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A108056
Numbers of the form (7^i)*(13^j).
9
1, 7, 13, 49, 91, 169, 343, 637, 1183, 2197, 2401, 4459, 8281, 15379, 16807, 28561, 31213, 57967, 107653, 117649, 199927, 218491, 371293, 405769, 753571, 823543, 1399489, 1529437, 2599051, 2840383, 4826809, 5274997, 5764801, 9796423, 10706059, 18193357, 19882681
OFFSET
1,2
LINKS
FORMULA
Sum_{n>=1} 1/a(n) = (7*13)/((7-1)*(13-1)) = 91/72. - Amiram Eldar, Sep 23 2020
a(n) ~ exp(sqrt(2*log(7)*log(13)*n)) / sqrt(91). - Vaclav Kotesovec, Sep 23 2020
MATHEMATICA
n = 10^7; Flatten[Table[7^i*13^j, {i, 0, Log[7, n]}, {j, 0, Log[13, n/7^i]}]] // Sort (* Amiram Eldar, Sep 23 2020 *)
PROG
(PARI) list(lim)=my(v=List(), N); for(n=0, log(lim)\log(13), N=13^n; while(N<=lim, listput(v, N); N*=7)); vecsort(Vec(v)) \\ Charles R Greathouse IV, Jun 28 2011
(Python)
from sympy import integer_log
def A108056(n):
def bisection(f, kmin=0, kmax=1):
while f(kmax) > kmax: kmax <<= 1
while kmax-kmin > 1:
kmid = kmax+kmin>>1
if f(kmid) <= kmid:
kmax = kmid
else:
kmin = kmid
return kmax
def f(x): return n+x-sum(integer_log(x//13**i, 7)[0]+1 for i in range(integer_log(x, 13)[0]+1))
return bisection(f, n, n) # Chai Wah Wu, Oct 22 2024
KEYWORD
nonn,easy
AUTHOR
Douglas Winston (douglas.winston(AT)srupc.com), Jun 02 2005
EXTENSIONS
More terms from Amiram Eldar, Sep 23 2020
STATUS
approved