A243846 Numbers for which the nozero power-sequence of n falls into a loop.
1, 366784, 14877, 531136, 29287878125, 13631616, 18916327, 1245376, 118971, 1, 24871, 1942272, 377414623, 361123756, 221285675921484375, 453559756, 16185473, 4136832, 113758939, 366784, 164961711, 3179798512, 131147731, 1841716224, 283439365914625, 118754727776
Offset: 1
A335505 Triangle read by rows, 0 <= k < n, n >= 1: T(n,k) is the eventual period of the sequence x(j) (or 0 if x(j) never enters a cycle) defined as follows: x(0) = 1 and for j > 1 x(j) is obtained from 5*x(j-1) by deleting all occurrences of the digit k in base n.
0, 1, 1, 6, 1, 1, 4, 1, 1, 2, 1, 1, 0, 0, 0, 6, 60, 6, 30, 2, 1, 48, 2, 3, 12, 0, 1, 6, 156, 14, 22, 2, 18, 1, 34, 78, 12, 36, 3, 48, 0, 1, 138, 198, 10, 684, 1, 1, 2, 20, 1, 2, 0, 22, 1872, 495, 2, 50, 315, 0, 1, 405, 245, 2780, 0, 1440
Offset: 1
Comments
T(1,0) = 0 is defined in order to make the triangle of numbers regular.
T(n,k) = 1 whenever k is a power of 5 and k > 1.
Examples
Triangle begins: n\k 0 1 2 3 4 5 6 7 8 9 ----------------------------------------------------- 1: 0 2: 1 1 3: 6 1 1 4: 4 1 1 2 5: 1 1 0 0 0 6: 6 60 6 30 2 1 7: 48 2 3 12 0 1 6 8: 156 14 22 2 18 1 34 78 9: 12 36 3 48 0 1 138 198 10 10: 684 1 1 2 20 1 2 0 22 1872 T(10,7) = 0 because A335506 never enters a cycle.
Links
- Pontus von Brömssen, Rows n = 1..32, flattened
Programs
-
Python
from sympy.ntheory.factor_ import digits from functools import reduce def drop(x,n,k): # Drop all digits k from x in base n. return reduce(lambda x,j:n*x+j if j!=k else x,digits(x, n)[1:],0) def cycle_length(n,k,m): # Brent's algorithm for finding cycle length. # Note: The function may hang if the sequence never enters a cycle. if (m,n,k)==(5,10,7): return 0 # A little cheating; see A335506. p=1 length=0 tortoise=hare=1 nz=0 while True: hare=drop(m*hare,n,k) while hare and hare%n==0: hare//=n nz+=1 # Keep track of the number of trailing zeros. length+=1 if tortoise==hare: break if p==length: tortoise=hare nz=0 p*=2 length=0 return length if not nz else 0 def A335505(n,k): return cycle_length(n,k,5) if n>1 else 0
A335502 Triangle read by rows, 0 <= k < n, n >= 1: T(n,k) is the eventual period of the sequence x(j) (or 0 if x(j) never enters a cycle) defined as follows: x(0) = 1 and for j > 1 x(j) is obtained from 2*x(j-1) by deleting all occurrences of the digit k in base n.
0, 1, 1, 4, 1, 1, 2, 1, 1, 0, 4, 1, 1, 3, 1, 12, 2, 1, 6, 1, 4, 78, 1, 1, 6, 1, 3, 6, 3, 1, 1, 0, 1, 0, 0, 0, 6, 1, 1, 18, 1, 4, 36, 4, 1, 36, 4, 1, 4, 1, 8, 4, 72, 1, 540, 100, 1, 1, 16, 1, 4, 17, 0, 1, 8, 4, 90, 2, 1, 12, 1, 4, 14, 6, 1, 4, 4, 240
Offset: 1
Comments
T(1,0) = 0 is defined in order to make the triangle of numbers regular.
One way of getting T(n,k) = 0 is to have x(j) = x(i)*n^e for some j > i and e > 0. For k < n <= 48, this is the only way to get T(n,k) = 0 (but see A335506 for another situation where the x-sequence is not periodic).
T(n,k) = 1 whenever k is a power of 2 and k > 1.
It seems that k = 0 and k = n-1 often lead to particularly long cycles.
Examples
Triangle begins: n\k 0 1 2 3 4 5 6 7 8 9 10 11 --------------------------------------------------- 1: 0 2: 1 1 3: 4 1 1 4: 2 1 1 0 5: 4 1 1 3 1 6: 12 2 1 6 1 4 7: 78 1 1 6 1 3 6 8: 3 1 1 0 1 0 0 0 9: 6 1 1 18 1 4 36 4 1 10: 36 4 1 4 1 8 4 72 1 540 11: 100 1 1 16 1 4 17 0 1 8 4 12: 90 2 1 12 1 4 14 6 1 4 4 240 For n = 10 and k = 5, the x-sequence starts 1, 2, 4, 8, 16, 32, 64, 128, 26, 2, and then repeats with a period of 8, so T(10,5) = 8. T(10,0) = 36, because A242350 eventually enters a cycle of length 36. For n=11 and k=7, the x-sequence starts (in base 11) 1, 2, 4, 8, 15, 2A, 59, 10. Disregarding trailing zeros, the sequence then repeats with period 7 and x(i+7*j) = x(i)*11^j for positive i and j. The x-sequence itself is therefore not eventually periodic, so T(11,7)=0.
Links
- Pontus von Brömssen, Rows n = 1..48, flattened
Programs
-
Python
from sympy.ntheory.factor_ import digits from functools import reduce def drop(x,n,k): # Drop all digits k from x in base n. return reduce(lambda x,j:n*x+j if j!=k else x,digits(x, n)[1:],0) def cycle_length(n,k,m): # Brent's algorithm for finding cycle length. # Note: The function may hang if the sequence never enters a cycle. if (m,n,k)==(5,10,7): return 0 # A little cheating; see A335506. p=1 length=0 tortoise=hare=1 nz=0 while True: hare=drop(m*hare,n,k) while hare and hare%n==0: hare//=n nz+=1 # Keep track of the number of trailing zeros. length+=1 if tortoise==hare: break if p==length: tortoise=hare nz=0 p*=2 length=0 return length if not nz else 0 def A335502(n,k): return cycle_length(n,k,2) if n>1 else 0
A335503 Triangle read by rows, 0 <= k < n, n >= 1: T(n,k) is the eventual period of the sequence x(j) (or 0 if x(j) never enters a cycle) defined as follows: x(0) = 1 and for j > 1 x(j) is obtained from 3*x(j-1) by deleting all occurrences of the digit k in base n.
0, 1, 1, 1, 1, 0, 12, 1, 2, 1, 28, 1, 0, 1, 2, 64, 1, 2, 1, 2, 4, 60, 4, 2, 1, 4, 0, 2, 54, 1, 2, 1, 62, 16, 2, 48, 2, 1, 0, 1, 0, 0, 0, 0, 0, 80, 40, 4, 1, 20, 2000, 60, 72, 4, 1, 40, 20, 5, 1, 85, 240, 5, 5, 20, 1, 320, 1260, 128, 2, 1, 272, 4, 2, 48, 68, 1, 20, 1440
Offset: 1
Comments
T(1,0) = 0 is defined in order to make the triangle of numbers regular.
T(n,k) = 1 whenever k is a power of 3 and k>1.
Examples
Triangle begins: n\k 0 1 2 3 4 5 6 7 8 9 10 11 --------------------------------------------------------------- 1: 0 2: 1 1 3: 1 1 0 4: 12 1 2 1 5: 28 1 0 1 2 6: 64 1 2 1 2 4 7: 60 4 2 1 4 0 2 8: 54 1 2 1 62 16 2 48 9: 2 1 0 1 0 0 0 0 0 10: 80 40 4 1 20 2000 60 72 4 1 11: 40 20 5 1 85 240 5 5 20 1 320 12: 1260 128 2 1 272 4 2 48 68 1 20 1440 T(10,0) = 80, because A243845 eventually enters a cycle of length 80.
Links
- Pontus von Brömssen, Rows n = 1..32, flattened
Programs
-
Python
from sympy.ntheory.factor_ import digits from functools import reduce def drop(x,n,k): # Drop all digits k from x in base n. return reduce(lambda x,j:n*x+j if j!=k else x,digits(x, n)[1:],0) def cycle_length(n,k,m): # Brent's algorithm for finding cycle length. # Note: The function may hang if the sequence never enters a cycle. if (m,n,k)==(5,10,7): return 0 # A little cheating; see A335506. p=1 length=0 tortoise=hare=1 nz=0 while True: hare=drop(m*hare,n,k) while hare and hare%n==0: hare//=n nz+=1 # Keep track of the number of trailing zeros. length+=1 if tortoise==hare: break if p==length: tortoise=hare nz=0 p*=2 length=0 return length if not nz else 0 def A335503(n,k): return cycle_length(n,k,3) if n>1 else 0
A335504 Triangle read by rows, 0 <= k < n, n >= 1: T(n,k) is the eventual period of the sequence x(j) (or 0 if x(j) never enters a cycle) defined as follows: x(0) = 1 and for j > 1 x(j) is obtained from 4*x(j-1) by deleting all occurrences of the digit k in base n.
0, 1, 1, 2, 1, 1, 1, 1, 0, 0, 2, 24, 2, 2, 1, 16, 18, 1, 6, 1, 42, 33, 1, 1, 15, 1, 24, 3, 3, 1, 1, 0, 1, 0, 0, 0, 3, 1, 195, 27, 1, 465, 147, 2, 6, 1002, 18, 4, 42, 1, 66, 2, 10, 10, 738, 1660, 25, 5, 180, 1, 2, 15, 35, 150, 4, 1490
Offset: 1
Comments
T(1,0) = 0 is defined in order to make the triangle of numbers regular.
T(n,k) = 1 whenever k is a power of 4 and k>1.
Examples
Triangle begins: n\k 0 1 2 3 4 5 6 7 8 9 ----------------------------------------------------- 1: 0 2: 1 1 3: 2 1 1 4: 1 1 0 0 5: 2 24 2 2 1 6: 16 18 1 6 1 42 7: 33 1 1 15 1 24 3 8: 3 1 1 0 1 0 0 0 9: 3 1 195 27 1 465 147 2 6 10: 1002 18 4 42 1 66 2 10 10 738
Links
- Pontus von Brömssen, Rows n = 1..32, flattened
Programs
-
Python
from sympy.ntheory.factor_ import digits from functools import reduce def drop(x,n,k): # Drop all digits k from x in base n. return reduce(lambda x,j:n*x+j if j!=k else x,digits(x, n)[1:],0) def cycle_length(n,k,m): # Brent's algorithm for finding cycle length. # Note: The function may hang if the sequence never enters a cycle. if (m,n,k)==(5,10,7): return 0 # A little cheating; see A335506. p=1 length=0 tortoise=hare=1 nz=0 while True: hare=drop(m*hare,n,k) while hare and hare%n==0: hare//=n nz+=1 # Keep track of the number of trailing zeros. length+=1 if tortoise==hare: break if p==length: tortoise=hare nz=0 p*=2 length=0 return length if not nz else 0 def A335504(n,k): return cycle_length(n,k,4) if n>1 else 0
A306567 a(n) is the largest value obtained by iterating x -> noz(x + n) starting from 0 (where noz(k) = A004719(k) omits the zeros from k).
9, 99, 27, 99, 96, 99, 63, 99, 81, 91, 99, 195, 94, 295, 93, 291, 113, 189, 171, 992, 159, 187, 187, 483, 988, 475, 153, 281, 181, 273, 279, 577, 297, 997, 567, 369, 333, 363, 351, 994, 219, 465, 357, 663, 459, 461, 423, 192, 441, 965, 399, 999, 437, 126, 551
Offset: 1
Comments
For any n > 0, a(n) is well defined:
- the set of zeroless numbers (A052382) contains arbitrarily large gaps,
- for example, for any k > 0, the interval I_k = [10^k..(10^(k+1)-1)/9-1] if free of zeroless numbers,
- let i be such that #I_i > n,
- let b_n be defined by b_n(0) = 0, and for any j > 0, b_n(j) = noz(b_n(j-1) + n),
- as b_n starts below 10^i and cannot cross the gap constituted by I_i,
- b_n is bounded (and eventually periodic), QED.
Examples
For n = 1: - noz(0 + 1) = 1, - noz(1 + 1) = 2, - noz(2 + 1) = 3, ... - noz(7 + 1) = 8, - noz(8 + 1) = 9, - noz(9 + 1) = noz(10) = 1, - hence a(1) = 9.
Links
- Rémy Sigrist, Table of n, a(n) for n = 1..10000
- Rémy Sigrist, PARI program for A306567
Programs
-
PARI
\\ See Links section.
Formula
Empirically, for any k >= 0:
- a( 10^k) = 9 * 10^k + (10^k-1)/9,
- a(2 * 10^k) = 99 * 10^k + 2 * (10^k-1)/9,
- a(3 * 10^k) = 27 * 10^k + 3 * (10^k-1)/9,
- a(4 * 10^k) = 99 * 10^k + 4 * (10^k-1)/9,
- a(5 * 10^k) = 96 * 10^k + 5 * (10^k-1)/9,
- a(6 * 10^k) = 99 * 10^k + 6 * (10^k-1)/9,
- a(7 * 10^k) = 63 * 10^k + 7 * (10^k-1)/9,
- a(8 * 10^k) = 99 * 10^k + 8 * (10^k-1)/9,
- a(9 * 10^k) = 81 * 10^k + 9 * (10^k-1)/9.
Comments
Examples
Links
Crossrefs
Programs
Mathematica
a[n_] := Block[{h = <||>, t = n}, While[! KeyExistsQ[h,t], h[t]=0; t = FromDigits@ Select[ IntegerDigits[n t], # > 0 &]]; t]; Array[a, 20] (* Giovanni Resta, May 20 2019 *)Formula
Extensions