A129317 -id:A129317 - cp's OEIS frontend

A129319 a(n) = sopfr(A129317(n)), where sopfr(x) is the sum of the prime factors of x, counting multiplicity.

5, 6, 8, 18, 15, 30, 34, 78, 29, 86, 115, 525, 50, 110, 33, 204, 286, 62, 506, 32, 185, 120, 35, 92, 180, 100, 390, 102, 624, 450, 44, 198, 455, 180, 294, 306, 45, 141, 882, 134, 650, 86, 708, 575, 116, 75, 150, 174, 152, 375, 160, 286, 242, 122, 584, 518, 1456, 448

Offset: 1

Walter Kehowski, Apr 09 2007

			a(6)=30 since sopfr(A129317(6))=sopfr(161)=sopfr(7*23)=30.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A001414, A006145, A039752, A129316, A129317, A129318, A129320.

A129316 Positive integers k such that sopfr(k) divides sopfr(k+1), where sopfr(k) is the sum of the prime factors of k, counting multiplicity.

Original entry on oeis.org

5, 8, 15, 77, 125, 160, 252, 496, 714, 948, 980, 1045, 1053, 1260, 1330, 1378, 1404, 1430, 1508, 1520, 1610, 1750, 1862, 1890, 2170, 2491, 2680, 2821, 3094, 3100, 3248, 3400, 3591, 3610, 3652, 3808, 4185, 4191, 4384, 4452, 4500, 4598, 4906, 5120, 5145

Offset: 1

Walter Kehowski, Apr 09 2007

A generalization of Ruth-Aaron pairs (A006145).

			a(6)=160 since sopfr(160)=sopfr(2^5*5)=10+5=15 and sopfr(161)=sopfr(7*23)=30.

Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Harvey P. Dale)

Cf. A001414, A006145, A039752, A129317, A129318, A129319.

sopf[n_]:=Total[Flatten[Table[#[[1]],{#[[2]]}]&/@FactorInteger[n]]]; Rest[ Flatten[Position[Partition[Table[sopf[n],{n,5200}],2,1],?(Divisible[#[[2]],#[[1]]]&),{1},Heads->False]]] (* _Harvey P. Dale, Jul 18 2013 *)

sopfr(k+1) mod sopfr(k) = 0.

A129318 a(n) = sopfr(A129316(n)), where sopfr(x) is the sum of the prime factors of x, counting multiplicity.

Original entry on oeis.org

5, 6, 8, 18, 15, 15, 17, 39, 29, 86, 23, 35, 25, 22, 33, 68, 26, 31, 46, 32, 37, 24, 35, 23, 45, 100, 78, 51, 39, 45, 44, 33, 35, 45, 98, 34, 45, 141, 147, 67, 25, 43, 236, 25, 29, 75, 150, 29, 76, 125, 160, 26, 121, 122, 73, 37, 52, 56, 40, 39, 54, 58, 239, 69, 36, 39

Offset: 1

Walter Kehowski, Apr 09 2007, corrected Apr 09 2007

			a(6)=15 since sopfr(A129316(6))=sopfr(160)=sopfr(2^5*5)=15.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A001414, A006145, A039752, A129316, A129317, A129319, A129320.

A129320 a(n) = A129319(n)/A129318(n).

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 2, 2, 1, 1, 5, 15, 2, 5, 1, 3, 11, 2, 11, 1, 5, 5, 1, 4, 4, 1, 5, 2, 16, 10, 1, 6, 13, 4, 3, 9, 1, 1, 6, 2, 26, 2, 3, 23, 4, 1, 1, 6, 2, 3, 1, 11, 2, 1, 8, 14, 28, 8, 1, 7, 1, 23, 8, 20, 7, 2

Offset: 1

Walter Kehowski, Apr 09 2007

			a(6)=2 since A129319(n)/A129318(n)=30/15=2.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A001414, A006145, A039752, A129316, A129317, A129318, A129319.

cp's OEIS Frontend

A129319 a(n) = sopfr(A129317(n)), where sopfr(x) is the sum of the prime factors of x, counting multiplicity.

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A129316 Positive integers k such that sopfr(k) divides sopfr(k+1), where sopfr(k) is the sum of the prime factors of k, counting multiplicity.

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A129318 a(n) = sopfr(A129316(n)), where sopfr(x) is the sum of the prime factors of x, counting multiplicity.

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A129320 a(n) = A129319(n)/A129318(n).

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