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 10 results.

A307614 Number of partitions of the n-th triangular number into consecutive positive triangular numbers.

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 2, 2, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 2, 1, 1, 2, 2, 1, 1, 1, 2, 1, 2, 1, 1, 2, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 2, 1
Offset: 1

Views

Author

Ilya Gutkovskiy, Apr 18 2019

Keywords

Examples

			A000217(4) = 10 = 1 + 3 + 6, so a(4) = 2.
		

Crossrefs

Formula

a(n) = [x^(n*(n+1)/2)] Sum_{i>=1} Sum_{j>=i} Product_{k=i..j} x^(k*(k+1)/2).

A299032 Number of ordered ways of writing n-th triangular number as a sum of n squares of positive integers.

Original entry on oeis.org

1, 1, 0, 3, 6, 0, 12, 106, 420, 2718, 18240, 120879, 694320, 5430438, 40668264, 300401818, 2369504386, 19928714475, 174151735920, 1543284732218, 14224347438876, 135649243229688, 1331658133954940, 13369350846412794, 138122850643702056, 1462610254141337590
Offset: 0

Views

Author

Ilya Gutkovskiy, Feb 01 2018

Keywords

Examples

			a(4) = 6 because fourth triangular number is 10 and we have [4, 4, 1, 1], [4, 1, 4, 1], [4, 1, 1, 4], [1, 4, 4, 1], [1, 4, 1, 4] and [1, 1, 4, 4].
		

Crossrefs

Programs

  • Maple
    b:= proc(n, t) option remember; local i; if n=0 then
          `if`(t=0, 1, 0) elif t<1 then 0 else 0;
          for i while i^2<=n do %+b(n-i^2, t-1) od; % fi
        end:
    a:= n-> b(n*(n+1)/2, n):
    seq(a(n), n=0..25);  # Alois P. Heinz, Feb 05 2018
  • Mathematica
    Table[SeriesCoefficient[(-1 + EllipticTheta[3, 0, x])^n/2^n, {x, 0, n (n + 1)/2}], {n, 0, 25}]

Formula

a(n) = [x^(n*(n+1)/2)] (Sum_{k>=1} x^(k^2))^n.

A338585 Number of partitions of the n-th triangular number into exactly n positive triangular numbers.

Original entry on oeis.org

1, 1, 0, 0, 1, 2, 3, 4, 9, 16, 29, 52, 92, 173, 307, 554, 1002, 1792, 3216, 5738, 10149, 17942, 31769, 55684, 97478, 170356, 295644, 512468, 886358, 1523779, 2614547, 4476152, 7627119, 12966642, 21988285, 37142199, 62591912, 105215149, 176266155, 294591431
Offset: 0

Views

Author

Ilya Gutkovskiy, Nov 08 2020

Keywords

Examples

			The 5th triangular number is 15 and 15 = 1 + 1 + 1 + 6 + 6 = 3 + 3 + 3 + 3 + 3, so a(5) = 2.
		

Crossrefs

Programs

  • Maple
    h:= proc(n) option remember; `if`(n<1, 0,
          `if`(issqr(8*n+1), n, h(n-1)))
        end:
    b:= proc(n, i, k) option remember; `if`(n=0, `if`(k=0, 1, 0),
          `if`(i*kn, 0, b(n, h(i-1), k)+b(n-i, h(min(n-i, i)), k-1)))
        end:
    a:= n-> (t-> b(t, h(t), n))(n*(n+1)/2):
    seq(a(n), n=0..42);  # Alois P. Heinz, Nov 10 2020
  • Mathematica
    h[n_] := h[n] = If[n < 1, 0, If[IntegerQ@Sqrt[8n+1], n, h[n-1]]];
    b[n_, i_, k_] := b[n, i, k] = If[n == 0, If[k == 0, 1, 0], If[i k < n || k > n, 0, b[n, h[i-1], k] + b[n-i, h[Min[n-i, i]], k-1]]];
    a[n_] := b[#, h[#], n]&[n(n+1)/2];
    a /@ Range[0, 42](* Jean-François Alcover, Nov 15 2020, after Alois P. Heinz *)
  • SageMath
    # Returns a list of length n, slow.
    def GeneralizedEulerTransform(n, a):
        R. = ZZ[[]]
        f = prod((1 - y*x^a(k) + O(x, y)^a(n)) for k in (1..n))
        coeffs = f.inverse().coefficients()
        coeff = lambda k: coeffs[x^a(k)*y^k] if x^a(k)*y^k in coeffs else 0
        return [coeff(k) for k in range(n)]
    def A338585List(n): return GeneralizedEulerTransform(n, lambda n: n*(n+1)/2)
    print(A338585List(12)) # Peter Luschny, Nov 12 2020

Formula

a(n) = [x^A000217(n) y^n] Product_{j>=1} 1 / (1 - y*x^A000217(j)).
a(n) = A319797(A000217(n),n).

A299031 Number of ordered ways of writing n-th triangular number as a sum of n squares of nonnegative integers.

Original entry on oeis.org

1, 1, 0, 3, 18, 60, 252, 1576, 10494, 64152, 458400, 3407019, 27713928, 225193982, 1980444648, 17626414158, 165796077562, 1593587604441, 15985672426992, 163422639872978, 1729188245991060, 18743981599820280, 208963405365941380, 2378065667103672024, 27742569814633730608
Offset: 0

Views

Author

Ilya Gutkovskiy, Feb 01 2018

Keywords

Examples

			a(3) = 3 because third triangular number is 6 and we have [4, 1, 1], [1, 4, 1] and [1, 1, 4].
		

Crossrefs

Programs

  • Mathematica
    Table[SeriesCoefficient[(1 + EllipticTheta[3, 0, x])^n/2^n, {x, 0, n (n + 1)/2}], {n, 0, 24}]

Formula

a(n) = [x^(n*(n+1)/2)] (Sum_{k>=0} x^(k^2))^n.

A331900 Number of compositions (ordered partitions) of the n-th triangular number into distinct triangular numbers.

Original entry on oeis.org

1, 1, 1, 1, 7, 1, 3, 13, 3, 55, 201, 159, 865, 1803, 7093, 43431, 14253, 22903, 130851, 120763, 1099693, 4527293, 4976767, 7516897, 14349685, 72866239, 81946383, 167841291, 897853735, 455799253, 946267825, 5054280915, 3941268001, 17066300985, 49111862599
Offset: 0

Views

Author

Ilya Gutkovskiy, Jan 31 2020

Keywords

Examples

			a(6) = 3 because we have [21], [15, 6] and [6, 15].
		

Crossrefs

Programs

  • Maple
    b:= proc(n, i, p) option remember; (t->
          `if`(t*(i+2)/3n, 0, b(n-t, i-1, p+1)))))((i*(i+1)/2))
        end:
    a:= n-> b(n*(n+1)/2, n, 0):
    seq(a(n), n=0..37);  # Alois P. Heinz, Jan 31 2020
  • Mathematica
    b[n_, i_, p_] := b[n, i, p] = With[{t = i(i+1)/2}, If[t(i+2)/3 < n, 0, If[n == 0, p!, b[n, i-1, p] + If[t > n, 0, b[n-t, i-1, p+1]]]]];
    a[n_] := b[n(n+1)/2, n, 0];
    a /@ Range[0, 37] (* Jean-François Alcover, Nov 17 2020, after Alois P. Heinz *)

Formula

a(n) = A331843(A000217(n)).

Extensions

More terms from Alois P. Heinz, Jan 31 2020

A335634 Number of ordered ways of writing the n-th n-gonal number as a sum of n nonzero n-gonal numbers.

Original entry on oeis.org

1, 1, 1, 0, 1, 30, 180, 700, 3780, 11844, 50610, 325820, 5803380, 126594910, 2114901789, 28282722650, 323420067880, 3190581939996, 29336527986960, 245438739897312, 1967485926594030, 16000631392009320, 184418174847183508, 4054670001158799616, 111835386569787369559
Offset: 0

Views

Author

Ilya Gutkovskiy, Oct 03 2020

Keywords

Examples

			a(4) = 1 because the fourth square is 16 and we have [4, 4, 4, 4].
		

Crossrefs

Programs

  • Mathematica
    Join[{1}, Table[SeriesCoefficient[Sum[x^(k (k (n - 2) - n + 4)/2), {k, 1, n}]^n, {x, 0, n (n^2 - 3 n + 4)/2}], {n, 1, 24}]]
  • PARI
    p(n,k) = {k * (k * (n - 2) - n + 4) / 2}
    a(n) = {my(m=p(n,n)); polcoef((sum(k=1, n, x^p(n,k)) + O(x*x^m))^n, m)} \\ Andrew Howroyd, Oct 03 2020

Formula

a(n) = [x^p(n,n)] (Sum_{k=1..n} x^p(n,k))^n, where p(n,k) = k * (k * (n - 2) - n + 4) / 2 is the k-th n-gonal number.

A338465 Number of ways to write 2*n as an ordered sum of n nonzero triangular numbers.

Original entry on oeis.org

1, 0, 2, 0, 6, 5, 20, 42, 70, 261, 297, 1430, 1584, 7293, 9634, 35945, 60150, 176596, 366401, 886977, 2150421, 4624410, 12205074, 25065216, 67616872, 139894305, 369551925, 793214982, 2011977414, 4517758504, 10992821055, 25669627965, 60531471286, 145112506352
Offset: 0

Views

Author

Ilya Gutkovskiy, Jan 31 2021

Keywords

Comments

Also number of ways to write n as an ordered sum of n nonnegative numbers one less than a triangular number.

Crossrefs

Programs

  • Mathematica
    Table[SeriesCoefficient[(EllipticTheta[2, 0, Sqrt[x]]/(2 x^(1/8)) - 1)^n, {x, 0, 2 n}], {n, 0, 33}]

Formula

a(n) = [x^(2*n)] (theta_2(sqrt(x)) / (2 * x^(1/8)) - 1)^n, where theta_2() is the Jacobi theta function.
a(n) = [x^n] (Sum_{k>=0} x^(k*(k + 3)/2))^n.

A336303 Number of ordered ways of writing the n-th n-gonal pyramidal number as a sum of n nonzero n-gonal pyramidal numbers.

Original entry on oeis.org

1, 1, 0, 0, 6, 0, 180, 630, 1120, 36288, 441000, 6579870, 59734620, 1252872192, 13668490836, 162131872695, 2971275208720, 52783774330940, 1334562954639156, 16933262255752698, 406499325562503480, 8838644883526856832, 190698441426122689290
Offset: 0

Views

Author

Ilya Gutkovskiy, Oct 04 2020

Keywords

Examples

			a(4) = 6 because the fourth square pyramidal number is 30 and we have [14, 14, 1, 1], [14, 1, 14, 1], [14, 1, 1, 14], [1, 14, 14, 1], [1, 14, 1, 14] and [1, 1, 14, 14].
		

Crossrefs

Programs

  • Mathematica
    Join[{1},Table[SeriesCoefficient[Sum[x^(k (k + 1) (k (n - 2) - n + 5)/6), {k, 1, n}]^n, {x, 0, n (n + 1) (n^2 - 3 n + 5)/6}], {n, 1, 22}]]

Formula

a(n) = [x^p(n,n)] (Sum_{k=1..n} x^p(n,k))^n, where p(n,k) = k * (k + 1) * (k * (n - 2) - n + 5) / 6 is the k-th n-gonal pyramidal number.

A338778 Number of ordered ways of writing n-th tetrahedral number as a sum of n positive tetrahedral numbers.

Original entry on oeis.org

1, 1, 0, 0, 0, 20, 195, 1890, 6286, 94584, 1065120, 12345432, 194450586, 2844976135, 44569913570, 740023110855, 13144353701940, 241663182769494, 4707408836458200, 95865898167054186, 2038122531703155798, 45103282424247100962, 1037559653596650520776, 24776005985596646165127
Offset: 0

Views

Author

Ilya Gutkovskiy, Nov 08 2020

Keywords

Examples

			The 5th tetrahedral number is 35 and 35 = 1 + 4 + 10 + 10 + 10 (20 permutations), so a(5) = 20.
		

Crossrefs

Formula

a(n) = [x^A000292(n)] (Sum_{j>=1} x^A000292(j))^n.

A319221 Number of ordered ways of writing n-th triangular number as a sum of n squares.

Original entry on oeis.org

1, 2, 0, 24, 144, 960, 4608, 74048, 859952, 9568800, 109975680, 1647979872, 23917274304, 358378620704, 5528847787008, 94307761212304, 1632598198916544, 29205907283227776, 538335591996965760, 10388234139989630128, 205386383159397554688, 4173254005731822569088
Offset: 0

Views

Author

Ilya Gutkovskiy, Sep 13 2018

Keywords

Crossrefs

Programs

  • Mathematica
    Table[SeriesCoefficient[EllipticTheta[3, 0, x]^n, {x, 0, n (n + 1)/2}], {n, 0, 21}]
    Join[{1}, Table[SquaresR[n, n (n + 1)/2], {n, 21}]]

Formula

a(n) = [x^(n*(n+1)/2)] theta_3(x)^n, where theta_3() is the Jacobi theta function.
a(n) = [x^(n*(n+1)/2)] (Sum_{k=-infinity..infinity} x^(k^2))^n.
Showing 1-10 of 10 results.