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.

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A298938 Number of ordered ways of writing n^3 as a sum of n squares of nonnegative integers.

Original entry on oeis.org

1, 1, 1, 4, 5, 686, 13942, 455988, 13617853, 454222894, 18323165948, 802161109047, 42149084452070, 2481730049781672, 157265294178424356, 10977302934685469078, 812821237985857557677, 64539935903231450294134, 5504599828399250884049308
Offset: 0

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Author

Ilya Gutkovskiy, Jan 29 2018

Keywords

Examples

			a(4) = 5 because we have [64, 0, 0, 0], [16, 16, 16, 16], [0, 64, 0, 0], [0, 0, 64, 0] and [0, 0, 0, 64].
		

Crossrefs

Programs

  • Mathematica
    Table[SeriesCoefficient[(1 + EllipticTheta[3, 0, x])^n/2^n, {x, 0, n^3}], {n, 0, 18}]

Formula

a(n) = [x^(n^3)] (Sum_{k>=0} x^(k^2))^n.

A298939 Number of ordered ways of writing n^3 as a sum of n squares of positive integers.

Original entry on oeis.org

1, 1, 1, 4, 1, 286, 7582, 202028, 6473625, 226029577, 8338249868, 391526193477, 19990594900630, 1159906506684446, 74890158861242740, 5119732406649036418, 380146984328280974281, 30198665638519565614034, 2555354508318427693497565
Offset: 0

Views

Author

Ilya Gutkovskiy, Jan 29 2018

Keywords

Examples

			a(3) = 4 because we have [25, 1, 1], [9, 9, 9], [1, 25, 1] and [1, 1, 25].
		

Crossrefs

Programs

  • Mathematica
    Table[SeriesCoefficient[(-1 + EllipticTheta[3, 0, x])^n/2^n, {x, 0, n^3}], {n, 0, 18}]

Formula

a(n) = [x^(n^3)] (Sum_{k>=1} x^(k^2))^n.

A331884 Number of compositions (ordered partitions) of n^2 into distinct squares.

Original entry on oeis.org

1, 1, 1, 1, 1, 3, 1, 7, 1, 31, 123, 151, 121, 897, 7351, 5415, 14881, 48705, 150583, 468973, 1013163, 1432471, 1730023, 50432107, 14925241, 125269841, 74592537, 241763479, 213156871, 895153173, 7716880623, 2681163865, 3190865761, 22501985413, 116279718801
Offset: 0

Views

Author

Ilya Gutkovskiy, Jan 30 2020

Keywords

Examples

			a(5) = 3 because we have [25], [16, 9] and [9, 16].
		

Crossrefs

Programs

  • Maple
    b:= proc(n, i, p) option remember;
          `if`(i*(i+1)*(2*i+1)/6n, 0, b(n-i^2, i-1, p+1))+b(n, i-1, p)))
        end:
    a:= n-> b(n^2, n, 0):
    seq(a(n), n=0..35);  # Alois P. Heinz, Jan 30 2020
  • Mathematica
    b[n_, i_, p_] := b[n, i, p] = If[i(i+1)(2i+1)/6 < n, 0, If[n == 0, p!, If[i^2 > n, 0, b[n - i^2, i - 1, p + 1]] + b[n, i - 1, p]]];
    a[n_] := b[n^2, n, 0];
    a /@ Range[0, 35] (* Jean-François Alcover, Nov 08 2020, after Alois P. Heinz *)

Formula

a(n) = A331844(A000290(n)).

Extensions

a(24)-a(34) from Alois P. Heinz, Jan 30 2020

A307608 Number of partitions of n^2 into consecutive positive squares.

Original entry on oeis.org

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

Views

Author

Ilya Gutkovskiy, Apr 18 2019

Keywords

Examples

			29^2 = 20^2 + 21^2, so a(29) = 2.
		

Crossrefs

Formula

a(n) = [x^(n^2)] Sum_{i>=1} Sum_{j>=i} Product_{k=i..j} x^(k^2).
a(n) = A296338(A000290(n)).
a(n) >= 2 for n in A097812.

A385963 a(n) is the maximum number of distinct positive integers whose sum of squares is equal to n^2.

Original entry on oeis.org

0, 1, 1, 1, 1, 2, 1, 3, 1, 4, 5, 5, 5, 6, 7, 7, 7, 8, 8, 9, 9, 9, 9, 11, 10, 11, 11, 11, 11, 11, 13, 12, 12, 13, 14, 13, 14, 14, 15, 15, 15, 16, 16, 16, 16, 17, 17, 17, 18, 18, 18, 18, 19, 19, 19, 19, 19, 19, 19, 20, 21, 21, 21, 21, 22, 22, 22, 22, 23, 22, 24, 24, 24, 24, 24, 25
Offset: 0

Views

Author

Gonzalo Martínez, Jul 13 2025

Keywords

Comments

An upper bound is a(n) <= r for the largest r with 1^2 + ... + r^2 <= n^2, and with equality only at n = 0,1,24, the latter being a(70) = 24 (see comments A001032).

Examples

			For n = 11, there are A030273(11) = 4 partitions of 11^2 into distinct squares: {11^2}, {2^2, 6^2, 9^2}, {1^2, 2^2, 4^2, 10^2}, {1^2, 2^2, 4^2, 6^2, 8^2}, where the largest cardinality of these sets is 5. Therefore, a(11) = 5.
		

Crossrefs

Programs

  • PARI
    a(n)=poldegree(polcoef(prod(k=1, n, 1 + y*x^(k^2), 1 + O(x^(n^2+1))), n^2)) \\ Andrew Howroyd, Jul 13 2025

Extensions

More terms from Andrew Howroyd, Jul 13 2025
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