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.

Showing 1-10 of 13 results. Next

A359557 a(1) = 1, a(2) = 2; for n > 2, a(n) is the smallest positive number which has not appeared such that all the distinct prime factors of a(n-2) + a(n-1) are factors of a(n).

Original entry on oeis.org

1, 2, 3, 5, 4, 6, 10, 8, 12, 20, 14, 34, 18, 26, 22, 24, 46, 70, 58, 16, 74, 30, 52, 82, 134, 36, 170, 206, 94, 60, 154, 214, 92, 102, 194, 148, 114, 262, 188, 90, 278, 138, 78, 42, 120, 48, 84, 66, 150, 54, 204, 258, 462, 180, 642, 822, 366, 132, 498, 210, 354, 282, 318, 240, 186, 426, 306
Offset: 1

Views

Author

Scott R. Shannon, Jan 05 2023

Keywords

Comments

All terms other than 3 and 5 are even. As a(114) = 510 and a(115) = 570 both contain 2 and 5 as prime factors, all subsequent terms are multiples of 10. Likewise after 1994 terms all terms contain 2, 3, 5, 7, 11 as factors, so all subsequent terms are multiples of 2*3*5*7*11 = 2310.
The terms grow rapidly in size, e.g., a(2459) = 28318290. The smallest number not to appear is 7.
a(n) = k*m such that k = A007947(a(n-2)+a(n-1)) and m >= 1 produces the smallest k*m != a(j), j < n. - Michael De Vlieger, Jan 07 2023

Examples

			a(5) = 4 as a(3) + a(4) = 3 + 5 = 8 which contains 2 as its only distinct prime factor, and 4 is the smallest unused number to contain 2 as a factor.
a(12) = 34 as a(10) + a(11) = 20 + 14 = 34 which contains 2 and 17 as distinct prime factors, and 34 is also the smallest unused number to contain 2 and 17 as factors.
a(13) = 18 as a(11) + a(12) = 14 + 34 = 48 which contains 2 and 3 as distinct prime factors, and 18 is the smallest unused number to contain 2 and 3 as factors.
		

Crossrefs

Programs

  • Mathematica
    nn = 2^7; c[] = False; q[] = 1; f[n_] := Times @@ FactorInteger[n][[All, 1]]; Array[Set[{a[#], c[#]}, {#, True}] &, 2]; Set[{i, j, k}, {a[1], a[2], f[a[1] + a[2]]}]; Do[m = q[k]; While[c[k m], m++]; m *= k; While[c[k q[k]], q[k]++]; Set[{a[n], c[m], i, j, k}, {m, True, j, m, f[j + m]}], {n, 3, nn}]; Array[a, nn] (* Michael De Vlieger, Jan 07 2023 *)
  • Python
    from math import prod
    from sympy import factorint
    from itertools import count, islice
    def agen():
        i, j, aset = 1, 2, {1, 2}; yield from [i, j]
        while True:
            m = prod(factorint(i+j))
            an = next(k*m for k in count(1) if m*k not in aset)
            i, j = j, an; aset.add(an); yield an
    print(list(islice(agen(), 61))) # Michael S. Branicky, Jan 16 2023

A355647 a(1) = 1, a(2) = 2; for n > 2, a(n) is the smallest positive number that has not yet appeared that has the same number of divisors as the sum a(n-2) + a(n-1).

Original entry on oeis.org

1, 2, 3, 5, 6, 7, 11, 12, 13, 4, 17, 8, 9, 19, 18, 23, 29, 20, 25, 28, 31, 37, 32, 10, 24, 14, 15, 41, 30, 43, 47, 60, 53, 59, 48, 61, 67, 40, 71, 21, 44, 22, 42, 64, 26, 72, 45, 50, 27, 33, 84, 52, 54, 34, 56, 90, 35, 38, 73, 39, 80, 46, 96, 51, 63, 66, 55, 49, 70, 57, 79, 78, 83, 58, 62, 120
Offset: 1

Views

Author

Scott R. Shannon, Jul 12 2022

Keywords

Comments

In the first 500000 terms the smallest numbers that have not appeared are 15625, 25600, 28561, 36864. It is unknown if these and all other numbers eventually appear. In the same range on eighty-two occasions a(n) equals the sum of the previous two terms, these values begin 3, 5, 17, 64, 90, 73, 120, 144, 192.
See A355648 for the fixed points.

Examples

			a(5) = 6 as a(3) + a(4) = 3 + 5 = 8 which has four divisors, and 6 is the smallest unused number that has four divisors.
		

Crossrefs

Programs

  • Python
    from sympy import divisor_count
    from itertools import count, islice
    def agen():
        anm1, an, mink, seen = 1, 2, 3, {1, 2}
        yield 1
        for n in count(2):
            yield an
            k, target = mink, divisor_count(anm1+an)
            while k in seen or divisor_count(k) != target: k += 1
            while mink in seen: mink += 1
            anm1, an = an, k
            seen.add(an)
    print(list(islice(agen(), 76))) # Michael S. Branicky, Jul 26 2022

A355636 a(1) = a(2) = 1; for n > 2, a(n) is the smallest positive number that has not yet appeared that has the same number of divisors as the sum a(n-2) + a(n-1) but does not equal the sum.

Original entry on oeis.org

1, 1, 3, 9, 18, 6, 30, 100, 24, 12, 196, 48, 20, 28, 80, 60, 72, 84, 90, 40, 42, 8, 32, 54, 10, 729, 2, 14, 81, 15, 108, 21, 22, 5, 26, 7, 27, 33, 96, 34, 56, 126, 66, 320, 35, 38, 11, 4, 39, 13, 44, 46, 132, 51, 55, 57, 162, 58, 140, 150, 70, 156, 62, 65, 17, 69, 74, 77, 19, 160, 23, 82, 78
Offset: 1

Views

Author

Scott R. Shannon, Jul 11 2022

Keywords

Comments

In the first 250000 terms the smallest numbers that have not appeared are 64, 1024, 11664, 15625. It is unknown if these and all other numbers eventually appear.
See A355637 for the fixed points.

Examples

			a(6) = 6 as a(4) + a(5) = 9 + 18 = 27 which has four divisors, and 6 is the smallest unused number that does not equal 27 and has four divisors.
		

Crossrefs

Programs

  • PARI
    listm(nn) = my(va = vector(nn)); va[1] = 1; va[2] = 1; my(m = Map()); mapput(m, 1, 1); for (n=3, nn, my(s=va[n-2]+va[n-1], d=numdiv(s), k=1, vs=Vec(va, n-1)); while (mapisdefined(m, k) || (k==s) || (numdiv(k)!=d), k++); va[n] = k; mapput(m, k, n);); va; \\ Michel Marcus, Jul 11 2022
    
  • Python
    from sympy import divisor_count
    from itertools import count, islice
    def agen():
        anm1, an, mink, seen = 1, 1, 2, {1}
        yield 1
        for n in count(2):
            yield an
            k, target, tsum = mink, divisor_count(anm1+an), anm1+an
            while k in seen or k == tsum or divisor_count(k) != target: k += 1
            while mink in seen: mink += 1
            anm1, an = an, k
            seen.add(an)
    print(list(islice(agen(), 73))) # Michael S. Branicky, Jul 26 2022

A355702 a(1) = 1, a(2) = 2; for n > 2, a(n) is the smallest positive number that has not yet appeared that has the same number of prime divisors as the sum a(n-2) + a(n-1).

Original entry on oeis.org

1, 2, 3, 5, 8, 7, 4, 11, 6, 13, 17, 12, 19, 23, 18, 29, 31, 16, 37, 41, 20, 43, 27, 28, 9, 47, 24, 53, 10, 30, 36, 42, 44, 14, 15, 59, 21, 32, 61, 22, 67, 71, 45, 50, 25, 52, 26, 63, 73, 40, 79, 33, 48, 54, 66, 72, 68, 56, 70, 60, 75, 81, 84, 76, 64, 88, 90, 34, 78, 80, 35, 38, 83, 39, 46, 49, 51
Offset: 1

Views

Author

Scott R. Shannon, Jul 14 2022

Keywords

Comments

In the first 500000 terms on seventeen occasions the sum of the previous two terms equals the next term, these terms being 3, 5, 8, 11, 100,... ,131072, 262144. It in unknown if there are infinitely many such terms. In the same range there are seventy-three fixed points; see A356017. The sequence is conjectured to be a permutation of the positive integers.

Examples

			a(4) = 5 as a(2) + a(3) = 2 + 3 = 5 which has one prime divisor, and 5 is the smallest unused number that has one prime divisor.
a(6) = 7 as a(4) + a(5) = 5 + 8 = 13 which has one prime divisor, and 7 is the smallest unused number that has one prime divisor.
a(7) = 4 as a(5) + a(6) = 8 + 7 = 15 which has two prime divisors, and 4 is the smallest unused number that has two prime divisors.
		

Crossrefs

A359256 a(1) = 1, a(2) = 2; for n > 2, a(n) is the smallest positive number which has not appeared such that all the distinct prime factors of a(n-1) + a(n) are factors of a(n).

Original entry on oeis.org

1, 2, 6, 3, 24, 8, 56, 42, 7, 336, 48, 16, 112, 84, 12, 4, 28, 21, 60, 15, 10, 22, 66, 30, 18, 9, 72, 36, 45, 80, 20, 5, 120, 40, 85, 204, 39, 78, 26, 38, 90, 35, 14, 50, 75, 150, 93, 186, 57, 114, 102, 34, 94, 162, 54, 27, 216, 108, 135, 240, 144, 99, 198, 44, 77, 266, 95, 380, 132, 110, 11
Offset: 1

Views

Author

Scott R. Shannon and Eric Angelini, Jan 05 2023

Keywords

Comments

The primes do not occur in their natural order, and for all terms studied if a(n) is a prime p, then a(n-1) = p(p-1) and a(n+1) = p(p^2-1). In the first 10000 terms the fixed points are 22, 165, 710, 1005, 9003, although it is likely more exist. The sequence is conjectured to be a permutation of the positive integers.

Examples

			a(3) = 6 as a(2) + 6 = 2 + 6 = 8 which has 2 as its only distinct prime factor, and 2 is a factor of 6.
a(8) = 42 as a(7) + 42 = 56 + 42 = 96 which has 2 and 3 as distinct prime factors,  and 2 and 3 are factors of 42.
a(10) = 336 as a(9) + 336 = 7 + 336 = 343 which has 7 as its only distinct prime factor, and 7 is a factor of 336. Note that 336 = 7(7^2-1).
		

Crossrefs

A353006 a(1) = 1, a(2) = 2; for n > 2, a(n) is the smallest positive number that has not appeared that shares a factor with (a(n-1) + a(n-2))*|a(n-1) - a(n-2)|.

Original entry on oeis.org

1, 2, 3, 5, 4, 6, 8, 7, 9, 10, 19, 12, 14, 13, 15, 16, 31, 18, 21, 24, 20, 11, 27, 22, 25, 30, 33, 28, 35, 36, 71, 40, 37, 39, 26, 45, 38, 42, 32, 34, 44, 46, 48, 47, 50, 51, 101, 52, 17, 23, 54, 49, 55, 56, 57, 113, 58, 60, 59, 63, 61, 62, 41, 66, 65, 131, 64, 67, 69, 68, 137, 70, 72, 74
Offset: 1

Views

Author

Scott R. Shannon, Apr 16 2022

Keywords

Comments

The sequences is conjectured to be a permutation of the positive integers. In the first 500000 terms there are 637 fixed points: 1, 2, 3, 6, 9, ..., 180611, 189383, 298097. As the terms slowly move away from the line a(n) = n as n increases it is likely no more exist.

Examples

			a(4) = 5 as (a(3)+a(2))*|a(3)-a(2)| = (3+2)*|3-2| = 5, and 5 is the smallest unused number that shares a factor with 5.
a(5) = 4 as (a(4)+a(3))*|a(4)-a(3)| = (5+3)*|5-3| = 16, and 4 is the smallest unused number that shares a factor with 16.
		

Crossrefs

Programs

A355649 a(1) = a(2) = 1; for n > 2, a(n) is the smallest positive number that has the same number of divisors as the sum a(n-2) + a(n-1).

Original entry on oeis.org

1, 1, 2, 2, 4, 6, 6, 12, 12, 24, 36, 60, 60, 120, 180, 180, 360, 360, 720, 840, 840, 1680, 2520, 2520, 5040, 7560, 10080, 10080, 20160, 27720, 27720, 55440, 83160, 110880, 110880, 221760, 332640, 554400, 554400, 1108800, 1441440, 1441440, 2882880, 4324320, 7207200, 7207200, 14414400, 21621600
Offset: 1

Views

Author

Scott R. Shannon, Jul 12 2022

Keywords

Examples

			a(7) = 6 as a(5) + a(6) = 4 + 6 = 10 which has four divisors, and 6 is the smallest positive number that has four divisors.
		

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

A363162 a(1) = 1, a(2) = 2; for n > 2, a(n) is the smallest positive number that has not yet appeared that has the same number of distinct prime divisors as a(n-2) + a(n-1).

Original entry on oeis.org

1, 2, 3, 4, 5, 7, 6, 8, 10, 12, 14, 15, 9, 18, 11, 13, 20, 21, 16, 17, 22, 24, 26, 28, 33, 19, 34, 23, 35, 36, 25, 27, 38, 39, 40, 29, 44, 31, 45, 46, 48, 50, 51, 32, 37, 52, 41, 54, 55, 43, 56, 57, 47, 58, 30, 62, 63, 49, 65, 42, 53, 68, 59, 61, 60, 64, 69, 72, 74, 75, 67, 76, 77, 80, 71, 73, 82
Offset: 1

Views

Author

Scott R. Shannon, Jul 06 2023

Keywords

Comments

The terms with different numbers of distinct prime divisors are concentrated along different lines in the graph; see the attached colored image. There are numerous fixed points in the first one million terms, although the last nonprime fixed point is a(n) = 3495. Beyond that there are thirty-one more fixed points all with prime values; it is likely more exist although this is unknown. The sequence is conjectured to be a permutation of the positive integers.

Examples

			a(4) = 4 as a(2) + a(3) = 2 + 3 = 5 which has one distinct prime divisor, and 4 is the smallest unused number that has one distinct prime divisor.
a(10) = 12 as a(8) + a(9) = 8 + 10 = 18 which has two distinct prime divisors, and 12 is the smallest unused number that has two distinct prime divisors.
		

Crossrefs

Programs

  • Mathematica
    nn = 120; c[] := False; f[x] := PrimeNu[x]; Array[Set[{a[#], c[#]}, {#, True}] &, 2]; i = 1; j = s = 2; u = 3; Do[k = u; s = f[i + j]; While[Or[c[k], f[k] != s], k++]; Set[{a[n], c[k], i, j}, {k, True, j, k}]; If[k == u, While[c[u], u++]], {n, 3, nn}]; Array[a, nn] (* Michael De Vlieger, Jul 08 2023 *)

A353075 a(1) = 1, a(2) = 2; for n > 2, a(n) is the smallest positive number that has not appeared that shares a factor with a(n-1) * a(n-2) + |a(n-1) - a(n-2)|.

Original entry on oeis.org

1, 2, 3, 7, 5, 37, 14, 541, 8101, 223, 23, 73, 13, 1009, 11, 12097, 46, 22, 4, 6, 8, 10, 12, 16, 18, 15, 9, 21, 24, 26, 20, 28, 30, 32, 34, 25, 859, 35, 17, 613, 69, 42841, 39, 1713601, 19, 92, 27, 2549, 38, 43, 33, 1429, 115, 44, 42, 36, 40, 48, 50, 52, 54, 45, 51, 57, 60, 49, 65, 55, 63
Offset: 1

Views

Author

Scott R. Shannon, Apr 22 2022

Keywords

Comments

The sequence produces numerous groupings of primes. For example a(3) to a(16) contains thirteen primes in fourteen terms, a(80) to a(102) contains fourteen primes in twenty-three terms. The sequences is conjectured to be a permutation of the positive integers.

Examples

			a(5) = 5 as a(4)*a(3)+|a(4)-a(3)| = 7*3+|7-3| = 25, and 5 is the smallest unused number that shares a factor with 25.
		

Crossrefs

Programs

Showing 1-10 of 13 results. Next