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A349184
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a(n) = n^n - Sum_{k=1..n-2} f_k(n), with f_k(n)=( floor( (n^n - Sum_{t=1..k-1} f_t(n))^(1/(n-k)) ) )^(n-k).
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1
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4, 2, 4, 16, 29, 29, 53, 25, 5, 47, 18, 95, 5, 18, 8, 168, 231, 72, 13, 487, 695, 216, 242, 14, 359, 377, 106, 240, 1429, 2840, 1469, 2085, 1276, 1197, 577, 7388, 1488, 2818, 2660, 4138, 4455, 448, 220, 2988, 19220, 459, 18020, 1453, 3514, 7682, 15159, 1821, 541, 38445
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OFFSET
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2,1
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COMMENTS
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a(n) is calculated starting from n^n and successively subtracting largest possible perfect powers, where those powers range from perfect (n-1)-th power down to perfect square.
At initial n^n, f_1(n) is the largest (n-1)-th power not exceeding n^n. Subtract it to leave remainder n^n - f_1(n). Then f_2(n) is the largest (n-2)-th perfect power not exceeding that remainder, so subtract it to remainder now n^n - f_1(n) - f_2(n). Continue this way through to subtracting f_{n-2}(n) which is the largest perfect square not exceeding the remainder at its point, and leaving result a(n) = n^n - f_1(n) - ... - f_{n-2}(n).
In other words: Subtract from n^n the next smaller power x^(n-1) with the largest x and the result >= 0. Repeat this power reduction and subtract it from the previous result, until x^2, leaving the term a(n). E.g., a(5) = 5^5 - 7^4 - 8^3 - 14^2 = 16.
The definitions f_k(n) and f_t(n) are the k-th and t-th function of the same function f(n).
It appears that log(a(n)) is of the order log(n)^2. (Possibly, log(a(n))/log(n)^2 converges to 1/2.) - Pontus von Brömssen, Nov 11 2021
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LINKS
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EXAMPLE
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a(5)=16, because
f_1 = ( floor( (5^5 )^(1/4) ) )^4 = 2401,
f_2 = ( floor( (5^5- f_1 )^(1/3) ) )^3 = 512,
f_3 = ( floor( (5^5-(f_1+f_2))^(1/2) ) )^2 = 196,
a(5) = 5^5 - (f_1 + f_2 + f_3) = 3125 -(2401 + 512 + 196) = 16.
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MATHEMATICA
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f[n_, k_] := Floor[(n^n - Sum[f[n, j], {j, k - 1}])^(1/(n - k))]^(n - k); Array[#^# - Sum[f[#, k], {k, # - 2}] &, 16, 2] (* Michael De Vlieger, Nov 10 2021 *)
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PROG
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(PARI) a(n) = {my(v=vector(n)); v[1] = n^n; for (i=2, n, v[i] = sqrtnint(v[1] - sum(t=1, i-1, v[t+1]), n-i+1)^(n-i+1); ); v[1] - sum(k=2, n-1, v[k]); } \\ Michel Marcus, Nov 10 2021
(Python)
from sympy import integer_nthroot as introot
from functools import reduce
return reduce(lambda x, e:x-introot(x, e)[0]**e, range(n-1, 1, -1), n**n) # Pontus von Brömssen, Nov 11 2021
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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