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 1-6 of 6 results.

A340159 a(n) is the smallest number m such that numbers m, m + 1, m + 2, ..., m + n - 1 have k, 2*k, 3*k, ..., n*k divisors respectively.

Original entry on oeis.org

1, 1, 61, 421, 211082, 11238341, 16788951482, 41126483642
Offset: 1

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Author

Jaroslav Krizek, Dec 29 2020

Keywords

Comments

a(n) is the smallest number m such that tau(m) = tau(m + 1)/2 = tau(m + 2)/3 = tau(m + 3)/4 = ... = tau(m + n - 1)/n, where tau(k) = the number of divisors of k (A000005).
Corresponding values of tau(a(n)): 1, 1, 2, 2, 4, 4, 4, ...
a(8) <= 41126483642. - David A. Corneth, Dec 31 2020
Any subsequent terms are > 10^11. - Lucas A. Brown, Mar 18 2024

Examples

			a(3) = 61 because 61, 62 and 63 have 2, 4, and 6 divisors respectively and there is no smaller number having this property.
		

Crossrefs

Cf. A294528 for similar sequence with primes.

Programs

  • PARI
    isok(m, n) = {my(k=numdiv(m)); for (i=1, n-1, if (numdiv(m+i) != (i+1)*k, return (0));); return(1);}
    a(n) = my(m=1); while(!isok(m, n), m++); m; \\ Michel Marcus, Dec 30 2020
    
  • Python
    # see LINKS

Extensions

a(7) from Jinyuan Wang, Dec 31 2020
a(8) from Lucas A. Brown, Mar 18 2024

A340158 Numbers m such that m, m + 1, m + 2, m + 3 and m + 4 have k, 2k, 3k, 4k and 5k divisors respectively.

Original entry on oeis.org

211082, 2364062, 2774165, 3379802, 3743573, 4390682, 5651042, 5845442, 6708578, 7326122, 7371482, 8566394, 8839202, 9056282, 10154642, 10301333, 10325621, 10446242, 10540202, 11238341, 11719562, 11978762, 12377282, 12871058, 13456202, 16840058, 16954562, 17155141
Offset: 1

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Author

Jaroslav Krizek, Dec 29 2020

Keywords

Comments

Numbers m such that tau(m) = tau(m + 1)/2 = tau(m + 2)/3 = tau(m + 3)/4 = tau(m + 4)/5, where tau(k) = the number of divisors of k (A000005).
Quintuples of [tau(a(n)), tau(a(n) + 1), tau(a(n) + 2), tau(a(n) + 3), tau(a(n) + 4)] = [tau(a(n)), 2*tau(a(n)), 3*tau(a(n)), 4*tau(a(n)), 5*tau(a(n))]: [4, 8, 12, 16, 20], [4, 8, 12, 16, 20], [4, 8, 12, 16, 20], [8, 16, 24, 32, 40], [4, 8, 12, 16, 20], [4, 8, 12, 16, 20], ...
Corresponding values of numbers k: 4, 4, 4, 8, 4, 4, 4, 4, 4, 8, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 8, 4, 4, 4, 4, 4, 4, 4, ...
1524085621 is the smallest prime term (see A294528).
Subsequence of A063446, A339778 and A340157.

Examples

			tau(211082) = 4, tau(211083) = 8, tau(211084) = 12, tau(211085) = 16, tau(211086) = 20.
		

Crossrefs

Programs

  • Magma
    [m: m in [1..10^6] | #Divisors(m) eq #Divisors(m + 1)/2 and #Divisors(m) eq #Divisors(m + 2)/3 and #Divisors(m) eq #Divisors(m + 3)/4 and #Divisors(m) eq #Divisors(m + 4)/5]
    
  • Mathematica
    Select[Range[5*10^6], Equal @@ (DivisorSigma[0, # + {0, 1, 2, 3, 4}]/{1, 2, 3, 4, 5}) &] (* Amiram Eldar, Dec 30 2020 *)
  • PARI
    isok(m) = my(k = numdiv(m)); (numdiv(m+1) == 2*k) && (numdiv(m+2) == 3*k) && (numdiv(m+3) == 4*k) && (numdiv(m+4) == 5*k); \\ Michel Marcus, Jan 16 2021

A340157 Numbers m such that numbers m, m + 1, m + 2 and m + 3 have k, 2k, 3k and 4k divisors respectively.

Original entry on oeis.org

421, 3013, 5029, 5223, 5245, 5893, 6487, 10533, 11911, 14677, 17173, 23077, 23573, 24613, 25141, 25213, 27637, 27973, 28357, 30661, 32407, 34117, 37477, 38282, 39751, 43495, 45973, 47365, 48423, 50821, 50965, 53413, 53989, 54421, 55141, 56103, 57877, 58165
Offset: 1

Views

Author

Jaroslav Krizek, Dec 29 2020

Keywords

Comments

Numbers m such that tau(m) = tau(m + 1)/2 = tau(m + 2)/3 = tau(m + 3)/4, where tau(k) = the number of divisors of k (A000005).
Quadruplets of [tau(a(n)), tau(a(n) + 1), tau(a(n) + 2), tau(a(n) + 3)] = [tau(a(n)), 2*tau(a(n)), 3*tau(a(n)), 4*tau(a(n))]: [2, 4, 6, 8], [4, 8, 12, 16], [4, 8, 12, 16], [4, 8, 12, 16], [4, 8, 12, 16], [4, 8, 12, 16], [4, 8, 12, 16], ...
Corresponding values of tau(a(n)): 2, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 2, 4, 4, 4, 4, 4, 4, 4, 4, 4, ...
Subsequence of A063446 and A339778. Supersequence of A340158.
Prime terms (primes p such that p, p + 1, p + 2 and p + 3 have 2, 4, 6 and 8 divisors respectively): 421, 30661, 50821, 54421, 130021, 195541, 423781, 635461, 1003381, 1577941, 1597381, 1883941, ...

Examples

			tau(421) = 2, tau(422) = 4, tau(423) = 6, tau(424) = 8.
		

Crossrefs

Programs

  • Magma
    [m: m in [1..10^5] | #Divisors(m) eq #Divisors(m + 1)/2 and #Divisors(m) eq #Divisors(m + 2)/3 and #Divisors(m) eq #Divisors(m + 3)/4]
    
  • Mathematica
    Select[Range[60000], Equal @@ (DivisorSigma[0, # + {0, 1, 2, 3}]/{1, 2, 3, 4}) &] (* Amiram Eldar, Dec 30 2020 *)
  • PARI
    isok(m, n=4) = {my(k=numdiv(m)); for (i=1, n-1, if (numdiv(m+i) != (i+1)*k, return (0));); return(1);} \\ Michel Marcus, Dec 30 2020

A341214 a(n) is the smallest prime p such that p, p - 1, p - 2, ..., p - n + 1 have 2, 4, 6, ..., 2*n divisors respectively.

Original entry on oeis.org

2, 7, 47, 1019, 55414379
Offset: 1

Views

Author

Jaroslav Krizek, Feb 07 2021

Keywords

Comments

a(n) is the smallest prime p such that tau(p) = tau(p - 1)/2 = tau(p - 2)/3 = ... = tau(p - n + 1)/n = 2, where tau(k) = the number of divisors of k (A000005).
No such prime p exists for n > 5, so a(5) is the final term. - Jon E. Schoenfield, Feb 07 2021

Examples

			a(4) = 1019 because 1016, 1017, 1018 and 1019 have 8, 6, 4, and 2 divisors respectively and there is no smaller prime having this property (see A340872).
		

Crossrefs

Cf. A341213 (similar sequence for natural numbers).

A340871 Primes p such that p, p + 1, p + 2 and p + 3 have 2, 4, 6 and 8 divisors respectively.

Original entry on oeis.org

421, 30661, 50821, 54421, 130021, 195541, 423781, 635461, 1003381, 1577941, 1597381, 1883941, 2070421, 2100661, 2162581, 2534821, 2585941, 2666581, 2851621, 3296581, 3658021, 3800581, 4657381, 4969141, 5739541, 5962741, 6188821, 6537301, 6556741, 7090261
Offset: 1

Views

Author

Jaroslav Krizek, Jan 24 2021

Keywords

Comments

Primes from A340157.
Term 1524085621 is the smallest prime p such that p, p + 1, p + 2, p + 3 and p + 4 have 2, 4, 6, 8 and 10 divisors respectively. No such run exists for any 6 consecutive integers with prime p with this property (see A294528).

Examples

			tau(421) = 2, tau (422) = 4, tau (423) = 6, tau (424) = 8.
		

Crossrefs

Programs

  • Magma
    [m: m in [1..10^7] | IsPrime(m) and #Divisors(m + 1) eq 4 and #Divisors(m + 2) eq 6 and #Divisors(m + 3) eq 8];
  • Mathematica
    Select[Range[10^6], DivisorSigma[0, # + {0, 1, 2, 3}] == {2, 4, 6, 8} &] (* Amiram Eldar, Jan 25 2021 *)
    Select[Prime[Range[490000]],DivisorSigma[0,#+{1,2,3}]=={4,6,8}&] (* Harvey P. Dale, Oct 02 2021 *)

A363335 Irregular table read by rows: T(n,k) is the smallest m that has 2*n divisors and is at the beginning of a run of exactly k consecutive integers whose number of divisors increases by 2, or -1 if no such m exists.

Original entry on oeis.org

2, 5, 61, 421, 1524085621, 10, 27, 187, 2545622, 12, 153, 35557, 363121, 223577456873, 44753756873
Offset: 1

Views

Author

Jon E. Schoenfield, May 29 2023

Keywords

Comments

Equivalently, T(n,k) is the smallest m such that tau(m+j) = 2*(n+j) for each j in 0..k-1 but not for j = -1 or j = k, where tau is the number-of-divisors function, A000005.
Row 1 is A294528.
No row can have more than 15 terms: if some row n were to have 16 (or more) terms, then T(n,16) would begin a run of 16 consecutive integers, two of which (let's call them x and y, with x < y) would be 4 times an odd number, and thus x and y would each have a number of divisors that is a multiple of 3, but this is impossible since y - x = 8 so tau(y) - tau(x) = 16.
Row 3 has no more than 11 terms.
T(3,7) <= 9392318479793089373.
T(3,8) <= 42121593968048969225.
T(3,9) and T(3,10) seem likely to exist.
T(3,11) cannot exist unless there is a solution to 3*p^4 + 5 = 8*q^4 with p and q prime; if T(3,11) exists, then it exceeds 10^300000.

Examples

			T(2,3) is the smallest m such that tau(m+j) = 2*(2+j) for each j in 0..2 but not for j = -1 or j = 3; i.e., tau(m) = 4, tau(m+1) = 6, and tau(m+2) = 8, but tau(m-1) != 2 and tau(m+3) != 10. The smallest such m is 187:
    m = 187 = 11*17 (which has 4 divisors),
    m+1 = 188 = 2^2*47 (which has 6 divisors), and
    m+2 = 189 = 3^3*7 (which has 8 divisors), but
    m-1 = 186 = 2*3*31 (which has 8 divisors, not 2), and
    m+3 = 190 = 2*5*19 (which has 8 divisors, not 10).
The first several rows of the table are as follows:
  Row n=1:  2, 5, 61, 421, 1524085621; (A294528)
  Row n=2:  10, 27, 187, 2545622;
  Row n=3:  12, 153, 35557, 363121, 223577456873, 44753756873, ...
  Row n=4:  24, 890, 1615, 795056874, 718511874, ...
  Row n=5:  48, 1424, 84281875, 1578123, ...
  Row n=6:  60, 1215, 53216, ...
  Row n=7:  192, 2624, ...
  Row n=8:  120, 6699, 31310, ...
  Row n=9:  180, 16928, ...
  ...
		

Crossrefs

Formula

T(n,k) = min_{j : A000005(m+j) = 2*(n+j) for j = 0..k-1 but not for j = -1 or j = k}.
Showing 1-6 of 6 results.