A005054 -id:A005054 - cp's OEIS frontend

A362556 Number of distinct n-digit suffixes generated by iteratively multiplying an integer by 8, where the initial integer is 1.

5, 21, 101, 502, 2502, 12502, 62503, 312503, 1562503, 7812504, 39062504, 195312504, 976562505, 4882812505, 24414062505, 122070312506, 610351562506, 3051757812506, 15258789062507, 76293945312507, 381469726562507

Offset: 1

Gil Moses, Apr 24 2023

			For n = 1, we begin with 1, iteratively multiply by 8 and count the number of terms before the last 1 digit begins to repeat. We obtain 1, 8, 64, 512, 4096, ... . The next term is 32768, which repeats the last 1 digit 8. Thus, the number of distinct terms is a(1) = 5.

Paolo Xausa, Table of n, a(n) for n = 1..1000
Wikipedia, Multiplicative order
Index entries for linear recurrences with constant coefficients, signature (6,-5,1,-6,5).

Cf. A001018, A005054.
Cf. A014391, A014392.
Cf. A362468 (with 4 as the multiplier).

A362556[n_]:=5^(n-1)4+Ceiling[n/3];Array[A362556,30] (* after Charles R Greathouse IV *) (* or *) LinearRecurrence[{6,-5,1,-6,5},{5,21,101,502,2502},30] (* Paolo Xausa, Nov 18 2023 *)

a(n)=4*5^(n-1)+ceil(n/3) \\ Charles R Greathouse IV, Apr 28 2023

def a(n):
     s, x, M = set(), 1, 10**n
     while x not in s: s.add(x); x = (x<<3)%M
     return len(s)

a(13)-a(21) from Charles R Greathouse IV, Apr 28 2023

A367631 Triangle read by rows: T(n,k) is the number of permutations of length n avoiding simultaneously the patterns 123 and 132 with the maximum number of non-overlapping descents equal k.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 0, 4, 0, 0, 0, 5, 3, 0, 0, 0, 2, 14, 0, 0, 0, 0, 0, 23, 9, 0, 0, 0, 0, 0, 16, 48, 0, 0, 0, 0, 0, 0, 4, 97, 27, 0, 0, 0, 0, 0, 0, 0, 94, 162, 0, 0, 0, 0, 0, 0, 0, 0, 44, 387, 81, 0, 0, 0, 0, 0, 0, 0, 0, 8, 476, 540, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 320, 1485, 243, 0, 0, 0, 0, 0, 0

Offset: 0

Tian Han, Nov 24 2023

Number of permutations of length n avoiding simultaneously the patterns 123 and 132 with the maximum number of non-overlapping descents equal k. A descent in a permutation a(1)a(2)...a(n) is position i such that a(i) > a(i+1).

			Triangle T(n,k) begins:
  1;
  1, 0;
  1, 1,  0;
  0, 4,  0,  0;
  0, 5,  3,  0,   0;
  0, 2, 14,  0,   0,    0;
  0, 0, 23,  9,   0,    0,   0;
  0, 0, 16, 48,   0,    0,   0, 0;
  0, 0,  4, 97,  27,    0,   0, 0, 0;
  0, 0,  0, 94, 162,    0,   0, 0, 0, 0;
  0, 0,  0, 44, 387,   81,   0, 0, 0, 0, 0;
  0, 0,  0,  8, 476,  540,   0, 0, 0, 0, 0, 0;
  0, 0,  0,  0, 320, 1485, 243, 0, 0, 0, 0, 0, 0;
  ...

Tian Han and Sergey Kitaev, Joint distributions of statistics over permutations avoiding two patterns of length 3, arXiv:2311.02974 [math.CO], 2023. See formula 7 at page 7.

Row sums give A011782.
Column sums give 3*A005054.
T(2n,n) gives A133494.
T(3n+2,n) gives A000079.
T(3n+1,n) gives A053220(n+1).

G.f.: (1 + x + x^2 - 2*x^2*z - x^3*z)/(1 - 3*x^2*z - 2*x^3*z).

A378771 a(n) is the least k such that the last k digits of m = A020666(n)^n contain all 10 possible digits (0 through 9).

Original entry on oeis.org

10, 10, 10, 11, 13, 11, 13, 17, 15, 16, 15, 15, 16, 18, 17, 15, 17, 15, 13, 16, 17, 15, 24, 17, 23, 16, 19, 20, 20, 22, 22, 25, 32, 17, 20, 23, 20, 19, 19, 23, 19, 14, 21, 19, 17, 25, 22, 17, 27, 19, 24, 13, 20, 28, 18, 33, 26, 18, 33, 22, 23, 20, 25, 25, 25, 22

Offset: 1

David A. Corneth, Dec 06 2024

A conjecture over at A020666 says that A020666(n) = 2 for n >= 189. Perhaps looking at last digits and some modular arithmetic leads to a proof.

A020666(189) = 2 and a(189) = 33 so the last 33 digits of 2^189 contain all 10 digits. Therefore for k = 189 + 4*5^(33-1) the last 33 digits of 2^k contain all 10 possible digits.

			a(4) = 11 since m = A020666(4)^4 = 763^4 = 338920744561. The last 11 digits contain all 10 possible digits (0 through 9) but the last 10 digits do not; 3 is missing in the last 10 digits.

David A. Corneth, Table of n, a(n) for n = 1..10000

Cf. A005054, A020666.

a(n) = {
	if(n == 1, return(10));
	my(i, todo = 10, v = vector(10), d);
	for(i = 2, oo,
		d = digits(i^n);
		if(#Set(d) == 10,
			forstep(i = #d, 1, -1,
				if(v[d[i]+1] == 0,
					v[d[i]+1] = 1;
					todo--;
					if(todo == 0,
						return(#d - i + 1))))));
}

A384853 Squared length of interior diagonal of n-th (U, V)-crossbox, where U = (1, 0, 1) and V = (0, 1, 0), as in Comments.

Original entry on oeis.org

1, 5, 9, 21, 57, 165, 489, 1461, 4377, 13125, 39369, 118101, 354297, 1062885, 3188649, 9565941, 28697817, 86093445, 258280329, 774840981, 2324522937, 6973568805, 20920706409, 62762119221, 188286357657, 564859072965, 1694577218889, 5083731656661

Offset: 1

Clark Kimberling, Jul 02 2025

nonn

Suppose that U and V are 3-dimensional vectors over the field of real numbers. Define f(1) = U, f(2) = V, f(3) = UxV, where x = cross product, and for n>=2, define f(n) = h(n - 1), g(n) = f(n - 1) + g(n - 1) - h(n - 1), h(n) = f(n) x g(n).
The parallelopiped having edge vectors f(n), g(n), h(n) is the n-th (U,V)-crossbox, with volume |f(n).(g(n) x h(n))|, where . = dot product, and interior diagonal length ||g(n)||. These two sequences, after removal of their first 2 terms, are given for selected U and V by the following table, except for the 3 initial terms:
    U             V         volume   squared diagonal length, ||g(n)||^2
(1, 0, 0)     (0, 1, 0)     A000079           A052548
(1, 0, 0)     (0, 1, 1)     A008776         3*A052919
(1, 0, 0)     (1, 0, 1)     A000351           A178676
(1, 0, 0)     (1, 1, 1)     A167747         5*A204061
(1, 0, 0)     (0, 2, 0)     A005054         4*A199215
(1, 0, 0)     (1, 2, 0)     A013731         8*A199552
(1, 0, 0)     (2, 1, 0)     A011557        10*A000533
(1, 0, 0)     (1, 1, 2)     A067403        18*A135423
(1, 0, 0)     (2, 1, 1)     A334603        11*A199750
(1, 0, 1)     (0, 1, 0)     A008776           this sequence
(1, 1, 0)     (0, 1, 1)     A081341         6*A199318
(1, 1, 0)     (1, 1, 1)     A270369         9*A199559
(1, 2, 3)     (3, 2, 1)   2*A009992   48 + 96*A009992

			Taking U = (1, 0, 1) and V = (0, 1, 0), successive edge vectors are given by
(f(n)) = ( (1, 0, 1), (-1,0,1), (-1,2,-1), (3,0,-3), (3,-6,3), ...)
(g(n)) = ( (0,1,0), (2,1,0), (2,-1,2), (-2,1,4), (-2,7,-2), (10,1,-8), ...)
(h(n)) = ( (-1.0,1), (-1,2,-1), (3,0,-3), (3,-6,3), (-9,0,9),...)
The successive volumes are (2, 6, 18, 54, 162, 486, 1458, 4374, 13122,...).
The lengths of diagonals of the first five crossboxes are 1, sqrt(5), 3, sqrt(21), sqrt(57), so the first five squared lengths are 1, 5, 9, 21, 57.

Index entries for linear recurrences with constant coefficients, signature (4,-3).

Cf. A000079, A052548, A008776.

f[1] = {1, 0, 1}; g[1] = {0, 1, 0}; h[1] = Cross[f[1], g[1]];
f[n_] := f[n] = h[n - 1];
g[n_] := g[n] = f[n - 1] + g[n - 1] - h[n - 1];
h[n_] := h[n] = Cross[f[n], g[n]];
v[n_] := f[n] . Cross[g[n], h[n]] (* signed volume of nth parallelopiped P(n) *)
d[n_] := Norm[g[n]] (* length of interior diagonal of P(n) *)
Column[Table[{f[n], g[n], h[n]}, {n, 1, 16}]]  (* edge vectors of P(n) *)
Table[v[n], {n, 1, 16}]  (* A008776 *)
u = Table[d[n]^2, {n, 1, 30}] (* A384853 *)
Join[{1},Table[1+2*(3^(n-1)+1),{n,40}]] (* or *) LinearRecurrence[{4,-3},{1,5,9},50] (* Harvey P. Dale, Jul 20 2025 *)

a(0) = 1, a(n) = 1 + 2 * (3^(n-1)+1) for n>=1.
a(n) = 4*a(n-1) - 3*a(n-2) for n>=4.
In general, suppose that U = (a,b,c) and V = (s,t,u), and let D = -(a^2 + b^2 + c^2 + s^2 + t^2 + u^2 + 2 (a s + b t + c u)). Then, linear recurrences are given for n>=3 by f(n) = D*f (n - 2), g(n) = g(n - 1) + D*g(n - 2) - D*g(n - 3), h(n) = D*h(n - 2). If w(n) denotes the volume of the n-th (U,V)-crossbox, then w(n) = D*w(n-1) for n>=2.

A309263 Terms of A140110 that are not divisible by 6.

Original entry on oeis.org

1, 2, 4, 8, 16, 20, 32, 64, 100, 128, 256, 272, 500, 512, 1024, 2048, 2500, 4096, 4624, 8192, 10100, 12500, 16384, 32768, 62500, 65536, 65792, 78608, 131072, 262144, 312500, 524288, 1020100, 1048576, 1336336, 1562500, 2097152, 4194304, 7812500, 8388608

Offset: 1

J. Lowell, Jul 19 2019

nonn

Includes all powers of 2.
Conjecture: The sequence includes all numbers of the form 4*5^n.
The number 10100 is a counterexample for: (a) Prime factorizations of numbers of this sequence will always have only 2's and Fermat primes. (b) No number in this sequence is divisible by more than one distinct odd prime.

			20 is in this sequence because it is in A140110 and is not divisible by 6.
24, which is in A140110, is not in this sequence because it is divisible by 6.

Cf. A140110, A000079, A005054.

isok(n) = {if(n%6 == 0, return(0)); my(d = divisors(n)); for (k=1, #d - 1, r = d[k+1]/d[k]; if(numerator(r) != denominator(r) + 1, return(0)); ); return(1); } \\ Jinyuan Wang, Aug 03 2019

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A362556 Number of distinct n-digit suffixes generated by iteratively multiplying an integer by 8, where the initial integer is 1.

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A367631 Triangle read by rows: T(n,k) is the number of permutations of length n avoiding simultaneously the patterns 123 and 132 with the maximum number of non-overlapping descents equal k.

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A378771 a(n) is the least k such that the last k digits of m = A020666(n)^n contain all 10 possible digits (0 through 9).

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A384853 Squared length of interior diagonal of n-th (U, V)-crossbox, where U = (1, 0, 1) and V = (0, 1, 0), as in Comments.

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A309263 Terms of A140110 that are not divisible by 6.

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