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A358395
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Odd numbers k such that sigma(k) + sigma(k+2) > 2*sigma(k+1); odd terms in A053228.
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2
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1125, 1573, 1953, 2205, 2385, 3465, 5185, 5353, 5773, 6433, 6613, 6825, 7245, 7425, 7665, 7693, 8505, 8925, 9133, 9205, 9405, 9945, 10393, 10773, 11473, 11653, 12285, 12493, 12705, 13473, 13585, 13725, 14025, 15013, 15145, 15433, 16065, 16245, 16905, 17253, 17325, 17953
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OFFSET
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1,1
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COMMENTS
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Odd numbers k such that A053223(k) > 0.
Terms congruent to 5 modulo 6 exist but must be very large: for example A053223(670173643268502741420822977335461337017377351999597045900203591953125) = 1311786588705365455963902347308393766941056366825184647502989937872.
A number m coprime to 2 and 3 such that sigma(m)/m >= 3 (m = A358412(3) = A358413(2) = 5^4*7^3*11^2*13^2*17*...*157 ~ 5.16404*10^66 is the smallest such number; see the link from Mercurial, the Spectre) produces a family of infinitely many terms congruent to 5 modulo 6 in this sequence, by Dirichlet's theorem on arithmetic progressions. Concretely, let k == 5 (mod 6), N(t) = t*k*(k+2) + (k+1)/6 for t >= 0, then:
(i) If sigma(k)/k >= 3. If N(t) is prime and 6*N(t)+1 is composite, then sigma(6*N(t)-1) >= 3*(6*N(t)-1), sigma(6*N(t)) = 12*(N(t)+1) and sigma(6*N(t)+1) >= 1+sqrt(6*N(t)+1)+(6*N(t)+1), so A053223(6*N(t)-1) >= sqrt(6*N(t)+1) - 25 >= sqrt(k+2) - 25 > 0.
(ii) If sigma(k+2)/(k+2) >= 3. If N(t) is prime and 6*N(t)-1 is composite, then sigma(6*N(t)+1) >= 3*(6*N(t)+1), sigma(6*N(t)) = 12*(N(t)+1) and sigma(6*N(t)-1) >= 1+sqrt(6*N(t)-1)+(6*N(t)-1), so A053223(6*N(t)-1) >= sqrt(6*N(t)-1) - 21 >= sqrt(k) - 21 > 0.
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LINKS
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EXAMPLE
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1125 is a term since sigma(1126) = 1692 is smaller than the average of sigma(1125) = 2028 and sigma(1127) = 1368.
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PROG
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(PARI) isA358395(n) = (n%2) && (sigma(n) + sigma(n+2) > 2*sigma(n+1))
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CROSSREFS
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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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