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 1-4 of 4 results.

A076425 Numbers n such that zero is never reached by iterating the mapping k -> abs(reverse(lpd(k))-reverse(gpf(k))). lpd(k) is the largest proper divisor and gpf(k) is the largest prime factor of k.

Original entry on oeis.org

2074, 2113, 2179, 2914, 3111, 4112, 4371, 4390, 4456, 4956, 4978, 5185, 5450, 5750, 6474, 6585, 6827, 7248, 7259, 7285, 7467, 8175, 8625, 8647, 9378, 9711, 9739, 10199, 10975, 11407, 11752, 12006, 12232, 12338, 12445, 12826, 13224, 13396
Offset: 1

Views

Author

Klaus Brockhaus, Oct 11 2002

Keywords

Comments

n such that A076423(n) = -1.

Examples

			For 4112 the mapping leads to a fixed point (cf. A076426): 4112 -> 5750 -> 5750 -> ...; for 2074 the mapping leads to a cycle: 2074 -> 7285 -> 7467 -> 9711 -> 7285 -> ...

Crossrefs

Cf. A076423, A076426.

Programs

  • PARI
    {stop=20; for(n=1,13600,c=1; b=1; k=n; while(b&&c1,v[a-1],1); p=0; while(z>0,d=divrem(z,10); z=d[1]; p=10*p+d[2]); z=if(k==1,1,vecmax(component(factor(k),1))); q=0; while(z>0,d=divrem(z,10); z=d[1]; q=10*q+d[2]); a=abs(p-q); if(a==0,b=0,k=a; c++)); if(a>0,print1(n,",")))}

A076426 Fixed points of the mapping k -> abs(reverse(lpd(k))-reverse(Lpf(k))). lpd(k) is the largest proper divisor and Lpf(k) is the largest prime factor of k.

Original entry on oeis.org

5750, 33866, 74841, 517250, 577750, 5538710, 51414250, 51454250, 51687250, 51727250, 51748250, 51858250, 52525250, 57515750, 57535750, 57575750, 57757750, 67597352, 841794296, 5120202250, 5120802250, 5121612250
Offset: 1

Views

Author

Klaus Brockhaus, Oct 11 2002

Keywords

Comments

Besides these fixed points (cycles of length 1) there are five cycles of length 2 ([9378, 9739], [518775, 522075], [5170250, 5197250], [5219475, 5249775], [5255750, 5755250]) and one cycle of length 3 ([7285, 7467, 9711]) below 8000000.

Examples

			lpd(5750) = 2875; Lpf(5750) = 23; 5782 - 32 = 5750.

Crossrefs

Cf. A076423, A076425.

Programs

  • PARI
    {for(n=1,34000,v=divisors(n); a=matsize(v)[2]; z=if(a>1,v[a-1],1); p=0; while(z>0,d=divrem(z,10); z=d[1]; p=10*p+d[2]); z=if(n==1,1,vecmax(component(factor(n),1))); q=0; while(z>0,d=divrem(z,10); z=d[1]; q=10*q+d[2]); if(abs(p-q)==n,print1(n,",")))}

Formula

abs(reverse(lpd(n))-reverse(Lpf(n))) = n.

Extensions

Offset corrected and a(7)-a(22) from Donovan Johnson, Aug 09 2010

A076424 Smallest number that requires n steps to reach 0 when iterating the mapping k -> abs(reverse(lpd(k))-reverse(Lpf(k))). lpd(k) is the largest proper divisor and Lpf(k) is the largest prime factor of k.

Original entry on oeis.org

1, 2, 3, 12, 31, 23, 56, 102, 193, 257, 570, 1129, 4970, 3229, 11551, 11969, 24232, 20094, 24103, 35996, 100090, 222284, 116269, 231488, 388768, 1751753, 2046872, 1140163, 1149979, 2156214, 3199384, 2971734, 7018074, 10163234, 13135933
Offset: 1

Views

Author

Klaus Brockhaus, Oct 11 2002

Keywords

Examples

			a(5) =31 since 31 requires 5 steps, but no m < 31 does. Although 23 < 31, 23 requires 6 steps.

Crossrefs

Cf. A074347, A076423.

Programs

  • PARI
    {m=36; z=19200000; v=listcreate(m); for(i=1,m,listinsert(v,-1,i)); for(n=1,z,c=1; b=1; k=n; while(b&&c<=m, d=divisors(k); i=matsize(d)[2]-1; z=if(i>0,d[i],1); p=0; while(z>0,d=divrem(z,10); z=d[1]; p=10*p+d[2]); z= if(k==1,1,vecmax(component(factor(k),1))); q=0; while(z>0,d=divrem(z,10); z=d[1]; q=10*q+d[2]); a= abs(p-q); if(a==0,b=0,k=a; c++)); if(a==0,if(v[c]<0,v[c]=n; print1([c,n])))); print(); for(i=1,m,print1(v[i],","))}

A175764 Number of iterations of the mapping k->f(k) to reach one of 2, 5, or 29, starting with k=n, and with f(k)=(k^2+4)/d, where d is the next-to-largest divisor of k^2+4, or -1 if the sequence never reaches one of the required values.

Original entry on oeis.org

1, 0, 9, 1, 0, 1, 2, 1, 1, 1, 1, 1, 8, 1, 2, 1, 4, 1, 1, 1, 1, 1, 9, 1, 5, 1, 3, 1, 0, 1, 1, 1, 3, 1, 2, 1, 6, 1, 1, 1, 1, 1, 5, 1, 2, 1, 10, 1, 1, 1, 1, 1, 1, 1, 9, 1, 10, 1, 1, 1, 1, 1, 1, 1, 2, 1, 6, 1, 1, 1, 1, 1, 10, 1, 9, 1, 5, 1, 1, 1, 1, 1, 2, 1, 2, 1, 6, 1, 1, 1, 1, 1, 5, 1, 2, 1, 3, 1, 1, 1, 1, 1, 11
Offset: 1

Views

Author

John W. Layman, Aug 30 2010

Keywords

Comments

It appears that the sequence always reaches 2, 5, or 29 for any initial value n. Is this easy to prove?
It appears that a(n) is 1 whenever n>29 and n mod 10 is one of {0,1,2,4,6,8,9}. This has been verified to n=5000. Also, it appears that a(n) is 9 whenever n mod 130 is one of {3,23,55,75,107,127}. This has also been verified to n=5000. Are these conjectures easy to prove?
From Antti Karttunen, May 19 2021: (Start)
Note that the first four terms of iteration 47017 -> 2210598293 -> 4886744813014513853 -> 23880274867524255960728999629928905613 are all primes (see also A231120), but then (4+(23880274867524255960728999629928905613^2)) is composite, and its smallest prime divisor is 1946761. Actually, a(23880274867524255960728999629928905613) = 2, thus a(47017) = 5.
(End)

Examples

			For n=3, we have 3 -> (3^2+4)/d = 13/1 -> (13^2+4)/d = 173/1 -> (173^2+4)/d = 29933/809 = 37, since the divisors of 29933 are {1,37,809,29933}. Continuing, we get the orbit {3,13,173,37,1373,1217,97,9413,89,5,29,5,29,...}, showing that 5 is reached after 9 steps, after which the orbit is periodic {...,5,29,5,29,...}. Thus a(3)=9.

Links

Crossrefs

Cf. A032742, A076423, A087717, A231120.

Programs

  • PARI
    A175764(n) = if(2==n||5==n||29==n,0,1+A175764(f(n)));
f(k) = { my(u=(4+(k^2)), ds=divisors(u)); (u/ds[#ds-1]); };
\\ Alternatively, "f" could be defined as:
f(k) = { my(u=(4+(k^2))); (u/A032742(u)); };
A032742(n) = if(1==n||isprime(n),1,forprime(p=2,n,if(!(n%p),return(n/p)))); \\ And not requiring full factorization when this is used. - Antti Karttunen, May 19 2021
Showing 1-4 of 4 results.

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