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.

Previous Showing 21-30 of 36 results. Next

A361125 Indices of records in A361103.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 9, 10, 11, 18, 22, 24, 28, 29, 38, 42, 47, 52, 54, 57, 59, 66, 68, 72, 75, 77, 80, 83, 86, 92, 94, 98, 104, 107, 114, 115, 118, 119, 121, 124, 127, 131, 133, 135, 138, 143, 149, 163, 165, 175, 181, 188, 197, 199, 202, 204, 206, 211, 213, 216, 218, 222
Offset: 1

Views

Author

Keywords

Crossrefs

A361126 a(n) = A361102(A361125(n)).

Original entry on oeis.org

1, 6, 10, 12, 14, 15, 22, 24, 26, 38, 44, 46, 52, 54, 66, 72, 78, 86, 88, 92, 94, 104, 106, 112, 116, 118, 122, 126, 132, 140, 142, 146, 154, 158, 166, 168, 172, 174, 176, 180, 184, 188, 190, 194, 198, 204, 210, 226, 230, 244, 250, 260, 272, 274, 278, 280, 284, 290, 292
Offset: 1

Views

Author

Keywords

Comments

These are the terms of C which are the slowest to appear in A360519.

A361327 a(n) is the greatest prime factor of A361321(n) with a(1) = 1.

Original entry on oeis.org

1, 3, 5, 7, 7, 11, 11, 7, 13, 13, 5, 11, 11, 7, 13, 13, 17, 17, 19, 19, 17, 17, 7, 5, 23, 23, 19, 19, 11, 23, 23, 13, 11, 29, 29, 7, 13, 31, 31, 23, 23, 37, 37, 11, 29, 29, 11, 41, 41, 31, 31, 13, 43, 43, 37, 37, 17, 47, 47, 53, 53, 59, 59, 61, 61, 13, 13, 67, 67, 71, 71, 73, 73, 79, 79, 11, 17
Offset: 1

Views

Author

Keywords

Examples

			A361321(3) = 10 = 2*5, so a(3) = 5.
		

Crossrefs

Cf. A361321, A000469, A361328 (least prime factor), A361329, A361329, A336957, A360519, A361120.

A361328 a(n) is the least prime factor of A361321(n) with a(1) = 1.

Original entry on oeis.org

1, 2, 2, 5, 3, 3, 2, 2, 7, 3, 3, 5, 7, 2, 2, 5, 5, 2, 2, 3, 3, 7, 2, 2, 3, 7, 7, 5, 2, 2, 13, 11, 2, 2, 5, 3, 2, 2, 5, 5, 2, 2, 5, 3, 3, 7, 2, 2, 3, 3, 7, 2, 2, 3, 3, 7, 2, 2, 3, 3, 2, 2, 3, 3, 2, 2, 3, 3, 2, 2, 3, 3, 2, 2, 3, 3, 11, 2, 2, 11, 11, 2, 2, 13, 13, 2, 2, 3, 3, 5, 7, 2, 2, 3, 5, 5
Offset: 1

Views

Author

Keywords

Examples

			A361321(3) = 10 = 2*5, so a(3) = 2.
		

Crossrefs

Cf. A361321, A000469, A361327 (greatest prime factor), A361329, A361329, A336957, A360519, A361119.

A361329 a(n) = gcd(A361321(n), A361321(n+1)).

Original entry on oeis.org

1, 2, 5, 7, 3, 11, 2, 7, 13, 3, 5, 11, 7, 2, 13, 5, 17, 2, 19, 3, 17, 7, 10, 3, 23, 7, 19, 5, 2, 23, 13, 11, 2, 29, 5, 3, 2, 31, 5, 23, 2, 37, 5, 3, 29, 7, 2, 41, 3, 31, 7, 2, 43, 3, 37, 7, 2, 47, 3, 53, 2, 59, 3, 61, 2, 65, 3, 67, 2, 71, 3, 73, 2, 79, 3, 11, 17, 2, 19, 11, 23, 2, 17, 13, 19, 2
Offset: 1

Views

Author

Keywords

Examples

			A361321(3) = 10, A361321(4) = 35, so a(3) = gcd(10,35) = 5;
		

Crossrefs

A361606 Lexicographically earliest infinite sequence of distinct positive numbers such that, for n > 3, a(n) shares a factor with a(n-1) and a(n-2) but not with a(n-1) + a(n-2).

Original entry on oeis.org

1, 6, 10, 15, 12, 20, 45, 18, 40, 75, 24, 50, 105, 14, 30, 21, 28, 36, 63, 56, 48, 147, 98, 54, 189, 70, 60, 231, 22, 42, 33, 44, 72, 99, 88, 78, 143, 66, 26, 39, 84, 52, 91, 112, 104, 455, 80, 126, 35, 90, 154, 55, 100, 132, 135, 110, 96, 165, 130, 102, 85, 120, 34, 51, 108, 68, 153, 114
Offset: 1

Views

Author

Scott R. Shannon, Mar 17 2023

Keywords

Comments

All terms must contain two or more distinct prime factors. If a(n) was a prime power then a(n+1) would contain the same prime factor, which in turn would imply that a(n) + a(n+1) is a multiple of the prime. But that would make finding a(n+2) impossible as any factor of a(n) would also be a factor of the sum.
To ensure the sequence is infinite a(n) must also contain a prime factor not in a(n-1). If this were not the case the sum a(n-1) + a(n) would be a multiple of the distinct prime factors of a(n), implying a(n+1) would not exist as any factor of a(n) would be a factor of the sum.
The last even term is a(114) = 210. As a(115) = 119 and a(116) = 255, the first occurrence of consecutive odd values, the resulting sum is even, so a(117) must be odd. This forces all subsequent terms to also be odd.
There is a concentration of terms at a(n) ~ 3.4*n. See the linked image. The only fixed point in the first 50000 terms is 14, although it is possible more exist.

Examples

			a(8) = 18 = 2*3*3 as a(6) = 20 = 2*2*5 and a(7) = 45 = 3*3*5 and a(6) + a(7) = 20 + 45 = 65 = 5*13. As the sum contains 5 as a factor a(8) cannot, but it must contain both 2 and 3 while containing a factor not in 45 = 3*3*5. The smallest unused number satisfying these conditions is 18.
		

Crossrefs

Programs

A362754 a(1) = 1, a(2) = 6; for n > 2, a(n) is the smallest positive number that has not yet appeared that shares a factor with a(n-1) and also contains as a factor the smallest prime that is not a factor of a(n-1).

Original entry on oeis.org

1, 6, 10, 12, 15, 18, 20, 24, 30, 14, 21, 28, 36, 40, 42, 35, 50, 45, 48, 60, 56, 54, 70, 63, 66, 55, 22, 33, 44, 72, 75, 78, 65, 26, 39, 52, 84, 80, 90, 98, 96, 100, 102, 85, 34, 51, 68, 108, 105, 110, 99, 88, 114, 95, 38, 57, 76, 120, 112, 126, 130, 117, 104, 132, 135, 138, 115, 46, 69, 92, 144
Offset: 1

Views

Author

Scott R. Shannon, May 02 2023

Keywords

Comments

No term can be a prime power as each term must contain at least two distinct prime factors. This make the sequence similar to A360519 and A361606. A close examination of the lines of concentrated terms, see the attached images, shows they have a slight downward curvature. In the first 250000 terms the only fixed points are 1, 69, 87, 116825, although it is possible more exist for very large values of n.

Examples

			a(3) = 10 as a(2) = 6 = 2*3, and 10 is the smallest unused number that shares a factor with 6 while also containing 5 as a prime factor, the smallest prime not a factor of 6.
a(4) = 12 as a(3) = 10 = 2*5, and 12 is the smallest unused number that shares a factor with 10 while also containing 3 as a prime factor, the smallest prime not a factor of 10.
		

Crossrefs

Programs

  • Mathematica
    nn = 120; c[] := False; m[] := 2; MapIndexed[Set[{a[First[#2]], c[#1]}, {#1, True}] &, {1, 6}]; j = a[2]; Do[q = 2; While[Divisible[j, q], q = NextPrime[q]]; k = m[q]; While[Or[c[#], PrimePowerQ[#], CoprimeQ[j, k]] &[q k], k++]; k *= q; While[c[m[q] q], m[q]++]; Set[{a[n], c[k], j}, {k, True, k}], {n, 3, nn}]; Array[a, nn] (* Michael De Vlieger, May 02 2023 *)

A368231 Lexicographically earliest infinite sequence of distinct positive numbers such that, for n>3, a(n) has a common factor with a(n-1) but not with a(n-2) or n.

Original entry on oeis.org

1, 15, 35, 77, 143, 65, 30, 21, 91, 221, 85, 55, 33, 39, 182, 133, 95, 115, 69, 51, 170, 145, 203, 119, 102, 45, 155, 341, 154, 161, 207, 57, 190, 185, 407, 187, 153, 63, 217, 403, 130, 205, 123, 87, 319, 209, 247, 299, 138, 93, 589, 323, 238, 259, 111, 75, 70, 287, 451, 253, 230, 195, 377
Offset: 1

Views

Author

Scott R. Shannon, Dec 18 2023

Keywords

Comments

This is a variation of the Enots Wolley sequence A336957 and A360519, with an additional restriction that no term a(n) can have a common factor with n. For the sequence to be infinite a(n) must always have a prime factor that is not a factor of a(n-1)*(n+1). See the examples below.
Other than no term being a prime or prime power, see A336957, no term can be an even number with only two distinct prime factors. Clearly no term a(2*k) can be even, so if we assume that a(2*k+1) = 2^n*p^m, with n and m>=1, then a(2*k) must have p as a factor. But as a(2*k+2) must share a factor with a(2*k+1) and cannot have 2 as a factor, it must also have p as a factor. However that is not allowed as a(n) cannot share a factor with a(n-2), so no term can be even with only two distinct prime factors. Therefore the smallest even number is a(7) = 30.

Examples

			a(2) = 15 as 15 is the smallest number that is not a prime power and does not have 2 as a factor.
a(3) = 35 as a(3) is chosen so it shares a factor with a(2) = 3*5 while not having 3 as a factor; it therefore must be a multiple of 5 while not being a power of 5. The smallest number meeting those criteria is 10, but a(2)*(3+1) = 15*4 = 60, and 10 has no prime factor not in 60, so choosing 10 would mean a(4) would not exist. The next smallest available number is 35.
a(4) = 77 as a(4) must be a multiple of 7 but not a power of 7, not a multiple of 2, 3 or 5, while having a prime factor not in 35*(4+1) = 165. The smallest number satisfying these criteria is 77.
		

Crossrefs

A361104 a(n) = k such that A361103(k-1) = n, or -1 if n never appears in A361103.

Original entry on oeis.org

1, 2, 3, 17, 9, 4, 8, 31, 15, 7, 5, 47, 64, 6, 21, 10, 96, 20, 11, 13, 57, 38, 14, 16, 79, 37, 18, 12, 160, 28, 22, 19, 61, 24, 26, 23, 131, 52, 27, 25, 41, 33, 46, 29, 77, 45, 42, 34, 54, 59, 36, 32, 68, 72, 44, 40, 104, 82, 50, 49, 75, 111, 51, 35, 98, 143, 63, 30, 85
Offset: 1

Views

Author

Keywords

Comments

Imagine the offset of A361103 is 1, and assume it really is a permutation of the natural numbers. In tabular form, it is
..1..2..3..4..5..6..7..8..9.10.11...
..1..2..3..6.11.14.10..7..5.16..19...
Then the inverse permutation would be
..1..2..3.17..9..4..8.31.15..7..5.47...
which is the present sequence.

Examples

			A361103(16) = 4, so a(4) = 17.
		

Crossrefs

Extensions

Edited by N. J. A. Sloane, Mar 04 2023

A361124 Records in A361103.

Original entry on oeis.org

1, 2, 3, 6, 11, 14, 16, 19, 28, 32, 36, 40, 44, 68, 72, 76, 80, 84, 88, 92, 96, 100, 104, 112, 116, 120, 123, 132, 136, 139, 144, 148, 156, 160, 164, 171, 172, 175, 180, 184, 188, 192, 196, 200, 216, 220, 228, 236, 244, 248, 256, 271, 272, 276, 280, 284, 288, 292, 296
Offset: 1

Views

Author

Keywords

Comments

These are the terms in C (A361102) that are the slowest to appear in A360519.

Crossrefs

Previous Showing 21-30 of 36 results. Next