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A369354
The number of terms beyond the first term for the series starting at k(1) = n and then iteratively finding the smallest k(i+1) such that k(i+1) - k(i) = sopfr(k(i+1) + k(i)), where i >= 1 and sopfr(m) is the sum of the primes dividing m, with repetition.
7
1, 0, 0, 1, 2, 0, 0, 0, 1, 1, 1, 0, 1, 0, 1, 0, 2, 0, 2, 0, 0, 0, 0, 1, 1, 1, 0, 1, 0, 1, 1, 0, 0, 0, 3, 0, 0, 0, 0, 2, 2, 0, 2, 0, 1, 1, 0, 0, 2, 0, 0, 0, 3, 0, 2, 0, 1, 0, 0, 0, 0, 1, 0, 0, 2, 0, 2, 1, 0, 1, 0, 0, 2, 1, 1, 2, 0, 0, 1, 1, 0, 1, 0, 0, 0, 0, 1, 0, 1, 4, 0, 1, 0, 0, 0, 0, 0, 0, 0
OFFSET
1,5
COMMENTS
In the first 500000 terms the largest value is a(443314) = 17. See the examples.
See A369357 for the numbers that terminate the series with lengths >= 1.
LINKS
EXAMPLE
a(1) = 1 as the series when starting at 1 is 1, 7, and there is no number k such that k - 7 = sopfr(k + 7).
a(2) = 0 as there is no number k such that k - 2 = sopfr(k + 2).
a(5) = 2 as the series when starting at 5 is 5, 13, 23, and there is no number k such that k - 23 = sopfr(k + 23).
a(443314) = 17 as the series when starting at 443314 is 443314, 454409, 454790, 455085, 456828, 474078, 474578, 482330, 482497, 483143, 486180, 486225, 534858, 535155, 535369, 540494, 580542, 581003, and there is no number k such that k - 581003 = sopfr(k + 581003).
KEYWORD
nonn
AUTHOR
Scott R. Shannon, Jan 25 2024
STATUS
approved