1. - _Michael Somos_, Dec 07 2018

References

  • Kerawala, S. M.; Poncelet Porism in Two Circles. Bull. Calcutta Math. Soc. 39, 85-105, 1947.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

Crossrefs

Cf. A027748, A124010.
Cf. A007434, A209229, A007434, A328407, A007814, A000265, A008683, A010724.

Programs

  • Haskell
    a002348 n = product (zipWith d ps es) * 4 ^ e0 `div` 8 where
   d p e = (p ^ 2 - 1) * p ^ e
   e0 = if even n then head $ a124010_row n else 0
   es = map ((* 2) . subtract 1) $
            if even n then tail $ a124010_row n else a124010_row n
   ps = if even n then tail $ a027748_row n else a027748_row n
-- Reinhard Zumkeller, Mar 18 2012
  • Mathematica
    Poncelet[ n_Integer /; n >= 3 ] := Module[ {p, a, i}, {p, a}=Transpose[ FactorInteger[ n ] ];
If[ p[[1]]==2, 4^a[[1]] Product[ p[[i]]^(2(a[[i]] - 1))(p[[i]]^2 - 1), {i, 2, Length[ p ]} ]/8, (* Else *) Product[ p[[i]]^(2(a[[i]] - 1))(p[[i]]^2 - 1), {i, Length[ p ]} ]/8 ] ]
  • PARI
    {a(n) = my(p, e); if( n<3, 0, p=factor(n)~; e=p[2,]; p=p[1,]; if( p[1]==2, 4^e[1], 1) * prod(i=1 + (p[1]==2), length(p), p[i]^(2*(e[i] - 1)) * (p[i]^2 - 1)) / 8)}; /* Michael Somos, Dec 09 1999 */

Formula

From Ridouane Oudra, Jul 19 2025: (Start)
a(n) = (1/8) * Sum_{k=0..A007814(n)} A007434(n/2^k).
a(n) = (1/8) * Sum_{d|n} A209229(d)*A007434(n/d).
a(n) = (1/8) * Sum_{d|n} A328407(d)*mu(n/d).
a(n) = (n^2/8) * Prod_{p|n, p prime > 2} (1 - 1/p^2).
a(n) = (1/8) * 4^A007814(n) * A007434(A000265(n)).
a(n) = A007434(n)/A010724(n). (End)

Extensions

Extended with Mathematica program by Eric W. Weisstein

A328408 G.f. A(x) satisfies: A(x) = A(x^2) + x * (1 + 4*x + x^2) / (1 - x)^4.

Original entry on oeis.org

1, 9, 27, 73, 125, 243, 343, 585, 729, 1125, 1331, 1971, 2197, 3087, 3375, 4681, 4913, 6561, 6859, 9125, 9261, 11979, 12167, 15795, 15625, 19773, 19683, 25039, 24389, 30375, 29791, 37449, 35937, 44217, 42875, 53217, 50653, 61731, 59319, 73125, 68921, 83349, 79507, 97163, 91125
Offset: 1

Views

Author

Ilya Gutkovskiy, Oct 14 2019

Keywords

Links

Crossrefs

Cf. A000578, A001511, A016755, A023872, A059376, A129527, A209229, A328407.

Programs

  • Magma
    [n eq 1 select 1 else IsOdd(n) select n^3 else Self(n div 2)+n^3 :n in [1..45]]; // Marius A. Burtea, Oct 15 2019
  • Mathematica
    nmax = 45; CoefficientList[Series[Sum[x^(2^k) (1 + 4 x^(2^k) + x^(2^(k + 1)))/(1 - x^(2^k))^4, {k, 0, Floor[Log[2, nmax]] + 1}], {x, 0, nmax}], x] // Rest
a[n_] := If[EvenQ[n], a[n/2] + n^3, n^3]; Table[a[n], {n, 1, 45}]
Table[DivisorSum[n, Boole[IntegerQ[Log[2, n/#]]] #^3 &], {n, 1, 45}]
f[p_, e_] :=p^(3*e); f[2, e_] := (8^(e+1)-1)/7; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Oct 23 2023 *)

Formula

G.f.: Sum_{k>=0} x^(2^k) * (1 + 4*x^(2^k) + x^(2^(k+1))) / (1 - x^(2^k))^4.
G.f.: (1/7) * Sum_{k>=1} J_3(2*k) * x^k / (1 - x^k), where J_3() is the Jordan function (A059376).
Dirichlet g.f.: zeta(s-3) / (1 - 2^(-s)).
a(2*n) = a(n) + 8*n^3, a(2*n+1) = (2*n + 1)^3.
a(n) = Sum_{d|n} A209229(n/d) * d^3.
Product_{n>=1} (1 + x^n)^a(n) = g.f. for A023872.
Sum_{k=1..n} a(k) ~ 4*n^4/15. - Vaclav Kotesovec, Oct 15 2019
Multiplicative with a(2^e) = (8^(e+1)-1)/7, and a(p^e) = p^(3*e) for an odd prime p. - Amiram Eldar, Oct 23 2023

A338045 G.f.: Sum_{k>=0} x^(2^k) / (1 - x^(2^k))^3.

Original entry on oeis.org

1, 4, 6, 14, 15, 27, 28, 50, 45, 70, 66, 105, 91, 133, 120, 186, 153, 216, 190, 280, 231, 319, 276, 405, 325, 442, 378, 539, 435, 585, 496, 714, 561, 748, 630, 882, 703, 931, 780, 1100, 861, 1134, 946, 1309, 1035, 1357, 1128, 1581, 1225, 1600, 1326, 1820, 1431, 1863, 1540
Offset: 1

Views

Author

Ilya Gutkovskiy, Oct 08 2020

Keywords

Crossrefs

Cf. A000217, A000294, A000384, A001511, A014105, A129527, A195094, A209229, A328407, A338046.

Programs

  • Mathematica
    nmax = 55; CoefficientList[Series[Sum[x^(2^k) /(1 - x^(2^k))^3, {k, 0, Floor[Log[2, nmax]] + 1}], {x, 0, nmax}], x] // Rest
a[n_] := If[EvenQ[n], a[n/2] + n (n + 1)/2, n (n + 1)/2]; Table[a[n], {n, 1, 55}]
Table[(1/2) DivisorSum[n, Boole[IntegerQ[Log[2, n/#]]] # (# + 1) &], {n, 1, 55}]

Formula

G.f. A(x) satisfies: A(x) = A(x^2) + x / (1 - x)^3.
a(2*n) = a(n) + A014105(n), a(2*n+1) = A000384(n+1).
a(n) = (1/2) * Sum_{d|n} A209229(n/d) * d * (d + 1).
a(n) = Sum_{d|n} A195094(d).
Product_{n>=1} (1 + x^n)^a(n) = g.f. for A000294.

A338046 G.f.: Sum_{k>=0} x^(2^k) / (1 - x^(2^k))^4.

Original entry on oeis.org

1, 5, 10, 25, 35, 66, 84, 145, 165, 255, 286, 430, 455, 644, 680, 961, 969, 1305, 1330, 1795, 1771, 2310, 2300, 3030, 2925, 3731, 3654, 4704, 4495, 5640, 5456, 6945, 6545, 8109, 7770, 9741, 9139, 11210, 10660, 13275, 12341, 15015, 14190, 17490, 16215, 19596, 18424, 22630
Offset: 1

Views

Author

Ilya Gutkovskiy, Oct 08 2020

Keywords

Crossrefs

Cf. A000292, A000335, A000447, A001511, A002492, A129527, A209229, A328407, A338045.

Programs

  • Mathematica
    nmax = 48; CoefficientList[Series[Sum[x^(2^k) /(1 - x^(2^k))^4, {k, 0, Floor[Log[2, nmax]] + 1}], {x, 0, nmax}], x] // Rest
a[n_] := If[EvenQ[n], a[n/2] + n (n + 1) (n + 2)/6, n (n + 1) (n + 2)/6]; Table[a[n], {n, 1, 48}]
Table[(1/6) DivisorSum[n, Boole[IntegerQ[Log[2, n/#]]] # (# + 1) (# + 2) &], {n, 1, 48}]

Formula

G.f. A(x) satisfies: A(x) = A(x^2) + x / (1 - x)^4.
a(2*n) = a(n) + A002492(n), a(2*n+1) = A000447(n+1).
a(n) = (1/6) * Sum_{d|n} A209229(n/d) * d * (d + 1) * (d + 2).
Product_{n>=1} (1 + x^n)^a(n) = g.f. for A000335.
Showing 1-4 of 4 results.

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