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 1-3 of 3 results.

A117581 For each successive prime p, the largest integer n such that both n and n-1 factor into primes less than or equal to p.

Original entry on oeis.org

2, 9, 81, 4375, 9801, 123201, 336141, 11859211, 11859211, 177182721, 1611308700, 3463200000, 63927525376, 421138799640, 1109496723126, 1453579866025, 20628591204481, 31887350832897, 31887350832897, 119089041053697, 2286831727304145, 9591468737351909376, 9591468737351909376, 9591468737351909376, 9591468737351909376, 9591468737351909376, 19316158377073923834001
Offset: 1

Views

Author

Gene Ward Smith, Mar 29 2006

Keywords

Comments

By a theorem of Størmer, the number of such integers is finite; moreover he provides an algorithm for finding the complete list.
Størmer came to this problem from music theory. Another way to formulate the statement of the theorem is that for any prime p, there are only a finite number of superparticular ratios R = n/(n-1) such that R factors into primes less than or equal to p. The numerator of the smallest such R for the i-th prime is the i-th element of the above sequence. For instance, 81/80, the syntonic comma, is the smallest 5-limit superparticular "comma", i.e., small ratio greater than one.
An effective abc conjecture (c < rad(abc)^2) would imply that a(21) = 2286831727304145 and a(22) = ... = a(26) = 9591468737351909376 and a(27) = ... = a(32) = 19316158377073923834001 and a(33) = 124225935845233319439174. - Lucas A. Brown, Oct 16 2022

Links

Crossrefs

Cf. A002071, A002072, A116486, A117582, A117583.

Formula

a(n) = A002072 + 1.

Extensions

Entry edited by N. J. A. Sloane, Apr 01 2006
Corrected and extended by Don Reble, Nov 21 2006
More terms from A002072 added by Amiram Eldar, Apr 13 2025

A137845 Logarithmically smooth numbers; numbers n whose largest prime factor is less than log(n).

Original entry on oeis.org

8, 16, 24, 27, 32, 36, 48, 54, 64, 72, 81, 96, 108, 128, 144, 150, 160, 162, 180, 192, 200, 216, 225, 240, 243, 250, 256, 270, 288, 300, 320, 324, 360, 375, 384, 400, 405, 432, 450, 480, 486, 500, 512, 540, 576, 600, 625, 640, 648, 675, 720, 729, 750, 768, 800
Offset: 1

Views

Author

T. D. Noe, Feb 14 2008

Keywords

Comments

The graph of this sequence has inflections when n first exceeds exp(prime(k)) for some k. See A051102. It appears that (2400, 2401) and (4374, 4375) are the only consecutive numbers in this sequence. See A116486 for a slightly different definition of logarithmically smooth.
The sequence is closed under multiplication, since if x,y are sequence terms, and a prime p divides x, then p is less than log(x), which is less than log(xy). - Richard Locke Peterson, Apr 12 2020
The Euler phi function of a(n) need not be logarithmically smooth, since phi(27)=18. This differs from k-smooth numbers. - Richard Locke Peterson, May 09 2020

Examples

			48 = 2^4 * 3, and log(48) = 3.8712... > 3. Hence 48 is in the sequence.
49 = 7^2 but log(49) = 3.89182... < 7, so 49 is not in the sequence.

Links

Crossrefs

Cf. A048098, A063539 (two versions of Sqrt-smooth numbers).
See also A333534.

Programs

  • Mathematica
    Select[Range[2,1000], FactorInteger[#][[-1,1]] < Log[#] &]
  • PARI
    sm(N, p)=if(p==2, return(powers(2, logint(N, 2)))); my(v=[], q=precprime(p-1), t=1); for(e=0, logint(N, p), v=concat(v, sm(N\t, q)*t); t*=p); Set(v)
smCapped(N, p, lim)=my(v=sm(N\1,p), i); i=setsearch(v,lim\=1,1); if(i==0, i=setsearch(v,lim)+1); v[i..#v]
list(lim)=if(lim<8,return([])); my(P=primes([2,log(lim\=1)\1]),v=[]); for(i=2,#P, v=concat(v,smCapped(exp(P[i]),P[i-1],exp(P[i-1])))); v=concat(v,smCapped(lim,P[#P],exp(P[#P]))); v \\ Charles R Greathouse IV, Apr 16 2020

A122463 a(n) is the smallest integer such that all its prime factors are <= its n-th root, and such that the equivalent limitation holds also for a(n)-1.

Original entry on oeis.org

3, 9, 2401, 2401, 5909761, 1611308700, 421138799640, 2286831727304145, 3948741978036988496
Offset: 1

Views

Author

Fred Schneider, Sep 09 2006

Keywords

Comments

Smooth Power Duos: Search for consecutive numbers a(n)-1 and a(n) such that the largest prime factor of a(n)-1 raised to the power n remains <= a(n)-1 and such that the largest prime factor of a(n) raised to the power n remains <= a(n): (A006530(a(n)))^n <= a(n) and (A006530(a(n)-1))^n <= a(n)-1.
The prime factorization for the a(n) and a(n)-1 are:
n=1: 3=3, 2=2. n=2: 9=3^2, 8= 2^3. n=3 or 4: 2401=7^4, 2400=2^5*3*5^2.
n=5: 5909761 = 11^2*13^2*17^2, 5909760 = 2^8*3^5*5*19.
n=6: 1611308700 = 2^2*3^6*5^2*23*31^2, 1611308699 = 7^4*11*13^2*19^2 .
n=7: 421138799640 = 2^3*3^5*5*13^4*37*41, 421138799639 = 17*19*23^2*31*43^3 .
n=8: 2286831727304145 = 3^15*5*7^3*19*67*73, 2286831727304144 = 2^4*17*23^2*37*41^2*59*61*71 .
n=9: 3948741978036988496 = 2^4*7^5*13*23*43*59^3*67*83, 3948741978036988495 = 5*11*17*31*97^2*101*103*109*113^2 .
Note: All numbers through 2^62 have been searched

Examples

			For n=7, a(7)= 421138799640 = 2^3*3^5*5*13^4*37*41 and a(7)-1 =421138799639 = 17*19*23^2*31*43^3 are solutions because 41 <= floor(421138799640^(1/7)) = 45 and 43 <= floor(421138799639^(1/7)) = 45.

Links

Crossrefs

Cf. A122464, A122465, A116486.

Programs

  • PARI
    isok1(k, n) = {r = floor(sqrtn(k, n)); f = factor(k); for (j=1, #f~, if (f[j, 1] > r, return (0));); return (1);}
a(n) = {i = 3; found = 0; while (! found, found = (isok1(i, n) && isok1(i-1, n)); if (! found, i++);); return (i);} \\ Michel Marcus, Jul 12 2013

Extensions

Edited by R. J. Mathar, Sep 02 2009
Showing 1-3 of 3 results.

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