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A350862
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Decimal expansion of Sum_{k>=1} (k^(1/k^(1 + 1/1111)) - 1).
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2
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1, 2, 3, 4, 3, 2, 1, 9, 8, 2, 8, 1, 3, 8, 9, 0, 9, 3, 3, 3, 6, 8, 6, 4, 2, 4, 4, 0, 0, 4, 8, 8, 7, 7, 4, 8, 6, 8, 2, 6, 9, 1, 2, 5, 8, 7, 7, 1, 5, 4, 8, 3, 8, 1, 2, 6, 2, 3, 5, 0, 2, 6, 6, 6, 4, 0, 7, 4, 2, 2, 6, 9, 9, 0, 2, 7, 0, 3, 0, 1, 1, 3, 8, 2, 7, 7, 9
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OFFSET
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7,2
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COMMENTS
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Conjecture: If x is a whole number greater than 1, the Sum_{k>=1} (k^(1/k^(1 + 1/sqrt(x))) - 1) = x + C, where C is a constant less than 1.
The above relation was tested for all 1 < x < 10^7.
This sum demonstrates this relationship: setting sqrt(x) = 1111 generates the sum 1111^2 + C or 1234321 + C. Another example would be Sum_{k>=1} (k^(1/k^(1 + 1/sqrt(1729))) - 1) = 1729.84841430674....
Evaluating the sum at larger x values converges slower and slower. Monotonically changing extrapolation methods such as Richardson's Extrapolation must be used to compute these values.
Since the output (x + C) will be the square of the input (sqrt(x)) plus a constant less than 1, this implies that Sum_{k>=1} (k^(1/k^(1 + 1/sqrt(x))) - 1) diverges as x tends to infinity, or simplified to Sum_{k>=1} (k^(1/k) - 1).
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LINKS
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EXAMPLE
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1234321.98281389093336864244004887748682691258771548...
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MATHEMATICA
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digits = 120; d = 1; j = 1; s = 0; While[Abs[d] > 10^(-digits - 5), d = (-1)^j/j!*Derivative[j][Zeta][(1 + 1/1111)*j]; s += d; j++]; RealDigits[s, 10, 120][[1]] (* Vaclav Kotesovec, Jun 18 2023 *)
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PROG
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(PARI) sumpos(k=1, k^(1/(k^(1 + 1/1111))) - 1).
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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