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Numbers n such that A051950(n+1) = 4. Numbers n such that A049820(n) - A049820(n+1) = 3. Sequence of starts of first run of n (n>=2) consecutive integers m_1, m_2, ..., m_n such that tau(m_k) - tau(m_k-1) = 4, for all k=n...2: 11, 458, 3013, ... (a(5) > 100000); example for n=4: tau(3013) = 4, tau(3014) = 8, tau(3015) = 12, tau(3016) = 16.

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Numbers n such that tau(n+1) - tau(n) = 4; , where tau(n) = the number of divisors of n (A000005).

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allocated for Jaroslav KrizekNumbers n such that tau(n+1) - tau(n) = 4; where tau(n) = the number of divisors of n (A000005).

11, 17, 19, 31, 39, 43, 55, 65, 67, 69, 77, 87, 97, 129, 134, 163, 175, 183, 185, 194, 207, 211, 221, 237, 241, 247, 249, 254, 265, 283, 295, 309, 321, 327, 331, 337, 343, 351, 365, 398, 404, 417, 437, 454, 458, 459, 469, 471, 473, 482, 493, 494, 497, 505, 517

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Numbers n such that A051950(n) = 4. Numbers n such that A049820(n) - A049820(n+1) = 3. Sequence of starts of first run of n (n>=2) consecutive integers m_1, m_2, ..., m_n such that tau(m_k) - tau(m_k-1) = 4, for all k=n...2: 11, 458, 3013, ... (a(5) > 100000); example for n=4: tau(3013) = 4, tau(3014) = 8, tau(3015) = 12, tau(3016) = 16.

Jaroslav Krizek, <a href="/A230654/b230654.txt">Table of n, a(n) for n = 1..4000</a>

19 is in sequence because tau(20) - tau(19) = 6 - 2 = 4.

Select[ Range[ 50000], DivisorSigma[0, # ] + 4 == DivisorSigma[0, # + 1] &]

Cf. A055927 (numbers n such that tau(n+1) - tau(n) = 1), A230115 (numbers n such that tau(n+1) - tau(n) = 2), A230653 (numbers n such that tau(n+1) - tau(n) = 3), A000005.

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Jaroslav Krizek, Nov 03 2013

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