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.

User: Rémi Eismann

Rémi Eismann's wiki page.

Rémi Eismann has authored 75 sequences. Here are the ten most recent ones:

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

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.
		

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

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.
		

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

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.
		

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

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.
		

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

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.
		

Programs

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

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.
		

Extensions

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

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

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.
		

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

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.
		

A184752 a(n) = largest k such that A014612(n+1) = A014612(n) + (A014612(n) mod k), or 0 if no such k exists.

Original entry on oeis.org

0, 0, 16, 13, 26, 26, 18, 40, 43, 40, 48, 41, 60, 64, 66, 65, 74, 74, 64, 86, 97, 96, 99, 100, 106, 112, 115, 110, 123, 120, 122, 129, 146, 143, 152, 144, 163, 160, 169, 170, 170, 173, 168, 178, 184, 186, 185, 183, 202, 202, 214
Offset: 1

Author

Rémi Eismann, Jan 21 2011

Keywords

Comments

From the definition, a(n) = A014612(n) - A114403(n) if A014612(n) - A114403(n) > A114403(n), 0 otherwise where A014612 are the 3-almost primes and A114403 are the gaps between 3-almost primes.

Examples

			For n = 1 we have A014612(1) = 8, A014612(2) = 12; there is no k such that 12 - 8 = 4 = (8 mod k), hence a(1) = 0.
For n = 3 we have A014612(3) = 18, A014612(4) = 20; 16 is the largest k such that 20 - 18 = 2 = (18 mod k), hence a(3) = 16.
For n = 21 we have A014612(21) = 98, A014612(22) = 99; 97 is the largest k such that 99 - 98 = 1 = (97 mod k), hence a(21) = 97.
		

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

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.