A005054 -id:A005054 - cp's OEIS frontend

A085388 First differences of n^k.

1, 1, 0, 1, 1, 0, 1, 2, 2, 0, 1, 3, 6, 4, 0, 1, 4, 12, 18, 8, 0, 1, 5, 20, 48, 54, 16, 0, 1, 6, 30, 100, 192, 162, 32, 0, 1, 7, 42, 180, 500, 768, 486, 64, 0, 1, 8, 56, 294, 1080, 2500, 3072, 1458, 128, 0, 1, 9, 72, 448, 2058, 6480, 12500, 12288, 4374, 256, 0, 1, 10, 90, 648

Offset: 1

Paul Barry, Jun 30 2003

T(n,k) is the number of k-digit numbers in base n; n,k >= 2. - Mohammed Yaseen, Nov 11 2022

			Rows begin
  1,   0,   0,   0,   0, ...
  1,   1,   2,   4,   8, ...
  1,   2,   6,  18,  54, ...
  1,   3,  12,  48, 192, ...
  1,   4,  20, 100, 500, ...

Rows include A011782, A025192, A002001, A005054, A052934, A055272, A055274, A055275, A052268, A055276.
Diagonals include A053506, A085389, A085390.
Row-wise binomial transform is A083064.

T(n,k) = (n-1)*n^(k-1) + 0^k/n. - Corrected by Mohammed Yaseen, Nov 11 2022

T(n,0) = 1; T(n,k) = n^k - n^(k-1) for k >= 1. - Mohammed Yaseen, Nov 11 2022

Offset corrected by Mohammed Yaseen, Nov 11 2022

A060460 Consider the final n decimal digits of 2^j for all values of j. They are periodic. Sequence gives position (or phase) of the maximal value seen in these n digits.

Original entry on oeis.org

3, 12, 53, 254, 1255, 6256, 31257, 156258, 781259, 3906260, 19531261, 97656262, 488281263, 2441406264, 12207031265, 61035156266, 305175781267, 1525878906268, 7629394531269, 38146972656270, 190734863281271

Offset: 1

Labos Elemer, Apr 09 2001

The last n digits of 2^a(n) are predictable if maximal values of periods are known.

			a(2) = 5*3-(3+4*0) = 15-3 = 12, etc...
For n=2, the last 2 digits of powers of 2 have the period {2,4,8,16,32,64,28,56,12,24,48,96,92,84,68,36,72,44,88,76,52,4,8,16,32} displayed in A000855. The maximum is 96 and it occurs at 2^12=4096. So a(2)=12.

Harvey P. Dale, Table of n, a(n) for n = 1..1000
Index entries for sequences related to final digits of numbers
Index entries for linear recurrences with constant coefficients, signature (7, -11, 5).

Cf. A000079, A000855, A005054, A060458.

nxt[{n_,a_,b_}]:={n+1,b,5b-(3+4(n-1))}; NestList[nxt,{2,3,12},20][[All,2]] (* or *) Table[2*5^(n-1)+n,{n,30}] (* or *) LinearRecurrence[{7,-11,5},{3,12,53},30] (* Harvey P. Dale, Aug 01 2021 *)

a(1) = 3, a(n) = 5*a(n-1)-(3+4*(n-2)).
a(n) = a(n) = 2*5^(n-1) + n.
G.f.: (-3 + 9 x - 2 x^2)/((-1 + x)^2 (-1 + 5 x)) - Harvey P. Dale, Aug 01 2021

Offset 1 (and formulas adapted) from Michel Marcus, Mar 25 2020

A196731 Expansion of g.f. (1-x)/(1-12*x).

Original entry on oeis.org

1, 11, 132, 1584, 19008, 228096, 2737152, 32845824, 394149888, 4729798656, 56757583872, 681091006464, 8173092077568, 98077104930816, 1176925259169792, 14123103110037504, 169477237320450048, 2033726847845400576, 24404722174144806912, 292856666089737682944, 3514279993076852195328

Offset: 0

Philippe Deléham, Oct 05 2011

Index entries for linear recurrences with constant coefficients, signature (12).

Cf. A002001, A005054, A025192, A052934, A055272, A055274, A055276, A193722.

a:= n-> ceil(11*12^(n-1)):
seq(a(n), n=0..20);  # Alois P. Heinz, Mar 25 2025

Join[{1},NestList[12#&,11,20]] (* Harvey P. Dale, Sep 19 2018 *)

a(n)=if(n,11*12^n--,1) \\ Charles R Greathouse IV, Oct 05 2011

a(n)=(11+!n)*12^(n-1)  \\ M. F. Hasler, Oct 05 2011

a(n) = Sum_{k=0..n} A193722(n,k)*9^(n-k).
a(n+1) = 12*a(n) for n > 0. - M. F. Hasler, Oct 05 2011
From Elmo R. Oliveira, Mar 18 2025: (Start)
a(n) = 11*12^(n-1) with a(0)=1.
E.g.f.: (11*exp(12*x) + 1)/12. (End)

More terms from Elmo R. Oliveira, Mar 25 2025

A243911 Least number k such that n^k ends in two identical digits, or 0 if no such number exists.

Original entry on oeis.org

18, 0, 9, 0, 0, 0, 6, 0, 2, 1, 2, 17, 3, 0, 0, 11, 0, 5, 2, 0, 1, 0, 0, 0, 0, 0, 14, 0, 2, 7, 0, 1, 7, 0, 0, 7, 2, 5, 2, 0, 3, 0, 1, 0, 0, 0, 9, 0, 2, 0, 18, 3, 9, 1, 0, 0, 6, 5, 2, 0, 2, 0, 3, 0, 1, 0, 0, 0, 2, 3, 7, 11, 0, 0, 0, 1, 14, 5, 2, 0, 0, 0, 7, 0, 0, 0, 1, 0, 2, 9, 3, 0, 11

Offset: 2

Derek Orr, Jun 14 2014

For all n > 1, the 2-digit ending of n^k repeats itself after a certain k-value. Thus a(n) = 0 is definite.

a(10*n) = 2 for all n > 0. Thus there are infinitely many nonzero entries. a(5^n) = 0 for all n > 0. Thus there are infinitely many zero entries.

			2^18 = 262144 ends in two of the same digit. Thus a(2) = 18.

Cf. A005054, A216099.

def b(n,p):
  lst = []
  count = 0
  lst1 = []
  for i in range(1,5**(n+2)):
    st = str(p**i)
    if len(st) >= n:
      if int(st[len(st)-n:len(st)]) not in lst:
        lst.append(int(st[len(st)-n:len(st)]))
        lst1.append(i)
      else:
        return len(lst)+min(lst1)
def a(p):
  for i in range(1,b(2,p)+2):
    st = str(p**i)
    if int(st[len(st)-2:len(st)])%11==0:
      return i
p = 2
while p < 100:
  if a(p):
    print(a(p),end=', ')
  else:
    print(0,end=', ')
  p += 1

A371887 a(1) = 1; for n > 1, a(n) is the smallest positive integer k such that the digits of 2^k contain 2^a(n-1) as a proper substring.

Original entry on oeis.org

1, 5, 15, 507

Offset: 1

Adam Vulic, Apr 11 2024

From David A. Corneth, Apr 11 2024: (Start)
This sequence is well defined as A030000 is well defined; every finite string of digits is contained in some power of 2.
An upper bound for a(n), n > 1, can be found by solving 2^k == 2^a(n-1) (mod 10^m) where m is the number of digits of 2^a(n-1) (cf. A034887). This gives a(n) <= k = a(n-1) + 4*5^(m-1) (cf. A005054). So a(5) <= 507 + 4*5^152, which is about 7*10^106. (End)

			a(2) is the smallest k > 0 such that the digits of 2^k contain 2^a(1) = 2^1 = 2 as a proper substring, so a(2) = 5. (2^5 = 32.)
a(3) is the smallest k > 0 such that the digits of 2^k contain 2^a(2) = 32 as a proper substring, so a(3) = 15. (2^15 = 32768.)

Brady Haran, Apocalyptic Numbers, Numberphile video, 2024.

Cf. A018856, A005054, A030000, A034887.

k = 0; Rest@ NestList[(While[SequenceCount[IntegerDigits[2^k], IntegerDigits[2^#]] == 0, k++]; k++; k - 1) &, 1, 4] (* Michael De Vlieger, Apr 19 2024 *)

A052675 Expansion of e.g.f. (1-x)/(1-5*x).

Original entry on oeis.org

1, 4, 40, 600, 12000, 300000, 9000000, 315000000, 12600000000, 567000000000, 28350000000000, 1559250000000000, 93555000000000000, 6081075000000000000, 425675250000000000000, 31925643750000000000000

Offset: 0

encyclopedia(AT)pommard.inria.fr, Jan 25 2000

G. C. Greubel, Table of n, a(n) for n = 0..350
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 623

Cf. A000142, A005054, A052562.

spec := [S,{S=Sequence(Prod(Sequence(Z),Union(Z,Z,Z,Z)))},labeled]: seq(combstruct[count](spec,size=n), n=0..20);

Table[(4/5)*(5^n*n! + Boole[n==0]/4), {n, 0, 50}] (* G. C. Greubel, Jun 12 2022 *)
With[{nn=20},CoefficientList[Series[(1-x)/(1-5x),{x,0,nn}],x] Range[0,nn]!] (* Harvey P. Dale, Jul 31 2023 *)

[4*factorial(n)*5^(n-1) + bool(n==0)/5 for n in (0..40)] # G. C. Greubel, Jun 12 2022

E.g.f.: (1 - x)/(1 - 5*x).
D-finite Recurrence: a(0)=1, a(1)=4, a(n) = 5*n*a(n-1).
a(n) = 4*5^(n-1)*n!, n>0.
a(n) = (4/5) * A052562(n).
a(n) = n!*A005054(n). - R. J. Mathar, Jun 03 2022
G.f.: (4/5)*(Hypergeometric2F0([1, 1], [], 5*x) + 1/4). - G. C. Greubel, Jun 12 2022

A181610 The last n digits of powers of 2 start cycling. a(n) is the number of zero-free terms in this cycle.

Original entry on oeis.org

4, 18, 81, 364, 1638, 7371, 33170, 149268, 671701, 3022653, 13601945, 61208743, 275439346, 1239477074, 5577646830, 25099410745, 112947348510, 508263067945, 2287183805359, 10292327123878, 46315472056678, 208419624257654, 937888309161430, 4220497391215744

Offset: 1

Tanya Khovanova, Jan 30 2011

			The last two digits of powers of two cycle through 20 terms (A000855): 04, 08, 16, 32, 64, 28, 56, 12, 24, 48, 96, 92, 84, 68, 36, 72, 44, 88, 76, 52. Out of those 18 do not contain a zero. Hence a(2) = 18.

Hiroaki Yamanouchi, Table of n, a(n) for n = 1..26

The corresponding cycle length is A005054. See A126605 for n=3.

f[n_] := Block[{c = 0, k = n, lmt = n + 4*5^(n - 1)},
While[k < lmt, m = PowerMod[2, k, 10^n];
  If[m >= 10^(n - 1) && !MemberQ[ IntegerDigits@ m, 0], c++ ]; k++ ]; c];
Array[ f, 11] (* Robert G. Wilson v, Jan 30 2011 *)

a(8)-a(11) from Robert G. Wilson v, Jan 30 2011

a(12)-a(24) from Hiroaki Yamanouchi, Mar 21 2015

A264872 Array read by antidiagonals: T(n,m) = 2^n*(1+2^n)^m; n,m >= 0.

Original entry on oeis.org

1, 2, 2, 4, 6, 4, 8, 18, 20, 8, 16, 54, 100, 72, 16, 32, 162, 500, 648, 272, 32, 64, 486, 2500, 5832, 4624, 1056, 64, 128, 1458, 12500, 52488, 78608, 34848, 4160, 128, 256, 4374, 62500, 472392, 1336336, 1149984, 270400, 16512, 256, 512, 13122, 312500

Offset: 0

R. J. Mathar, Nov 27 2015

Start with an n X m rectangle and cut it vertically along any set of the m-1 separators. There are binomial(m-1,c) ways of doing this with 0 <= c < m cuts. Inside each of these 1+c regions cut vertically, for which there are 2^(n-1) choices. The total number of ways of dissecting the rectangle into rectangles in this way is Sum_{c=0..m-1} binomial(m-1,c) 2^((1+c)(n-1)) = 2^(n-1)*(1+2^(n-1))^(m-1) = T(n-1,m-1).

The symmetrized version of the array is S(n,m) = T(n,m) + T(m,n) - 2^(m+n) <= A116694(n,m), which counts tilings that start with guillotine cuts either horizontally or vertically, avoiding double counting of the tilings where the order of the cuts does not matter. - R. J. Mathar, Nov 29 2015

			   1,    2,     4,       8,       16,         32, ...
   2,    6,    18,      54,      162,        486, ...
   4,   20,   100,     500,     2500,      12500, ...
   8,   72,   648,    5832,    52488,     472392, ...
  16,  272,  4624,   78608,  1336336,   22717712, ...
  32, 1056, 34848, 1149984, 37949472, 1252332576, ...
.
The symmetrized version S(n,m) starts
   1,    2,     4,       8,       16,         32, ...
   2,    8,    30,     110,      402,       1478, ...
   4,   30,   184,    1116,     7060,      47220, ...
   8,  110,  1116,   11600,   130968,    1622120, ...
  16,  402,  7060,  130968,  2672416,   60666672, ...
  32, 1478, 47220, 1622120, 60666672, 2504664128, ...

R. J. Mathar, Counting 2-way monotonic terrace forms over rectangular landscapes, vixra:1511.0225 (2015), Section 6.

Cf. A000079 (row and column 0), A008776 (row 1), A005054 (row 2), A055275 (row 3), A063376 (column 1).

A264872 := proc(n,m)
    2^n*(1+2^n)^m ;
end proc:
seq(seq(A264872(n,d-n),n=0..d),d=0..12) ; # R. J. Mathar, Aug 14 2024

Table[2^(n - m) (1 + 2^(n - m))^m, {n, 9}, {m, 0, n}] // Flatten (* Michael De Vlieger, Nov 27 2015 *)

T(n,m) = 2^n*A264871(n,m).

T(n,m) <= A116694(n+1,m+1).

A308807 a(n) = 4*5^(n-1) + n.

Original entry on oeis.org

5, 22, 103, 504, 2505, 12506, 62507, 312508, 1562509, 7812510, 39062511, 195312512, 976562513, 4882812514, 24414062515, 122070312516, 610351562517, 3051757812518, 15258789062519, 76293945312520, 381469726562521, 1907348632812522, 9536743164062523

Offset: 1

Clive Tooth, Jun 25 2019

The last n decimal digits of 2^a(n) form the number 2^n.

			a(1) = 5, 2^5 = 32, the last digit of 32 is 2, which is 2^1.
a(2) = 22, 2^22 = 4194304, the last 2 digits of 4194304 are 04, which is 2^2.

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (7,-11,5).

seq(4*5^(n-1) + n, n=1..30); # Robert Israel, Jun 28 2019

Table[4*5^(n-1)+n,{n,30}] (* or *) LinearRecurrence[{7,-11,5},{5,22,103},30] (* Harvey P. Dale, Jun 27 2020 *)

Vec(x*(5 - 13*x + 4*x^2) / ((1 - x)^2*(1 - 5*x)) + O(x^25)) \\ Colin Barker, Jun 29 2019

a(n) = A005054(n) + n.
From Colin Barker, Jun 26 2019: (Start)
G.f.: x*(5 - 13*x + 4*x^2) / ((1 - x)^2*(1 - 5*x)).
a(n) = 7*a(n-1) - 11*a(n-2) + 5*a(n-3) for n>3.
(End)
Conjectures confirmed by Robert Israel, Jun 28 2019

A352301 a(n) is the n-th-to-last digit of A007013(n + 1).

Original entry on oeis.org

7, 2, 7, 3, 7, 6, 7, 5, 8, 2, 3, 6, 8, 0, 3, 0, 5, 1, 6, 0, 0, 4, 4, 2, 5, 3, 1, 0, 0, 0, 1, 8, 4, 2, 3, 1, 7, 2, 8, 6, 1, 9, 9, 0, 3, 6, 9, 7, 4, 1, 1, 9, 5, 4, 8, 7, 6, 7, 2, 2, 2, 7, 5, 7, 6, 5, 5, 6, 0, 0, 3, 9, 8, 7, 6, 3, 8, 2, 3, 4, 7, 2, 2, 4, 4, 7

Offset: 1

Davis Smith, Mar 11 2022

Although the sequences for the most significant digits of Mersenne numbers, A000225, are not cyclic (the most significant is not cyclic, the second most is not, etc.), the sequences for the least significant digits are. For example, if p == 3 (mod 4), then A000225(p) == 7 (mod 10). Since A007013(n + 1) = A000225(A007013(n)) and A007013(1) == 3 (mod 4), all subsequent values will be congruent to 7 (mod 10). Similarly, if p == 7 (mod 20), A000225(p) == 27 (mod 100). In general, if p == x (mod A005054(n)), then A000225(p) == A000225(x) (mod 10^n).

There are many primes of the form Sum_{i=1..n} a(i)*10^(i - 1). The largest known is for n = 7032 (it is 7032 digits long).

Davis Smith, Table of n, a(n) for n = 1..7032

Cf. A000225, A005054, A007013.

PARI
```
A352301_vec(n)=my(N=7,m=0);while(m
				
```

a(n) = floor(A007013(n + 1)/(10^(n - 1))) (mod 10).

A007013(n + 1) == Sum_{i=1..n} a(i)*10^(i - 1) (mod 10^n).

cp's OEIS Frontend

A085388 First differences of n^k.

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A060460 Consider the final n decimal digits of 2^j for all values of j. They are periodic. Sequence gives position (or phase) of the maximal value seen in these n digits.

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A196731 Expansion of g.f. (1-x)/(1-12*x).

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A243911 Least number k such that n^k ends in two identical digits, or 0 if no such number exists.

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A371887 a(1) = 1; for n > 1, a(n) is the smallest positive integer k such that the digits of 2^k contain 2^a(n-1) as a proper substring.

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A052675 Expansion of e.g.f. (1-x)/(1-5*x).

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A181610 The last n digits of powers of 2 start cycling. a(n) is the number of zero-free terms in this cycle.

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A264872 Array read by antidiagonals: T(n,m) = 2^n*(1+2^n)^m; n,m >= 0.

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