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A365020
Solutions k to the exponential Diophantine equation k^5 = Sum_{i=1..7} y_i^5 with positive y_i.
1
23, 26, 30, 34, 35, 38, 40, 41, 42, 46, 49, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100, 101, 102, 103, 104, 105
OFFSET
1,1
COMMENTS
This includes primitive solutions (where all y_i are coprime) and non-primitive solutions. There are k that have both types, e.g., 52^5 = 6^5 + 10^5 + 10^5 + 22^5 + 38^5 + 42^5 + 44^5 is non-primitive and 52^5 = 17^5 + 18^5 + 21^5 + 27^5 +31^5 + 44^5 + 44^5 is primitive. - R. J. Mathar, Aug 17 2023
Conjecture: sequence contains all integers >= 78 and a(n) = n+42 for n >= 36. - Chai Wah Wu, Aug 17 2023
EXAMPLE
Solutions [y_1,...,y_7]: k are [1, 7, 8, 14, 15, 18, 20]: 23, [3, 5, 5, 11, 19, 21, 22]: 26, [20, 23, 31, 32, 35, 41, 69]: 71, [5, 7, 12, 39, 45, 46, 68]: 72, [7, 19, 19, 26, 28, 56, 67]: 72, so 23, 26, 71 and 72 are in the sequence.
PROG
(Python)
from itertools import count, islice
from sympy import integer_nthroot
def A365020_gen(startvalue=1): # generator of terms >= startvalue
for n in count(max(startvalue, 1)):
n5, flag = n**5, False
for i1 in range(1, n):
i15=i1**5
for i2 in range(i1, n):
i25 = i15+i2**5
if i25>=n5: break
for i3 in range(i2, n):
i35 = i25+i3**5
if i35>=n5: break
for i4 in range(i3, n):
i45 = i35+i4**5
if i45>=n5: break
for i5 in range(i4, n):
i55 = i45+i5**5
if i55>=n5: break
for i6 in range(i5, n):
i65 = i6**5
i75 = n5-i55-i65
if i75<i65: break
if integer_nthroot(i75, 5)[1]:
yield n
flag = True
break
if flag: break
if flag: break
if flag: break
if flag: break
if flag: break
A365020_list = list(islice(A365020_gen(), 4)) # Chai Wah Wu, Aug 17 2023
CROSSREFS
Sequence in context: A359510 A345489 A025059 * A316724 A160774 A163142
KEYWORD
nonn
AUTHOR
R. J. Mathar, Aug 16 2023
STATUS
approved