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A389479
a(n) is the Frobenius number for the set { binomial(m,k), k=1..m-1 } where m = A024619(n).
1
49, 1043, 989, 20669, 12907, 99007, 67031, 700319, 7054529, 750397, 124807499, 7125065, 578549, 1935378079, 37337700047, 41645613, 188360339, 636214749065, 97469123567, 79628847, 161893747, 76693527973, 1777241641, 181135476008189, 2255226492893, 505642434227223
OFFSET
1,1
COMMENTS
Largest number not writable as Sum_{k=1..m-1} c_k*binomial(m,k) for c_k >= 0, where m = A024619(n) (m is not a power of a prime).
LINKS
Thomas F. Bloom, Erdős Problem #435
WonTae Hwang and Kyunghwan Song, The Frobenius problem for Numerical Semigroups generated by binomial coefficients, arXiv:2412.17882 [math.NT], 2024-2025. (See Theorem 0.1(1)(b), p. 2).
FORMULA
If m = Product_{k=1..c} p_k^b_k, then a(n) = -m + Sum_{k=1..c} Sum_{d=1..b_k} binomial(m,p_k^d)*(p_k-1) where m=A024619(n).
EXAMPLE
A024619(1)=6 so a(1) = -6 + (2-1)*binomial(6,2) + (3-1)*binomial(6,3) = 49.
PROG
(MATLAB)
for N=1:50,
F=factor(N); if max(F)==min(F), aN=0; continue; end;
aN=-N;
for pk=2:max(F), bk=sum(F==pk);
for d=1:bk, aN=aN+(pk-1)*nchoosek(N, pk^d); end;
end; disp(aN);
end;
(Python)
from sympy import binomial, factorint
def a(m): return 0 if len(f:=factorint(m))<2 else -m+sum((p-1)*binomial(m, p**j) for p, k in f.items() for j in range(1, k+1))
print([v for m in range(51) if (v:=a(m))]) # David Radcliffe, Oct 05 2025
(Python)
from math import comb
from sympy import primepi, integer_nthroot, factorint
def A389479(n):
def f(x): return int(n+1+sum(primepi(integer_nthroot(x, k)[0]) for k in range(1, x.bit_length())))
m, k = n, f(n)
while m != k:
m, k = k, f(k)
return -m+sum(sum(comb(m, p**d) for d in range(1, e+1))*(p-1) for p, e in factorint(m).items()) # Chai Wah Wu, Oct 11 2025
CROSSREFS
Cf. A024619.
Sequence in context: A284642 A304046 A380927 * A264877 A173246 A017765
KEYWORD
nonn,tabl,easy
AUTHOR
Michael R Peake, Oct 05 2025
STATUS
approved