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 1-5 of 5 results.

A309083 a(n) = n - floor(n/2^4) + floor(n/3^4) - floor(n/4^4) + ...

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71
Offset: 1

Views

Author

Ilya Gutkovskiy, Jul 11 2019

Keywords

Links

Crossrefs

Cf. A013938, A059851, A063775, A309081, A309082.

Programs

  • Mathematica
    Table[Sum[(-1)^(k + 1) Floor[n/k^4], {k, 1, n}], {n, 1, 75}]
nmax = 75; CoefficientList[Series[1/(1 - x) Sum[(-1)^(k + 1) x^(k^4)/(1 - x^(k^4)), {k, 1, Floor[nmax^(1/4)] + 1}], {x, 0, nmax}], x] // Rest
Table[Sum[Boole[IntegerQ[d^(1/4)] && OddQ[d]], {d, Divisors[n]}] - Sum[Boole[IntegerQ[d^(1/4)] && EvenQ[d]], {d, Divisors[n]}], {n, 1, 75}] // Accumulate

Formula

G.f.: (1/(1 - x)) * Sum_{k>=1} (-1)^(k+1) * x^(k^4)/(1 - x^(k^4)).
a(n) ~ 7*zeta(4)*n/8 = 7*Pi^4*n/720. - Vaclav Kotesovec, Oct 12 2019

A309127 a(n) = n + 2^4 * floor(n/2^4) + 3^4 * floor(n/3^4) + 4^4 * floor(n/4^4) + ...

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 96, 97, 98, 99, 100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 128, 129, 130, 131, 132, 133, 134, 135, 136, 137, 138, 139
Offset: 1

Views

Author

Ilya Gutkovskiy, Jul 13 2019

Keywords

Comments

Partial sums of A300909.

Links

Crossrefs

Cf. A013938, A024916, A300909, A309125, A309126.

Programs

  • Mathematica
    Table[Sum[k^4 Floor[n/k^4], {k, 1, n}], {n, 1, 75}]
nmax = 75; CoefficientList[Series[1/(1 - x) Sum[k^4 x^(k^4)/(1 - x^(k^4)), {k, 1, Floor[nmax^(1/4)] + 1}], {x, 0, nmax}], x] // Rest
  • PARI
    a(n) = sum(k=1, n, k^4*(n\k^4)); \\ Seiichi Manyama, Aug 30 2021

Formula

G.f.: (1/(1 - x)) * Sum_{k>=1} k^4 * x^(k^4)/(1 - x^(k^4)).
a(n) ~ zeta(5/4)*n^(5/4)/5 - n/2. - Vaclav Kotesovec, Aug 30 2021

A306533 Square array A(n,k), n >= 1, k >= 0, read by antidiagonals: A(n,k) = Sum_{j=1..n} floor(n/j^k).

Original entry on oeis.org

1, 1, 4, 1, 3, 9, 1, 2, 5, 16, 1, 2, 3, 8, 25, 1, 2, 3, 5, 10, 36, 1, 2, 3, 4, 6, 14, 49, 1, 2, 3, 4, 5, 7, 16, 64, 1, 2, 3, 4, 5, 6, 8, 20, 81, 1, 2, 3, 4, 5, 6, 7, 10, 23, 100, 1, 2, 3, 4, 5, 6, 7, 9, 12, 27, 121, 1, 2, 3, 4, 5, 6, 7, 8, 10, 13, 29, 144, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 14, 35, 169
Offset: 1

Views

Author

Ilya Gutkovskiy, Feb 22 2019

Keywords

Examples

			Square array begins:
   1,   1,  1,  1,  1,  1,  ...
   4,   3,  2,  2,  2,  2,  ...
   9,   5,  3,  3,  3,  3,  ...
  16,   8,  5,  4,  4,  4,  ...
  25,  10,  6,  5,  5,  5,  ...
  36,  14,  7,  6,  6,  6,  ...

Crossrefs

Columns k=0..4 give A000290, A006218, A013936, A013937, A013938.
Cf. A306534.

Programs

  • Mathematica
    Table[Function[k, Sum[Floor[n/j^k], {j, 1, n}]][i - n], {i, 0, 12}, {n, 1, i}] // Flatten

Formula

G.f. of column k (for k > 0): (1/(1 - x)) * Sum_{j>=1} x^(j^k)/(1 - x^(j^k)).

A347527 Partial sums of A347526.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 17, 18, 20, 21, 23, 24, 26, 27, 29, 30, 32, 33, 35, 36, 38, 39, 41, 42, 44, 45, 47, 48, 50, 51, 53, 54, 56, 57, 59, 60, 62, 63, 65, 66, 68, 69, 71, 72, 74, 75, 77, 78, 80, 81, 83, 84, 86, 87, 89, 90, 92, 93, 95, 96, 98, 99, 101, 102
Offset: 1

Views

Author

Seiichi Manyama, Sep 05 2021

Keywords

Links

Crossrefs

Cf. A006218, A013938, A094820, A347517, A347526.

Programs

  • Mathematica
    f[n_] := DivisorSum[n, 1 &, # <= n^(1/4) &]; Accumulate @ Array[f, 100] (* Amiram Eldar, Sep 05 2021 *)
  • PARI
    a(n) = sum(k=1, n, sumdiv(k, d, d^4<=k));
  • PARI
    N=99; x='x+O('x^N); Vec(sum(k=1, N^(1/4), x^k^4/(1-x^k))/(1-x))

Formula

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

A069470 a(n) = Sum_{k>=1} floor(n/(k*(k+1)/2)).

Original entry on oeis.org

0, 1, 2, 4, 5, 6, 9, 10, 11, 13, 15, 16, 19, 20, 21, 24, 25, 26, 29, 30, 32, 35, 36, 37, 40, 41, 42, 44, 46, 47, 52, 53, 54, 56, 57, 58, 62, 63, 64, 66, 68, 69, 73, 74, 75, 79, 80, 81, 84, 85, 87, 89, 90, 91, 94, 96, 98, 100, 101, 102, 107, 108, 109, 112, 113, 114, 118
Offset: 0

Views

Author

Henry Bottomley, Mar 25 2002

Keywords

Comments

The summation has floor(1/2 + sqrt(2*n)) = A002024(n) nonzero terms. - Enrique Pérez Herrero, Apr 05 2010

Examples

			a(11) = floor(11/1) + floor(11/3) + floor(11/6) + floor(11/10) + floor(11/15) + ... = 11 + 3 + 1 + 1 + 0 + ... = 16.

Links

Crossrefs

Cf. A000217, A006218, A007862, A013936, A013937, A013938, A013939.
Cf. A002024, A004275. - Enrique Pérez Herrero, Apr 05 2010

Programs

  • Magma
    [(&+[Floor(n/(k*(k+1)/2)): k in [1..100]]): n in [0..30]]; // G. C. Greubel, May 23 2018
  • Mathematica
    A069470[n_]:=Sum[Floor[(2*n)/(k*(1 + k))], {k, 1, Floor[1/2 + Sqrt[2*n]]}] (* Enrique Pérez Herrero, Apr 05 2010 *)
  • PARI
    for(n=0, 30, print1(sum(k=1, 100, floor(n/(k*(k+1)/2))), ", ")) \\ G. C. Greubel, May 23 2018

Formula

a(n) = a(n-1) + A007862(n).
It appears that limit((sum(floor((1/2)*n/(k*(k+1))), k=1..n))/n, n=infinity) = 1/2. - Stephen Crowley, Aug 12 2009
From Enrique Pérez Herrero, Apr 05 2010: (Start)
a(n) <= floor((2*n^2)/(1 + n)) = A004275(n).
a(n) <= floor((2*n*floor((1 + 2*sqrt(2*n))/2))/(1+floor((1+2*sqrt(2*n))/2))). (End)
G.f.: (1/(1 - x)) * Sum_{k>=1} x^(k*(k+1)/2)/(1 - x^(k*(k+1)/2)). - Ilya Gutkovskiy, Jul 11 2019
Showing 1-5 of 5 results.

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