cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A184753 a(n) = A184752(n)/A130650(n) unless A130650(n) = 0 in which case a(n) = 0.

Original entry on oeis.org

0, 0, 4, 1, 13, 2, 1, 10, 1, 5, 16, 1, 15, 16, 22, 5, 37, 2, 4, 2, 1, 24, 11, 10, 2, 28, 23, 11, 41, 20, 2, 3, 73, 13, 76, 12, 1, 20, 13, 85, 34, 1, 21, 2, 46, 62, 5, 3, 2, 2, 2, 1, 2, 78, 39, 80, 81, 122, 3, 63, 51, 32, 88, 1, 1, 1, 69, 70
Offset: 1

Views

Author

Rémi Eismann, Jan 21 2011

Keywords

Comments

a(n) is the "level" of 3-almost primes.
The decomposition of 3-almost primes into weight * level + gap is A014612(n) = A130650(n) * a(n) + A114403(n) if a(n) > 0.
a(n) = A014612(n) - A114403(n) if A014612(n) - A114403(n) > A114403(n), 0 otherwise.

Examples

			For n = 1 we have A130650(1) = 0, hence a(1) = 0.
For n = 3 we have A184752(3)/A130650(3)= 16 / 4 = 4; hence a(3) = 4.
For n = 21 we have A184752(21)/A130650(21)= 97 / 97 = 28; hence a(21) = 1.

Links

Crossrefs

Cf. A014612, A114403, A130650, A184752, A117078, A117563, A001223, A118534.

A184827 a(n) = largest k such that A000959(n+1) = A000959(n) + (A000959(n) mod k), or 0 if no such k exists.

Original entry on oeis.org

0, 0, 5, 5, 11, 9, 17, 19, 29, 29, 31, 37, 47, 39, 59, 65, 65, 71, 71, 71, 81, 87, 93, 99, 107, 103, 125, 125, 131, 129, 131, 143, 155, 157, 167, 153, 185, 191, 189, 197, 199, 203, 215, 215, 227, 233, 233, 223, 257, 255, 261, 263
Offset: 1

Views

Author

Rémi Eismann, Jan 23 2011

Keywords

Comments

From the definition, a(n) = A000959(n) - A031883(n) if A000959(n) - A031883(n) > A031883(n), 0 otherwise where A000959 are the lucky numbers and A031883 are the gaps between lucky numbers.

Examples

			For n = 1 we have A000959(1) = 1, A000959(2) = 3; there is no k such that 3 - 1 = 2 = (1 mod k), hence a(1) = 0.
For n = 3 we have A000959(3) = 7, A000959(4) = 9; 5 is the largest k such that 9 - 7 = 2 = (7 mod k), hence a(3) = 5; a(3) = 7 -2 = 5.
For n = 24 we have A000959(24) = 105, A000959(25) = 111; 99 is the largest k such that 111 - 105 = 6 = (105 mod k), hence a(24) = 99; a(24) = 105 - 6 = 99.

Links

Crossrefs

Cf. A000959, A031883, A130889, A184828, A117078, A117563, A001223, A118534.

A184828 a(n) = A184827(n)/A130889(n) unless A130889(n) = 0 in which case a(n) = 0.

Original entry on oeis.org

0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 13, 13, 1, 1, 1, 9, 3, 3, 11, 1, 1, 25, 25, 1, 3, 1, 13, 31, 1, 1, 3, 37, 1, 27, 1, 1, 7, 43, 5, 1, 1, 1, 1, 1, 17, 29, 1, 1, 1, 1, 3, 23, 5, 1, 45, 19, 19, 7, 31, 1, 5, 1, 1, 1, 43, 1, 31, 1, 5, 85, 85, 5, 1, 11, 43, 3
Offset: 1

Views

Author

Rémi Eismann, Jan 23 2011

Keywords

Comments

a(n) is the "level" of lucky numbers.
The decomposition of lucky numbers into weight * level + gap is A000959(n) = A130889(n) * a(n) + A031883(n) if a(n) > 0.
A184827(n) = A000959(n) - A031883(n) if A000959(n) - A031883(n) > A031883(n), 0 otherwise.

Examples

			For n = 1 we have A130889(1) = 0, hence a(1) = 0.
For n = 3 we have A184752(3)/A130889(3)= 5 / 5 = 1; hence a(3) = 1.
For n = 24 we have A184752(24)/A130889(24)= 99 / 9 = 11; hence a(24) = 11.

Links

Crossrefs

Cf. A000959, A031883, A130889, A184827, A117078, A117563, A001223, A118534.

A184829 a(n) = smallest k such that A000961(n+1) = A000961(n) + (A000961(n) mod k), or 0 if no such k exists.

Original entry on oeis.org

2, 0, 2, 3, 3, 2, 7, 7, 3, 5, 3, 3, 5, 3, 23, 5, 3, 2, 9, 11, 3, 13, 3, 5, 47, 3, 29, 61, 7, 3, 67, 7, 79, 7, 9, 31, 3, 9, 3, 5, 15, 9, 3, 2, 5, 25, 3, 43, 3, 29, 151, 53, 3, 5, 167, 3, 19, 3, 7, 3, 17, 199, 73, 3, 5, 227, 3, 239, 47, 6, 3, 251, 257, 3, 53, 7, 3, 277, 5
Offset: 1

Views

Author

Rémi Eismann, Jan 23 2011

Keywords

Comments

a(n) is the "weight" of prime powers.
The decomposition of prime powers into weight*level + gap is A000961(n) = a(n)*A184831(n) + A057820(n) if n > 2 and a(n) > 0. [amended by Jason Yuen, Oct 17 2024]

Examples

			For n = 1 we have A000961(1) = 1, A000961(2) = 2; 2 is the smallest k such that 2 = 1 + (1 mod k), hence a(1) = 2.
For n = 3 we have A000961(3) = 3, A000961(4) = 4; 2 is the smallest k such that 4 = 3 + (3 mod k), hence a(3) = 2.
For n = 24 we have A000961(24) = 49, A000961(25) = 53; 5 is the smallest k such that 53 = 49 + (49 mod k), hence a(24) = 5.

Links

Crossrefs

Cf. A000961, A057820, A184831, A184830, A117078, A117563, A001223, A118534.

Extensions

a(1) corrected by Jason Yuen, Oct 17 2024

A184830 a(n) = largest k such that A000961(n+1) = A000961(n) + (A000961(n) mod k), or 0 if no such k exists.

Original entry on oeis.org

0, 0, 2, 3, 3, 6, 7, 7, 9, 10, 15, 15, 15, 21, 23, 25, 27, 30, 27, 33, 39, 39, 45, 45, 47, 57, 58, 61, 63, 69, 67, 77, 79, 77, 81, 93, 99, 99, 105, 105, 105, 117, 123, 126, 125, 125, 135, 129, 147, 145, 151, 159, 165, 165, 167, 177, 171, 189, 189, 195
Offset: 1

Views

Author

Rémi Eismann, Jan 23 2011

Keywords

Comments

From the definition, a(n) = A000961(n) - A057820(n) if A000961(n) - A057820(n) > A057820(n), 0 otherwise where A000961 are the prime powers and A057820 are the gaps between prime powers.

Examples

			For n = 1 we have A000961(1) = 1, A000961(2) = 2; for all k >= 2, 2 = 1 + (1 mod k), hence the largest k does not exist and a(1) = 0.
For n = 3 we have A000961(3) = 3, A000961(4) = 4; 2 is the largest k such that 4 = 3 + (3 mod k), hence a(3) = 2.
For n = 24 we have A000961(24) = 49, A000961(25) = 53; 45 is the largest k such that 53 = 49 + (49 mod k), hence a(24) = 45.

Links

Crossrefs

Cf. A000961, A057820, A184829, A184831, A117078, A117563, A001223, A118534.

Programs

A184831 a(n) = A184830(n)/A184829(n) unless A184829(n) = 0 in which case a(n) = 0.

Original entry on oeis.org

0, 0, 1, 1, 1, 3, 1, 1, 3, 2, 5, 5, 3, 7, 1, 5, 9, 15, 3, 3, 13, 3, 15, 9, 1, 19, 2, 1, 9, 23, 1, 11, 1, 11, 9, 3, 33, 11, 35, 21, 7, 13, 41, 63, 25, 5, 45, 3, 49, 5, 1, 3, 55, 33, 1, 59, 9, 63, 27, 65, 11, 1, 3, 75, 45, 1, 79, 1, 5, 41, 85, 1, 1, 89, 5, 39, 93, 1, 57, 9
Offset: 1

Views

Author

Rémi Eismann, Jan 23 2011

Keywords

Comments

a(n) is the "level" of prime powers.
The decomposition of prime powers into weight * level + gap is A000961(n) = A184829(n) * a(n) + A057820(n) if a(n) > 0.
A184830(n) = A000961(n) - A057820(n) if A000961(n) - A057820(n) > A057820(n), 0 otherwise.

Examples

			For n = 1 we have A184829(1) = 0, hence a(1) = 0.
For n = 3 we have A184830(3)/A184829(3)= 2 / 2 = 1; hence a(3) = 1.
For n = 24 we have A184830(24)/A184829(24)= 45 / 5 = 9; hence a(24) = 9.

Links

Crossrefs

Cf. A000961, A057820, A184829, A184830, A117078, A117563, A001223, A118534.

A184832 a(n) = smallest k such that A005117(n+1) = A005117(n) + (A005117(n) mod k), or 0 if no such k exists.

Original entry on oeis.org

0, 0, 0, 2, 5, 4, 3, 3, 2, 13, 13, 3, 17, 2, 3, 4, 23, 2, 29, 29, 2, 3, 3, 2, 37, 37, 2, 41, 4, 3, 43, 7, 3, 53, 2, 3, 3, 2, 59, 2, 5, 5, 2, 3, 3, 2, 71, 2, 7, 4, 3, 3, 2, 5, 5, 3, 89, 2, 3, 3, 31, 2, 101, 101, 2, 3, 3, 2, 109, 109, 2, 113, 4, 3, 4, 11, 7, 5, 2, 3, 3, 2
Offset: 1

Views

Author

Rémi Eismann, Jan 23 2011

Keywords

Comments

a(n) is the "weight" of squarefree numbers.
The decomposition of squarefree numbers into weight * level + gap is A005117(n) = a(n) * A184834(n) + A076259(n) if a(n) > 0.

Examples

			For n = 1 we have A005117(1) = 1, A005117(2) = 2; there is no k such that 2 - 1 = 1 = (1 mod k), hence a(1) = 0.
For n = 4 we have A005117(4) = 5, A005117(5) = 6; 2 is the smallest k such that 6 - 5 = 1 = (5 mod k), hence a(4) = 2.
For n = 23 we have A005117(23) = 35, A005117(24) = 37; 3 is the smallest k such that 37 - 35 = 2 = (35 mod k), hence a(23) = 3.

Links

Crossrefs

Cf. A005117, A076259, A184834, A184833, A117078, A117563, A001223, A118534.

A184833 a(n) = largest k such that A005117(n+1) = A005117(n) + (A005117(n) mod k), or 0 if no such k exists.

Original entry on oeis.org

0, 0, 0, 4, 5, 4, 9, 9, 12, 13, 13, 15, 17, 20, 21, 20, 23, 28, 29, 29, 32, 33, 33, 36, 37, 37, 40, 41, 40, 45, 43, 49, 51, 53, 56, 57, 57, 60, 59, 64, 65, 65, 68, 69, 69, 72, 71, 76, 77, 76, 81, 81, 84, 85, 85, 87, 89, 92, 93, 93, 93, 100, 101, 101, 104, 105
Offset: 1

Views

Author

Rémi Eismann, Jan 23 2011

Keywords

Comments

From the definition, a(n) = A005117(n) - A076259(n) if A005117(n) - A076259(n) > A076259(n), 0 otherwise where A005117 are the squarefree numbers and A076259 are the gaps between squarefree numbers.

Examples

			For n = 1 we have A005117(1) = 1, A005117(2) = 2; there is no k such that 2 - 1 = 1 = (1 mod k), hence a(1) = 0.
For n = 4 we have A005117(4) = 5, A005117(5) = 6; 4 is the largest k such that 6 - 5 = 1 = (5 mod k), hence a(4) = 2; a(3) = 5 - 1 = 4.
For n = 23 we have A005117(23) = 35, A005117(24) = 37; 33 is the largest k such that 37 - 35 = 2 = (35 mod k), hence a(23) = 33; a(24) = 35 - 2 = 33.

Links

Crossrefs

Cf. A005117, A076259, A184832, A184834, A117078, A117563, A001223, A118534.

A184834 a(n) = A184833(n)/A184832(n) unless A184832(n) = 0 in which case a(n) = 0.

Original entry on oeis.org

0, 0, 0, 2, 1, 1, 3, 3, 6, 1, 1, 5, 1, 10, 7, 5, 1, 14, 1, 1, 16, 11, 11, 18, 1, 1, 20, 1, 10, 15, 1, 7, 17, 1, 28, 19, 19, 30, 1, 32, 13, 13, 34, 23, 23, 36, 1, 38, 11, 19, 27, 27, 42, 17, 17, 29, 1, 46, 31, 31, 3, 50, 1, 1, 52, 35, 35, 54, 1, 1, 56, 1, 28, 39
Offset: 1

Views

Author

Rémi Eismann, Jan 23 2011

Keywords

Comments

a(n) is the "level" of squarefree numbers.
The decomposition of squarefree numbers into weight * level + gap is A005117(n) = A184832(n) * a(n) + A076259(n) if a(n) > 0.
A184833(n) = A005117(n) - A076259(n) if A005117(n) - A076259(n) > A076259(n), 0 otherwise.

Examples

			For n = 1 we have A184832(1) = 0, hence a(1) = 0.
For n = 4 we have A184833(4)/A184832(4)= 4 / 2 = 1; hence a(4) = 2.
For n = 23 we have A184833(23)/A184832(23)= 33 / 3 = 11; hence a(23) = 11.

Links

Crossrefs

Cf. A005117, A076259, A184832, A184833, A117078, A117563, A001223, A118534.

A118144 Numbers of prime factors of l, where l is defined in A118534.

Original entry on oeis.org

0, 0, 1, 0, 2, 2, 2, 2, 1, 3, 2, 2, 2, 2, 1, 1, 2, 2, 3, 2, 1, 3, 2, 4, 2, 3, 3, 3, 3, 3, 2, 3, 4, 2, 3, 2, 1, 2, 2, 1, 2, 3, 4, 4, 3, 2, 1, 2, 4, 4, 1, 2
Offset: 1

Views

Author

Rémi Eismann and Fabien Sibenaler, May 14 2006, Feb 14 2008

Keywords

Comments

a(n) = 0 only for n = 1, 2 and 4.

Examples

			For a(1), l=0 thus a(1)=0,
for a(3), l=3 thus a(3)=1,
for a(8), l=15=3*5 thus a(8)=2,
for a(24), l=81=3*3*3*3 thus a(24)=4.

Links

Crossrefs

Cf. A118534, A117563, A118123, A117078.

Formula

a(n) = numbers of factors of l, largest l such that prime(n+1) = prime(n) + (prime(n) mod l), or 0 if no such l exists.
Previous Showing 51-60 of 77 results. Next

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