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 1-6 of 6 results.

A168656 Number of partitions of n such that the smallest part is divisible by the number of parts.

Original entry on oeis.org

1, 1, 1, 2, 2, 2, 2, 3, 4, 4, 5, 6, 7, 7, 8, 10, 11, 13, 15, 18, 20, 23, 25, 29, 33, 36, 41, 47, 53, 58, 66, 74, 83, 92, 103, 116, 130, 144, 160, 179, 199, 219, 243, 269, 298, 328, 362, 399, 441, 484, 533, 586, 645, 708, 778, 854, 937, 1026, 1124, 1230, 1347, 1470, 1607, 1756, 1917, 2089
Offset: 1

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Author

Vladeta Jovovic, Dec 01 2009, Dec 04 2009

Keywords

Crossrefs

Programs

  • Mathematica
    nmax = 100; Rest[CoefficientList[Series[Sum[x^(k^2)/((1 - x^(k^2))*Product[1 - x^j, {j, 1, k-1}]), {k, 1, Sqrt[nmax]}], {x, 0, nmax}], x]] (* Vaclav Kotesovec, Oct 16 2024 *)
    Table[Count[IntegerPartitions[n],?(Mod[#[[-1]],Length[#]]==0&)],{n,70}] (* _Harvey P. Dale, Dec 22 2024 *)
  • PARI
    N=100; x='x+O('x^N);
    Vec( sum(k=1,sqrtint(N), x^(k^2)/(1-x^(k^2)) / prod(i=1,k-1, 1-x^i) ) )

Formula

G.f.: Sum_{k>=1} x^(k^2)/((1-x^(k^2)) * Product_{i=1..k-1} (1-x^i)).
a(n) ~ c * exp(2*Pi*sqrt(n/15)) / n^(3/4), where c = 1 / (2 * 3^(1/4) * sqrt(5) * phi^(3/2)) = 0.08255116908... and phi = A001622 = (1+sqrt(5))/2 is the golden ratio. - Vaclav Kotesovec, Oct 17 2024

A213180 Sum over all partitions lambda of n of Sum_{p:lambda} p^m(p,lambda), where m(p,lambda) is the multiplicity of part p in lambda.

Original entry on oeis.org

0, 1, 3, 7, 16, 28, 59, 91, 170, 269, 450, 655, 1162, 1602, 2527, 3793, 5805, 8034, 12660, 17131, 26484, 37384, 53738, 73504, 114683, 153613, 221225, 313339, 453769, 609179, 927968, 1223909, 1804710, 2522264, 3539835, 4855420, 7439870, 9765555, 14009545
Offset: 0

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Author

Alois P. Heinz, Feb 27 2013

Keywords

Examples

			a(6) = 59: (1^6) + (2+1^4) + (2^2+1^2) + (2^3) + (3+1^3) + (3+2+1) + (3^2) + (4+1^2) + (4+2) + (5+1) + (6) = 1+3+5+8+4+6+9+5+6+6+6 = 59.
		

Crossrefs

Cf. A000070 (Sum 1), A006128 (Sum m), A014153 (Sum p), A024786 (Sum floor(1/m)), A066183 (Sum p^2*m), A066186 (Sum p*m), A073336 (Sum floor(m/p)), A116646 (Sum delta(m,2)), A117524 (Sum delta(m,3)), A103628 (Sum delta(m,1)*p), A117525 (Sum delta(m,2)*p), A197126, A213191.

Programs

  • Maple
    b:= proc(n, p) option remember; `if`(n=0, [1, 0], `if`(p<1, [0, 0],
          add((l->`if`(m=0, l, l+[0, l[1]*p^m]))(b(n-p*m, p-1)), m=0..n/p)))
        end:
    a:= n-> b(n, n)[2]:
    seq(a(n), n=0..40);
  • Mathematica
    b[n_, p_] := b[n, p] = If[n==0, {1, 0}, If[p<1, {0, 0}, Sum[Function[l, If[m==0, l, l+{0, l[[1]]*p^m}]][b[n-p*m, p-1]], {m, 0, n/p}]]]; a[n_] := b[n, n][[2]]; Table[a[n], {n, 0, 40}] (* Jean-François Alcover, Feb 15 2017, translated from Maple *)

Formula

From Vaclav Kotesovec, May 24 2018: (Start)
a(n) ~ c * 3^(n/3), where
c = 5.0144820680945600131204662934686439430547... if mod(n,3)=0
c = 4.6144523178014379613985400559486878971522... if mod(n,3)=1
c = 4.5237761454818383598444208605033385016299... if mod(n,3)=2
(End)

A281573 Expansion of Sum_{i>=1} mu(i)^2*x^i/(1 - x^i) / Product_{j>=1} (1 - x^j), where mu() is the Moebius function (A008683).

Original entry on oeis.org

1, 3, 6, 11, 19, 33, 51, 79, 118, 176, 252, 362, 505, 705, 965, 1314, 1765, 2365, 3127, 4124, 5387, 7012, 9052, 11653, 14893, 18982, 24048, 30378, 38176, 47857, 59704, 74302, 92099, 113879, 140300, 172463, 211297, 258325, 314887, 383037, 464684, 562653, 679566, 819269, 985449, 1183242, 1417738, 1695886
Offset: 1

Views

Author

Ilya Gutkovskiy, Jan 24 2017

Keywords

Comments

Total number of squarefree parts in all partitions of n.
Convolution of A000041 and A034444.

Examples

			a(5) = 19 because we have [5], [4, 1], [3, 2], [3, 1, 1], [2, 2, 1], [2, 1, 1, 1], [1, 1, 1, 1, 1] and 1 + 1 + 2 + 3 + 3 + 4 + 5 = 19.
		

Crossrefs

Programs

  • Mathematica
    nmax = 48; Rest[CoefficientList[Series[Sum[MoebiusMu[i]^2 x^i/(1 - x^i), {i, 1, nmax}]/Product[1 - x^j, {j, 1, nmax}], {x, 0, nmax}], x]]

Formula

G.f.: Sum_{i>=1} mu(i)^2*x^i/(1 - x^i) / Product_{j>=1} (1 - x^j).

A309535 Total number of square parts in all compositions of n.

Original entry on oeis.org

0, 1, 2, 5, 13, 30, 69, 156, 348, 769, 1682, 3653, 7884, 16924, 36160, 76944, 163137, 344770, 726533, 1527052, 3202076, 6700096, 13992080, 29167936, 60703424, 126141953, 261754114, 542448645, 1122778124, 2321317916, 4794159168, 9891365008, 20388823360
Offset: 0

Views

Author

Alois P. Heinz, Aug 06 2019

Keywords

Examples

			a(4) = 13: (1)(1)(1)(1), (1)(1)2, (1)2(1), 2(1)(1), 22, (1)3, 3(1), (4).
		

Crossrefs

Programs

  • Maple
    a:= proc(n) option remember; add(a(n-j)+
          `if`(issqr(j), ceil(2^(n-j-1)), 0), j=1..n)
        end:
    seq(a(n), n=0..33);
  • Mathematica
    CoefficientList[Series[(EllipticTheta[3, 0, x]-1)*(1-x)^2/(2*(1-2*x)^2), {x, 0, 30}], x] (* Vaclav Kotesovec, Aug 18 2019 *)
    Table[Sum[If[k == n, 1, (2^(n - k - 2)*(3 + n - k))] * If[IntegerQ[Sqrt[k]], 1, 0], {k, 1, n}], {n, 0, 30}] (* Vaclav Kotesovec, Aug 18 2019 *)

Formula

G.f.: Sum_{k>=1} x^(k^2)*(1-x)^2/(1-2*x)^2.
a(n) ~ c * 2^n * n, where c = (EllipticTheta[3, 0, 1/2] - 1)/8 = 0.1411171034014846448336823185681189155765645674... - Vaclav Kotesovec, Aug 18 2019, updated Mar 17 2024
a(n) = Sum_{k=1..A000196(n)} A045623(n-k^2). - Gregory L. Simay, Jun 07 2021

A263233 Triangle read by rows: T(n,k) is the number of partitions of n having k perfect square parts (0<=k<=n).

Original entry on oeis.org

1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 2, 1, 0, 1, 2, 1, 2, 1, 0, 1, 3, 3, 1, 2, 1, 0, 1, 3, 4, 3, 1, 2, 1, 0, 1, 5, 4, 5, 3, 1, 2, 1, 0, 1, 5, 8, 4, 5, 3, 1, 2, 1, 0, 1, 8, 8, 9, 4, 5, 3, 1, 2, 1, 0, 1, 9, 12, 9, 9, 4, 5, 3, 1, 2, 1, 0, 1, 13, 15, 13, 10, 9, 4, 5, 3, 1, 2, 1, 0, 1
Offset: 0

Views

Author

Emeric Deutsch, Nov 12 2015

Keywords

Comments

Sum of entries in row n = A000041(n) = number of partitions of n.
T(n,0) = A087153(n).
Sum_{k=0..n}k*T(n,k) = A073336(n) = total number of square parts in all partitions of n.

Examples

			T(8,2) = 5 because we have [6,1,1], [4,4], [4,3,1], [3,3,1,1], [2,2,2,1,1] (the partitions of 8 that have 2 perfect square parts).
Triangle starts:
  1;
  0, 1;
  1, 0, 1;
  1, 1, 0, 1;
  1, 2, 1, 0, 1;
  2, 1, 2, 1, 0, 1;
  3, 3, 1, 2, 1, 0, 1;
  3, 4, 3, 1, 2, 1, 0, 1;
  5, 4, 5, 3, 1, 2, 1, 0, 1;
		

Crossrefs

Programs

  • Maple
    h:= proc(i) options operator, arrow: i^2 end proc: g := product((1-x^h(i))/((1-x^i)*(1-t*x^h(i))), i = 1 .. 80): gser := simplify(series(g, x = 0, 30)): for n from 0 to 18 do P[n] := sort(coeff(gser, x, n)) end do: for n from 0 to 18 do seq(coeff(P[n], t, j), j = 0 .. n) end do; # yields sequence in triangular form.
  • Mathematica
    Needs["Combinatorica`"]; Table[Count[Replace[#, n_ /; ! IntegerQ@ Sqrt@ n -> Nothing, {1}] & /@ Combinatorica`Partitions@ n, w_ /; Length@ w == k], {n, 0, 12}, {k, 0, n}] // Flatten (* Michael De Vlieger, Dec 19 2015 *)

Formula

G.f.: Product_{i>=1}(1-x^h(i))/((1-x^i)*(1-t*x^h(i))), where h(i) = i^2.

A342228 Total sum of parts which are squares in all partitions of n.

Original entry on oeis.org

0, 1, 2, 4, 11, 16, 27, 42, 69, 108, 158, 229, 334, 469, 656, 903, 1255, 1685, 2283, 3032, 4033, 5290, 6936, 8986, 11650, 14969, 19172, 24402, 30998, 39110, 49260, 61712, 77155, 96000, 119209, 147394, 181958, 223713, 274533, 335792, 409980, 498981, 606273, 734572
Offset: 0

Views

Author

Ilya Gutkovskiy, Mar 06 2021

Keywords

Examples

			For n = 4 we have:
---------------------------------
Partitions        Sum of parts
.              which are squares
---------------------------------
4 ................... 4
3 + 1 ............... 1
2 + 2 ............... 0
2 + 1 + 1 ........... 2
1 + 1 + 1 + 1 ....... 4
---------------------------------
Total .............. 11
So a(4) = 11.
		

Crossrefs

Programs

  • Mathematica
    nmax = 43; CoefficientList[Series[Sum[k^2 x^(k^2)/(1 - x^(k^2)), {k, 1, Floor[nmax^(1/2)] + 1}]/Product[(1 - x^j), {j, 1, nmax}], {x, 0, nmax}], x]
    Table[Sum[DivisorSum[k, # &, IntegerQ[#^(1/2)] &] PartitionsP[n - k], {k, 1, n}], {n, 0, 43}]

Formula

G.f.: Sum_{k>=1} k^2*x^(k^2)/(1 - x^(k^2)) / Product_{j>=1} (1 - x^j).
a(n) = Sum_{k=1..n} A035316(k) * A000041(n-k).
Showing 1-6 of 6 results.