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 1-2 of 2 results.

A369983 Maximum of the absolute value of the coefficients of (1 - x)^3 * (1 - x^2)^3 * (1 - x^3)^3 * ... * (1 - x^n)^3.

Original entry on oeis.org

1, 3, 8, 15, 44, 50, 117, 186, 356, 561, 972, 1761, 3508, 5789, 10470, 19023, 35580, 62388, 113418, 205653, 376496, 674085, 1226181, 2211462, 4056220, 7287672, 13261764, 24005627, 43800562, 79033269, 143513301, 260061408, 473603594, 855436899, 1553736558, 2813222766
Offset: 0

Views

Author

Ilya Gutkovskiy, Feb 07 2024

Keywords

Crossrefs

Programs

  • Mathematica
    Table[Max[Abs[CoefficientList[Product[(1 - x^k)^3, {k, 1, n}], x]]], {n, 0, 35}]
  • PARI
    a(n) = vecmax(apply(abs, Vec(prod(i=1, n, (1-x^i)^3)))); \\ Michel Marcus, Feb 07 2024
    
  • Python
    from collections import Counter
    def A369983(n):
        c = {0:1}
        for k in range(1,n+1):
            d = Counter(c)
            for j in c:
                a = c[j]
                d[j+k] -= 3*a
                d[j+2*k] += 3*a
                d[j+3*k] -= a
            c = d
        return max(map(abs,c.values())) # Chai Wah Wu, Feb 07 2024

A369879 Number of different coefficient values in expansion of Product_{k=1..n} (1-x^k).

Original entry on oeis.org

1, 2, 2, 3, 4, 3, 4, 5, 5, 5, 5, 5, 6, 7, 7, 9, 9, 11, 12, 15, 16, 17, 21, 23, 25, 33, 35, 41, 43, 53, 55, 77, 74, 97, 85, 135, 111, 175, 131, 223, 157, 269, 187, 315, 219, 375, 245, 437, 280, 505, 320, 593, 357, 671, 398, 761, 445, 825, 489, 933, 529, 1039, 576, 1119, 627, 1211, 682, 1309, 730, 1415
Offset: 0

Views

Author

Seiichi Manyama, Feb 04 2024

Keywords

Crossrefs

Programs

  • PARI
    a(n) = #Set(Vec(prod(k=1, n, 1-x^k)));
    
  • Python
    from collections import Counter
    def A369879(n):
        c = {0:1}
        for k in range(1,n+1):
            d = Counter(c)
            for j in c:
                d[j+k] -= c[j]
            c = d
        return len(set(c.values()))+int(max(c)+1>len(c)) # Chai Wah Wu, Feb 04 2024
Showing 1-2 of 2 results.