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.

A302998 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals: A(n,k) = [x^(n^2)] (1 + theta_3(x))^k/(2^k*(1 - x)), where theta_3() is the Jacobi theta function.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 6, 4, 1, 1, 5, 11, 11, 5, 1, 1, 6, 20, 29, 17, 6, 1, 1, 7, 36, 70, 54, 26, 7, 1, 1, 8, 63, 157, 165, 99, 35, 8, 1, 1, 9, 106, 337, 482, 357, 163, 45, 9, 1, 1, 10, 171, 702, 1319, 1203, 688, 239, 58, 10, 1, 1, 11, 265, 1420, 3390, 3819, 2673, 1154, 344, 73, 11, 1
Offset: 0

Views

Author

Ilya Gutkovskiy, Apr 17 2018

Keywords

Comments

A(n,k) is the number of nonnegative solutions to (x_1)^2 + (x_2)^2 + ... + (x_k)^2 <= n^2.

Examples

			Square array begins:
  1,  1,   1,   1,    1,     1,  ...
  1,  2,   3,   4,    5,     6,  ...
  1,  3,   6,  11,   20,    36,  ...
  1,  4,  11,  29,   70,   157,  ...
  1,  5,  17,  54,  165,   482,  ...
  1,  6,  26,  99,  357,  1203,  ...
		

Crossrefs

Programs

  • Mathematica
    Table[Function[k, SeriesCoefficient[(1 + EllipticTheta[3, 0, x])^k/(2^k (1 - x)), {x, 0, n^2}]][j - n], {j, 0, 11}, {n, 0, j}] // Flatten
    Table[Function[k, SeriesCoefficient[1/(1 - x) Sum[x^i^2, {i, 0, n}]^k, {x, 0, n^2}]][j - n], {j, 0, 11}, {n, 0, j}] // Flatten
  • PARI
    T(n,k)={if(k==0, 1, polcoef(((sum(j=0, n, x^(j^2)) + O(x*x^(n^2)))^k)/(1-x), n^2))} \\ Andrew Howroyd, Sep 14 2019

Formula

A(n,k) = [x^(n^2)] (1/(1 - x))*(Sum_{j>=0} x^(j^2))^k.

A341423 Number of positive solutions to (x_1)^2 + (x_2)^2 + (x_3)^2 + (x_4)^2 <= n^2.

Original entry on oeis.org

1, 5, 32, 94, 219, 437, 804, 1362, 2177, 3271, 4768, 6708, 9227, 12381, 16254, 20954, 26707, 33461, 41480, 50884, 61703, 74183, 88606, 104862, 123481, 144241, 167604, 193648, 222799, 254731, 290244, 329512, 372545, 419661, 470822, 526646, 587481, 653505
Offset: 2

Views

Author

Ilya Gutkovskiy, Feb 11 2021

Keywords

Crossrefs

Programs

  • Maple
    b:= proc(n, k) option remember; `if`(k=0, 1, `if`(n=0, 0,
          add((s->`if`(s>n, 0, b(n-s, k-1)))(j^2), j=1..isqrt(n))))
        end:
    a:= n-> b(n^2, 4):
    seq(a(n), n=2..39);  # Alois P. Heinz, Feb 11 2021
  • Mathematica
    Table[SeriesCoefficient[(EllipticTheta[3, 0, x] - 1)^4/(16 (1 - x)), {x, 0, n^2}], {n, 2, 39}]

Formula

a(n) is the coefficient of x^(n^2) in expansion of (theta_3(x) - 1)^4 / (16 * (1 - x)).

A302863 a(n) = [x^(n^2)] (1 + theta_3(x))^n/(2^n*(1 - x)), where theta_3() is the Jacobi theta function.

Original entry on oeis.org

1, 2, 6, 29, 165, 1203, 9763, 83877, 793049, 7903501, 83570177, 933697153, 10905583809, 133352809334, 1695473999478, 22354920990148, 305096197935075, 4296142551821184, 62336908825014452, 930284705538262688, 14255992611680074754, 224065160215526683317, 3607018540134004189466
Offset: 0

Views

Author

Ilya Gutkovskiy, Apr 14 2018

Keywords

Comments

a(n) = number of nonnegative solutions to (x_1)^2 + (x_2)^2 + ... + (x_n)^2 <= n^2.

Crossrefs

Programs

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

A349611 Number of solutions to x^2 + y^2 + z^2 + w^2 <= n^2, where x, y, z, w are positive odd integers.

Original entry on oeis.org

0, 0, 1, 1, 5, 11, 32, 44, 82, 120, 207, 277, 405, 541, 768, 966, 1272, 1592, 2087, 2489, 3103, 3719, 4588, 5348, 6386, 7522, 8891, 10175, 11909, 13623, 15818, 17742, 20278, 22720, 25923, 28917, 32361, 36031, 40368, 44488, 49400, 54358, 60377, 65835, 72341
Offset: 0

Views

Author

Ilya Gutkovskiy, Nov 23 2021

Keywords

Examples

			a(4) = 5 since there are solutions (1,1,1,1), (3,1,1,1), (1,3,1,1), (1,1,3,1), (1,1,1,3).
		

Crossrefs

Programs

  • Maple
    N:= 100: # for a(0) .. a(N)
    F:= add(x^(k^2),k = 1 ... N,2):
    F:= expand(F^4):
    L:= ListTools:-PartialSums([seq](coeff(F,x,n),n=0..N^2)):
    L[[seq(n^2+1,n=0..N)]]; # Robert Israel, Dec 21 2023
  • Mathematica
    Table[SeriesCoefficient[EllipticTheta[2, 0, x^4]^4/(16 (1 - x)), {x, 0, n^2}], {n, 0, 44}]

Formula

a(n) = [x^(n^2)] theta_2(x^4)^4 / (16 * (1 - x)).
Showing 1-4 of 4 results.