1,3

For n = 1, we have phi(x_1) = k*phi(x_1), thus A(1, k) = 0 iff k >= 2.

For n >= 2, if phi(Product_{i=1..n} x_i) = Sum_{i=1..n} k*phi(x_i) and phi(x_1) <= phi(x_2) <= ... <= phi(x_n), then phi(x_(n-1)) <= n*k and phi(x_n) <= k*(n-1)*phi(x_(n-1)). It implies that the equation has finite solutions iff n >= 2 or k >= 2.

Table of n, a(n) for n=1..15.

Shi Baohuai and Pan Xiaowei, On the arithmetic functional equation phi(x_1*...*x_(n-1)*x_n) = m*(phi(x_1) + ... + phi(x_(n-1)) + phi(x_n)), Mathematics Practice and Understanding, 2014, Issue 24, Pages 307-310.

The square array A(n,k) begins:

-1,  0,   0,   0,    0,   ...

3,   9,   35,  33,   17,  ...

15,  39,  31,  138,  57,  ...

118, 463, 558, 1080, 732, ...

...

Cf. A000010, A057635, A336385.

sign,tabl,more,new

Jinyuan Wang, Aug 10 2020

approved

0,3

This morphism is uniform (each expansion the same length) and cyclic (the expansions of 1 and 2 are the same as the expansion of 0 but each term +1 and +2 (mod 3) respectively).  Leech chooses this in order to reduce the cases which need to be considered to prove the word is squarefree (no repeat X X for a block X of any length).

a(n) can be calculated by writing n in base 13 and summing expansion terms, mod 3, for each digit (formula below).  The pattern of the expansion terms can be illustrated as follows, with negatives where term 0 followed by 2 can be taken as a downwards step before mod 3.

  Term:    0,1,2,1,0,2,1,2,0,1,2,1,0  (by base 13 digit)

  Pattern:     2               2

             1   1           1   1

           0       0       0       0

                    -1  -1

                      -2

Leech calls the three expansion blocks A,B,C.  The sequence here expands 0,1,2 -> A,B,C respectively.  Leech allows for a different order, so for example 0,1,2 -> B,A,C.  The effect is a re-mapping of symbols (0,1,2 -> 1,0,2 before expansion) and so remains squarefree.

Leech notes the expansions are symmetric (the same read forward or reversed), and in particular that taking the middle as the zero point can form symmetric bi-directional squarefree words.

Leech makes a final remark that the morphism was constructed by trial assuming symmetry and cyclic relationships and that it is the shortest such.  The type of computer search made by Zolotov confirms it is the shortest, and that it is the sole order 13, up to re-mappings of the symbols.  This is so even with non-symmetrics allowed, since the shortest uniform cyclic non-symmetrics are at order 18.

Table of n, a(n) for n=0..86.

John Leech, A Problem on Strings of Beads, The Mathematical Gazette, volume 41, number 338, December 1957, item 2726, pages 277-278.

Boris Zolotov, Another Solution to the Thue Problem of Non-Repeating Words, arXiv:1505.00019 [math.CO], 2015.  (Section 2 morphism 3, then section 5 result 8 and proofs in section 9.)

Index entries for sequences that are fixed points of mappings

Index entries for sequences related to squarefree words

Fixed point of the morphism, starting from 0,

  0 -> 0,1,2,1,0,2,1,2,0,1,2,1,0  [Leech]

  1 -> 1,2,0,2,1,0,2,0,1,2,0,2,1

  2 -> 2,0,1,0,2,1,0,1,2,0,1,0,2

a(n) = (Sum_{d each base 13 digit of n} t(d)) mod 3, where t(d) = 0,1,2,1,0,2,1,2,0,1,2,1,0 according as d=0 to 12 respectively.

(PARI) my(table=[0, 1, 2, 1, 0, 2, 1, 2, 0, 1, 2, 1, 0]); a(n) = my(v=digits(n, #table)); sum(i=1, #v, table[v[i]+1])%3;

Cf. A170823.

base,nonn,new

Kevin Ryde, Aug 11 2020

approved

1

The R5 dragon curve is a segment replacement where each segment expands to

   -1  4---E

       |

   -1  3---2  +1

           |

       S---1  +1

S and E are the start and end points of the existing segment.  New points 1,2,3,4 are inserted between them.  The new turns are +1,+1, -1,-1, where +1 is left and -1 is right.  The directions of the first and last segments mean existing turns at S and E are unchanged.

  Existing:       +    +    -     ...

  Additional: ++-- ++-- ++-- ++-- ...

The curve is drawn by a unit step forward, turn left a(1)*90 degrees, another unit step forward, turn left a(2)*90 degrees, and so on.  (The same way as Joerg Arndt in A175337.)

It's convenient to number points in the curve starting n=0 at the origin, so an expansion level is points 0 to 5^k inclusive.  The first turn is then at n=1.  The new turns on each expansion are at n == 1,2,3,4 (mod 5) and the existing turns become n == 0 (mod 5).  So a(n) is determined by removing base 5 low 0 digits until reaching a digit 1,2,3,4 (formula below).

The segment expansion is symmetric in 180 degree rotation.  Or equivalently the new turns +1,+1, -1,-1 are unchanged by flip +1 <-> -1 and read last to first.  This means expansions can equally well be considered as unfoldings in the manner of Dekking's folding product (DDUU)*.  Each "downward" fold D is +1 and each "upward" fold U is -1.

Table of n, a(n) for n=1..75.

Joerg Arndt, Matters Computational (The Fxtbook), section 1.31.5 pages 95-100.

Béla Bollobás, The Art of Mathematics: Coffee Time in Memphis, Cambridge University Press, 2006, section 95, pages 226-228.  Signs + + - - etc are the present sequence.  See A170823 for scanned annotated pages.

F. M. Dekking, Paperfolding Morphisms, Planefilling Curves, and Fractal Tiles, Theoretical Computer Science, volume 414, issue 1, January 2012, pages 20-37.  Also arXiv:1011.5788 [math.CO], 2010-2011.  Example DDUU briefly at start of section 4 and end of section 10.

Donald E. Knuth, Selected Papers on Fun and Games, CSLI Lecture Notes Number 192, CSLI Publications, 2010, ISBN 978-1-57586-585-0, pages 603-614 addendum to reprint of Number Representations and Dragon Curves.  Example (DDUU)*3 drawn page 607.

Kevin Ryde, Iterations of the R5 Dragon Curve, section Turn.

Index entries for sequences that are fixed points of mappings

a(n) = 1 if A277543(n) = 1 or 2, or a(n) = -1 otherwise, where A277543(n) is the lowest non-0 digit of n written in base 5.

a(n) = 1 or -1 according as A175337(n-1) = 0 or 1 respectively.

Morphism 1 -> 1,1,-1,-1,1 and -1 -> 1,1,-1,-1,-1 starting from 1.

G.f.: Sum_{k>=0} (x^(5^k) + x^(2*5^k) - x^(3*5^k) - x^(4*5^k)) / (1 - x^(5^(k+1))).

(PARI) a(n) = my(r); until(r, [n, r]=divrem(n, 5)); if(r<=2, 1, -1);

Cf. A175337 (as 0,1), A170823 (partial sums mod 3)

base,sign,new

Kevin Ryde, Aug 11 2020

approved

1,5

Hugo Pfoertner, Table of n, a(n) for n = 1..10000

a(1) = 0; a(n+1) = a(n) + A061398(n-1) for n>1.

a(1) = a(2) = a(3) = 0 because the only composite number < 5 is the square 4.

a(4) = 1: 6 is the first squarefree composite number < prime(4) = 7.

(PARI) m=0; pp=0; forprime(p=2, 320, forcomposite(c=pp, p, if(issquarefree(c), m++)); print1(m, ", "); pp=p)

Cf. A061398, A120944.

nonn,new

Hugo Pfoertner, Aug 11 2020

approved

1,2

Subsequence of A336826.

Seiichi Manyama, Table of n, a(n) for n = 1..500

192 = 24 * (2*4) = 32 * (3*2).

549504 = 1696 * (1*6*9*6) = 2862 * (2*8*6*2) = 3392 * (3*3*9*2) = 3816 * (3*8*1*6).

1798848 = 6246 * (6*2*4*6) = 12492 * (1*2*4*9*2) = 33312 * (3*3*3*1*2).

digprod[n_] := n * Times @@ IntegerDigits[n]; seqQ[0] = True; seqQ[n_] := DivisorSum[n, Boole[digprod[#] == n] &] > 1; Select[Range[0, 2 * 10^5], seqQ] (* Amiram Eldar, Aug 08 2020 *)

Cf. A098736, A336826, A336876.

nonn,base,new

Seiichi Manyama, Aug 08 2020

approved

0,2

Table of n, a(n) for n=0..19.

E.g.f.: 2 * exp(x) / (3 - exp(2*x)).

a(n) = Sum_{k=0..n} binomial(n,k) * A122704(k).

a(n) = Sum_{k=0..n} (-1)^(n-k) * binomial(n,k) * A123227(k).

Table[2^(n + 1) HurwitzLerchPhi[1/3, -n, 1/2]/3, {n, 0, 19}]

nmax = 19; CoefficientList[Series[2 Exp[x]/(3 - Exp[2 x]), {x, 0, nmax}], x] Range[0, nmax]!

Cf. A080253, A122704, A123227.

nonn,new

Ilya Gutkovskiy, Aug 11 2020

approved

0,2

Table of n, a(n) for n=0..17.

a(n) = n! * [x^n] exp(n*x) / (2 - exp(2*x)).

a(n) = Sum_{k=0..n} binomial(n,k) * n^k * A216794(n-k).

Table[2^(n - 1) HurwitzLerchPhi[1/2, -n, n/2], {n, 0, 17}]

Table[n! SeriesCoefficient[Exp[n x]/(2 - Exp[2 x]), {x, 0, n}], {n, 0, 17}]

Cf. A080253, A162314, A216794, A292916.

nonn,new

Ilya Gutkovskiy, Aug 11 2020

approved

0,2

Table of n, a(n) for n=0..18.

a(n) = n! * [x^n] exp(n*x + (exp(2*x) - 1) / 2).

a(n) = Sum_{k=0..n} binomial(n,k) * n^(n-k) * A004211(k).

Table[n! SeriesCoefficient[Exp[n x + (Exp[2 x] - 1)/2], {x, 0, n}], {n, 0, 18}]

Unprotect[Power]; 0^0 = 1; Table[Sum[Binomial[n, k] n^(n - k) 2^k BellB[k, 1/2], {k, 0, n}], {n, 0, 18}]

Cf. A004211, A007405, A134980, A337010, A337011.

nonn,new

Ilya Gutkovskiy, Aug 11 2020

approved

0,2

Table of n, a(n) for n=0..20.

E.g.f.: exp(4*x + (exp(2*x) - 1) / 2).

a(0) = 1; a(n) = 5 * a(n-1) + Sum_{k=2..n} binomial(n-1,k-1) * 2^(k-1) * a(n-k).

a(n) = Sum_{k=0..n} binomial(n,k) * 4^(n-k) * A004211(k).

nmax = 20; CoefficientList[Series[Exp[4 x + (Exp[2 x] - 1)/2], {x, 0, nmax}], x] Range[0, nmax]!

a[0] = 1; a[n_] := a[n] = 5 a[n - 1] + Sum[Binomial[n - 1, k - 1] 2^(k - 1) a[n - k], {k, 2, n}]; Table[a[n], {n, 0, 20}]

Cf. A004211, A007405, A045379, A337010.

nonn,new

Ilya Gutkovskiy, Aug 11 2020

approved

0,2

Table of n, a(n) for n=0..20.

E.g.f.: exp(3*x + (exp(2*x) - 1) / 2).

a(0) = 1; a(n) = 4 * a(n-1) + Sum_{k=2..n} binomial(n-1,k-1) * 2^(k-1) * a(n-k).

a(n) = Sum_{k=0..n} binomial(n,k) * A004211(k+1).

a(n) = Sum_{k=0..n} binomial(n,k) * 3^(n-k) * A004211(k).

nmax = 20; CoefficientList[Series[Exp[3 x + (Exp[2 x] - 1)/2], {x, 0, nmax}], x] Range[0, nmax]!

a[0] = 1; a[n_] := a[n] = 4 a[n - 1] + Sum[Binomial[n - 1, k - 1] 2^(k - 1) a[n - k], {k, 2, n}]; Table[a[n], {n, 0, 20}]

Cf. A004211, A005494, A007405, A337011.

nonn,new

Ilya Gutkovskiy, Aug 11 2020

approved