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A124980
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Smallest strictly positive number decomposable in n different ways as a sum of two squares.
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5
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1, 25, 325, 1105, 4225, 5525, 203125, 27625, 71825, 138125, 2640625, 160225, 17850625, 1221025, 1795625, 801125, 1650390625, 2082925, 49591064453125, 4005625, 44890625, 2158203125, 30525625, 5928325, 303460625, 53955078125, 35409725, 100140625
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OFFSET
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1,2
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COMMENTS
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The number must be strictly positive, but one of the squares may be zero, as we see from a(1) = 1 = 1^2 + 0^2 and a(2) = 25 = 3^2 + 4^2 = 5^0 + 0^2. - M. F. Hasler, Jul 07 2024
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LINKS
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FORMULA
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EXAMPLE
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a(3) = 325 is decomposable in 3 ways: 15^2 + 10^2 = 17^2 + 6^2 = 18^2 + 1^2.
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PROG
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(PARI)
PD(n, L=n, D=Vecrev(divisors(n)[^1])) = { if(n>1, concat(vector(#D, i, if(D[i] > L, [], D[i] < n, [concat(D[i], P) | P <- PD(n/D[i], D[i])], [[n]]))), [[]])}
apply( {A124980(n)=vecmin([prod(i=1, #a, A002144(i)^(a[i]-1)) | a<-concat([PD(n*2, n), PD(n*2-1)])])}, [1..44]) \\ M. F. Hasler, Jul 07 2024
(Python)
from sympy import divisors, isprime, prod
def PD(n, L=None): return [[]] if n==1 else [
[d]+P for d in divisors(n)[:0:-1] if d <= (L or n) for P in PD(n//d, d)]
A2144=lambda upto=999: filter(isprime, range(5, upto, 4))
return min(prod(a**(f-1) for a, f in zip(A2144(), P))
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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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