A243977 -id:A243977 - cp's OEIS frontend

A244187 Numbers n not divisible by 10 such that, for some k, n^k can end in a repdigit of any length.

53, 71, 77, 133, 177, 213, 231, 237, 273, 311, 317, 373, 391, 397, 453, 471, 477, 497, 533, 613, 631, 637, 711, 717, 773, 791, 797, 817, 853, 871, 877, 913, 933, 1013, 1031, 1037, 1111, 1117, 1137, 1173, 1191, 1197, 1233, 1253, 1271, 1277, 1333, 1413, 1431, 1437, 1511, 1517, 1553, 1573, 1591, 1597, 1653, 1671, 1677, 1733, 1777, 1813, 1831, 1837, 1873, 1911, 1917, 1973, 1991, 1997

Offset: 1

Derek Orr, Jun 21 2014

Trivially, (10*m)^k can end in any number of zeros. Thus numbers divisible by 10 are not included in this sequence.
This sequence contains only numbers that end in a 1, 3, or 7.
Those that end in 1 are congruent to {71, 231, 311, 391} (mod 400).
Those that end in 3 are congruent to {53, 133, 213, 373} (mod 400) or are a subsequence of 273 mod 320 (for n < 2000, only 593 == 273 (mod 320) but it is not a term of this sequence).
Those that end in 7 are congruent to {77, 237, 317, 397} (mod 400) or are a subsequence of 177 (mod 320) (for n < 2000, only 1457 == 177 (mod 320) but it is not a term of this sequence).
If a(n) ends in a 1, then a(n+1) ends in a 7 and a(n+1) = a(n)+6. Note that the converse is not true.
Fix an M large enough. There exist infinitely many k such that a(n)^k ends in M of the same digit. Those k values can be given as X mod Y for some X and Y. Take two consecutive powers k and k' that satisfy this. Let the digit immediately before the repeating digit in a(n)^k be called D and the digit immediately before the repeating digit in a(n)^k' be called D'. D - D' will always be odd for large enough M. Example: M = 4 and a(n) = 71. 71^k ends in 4 ones for k = 13, 263, 513, 763, ... == 13 (mod 250). Taking k = 13 and k' = 263, 71^13 ends in 31111 and 71^263 ends in 61111. Thus, D = 3 and D' = 6, and we see 3 - 6 is odd.

			53^9 ends in 33. 53^29 ends in 333. 53^529 ends in 3333. 53^4529 ends in 33333. 53^49529 ends in 333333. This can continue for any number of 3's, where the exponent of 53 is some function based on how many 3's are at the end. We see f(2) = 9, f(3) = 29, f(4) = 529, f(5) = 4529, f(6) = 49529, and so on. Thus 53 is a term of this sequence.

Cf. A243977.

a1(n,p,m)=hh=0;for(i=1,10^p,st=(n^i)%10^p;w=digits(st);sb=(n^i)%10^(p+1);if(#w==p&&vecmin(w)==vecmax(w),hh++);if(hh==m,return((sb-st)/10^p)))
a2(n,p)=for(i=1,10^p,v=digits((n^i)%10^p);if(#v==p&&vecmin(v)==vecmax(v),return(vecmin(v))))
seq(n)={p=3;while(p<10,if(!a2(n,p),return(0));aa=a1(n,p,3);di1=aa-a1(n,p,2);di2=aa-a2(n,p);if(di1==Mod(0,2)&&di2==Mod(1,2),return(0));if(di1==Mod(0,2)&&di2==Mod(0,2),p++);if(di1==Mod(1,2),return(1)))}
n=1;while(n<2000,if(seq(n),print1(n,", "));n++)

A244188 Numbers n such that the digit that repeats the most at the end of n^k for some k is not the last digit of n.

Original entry on oeis.org

2, 12, 22, 33, 37, 42, 52, 62, 72, 73, 77, 92, 102, 112, 113, 117, 122, 142, 152, 153, 162, 172, 192, 197, 202, 212, 222, 233, 237, 242, 252, 262, 272, 273, 277, 292, 302, 312, 313, 317, 322, 342, 352, 353, 362, 372, 392, 397, 402, 412, 422, 433, 437, 442, 452, 462, 472

Offset: 1

Derek Orr, Jun 22 2014

a(n) ends in either a 2, 3, or 7.
If a(n) ends in a 2, the digit that repeats itself the most at the end of a(n)^k is 8.
If a(n) ends in a 3, the digit that repeats itself the most at the end of a(n)^k is 7.
If a(n) ends in a 7, the digit that repeats itself the most at the end of a(n)^k is 3.
The numbers that end in 2 are congruent to {2, 12, 22, 42} mod 50.
The numbers that end in 3 are congruent to {33, 73, 113, 153} mod 200.
The numbers that end in 7 are congruent to {37, 77, 117, 197} mod 200.

			2^k ends in 2 of the same digit for k = 18 mod 20 (last digit is 4) and 19 mod 20 (last digit is 8). 2^k ends in 3 of the same digit for k = 39 mod 100 (last digit is 8). Since 8 is the only possibility, 8 must remain the only possibility for any larger run of identical digits. Since 8 is not the last digit of 2, then 2 is a member of this sequence.
33^k ends in 2 of the same digit for k = 1 mod 20 (last digit is 3) and 7 mod 20 (last digit is 7). 33^k ends in 3 of the same digit for k = 87 mod 100 (last digit is 7). Since 7 is the only possibility, 7 must remain the only possibility for any larger run of identical digits. Since 7 is not the last digit of 33, 33 is a member of this sequence.

Cf. A243977, A244187, A243911, A243912.

seq(n)=for(m=2,6,cc=0;for(i=10^(m-1),10^m,st1=(n^i)%10^m;b="";for(j=1,m,b=concat(b,"1"));if(st1%eval(b)==0,for(d=i+1,10^m,sb1=(n^d)%10^m;if(sb1%eval(b)==0,if(sb1%10==st1%10,return(st1%10));if(sb1%10!=st1%10,cc++;break)))));if(cc==0,return(n%10)))
n=1;while(n<1000,if(seq(n)!=n%10,print1(n,", "));n++)

A244364 a(n) is the least integer m > 1 such that n is the largest number of identical digits that can end m^k for positive integer k.

Original entry on oeis.org

3, 4, 2, 33, 17, 319, 639, 1279, 2559, 5119, 10239, 20479, 40959, 81919, 163839, 327679, 655359, 1310719, 2621439, 5242879, 10485759, 20971519, 41943039, 83886079, 167772159, 335544319, 671088639

Offset: 1

Derek Orr, Jun 26 2014

It is unknown if a(n) ends in a 9 for all n > 5.

a(n) <= 5*2^n-1 because (5*2^n-1)^(5^(n-1)) = -1 (mod 10^n), (5*2^n-1)^k != -1 (mod 10^(n+1)), and (5*2^n-1)^k != 11 (mod 100). - Hiroaki Yamanouchi, Jul 11 2014

			33^k ends in 4 identical digits (let k == 187 mod 500) and never ends in more than 4 identical digits. Since this is not true for 2 <= m <= 32, a(4) = 33.

Cf. A243977.

a(n,p)=for(c=0,10^p,st=Str(n^c);if(#st>p,jo=(eval(st)%(10^p));if(jo==1,return(c));if(ispower(jo,,&k),if(ispower(n)!=0,if(jo^(1/(ispower(jo,,&k)))==(n^(1/(ispower(n)))),return(c)));if(ispower(n)==0,if(jo==n^(ispower(jo,,&k)),return(c))))))
hup(x)={m=1;for(i=2,100,f=0;for(j=m,a(x,i),dt=(x^j)%(10^i);b="";for(w=1,i,b=concat(b,"1"));if(dt%eval(b)==0,f++;r=j;break));if(f==0,return(i-1));m=r)}
n=1;while(n<100,for(x=2,10^3,if(hup(x)==n,print1(x,", ");break));n++)

Conjecture for n>5: a(n) = 10*2^(n-1)-1, k = (2r+1)*5^(n-2), r>=0. - Lars Blomberg, Jul 08 2014

a(9)-a(27) from Lars Blomberg, Jul 08 2014

Definition and example edited by Robert Israel, Jul 10 2014

cp's OEIS Frontend

A244187 Numbers n not divisible by 10 such that, for some k, n^k can end in a repdigit of any length.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

PARI

A244188 Numbers n such that the digit that repeats the most at the end of n^k for some k is not the last digit of n.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

PARI

A244364 a(n) is the least integer m > 1 such that n is the largest number of identical digits that can end m^k for positive integer k.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

PARI

Formula

Extensions