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A167219
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Numbers k such that there exists a positive integer B for which k = Sum_{i=0..m} (B^i)*a_i where the a_i are defined by k = Product_{i=0..m} prime(i+1)^a_i.
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3
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3, 6, 10, 12, 24, 27, 36, 48, 96, 100, 144, 175, 192, 216, 273, 384, 486, 576, 768, 972, 1296, 1536, 1728, 2304, 3072, 3125, 6144, 9216, 12288, 13824, 17496, 19683, 20736, 24576, 36864, 46656, 49152, 62208, 69984, 98304, 110592, 147456, 196608, 331776, 393216, 589824
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OFFSET
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1,1
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COMMENTS
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Previous name: Numbers k such that there exists a solution to (p_m ^ a_m)*(p_m-1 ^ a_m-1)*...*(3^a_1)*(2^a_0) = (B^m)*a_m + (B^m-1)*a_m-1 + ... + (B^1)*a_1 + (B^0)*a_0 where k = (p_m ^ a_m)*(p_m-1 ^ a_m-1)*...*(3^a_1)*(2^a_0); a_m >= 1; a_(i<m) >= 0; p_0, p_1, ..., p_m are prime numbers; a_0, a_1, ..., a_m, B are integers.
B is the base in which we can express k as Sum_{i=0..m} B^i * a_i. B may also be seen as the variable in a polynomial, and k is then also an encoding of the polynomial (defined by the product of primes formula).
For k = (2^r)*3 we have B = (2^r)*3 - r.
A167221(n) is the smallest positive integer that yields a solution for k = a(n).
Negative B's can be obtained when the polynomial is an even function. This happens for instance when for k = 10, 100, 3125, ... - Michel Marcus, Aug 10 2022
Positive integers k such that k is a fixed point of a completely additive function f_B:N+ -> Z, B > 0, where f_B(prime(i+1)) = B^i for all i >= 0. Equivalently, since row B of A104244 is f_B, {a(n)} lists the columns of A104244 that contain their own column number.
If we require B to be negative instead, the sequence appears to start 10, 100, 3125, 1799875, 65610000, ... . Of these, 1799875 = 5^3 * 7 * 11^2 * 17 is the only k with only negative solutions (B = -11); the solutions for 65610000 are {4049, -4051}.
(End)
If p is the (k+1)-th prime and p is congruent to 1 modulo k, then p^p is a term with p^((p-1)/k) a solution for B. The list of such primes starts 3, 5, 7, 31, 97, 101, 331, ... . I suspect this list is infinite, meaning the greatest prime factor of the terms would be unbounded. - Peter Munn, Aug 15 2022
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LINKS
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EXAMPLE
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For k = 10 = 2^1 * 3^0 * 5^1, k = B^0 * 1 + B^1 * 0 + B^2 * 1, so we have to solve the equation 10 = 1 + B^2 for a positive integer B, B = 3. But B=-3 works too. Thus 10 is a term.
For k = 12 = 2^2 * 3^1, k = B^0 * 2 + B^1 * 1, so we have to solve the equation 12 = 2 + B for a positive integer B. B = 10. Thus 12 is a term.
For k = 21 = 2^0 * 3^1 * 5^0 * 7^1, k = B^0 * 0 + B^1 * 1 + B^2 * 0 + B^3 * 1, so we have to solve the equation 21 = B + B^3 for an integer B. No such B exists, so 21 is not a term of this sequence.
In other words:
10 is a term because 10 = 5^1 * 3^0 * 2^1 and 101 in base 3 is 10.
12 is a term because 12 = 3^1 * 2^2 and 12 in base 10 is 12. (End)
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PROG
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(PARI) isok(k) = if (k>1, my(f=factor(k), v=primes(primepi(vecmax(f[, 1])))); my(p=sum(i=1, #v, 'x^(i-1)*valuation(k, v[i]))); p -= k; my(c=-polcoef(p, 0)); my(q=(p+c)/x); my(d=divisors(c)); for (k=1, #d, if(subst(q, x, d[k]) == c/d[k], return(1)); ); ); \\ Michel Marcus, Aug 08 2022
(Python)
from sympy import divisors, factorint, sieve
def ok(n):
if n < 2: return False
f = factorint(n)
a = [f[pi] if pi in f else 0 for pi in sieve.primerange(2, max(f)+1)]
for B in range(1, n+1):
polyB = sum(B**i*ai for i, ai in enumerate(a) if ai > 0)
if polyB == n: return True
elif polyB > n: return False
return False
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CROSSREFS
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A206284 describes the polynomial encoding used here.
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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Incorrect term 71 removed, new name and more terms from Michel Marcus, Aug 08 2022
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STATUS
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approved
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