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A332101
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Least m such that m^n <= Sum_{k<m} k^n.
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6
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2, 3, 5, 6, 8, 9, 11, 12, 14, 15, 16, 18, 19, 21, 22, 24, 25, 27, 28, 29, 31, 32, 34, 35, 37, 38, 40, 41, 42, 44, 45, 47, 48, 50, 51, 52, 54, 55, 57, 58, 60, 61, 63, 64, 65, 67, 68, 70, 71, 73, 74, 76, 77, 78, 80, 81, 83, 84, 86, 87, 89, 90, 91, 93, 94, 96, 97
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OFFSET
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0,1
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COMMENTS
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In a list (1^n, 2^n, 3^n, ...) (rows of table A051128 or A051129), a(n) is the index of the first term less than or equal to the sum of all earlier terms, cf. example.
Obviously a lower bound for any s solution to s^n = Sum_{x in S} x^n, S subset of {1, ..., s-1}, cf. A030052.
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LINKS
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FORMULA
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a(n) = round(n / log(2)) + 2. (Conjectured; verified up to 10^4, in particular for 3525/log(2) = 5085.500019... and 7844/log(2) ~ 11316.49990...)
a(n) = A078607(n) + 2 for almost all n > 1. (n = 777451915729368 might be an exception to this equality or the above one.) - M. F. Hasler, May 08 2020
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EXAMPLE
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For n = 0, m^0 > Sum_{0 < k < m} k^0 = 0 for m = 0, 1 (empty sums), but 2^0 = Sum_{0 < k < 2} k^0 = 1, so a(0) = 2.
For n = 1, 1^1 > Sum_{0 < k < 1} k^1 = 0 (empty sum) and 2^1 > Sum_{0 < k < 2} k^1 = 1, but 3^1 <= Sum_{0 < k < 3} k^1 = 1 + 2, so a(1) = 3.
To find a(n) one can add up terms in row n of the table k^n until the sum equals or exceeds the next term, whose column number k is then a(n):
n |k: 1 2 3 4 5 6 Comment
--+---------------------------------------------------------------
1 | 1 2 3 1 < 2 but 1 + 2 >= 3, so a(1) = 3.
2 | 1 4 9 16 25 1 + 4 + 9 + 16 > 25, and a(2) = 5.
3 | 1 8 27 64 125 216 1 + 8 + 27 + 64 + 125 > 216: a(3) = 6.
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MATHEMATICA
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Table[Block[{m = 1, s = 0}, While[m^n > s, s = s + m^n; m++]; m], {n, 0, 66}] (* Michael De Vlieger, Apr 30 2020 *)
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PROG
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(PARI) apply( A332101(n, s)=for(m=1, oo, s<m^n||return(m); s+=m^n), [0..66])
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CROSSREFS
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Cf. A078607, A332097 (maximum of E(s), cf comments), A030052 (least k such that k^n = sum of distinct n-th powers), A332065 (all k such that k^n is a sum of distinct n-th powers).
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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