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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A184219 a(n) = A184218(n)/A130703(n) unless A130703(n) = 0 in which case a(n) = 0.

Original entry on oeis.org

0, 0, 0, 0, 1, 1, 2, 1, 1, 2, 3, 1, 1, 5, 4, 1, 5, 4, 5, 7, 1, 5, 9, 5, 1, 9, 10, 1, 9, 7, 8, 11, 1, 14, 11, 1, 7, 13, 10, 1, 13, 10, 11, 15, 1, 11, 20, 7, 5, 18, 14, 5, 17, 22, 14, 19, 11, 14, 19, 1, 1, 27, 16, 13, 27, 16, 26, 23
Offset: 1

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Author

Rémi Eismann, Jan 10 2011

Keywords

Comments

a(n) is the "level" of triangular numbers.
The decomposition of triangular numbers into weight * level + gap is A000217(n) = A130703(n) * a(n) + (n + 1) if a(n) > 0.
A184218(n) = A000217(n) - (n + 1) if A000217(n) - (n + 1) > (n + 1), 0 otherwise.

Examples

			For n = 3 we have A130703(3) = 0, hence a(3) = 0.
For n = 5 we have A184218(5)/A130703(5) = 9 / 9 = 1, hence a(5) = 1.
For n = 24 we have A184218(24)/A130703(24) = 275 / 55 = 5, hence a(24) = 5.
		

Crossrefs

A184221 a(n) = A184220(n)/A133150(n) unless A133150(n) = 0 in which case a(n) = 0.

Original entry on oeis.org

0, 0, 0, 0, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 7, 7, 1, 2, 1, 2, 1, 7, 7, 2, 1, 2, 1, 2, 7, 14, 1, 2, 1, 2, 1, 14, 7, 17, 1, 2, 1, 2, 17, 14, 1, 2, 1, 2, 23, 14, 7, 2, 1, 2, 17, 2, 7, 14, 1, 17, 1, 23, 1, 14, 7, 2, 1, 31, 1, 2, 7, 34, 1, 2, 1
Offset: 1

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Author

Rémi Eismann, Jan 10 2011

Keywords

Comments

a(n) is the "level" of squares (A000290).
The decomposition of squares into weight * level + gap is A000217(n) = A133150(n) * a(n) + A005408(n) if a(n) > 0.
A184220(n) = A000290(n) - A005408(n) if A000217(n) - A005408(n) > A005408(n), 0 otherwise.

Examples

			For n = 3 we have A133150(3) = 0, hence a(3) = 0.
For n = 5 we have A184220(5)/A133150(5) = 14 / 14 = 1, hence a(5) = 1.
For n = 25 we have A184220(25)/A133150(25) = 574 / 82 = 5, hence a(25) = 7.
		

Crossrefs

A184728 a(n) = largest k such that A001358(n+1) = A001358(n) + (A001358(n) mod k), or 0 if no such k exists.

Original entry on oeis.org

0, 0, 8, 6, 13, 9, 20, 19, 24, 19, 32, 33, 32, 37, 32, 43, 47, 47, 53, 56, 54, 59, 61, 64, 71, 72, 79, 84, 85, 83, 89, 92, 93, 84, 101, 107, 112, 117, 117, 120, 121, 117, 125, 132, 127, 140, 141, 141, 144, 137, 152, 157, 157
Offset: 1

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Author

Rémi Eismann, Jan 20 2011

Keywords

Comments

a(n) = A001358(n) - A065516(n) if A001358(n) - A065516(n) > A065516(n), 0 otherwise.
A001358(n): semiprimes; A065516(n): first difference of semiprimes.

Examples

			For n = 1 we have A001358(n) = 4, A001358(n+1) = 6; there is no k such that 6 - 4 = 2 = (4 mod k), hence a(1) = 0.
For n = 3 we have A001358(n) = 9, A001358(n+1) = 10; 8 is the largest k such that 10 - 9 = 1 = (9 mod k), hence a(3) = 8; a(3) = A001358(3) - A065516(3) = 8.
For n = 20 we have A001358(n) = 57, A001358(n+1) = 58; 56 is the largest k such that 58 - 57 = 1 = (57 mod k), hence a(20) = 56; a(20) = A001358(20) - A065516(20) = 56.
		

Crossrefs

A184729 a(n) = A184728(n)/A130533(n) unless A130533(n) = 0 in which case a(n) = 0.

Original entry on oeis.org

0, 0, 4, 1, 1, 1, 10, 1, 12, 1, 16, 11, 8, 1, 4, 1, 1, 1, 1, 28, 9, 1, 1, 8, 1, 12, 1, 42, 17, 1, 1, 46, 31, 7, 1, 1, 28, 39, 39, 60, 11, 13, 25, 66, 1, 70, 47, 47, 72, 1, 38, 1, 1, 26, 1, 7, 88, 1, 1, 61, 20, 17, 100, 67, 67, 102, 29, 41, 106
Offset: 1

Views

Author

Rémi Eismann, Jan 20 2011

Keywords

Comments

a(n) is the "level" of semiprimes.
The decomposition of semiprimes into weight * level + gap is A001358(n) = A130533(n) * a(n) + A065516(n) if a(n) > 0.
A184728(n) = A001358(n) - A065516(n) if A001358(n) - A065516(n) > A065516(n), 0 otherwise.

Examples

			For n = 1 we have A130533(1) = 0, hence a(1) = 0.
For n = 3 we have A184728(3)/A130533(3)= 8 / 2 = 4; hence a(3) = 4.
For n = 20 we have A184728(20)/A130533(20)= 56 / 2 = 28; hence a(20) = 28.
		

Crossrefs

A184750 a(n) = largest k such that A000326(n+1) = A000326(n) + (A000326(n) mod k), or 0 if no such k exists.

Original entry on oeis.org

0, 0, 0, 0, 19, 32, 48, 67, 89, 114, 142, 173, 207, 244, 284, 327, 373, 422, 474, 529, 587, 648, 712, 779, 849, 922, 998, 1077, 1159, 1244, 1332, 1423, 1517, 1614, 1714, 1817, 1923, 2032, 2144, 2259, 2377, 2498, 2622, 2749
Offset: 1

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Author

Rémi Eismann, Jan 21 2011

Keywords

Comments

From the definition, a(n) = A000326(n) - A016777(n) if A000326(n) - A016777(n) > A016777(n), 0 otherwise, where A000326 are the pentagonal numbers and A016777 are the gaps between pentagonal numbers: 3n + 1.

Examples

			For n = 3 we have A000326(3) = 12, A000326(4) = 22; there is no k such that 22 - 12 = 10 = (12 mod k), hence a(3) = 0.
For n = 5 we have A000326(5) = 35, A000326(6) = 51; 19 is the largest k such that 51 - 35 = 16 = (35 mod k), hence a(5) = 19; a(5) = (75-35-2)/2 = 19.
For n = 25 we have A000326(25) = 925, A000326(26) = 1001; 849 is the largest k such that 1001 - 925 = 76 = (925 mod k), hence a(25) = 849; a(25) = (1875-175-2)/2 = 849.
		

Crossrefs

Programs

  • Maple
    A184750:=n->(3*n^2 - 7*n - 2)*signum(floor(n/5))/2; seq(A184750(n), n=1..50); # Wesley Ivan Hurt, Apr 05 2014
  • Mathematica
    Table[(3 n^2 - 7 n - 2) Sign[Floor[n/5]]/2, {n, 50}] (* Wesley Ivan Hurt, Apr 05 2014 *)
  • PARI
    concat([0,0,0,0], Vec(-x^5*(9*x^2-25*x+19)/(x-1)^3 + O(x^100))) \\ Colin Barker, Apr 05 2014

Formula

a(n) = (3n^2-7n-2)/2 for n >= 5 and a(n) = 0 for n <= 4.
a(n) = 3*a(n-1)-3*a(n-2)+a(n-3) for n>7. G.f.: x^5*(9*x^2-25*x+19) / (1-x)^3. - Colin Barker, Apr 05 2014
a(n) = A000326(n) - A016777(n), n>=5, (see a comment above). - Wolfdieter Lang, Apr 19 2014

Extensions

Edited - Wolfdieter Lang, Apr 19 2014

A184751 a(n) = A184750(n)/A133151(n) unless A133151(n) = 0 in which case a(n) = 0.

Original entry on oeis.org

0, 0, 0, 0, 1, 1, 2, 1, 1, 3, 2, 1, 3, 4, 4, 3, 1, 2, 6, 1, 1, 9, 8, 1, 3, 2, 2, 3, 1, 4, 12, 1, 1, 6, 2, 1, 3, 16, 16, 9, 1, 2, 19, 1, 1, 12, 4, 19, 9, 2, 2, 3, 1, 8, 24, 1, 1, 18, 23, 1, 3, 19, 4, 3, 1, 23, 19, 1, 1, 32, 16, 1, 3, 2, 2, 27, 1, 4, 12, 1, 19, 23
Offset: 1

Views

Author

Rémi Eismann, Jan 21 2011

Keywords

Comments

a(n) is the "level" of pentagonal numbers (A000326).
The decomposition of pentagonal numbers into weight * level + gap is A000326(n) = A133151(n) * a(n) + A016777(n) if a(n) > 0.
A184750(n) = A000326(n) - A016777(n) if A000326(n) - A016777(n) > A016777(n), 0 otherwise.

Examples

			For n = 3 we have A133151(3) = 0, hence a(3) = 0.
For n = 5 we have A184750(5)/A133151(5) = 19 / 19 = 1, hence a(5) = 1.
For n = 25 we have A184750(25)/A133151(25) = 849 / 283 = 5, hence a(25) = 3.
		

Crossrefs

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.
		

Crossrefs

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.
		

Crossrefs

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.
		

Crossrefs

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.
		

Crossrefs

Extensions

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