A175046 -id:A175046 - cp's OEIS frontend

A175048 Write n in binary, then increase each run of 1's by one 1. a(n) is the decimal equivalent of the result.

3, 6, 7, 12, 27, 14, 15, 24, 51, 54, 55, 28, 59, 30, 31, 48, 99, 102, 103, 108, 219, 110, 111, 56, 115, 118, 119, 60, 123, 62, 63, 96, 195, 198, 199, 204, 411, 206, 207, 216, 435, 438, 439, 220, 443, 222, 223, 112, 227, 230, 231, 236, 475, 238, 239, 120, 243, 246

Offset: 1

Leroy Quet, Dec 02 2009

			12 in binary is 1100. Increase each run of 1 by one digit to get 11100, which is 28 in decimal. So a(12) = 28.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Cf. A175046, A175047, A348710.

Cf. A030308, A007088.

import Data.List (group)
a175048 = foldr (\b v -> 2 * v + b) 0 . concatMap
   (\bs@(b:_) -> if b == 1 then 1 : bs else bs) . group . a030308_row
-- Reinhard Zumkeller, Jul 05 2013

Table[FromDigits[Flatten[If[MemberQ[#,1],Join[{1},#],#]&/@ Split[ IntegerDigits[ n,2]]],2],{n,60}] (* Harvey P. Dale, Oct 10 2013 *)

def a(n): return int(("0"+bin(n)[2:]).replace("01", "011"), 2)
print([a(n) for n in range(1, 61)]) # Michael S. Branicky, Jul 27 2022

a(2^n) = 3*2^n. a(4n) = 2*a(2n), a(4n+1) = 4*a(2n)+3, a(4n+2) = 2*a(2n+1), a(4n+3) = 2*a(2n+1)+1. - Chai Wah Wu, Nov 21 2018

Extended by Ray Chandler, Dec 18 2009

A320037 Write n in binary, then modify each run of 0's by appending one 1, and modify each run of 1's by appending one 0. a(n) is the decimal equivalent of the result.

Original entry on oeis.org

2, 9, 6, 17, 38, 25, 14, 33, 70, 153, 78, 49, 102, 57, 30, 65, 134, 281, 142, 305, 614, 313, 158, 97, 198, 409, 206, 113, 230, 121, 62, 129, 262, 537, 270, 561, 1126, 569, 286, 609, 1222, 2457, 1230, 625, 1254, 633, 318, 193, 390, 793, 398, 817, 1638, 825, 414

Offset: 1

Chai Wah Wu, Oct 04 2018

A variation of A175046. Indices of record values are given by A319423.
From Chai Wah Wu, Nov 21 2018: (Start)
Let f(k) = Sum_{i=2^k..2^(k+1)-1} a(i), i.e., the sum ranges over all numbers with a (k+1)-bit binary expansion. Thus f(0) = a(1) = 2 and f(1) = a(2) + a(3) = 15.
Then f(k) = 16*6^(k-1) - 2^(k-1) for k > 0.
Proof: looking at the last 2 bits of n, it is easy to see that a(4n) = 2a(2n)-1, a(4n+1) = 4a(2n)+2, a(4n+2) = 4a(2n+1)+1 and a(4n+3) = 2a(2n+1)+2. By summing over the recurrence relations for a(n), we get f(k+2) = Sum_{i=2^k..2^(k+1)-1} (f(4i) + f(4i+1) + f(4i+2) + f(4i+3)) =  Sum_{i=2^k..2^(k+1)-1} (6a(2i) + 6a(2i+1) + 4) = 6*f(k+1) + 2^(k+2). Solving this first order recurrence relation with the initial condition f(1) = 15 shows that f(k) = 16*6^(k-1) - 2^(k-1) for k > 0.
(End)

			6 in binary is 110. Modify each run by appending the opposite digit to get 11001, which is 25 in decimal. So a(6) = 25.

Chai Wah Wu, Table of n, a(n) for n = 1..10000
Chai Wah Wu, Record values in appending and prepending bitstrings to runs of binary digits, arXiv:1810.02293 [math.NT], 2018.

Cf. A175046, A319423, A320038.

from re import split
def A320037(n):
    return int(''.join(d+'0' if '1' in d else d+'1' for d in split('(0+)|(1+)',bin(n)[2:]) if d != '' and d != None),2)

a(4n) = 2a(2n)-1, a(4n+1) = 4a(2n)+2, a(4n+2) = 4a(2n+1)+1 and a(4n+3) = 2a(2n+1)+2. - Chai Wah Wu, Nov 21 2018

A320038 Write n in binary, then modify each run of 0's by prepending one 1, and modify each run of 1's by prepending one 0. a(n) is the decimal equivalent of the result.

Original entry on oeis.org

1, 6, 3, 12, 25, 14, 7, 24, 49, 102, 51, 28, 57, 30, 15, 48, 97, 198, 99, 204, 409, 206, 103, 56, 113, 230, 115, 60, 121, 62, 31, 96, 193, 390, 195, 396, 793, 398, 199, 408, 817, 1638, 819, 412, 825, 414, 207, 112, 225, 454, 227, 460, 921, 462, 231, 120, 241

Offset: 1

Chai Wah Wu, Oct 04 2018

nonn

A variation of A175046. Indices of record values are given by A319423.
From Chai Wah Wu, Nov 25 2018: (Start)
Let f(k) = Sum_{i=2^k..2^(k+1)-1} a(i), i.e., the sum ranges over all numbers with a (k+1)-bit binary expansion. Thus f(0) = a(1) = 1 and f(1) = a(2) + a(3) = 9.
Then f(k) = 10*6^(k-1) - 2^(k-1) for k > 0.
Proof: looking at the last 2 bits of n, it is easy to see that a(4n) = 2*a(2n), a(4n+1) = 4*a(2n)+1, a(4n+2) = 4*a(2n+1)+2 and a(4n+3) = 2*a(2n+1)+1. By summing over the recurrence relations for a(n), we get f(k+2) = Sum_{i=2^k..2^(k+1)-1} (f(4i) + f(4i+1) + f(4i+2) + f(4i+3)) =  Sum_{i=2^k..2^(k+1)-1} (6a(2i) + 6a(2i+1) + 4) = 6*f(k+1) + 2^(k+2). Solving this first-order recurrence relation with the initial condition f(1) = 9 shows that f(k) = 10*6^(k-1) - 2^(k-1) for k > 0.
(End)

			6 in binary is 110. Modify each run by prepending the opposite digit to get 01110, which is 14 in decimal. So a(6) = 14.

Chai Wah Wu, Table of n, a(n) for n = 1..10000
Chai Wah Wu, Record values in appending and prepending bitstrings to runs of binary digits, arXiv:1810.02293 [math.NT], 2018.

Cf. A175046, A319423, A320037.

a[n_] := Split[IntegerDigits[n, 2]] /. {a0:{(0)...} :> Prepend[a0, 1], a1:{(1)...} :> Prepend[a1, 0]} // Flatten // FromDigits[#, 2]&;
Array[a, 60] (* Jean-François Alcover, Dec 02 2018 *)

from re import split
def A320038(n):
    return int(''.join('0'+d if '1' in d else '1'+d for d in split('(0+)|(1+)',bin(n)[2:]) if d != '' and d != None),2)

a(n) = floor(A175046(n)/2).

a(4n) = 2*a(2n), a(4n+1) = 4*a(2n)+1, a(4n+2) = 4*a(2n+1)+2 and a(4n+3) = 2*a(2n+1)+1. - Chai Wah Wu, Nov 25 2018

A320039 Write n in binary, then modify each run of 0's and each run of 1's by appending a 1. a(n) is the decimal equivalent of the result.

Original entry on oeis.org

3, 13, 7, 25, 55, 29, 15, 49, 103, 221, 111, 57, 119, 61, 31, 97, 199, 413, 207, 441, 887, 445, 223, 113, 231, 477, 239, 121, 247, 125, 63, 193, 391, 797, 399, 825, 1655, 829, 415, 881, 1767, 3549, 1775, 889, 1783, 893, 447, 225, 455, 925, 463, 953, 1911, 957

Offset: 1

Chai Wah Wu, Oct 05 2018

A variation of A175046. Indices of record values are given by A319423.
From Chai Wah Wu, Nov 21 2018: (Start)
Let f(k) = Sum_{i=2^k..2^(k+1)-1} a(i), i.e., the sum ranges over all numbers with a (k+1)-bit binary expansion. Thus f(0) = a(1) = 3 and f(1) = a(2) + a(3) = 20.
Then f(k) = 21*6^(k-1) - 2^(k-1) for k >= 0.
Proof: the equation for f is true for k = 0. Looking at the last 2 bits of n, it is easy to see that a(4n) = 2*a(2n)-1, a(4n+1) = 4*a(2n)+3, a(4n+2) = 4*a(2n+1)+1 and a(4n+3) = 2*a(2n+1)+1. By summing over the recurrence relations for a(n), we get f(k+2) = Sum_{i=2^k..2^(k+1)-1} (f(4i) + f(4i+1) + f(4i+2) + f(4i+3)) =  Sum_{i=2^k..2^(k+1)-1} (6a(2i) + 6a(2i+1) + 4) = 6*f(k+1) + 2^(k+2). Solving this first-order recurrence relation with the initial condition f(1) = 20 shows that f(k) = 21*6^(k-1) - 2^(k-1) for k > 0.
(End)

			6 in binary is 110. Modify each run by appending a 1 to get 11101, which is 29 in decimal. So a(6) = 29.

Chai Wah Wu, Table of n, a(n) for n = 1..10000
Chai Wah Wu, Record values in appending and prepending bitstrings to runs of binary digits, arXiv:1810.02293 [math.NT], 2018.

Cf. A175046, A319423, A320037, A320038.

Array[FromDigits[Flatten@ Map[Append[#, 1] &, Split@ IntegerDigits[#, 2]], 2] &, 54] (* Michael De Vlieger, Nov 23 2018 *)

from re import split
def A320039(n):
    return int(''.join(d+'1' for d in split('(0+)|(1+)',bin(n)[2:]) if d != '' and d != None),2)

a(4n) = 2*a(2n)-1, a(4n+1) = 4*a(2n)+3, a(4n+2) = 4*a(2n+1)+1 and a(4n+3) = 2*a(2n+1)+1. - Chai Wah Wu, Nov 21 2018

A266150 Take the binary representation of n, increase each run of 0's by one 0 if the length of run is odd, otherwise, if length of run is even, remove one 0. a(n) is the decimal equivalent of the result.

Original entry on oeis.org

0, 1, 4, 3, 2, 9, 12, 7, 16, 5, 36, 19, 6, 25, 28, 15, 8, 33, 20, 11, 18, 73, 76, 39, 48, 13, 100, 51, 14, 57, 60, 31, 64, 17, 132, 67, 10, 41, 44, 23, 144, 37, 292, 147, 38, 153, 156, 79, 24, 97, 52, 27, 50, 201, 204, 103, 112, 29, 228, 115, 30, 121, 124, 63, 32

Offset: 0

Alex Ratushnyak, Dec 21 2015

This is a self-inverse permutation of the positive integers.

			a(4) = 2 since 4 = 100 binary -> 10 = 2 decimal.
a(5) = 9 since 5 = 101 binary -> 1001 = 9 decimal.
a(6) = 12 since 6 = 110 binary -> 1100 = 12 decimal.

Cf. A007088, A084483, A162853, A175046, A175047, A175048, A266151.

Table[FromDigits[#, 2] &@ Flatten[If[First@ # == 0, If[OddQ@ Length@ #, Append[IntegerDigits@ #, 0], Most@ IntegerDigits@ #], #] & /@ Split@ IntegerDigits[n, 2]], {n, 64}] (* Michael De Vlieger, Dec 22 2015 *)

a(n) = if (n==0, 0, my (b=n%2, r=valuation(n+b, 2), rr=if (b, r, r%2, r+1, r-1)); (a(n\2^r)+b)*2^rr-b) \\ Rémy Sigrist, Jan 20 2019

a(0) = 0 prepended by Rémy Sigrist, Jan 20 2019

A266151 Take the binary representation of n, increase each run of 1's by one 1 if the length of run is odd, otherwise, if length of run is even, remove one 1. a(n) is the decimal equivalent of the result.

Original entry on oeis.org

0, 3, 6, 1, 12, 27, 2, 15, 24, 51, 54, 13, 4, 11, 30, 7, 48, 99, 102, 25, 108, 219, 26, 111, 8, 19, 22, 5, 60, 123, 14, 63, 96, 195, 198, 49, 204, 411, 50, 207, 216, 435, 438, 109, 52, 107, 222, 55, 16, 35, 38, 9, 44, 91, 10, 47, 120, 243, 246, 61, 28, 59, 126, 31

Offset: 0

Alex Ratushnyak, Dec 21 2015

This is a self-inverse permutation of the positive integers.

			a(4) = 12 since 4 = 100 binary -> 1100 = 12 decimal,
a(5) = 27 since 5 = 101 binary -> 110011 = 27 decimal,
a(6) = 2 since 6 = 110 binary -> 10 = 2 decimal.

Cf. A007088, A084483, A162853, A175046, A175047, A175048, A266150.

Table[FromDigits[#, 2] &@ Flatten[If[First@ # == 1, If[OddQ@ Length@ #, Append[IntegerDigits@ #, 1], Most@ IntegerDigits@ #], #] & /@ Split@ IntegerDigits[n, 2]], {n, 63}] (* Michael De Vlieger, Dec 22 2015 *)

a(n) = if (n==0, 0, my (b=n%2, r=valuation(n+b, 2), rr=if (b==0, r, r%2, r+1, r-1)); (a(n\2^r)+b)*2^rr-b) \\ Rémy Sigrist, Jan 20 2019

a(0) = 0 prepended by Rémy Sigrist, Jan 20 2019

A319419 In binary expansion of n, delete one symbol from each run. Set a(n)=-1 if the result is the empty string.

Original entry on oeis.org

-1, -1, -1, 1, 0, -1, 1, 3, 0, 0, -1, 1, 2, 1, 3, 7, 0, 0, 0, 1, 0, -1, 1, 3, 4, 2, 1, 3, 6, 3, 7, 15, 0, 0, 0, 1, 0, 0, 1, 3, 0, 0, -1, 1, 2, 1, 3, 7, 8, 4, 2, 5, 2, 1, 3, 7, 12, 6, 3, 7, 14, 7, 15, 31, 0, 0, 0, 1, 0, 0, 1, 3, 0, 0, 0, 1, 2, 1, 3, 7, 0, 0

Offset: 0

N. J. A. Sloane, Sep 21 2018

If the binary expansion of n is 1^b 0^c 1^d 0^e ..., then a(n) is the number whose binary expansion is 1^(b-1) 0^(c-1) 1^(d-1) 0^(e-1) .... Leading 0's are omitted, and if the result is the empty string, here we set a(n) = -1. See A318921 for a version which represents the empty string by 0.
Lenormand refers to this operation as planing ("raboter") the runs (or blocks) of the binary expansion.
A175046 expands the runs in a similar way, and a(A175046(n)) = A001477(n). - Andrew Weimholt, Sep 08 2018 (Comment copied from A318921.)
a(n) = -1 iff n in A000975.

			n / "planed" string / a(n)
0 e -1 (e = empty string)
1 e -1
10 e -1
11 1 1
100 0 0
101 e -1
110 1 1
111 11 3
1000 00 0
1001 0 0
1010 e -1
1011 1 1
1100 10 2
1101 1 1
1110 11 3
1111 111 7
10000 000 0
...

N. J. A. Sloane, Table of n, a(n) for n = 0..16384
Claude Lenormand, Deux transformations sur les mots, Preprint, 5 pages, Nov 17 2003. Apparently unpublished. This is a scanned copy of the version that the author sent to me in 2003.
N. J. A. Sloane, Coordination Sequences, Planing Numbers, and Other Recent Sequences (II), Experimental Mathematics Seminar, Rutgers University, Jan 31 2019, Part I, Part 2, Slides. (Mentions this sequence)

Cf. A000975, A175046, A318921.

r:=proc(n) local t1, t2, L, len, i, j, k, b1;
if n <= 2 then return(-1); fi;
b1:=[]; t1:=convert(n, base, 2); L:=nops(t1); p:=1; len:=1;
for i from 2 to L do
t2:=t1[L+1-i];
if (t2=p) and (i1 then for j from 1 to len-1 do b1:=[op(b1), p]; od: fi;
   p:=t2; len:=1;
fi;               od;
if nops(b1)=0 then return(-1);
else k:=b1[1];
for i from 2 to nops(b1) do k:=2*k+b1[i]; od;
return(k);
fi;
end;
[seq(r(n), n=0..120)];

from re import split
def A319419(n):
    s = ''.join(d[:-1] for d in split('(0+)|(1+)',bin(n)[2:]) if d not in {'','0','1',None})
    return -1 if s == '' else int(s,2) # Chai Wah Wu, Sep 24 2018

from itertools import groupby
def a(n):
    s = "".join(k*(len(list(g))-1) for k, g in groupby(bin(n)[2:]))
    return int(s, 2) if s != "" else -1
print([a(n) for n in range(82)]) # Michael S. Branicky, Jun 01 2025

A320261 Write n in binary, then modify each run of 0's and each run of 1's by prepending a 1. a(n) is the decimal equivalent of the result.

Original entry on oeis.org

3, 14, 7, 28, 59, 30, 15, 56, 115, 238, 119, 60, 123, 62, 31, 112, 227, 462, 231, 476, 955, 478, 239, 120, 243, 494, 247, 124, 251, 126, 63, 224, 451, 910, 455, 924, 1851, 926, 463, 952, 1907, 3822, 1911, 956, 1915, 958, 479, 240, 483, 974, 487, 988, 1979, 990

Offset: 1

Chai Wah Wu, Oct 08 2018

A variation of A175046. Indices of record values are given by A319423.
From Chai Wah Wu, Nov 25 2018 : (Start)
Let f(k) = Sum_{i=2^k..2^(k+1)-1} a(i), i.e., the sum ranges over all numbers with a (k+1)-bit binary expansion. Thus f(0) = a(1) = 3 and f(1) = a(2) + a(3) = 21, and f(2) = a(4)+a(5)+a(6)+a(7) = 132.
Then f(k) = 135*6^(k-2) - 3*2^(k-2) for k >= 0.
Proof: the formula is true for k = 0. Looking at the last 2 bits of n, it is easy to see that a(4n) = 2*a(2n), a(4n+1) = 4*a(2n)+3, a(4n+2) = 4*a(2n+1)+2 and a(4n+3) = 2*a(2n+1)+1. By summing over the recurrence relations for a(n), we get f(k+2) = Sum_{i=2^k..2^(k+1)-1} (f(4i) + f(4i+1) + f(4i+2) + f(4i+3)) =  Sum_{i=2^k..2^(k+1)-1} (6a(2i) + 6a(2i+1) + 6) = 6*f(k+1) + 3*2^(k+1). Solving this first-order recurrence relation with the initial condition f(1) = 21 shows that f(k) = 135*6^(k-2) - 3*2^(k-2) for k > 0.
(End)

			6 in binary is 110. Modify each run by prepending a 1 to get 11110, which is 30 in decimal. So a(6) = 30.

Chai Wah Wu, Table of n, a(n) for n = 1..10000
Chai Wah Wu, Record values in appending and prepending bitstrings to runs of binary digits, arXiv:1810.02293 [math.NT], 2018.

Cf. A175046, A319423, A320037, A320038, A320039.

Table[FromDigits[Flatten[Join[{1},#]&/@Split[IntegerDigits[n,2]]],2],{n,60}] (* Harvey P. Dale, Apr 26 2019 *)

from re import split
def A320261(n):
    return int(''.join('1'+d for d in split('(0+)|(1+)',bin(n)[2:]) if d != '' and d != None),2)

a(4n) = 2*a(2n), a(4n+1) = 4*a(2n)+3, a(4n+2) = 4*a(2n+1)+2 and a(4n+3) = 2*a(2n+1)+1. - Chai Wah Wu, Nov 25 2018

A321003 a(n) = 2^n(43^n-1).

Original entry on oeis.org

3, 22, 140, 856, 5168, 31072, 186560, 1119616, 6718208, 40310272, 241863680, 1451186176, 8707125248, 52242767872, 313456640000, 1880739905536, 11284439564288, 67706637647872, 406239826411520, 2437438959517696, 14624633759203328, 87747802559414272

Offset: 0

N. J. A. Sloane, Nov 01 2018

Conjectured to be the sum of A175046(i) for 1 <= i < 2^(n+1).

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (8,-12).

Cf. A175046, A171498, A321002.

a := n -> 2^n*(4*3^n-1):
seq(a(n), n = 0 .. 25); # Stefano Spezia, Nov 02 2018

a[n_]:=2^n*(4*3^n-1); Array[a, 25, 0] (* or *)
CoefficientList[Series[-E^(2 x) + 4 E^(6 x), {x, 0, 25}], x]*Table[k!, {k, 0, 25}] (* Stefano Spezia, Nov 02 2018 *)

Vec((3 - 2*x) / ((1 - 2*x)*(1 - 6*x)) + O(x^25)) \\ Colin Barker, Nov 02 2018

a(n) = 2^n*(4*3^n-1); \\ Michel Marcus, Nov 02 2018

From Colin Barker, Nov 02 2018: (Start)
G.f.: (3 - 2*x) / ((1 - 2*x)*(1 - 6*x)).
a(n) = 8*a(n-1) - 12*a(n-2) for n>1.
(End)
E.g.f.: -exp(2*x)+4*exp(6*x). - Stefano Spezia, Nov 02 2018

A321536 Write n in base 10, lengthen all the runs of successive identical digits by 1.

Original entry on oeis.org

0, 11, 22, 33, 44, 55, 66, 77, 88, 99, 1100, 111, 1122, 1133, 1144, 1155, 1166, 1177, 1188, 1199, 2200, 2211, 222, 2233, 2244, 2255, 2266, 2277, 2288, 2299, 3300, 3311, 3322, 333, 3344, 3355, 3366, 3377, 3388, 3399, 4400, 4411, 4422, 4433, 444, 4455, 4466

Offset: 0

N. J. A. Sloane, Nov 13 2018

A321537(a(n)) = n. Compare A321538.

			10 -> 1100, so a(10)=1100; 11->111, so a(11)=111.

Giovanni Resta, Table of n, a(n) for n = 0..10000

A base-10 analog of A175046.

Cf. A321537, A321538.

a[n_] := FromDigits@ Flatten[ Append[ #, Last@#] & /@ Split@ IntegerDigits[ n]]; a /@ Range[0, 46] (* Giovanni Resta, Nov 13 2018 *)

a(n)={my(v=digits(n)); my(L=List()); for(i=1, #v, my(t=v[i]); if(i==1 || t<>v[i-1], listput(L,t)); listput(L,t)); fromdigits(Vec(L))} \\ Andrew Howroyd, Nov 13 2018

from re import split
def A321536(n):
    return int(''.join(d+d[0] for d in split('(0+)|(1+)|(2+)|(3+)|(4+)|(5+)|(6+)|(7+)|(8+)|(9+)',str(n)) if d != '' and d != None)) # Chai Wah Wu, Nov 13 2018

More terms from Giovanni Resta, Nov 13 2018

cp's OEIS Frontend

A175048 Write n in binary, then increase each run of 1's by one 1. a(n) is the decimal equivalent of the result.

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A320037 Write n in binary, then modify each run of 0's by appending one 1, and modify each run of 1's by appending one 0. a(n) is the decimal equivalent of the result.

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A320038 Write n in binary, then modify each run of 0's by prepending one 1, and modify each run of 1's by prepending one 0. a(n) is the decimal equivalent of the result.

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A320039 Write n in binary, then modify each run of 0's and each run of 1's by appending a 1. a(n) is the decimal equivalent of the result.

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A266150 Take the binary representation of n, increase each run of 0's by one 0 if the length of run is odd, otherwise, if length of run is even, remove one 0. a(n) is the decimal equivalent of the result.

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A266151 Take the binary representation of n, increase each run of 1's by one 1 if the length of run is odd, otherwise, if length of run is even, remove one 1. a(n) is the decimal equivalent of the result.

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A319419 In binary expansion of n, delete one symbol from each run. Set a(n)=-1 if the result is the empty string.

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A321003 a(n) = 2^n(43^n-1).