A359328 Maximal coefficient of x^2*(x^2 + x^3)*(x^2 + x^3 + x^5)*...*(x^2 + x^3 + x^5 + ... + x^prime(n)).

1, 1, 1, 2, 4, 12, 46, 251, 1576, 11578, 94933, 875134, 8900088, 99276703, 1214131109, 16107824706, 229757728186, 3499486564517, 56862172844198, 980725126968577, 17899265342632635, 345197504845310134, 7005723403640260805, 149261757412790940113, 3329108788695272565243

Offset

0,4

Comments

Excluding the term 1 from A326178, the exponents of the product x^0*x^2*(x^2 + x^3)*(x^2 + x^3 + x^5)*...*(x^2 + x^3 + x^5 + ... + x^prime(n)) are given by all the other terms of A326178.

a(n) is the number of compositions of the terms of the n-th row of A359337 into n prime parts less than or equal to prime(n), with the first part equal to 2, the second part less than or equal to 3, ..., and the n-th part less than or equal to prime(n).

Links

Table of n, a(n) for n=0..24.

Index entries for sequences related to compositions

Mathematica

Table[Max[CoefficientList[Product[Sum[x^Prime[i], {i, k}], {k, n}], x]], {n, 0, 24}]

Prog

(PARI) a(n) = vecmax(Vec(prod(k=1, n, sum(i=1, k, x^prime(i))))); \\ Michel Marcus, Dec 27 2022
(Python)
from collections import Counter
from sympy import prime, primerange
def A359328(n):
    if n == 0: return 1
    c, p = {0:1}, list(primerange(prime(n)+1))
    for k in range(1, n+1):
        d = Counter()
        for j in c:
            a = c[j]
            for i in p[:k]:
                d[j+i] += a
        c = d
    return max(c.values()) # Chai Wah Wu, Feb 01 2024

Crossrefs

Cf. A000040, A326178, A350457.

Cf. A359337 (corresponding exponents), A359338 (minimal corresponding exponent), A359339 (maximal corresponding exponent).

Sequence in context: A165901 A074449 A364622 * A332235 A131387 A207647

Adjacent sequences: A359325 A359326 A359327 * A359329 A359330 A359331

Keyword

nonn

Author

Stefano Spezia, Dec 26 2022

Status

approved