A243657 -id:A243657 - cp's OEIS frontend

A243658 a(0)=0; thereafter a(n) = noz(n+a(n-1)), where noz(n) = A004719(n).

0, 1, 3, 6, 1, 6, 12, 19, 27, 36, 46, 57, 69, 82, 96, 111, 127, 144, 162, 181, 21, 42, 64, 87, 111, 136, 162, 189, 217, 246, 276, 37, 69, 12, 46, 81, 117, 154, 192, 231, 271, 312, 354, 397, 441, 486, 532, 579, 627, 676, 726, 777, 829, 882, 936, 991, 147, 24, 82, 141, 21, 82, 144, 27, 91, 156, 222, 289

Offset: 0

N. J. A. Sloane, Jun 11 2014

Zeroless analog of triangular numbers.

N. J. A. Sloane, Table of n, a(n) for n = 0..5000

Cf. A004719, A242350, A243063, A243657.

Row n = 3 of A373169.

noz:=proc(n) local a,t1,i,j; a:=0; t1:=convert(n,base,10); for i from 1 to nops(t1) do j:=t1[nops(t1)+1-i]; if j <> 0 then a := 10*a+j; fi; od: a; end;
t1:=[0]; for n from 1 to 50 do t1:=[op(t1),noz(n+t1[n])]; od: t1;

noz[n_] := FromDigits[DeleteCases[IntegerDigits[n], 0]];
Block[{n = 0}, NestList[noz[++n+#] &, 0, 100]] (* Paolo Xausa, Apr 17 2024 *)

from itertools import count, islice
def noz(n): return int(str(n).replace("0", ""))
def agen(): # generator of terms
    yield (an:=0)
    yield from (an:=noz(n+an) for n in count(1))
print(list(islice(agen(), 68))) # Michael S. Branicky, Jul 02 2024

A242350 Multiply a(n-1) by 2 and drop all 0's.

Original entry on oeis.org

1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 124, 248, 496, 992, 1984, 3968, 7936, 15872, 31744, 63488, 126976, 253952, 5794, 11588, 23176, 46352, 9274, 18548, 3796, 7592, 15184, 3368, 6736, 13472, 26944, 53888, 17776, 35552, 7114, 14228, 28456, 56912, 113824

Offset: 1

J. Lowell, May 11 2014

Sequence enters a loop having period 36 at index 491: a(491) = a(527) = 366784, min and max being 28714 and 11772544. Starting with 3 instead of 1 gives another cycle. - Tom Edgar and Michel Marcus, May 13 2014

Zeroless analog of powers of 2. - N. J. A. Sloane, Jun 11 2014

			Term after 512 is 124 because 512*2=1024, and 1024 becomes 124 if all 0's are taken out.

Michel Marcus, Table of n, a(n) for n = 1..600
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1).

Cf. A052382, A004719, A243063, A243657, A243658.

NestList[FromDigits[Select[IntegerDigits[2 #],#!=0&]]&,1,50] (* Harvey P. Dale, Oct 22 2018 *)

dropz(n)=d = digits(n); s = 0; for (i=1, #d, if (d[i], s = 10*s + d[i]);); s;
lista(nn) = a = 1; for (i=1, nn, print1(a, ", "); a = dropz(2*a);) \\ Michel Marcus, May 12 2014

More terms from Michel Marcus, May 12 2014

A243063 Numbers generated by a Fibonacci-like sequence in which zeros are suppressed.

Original entry on oeis.org

1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 61, 438, 499, 937, 1436, 2373, 389, 2762, 3151, 5913, 964, 6877, 7841, 14718, 22559, 37277, 59836, 97113, 156949, 25462, 182411, 27873, 21284, 49157, 7441, 56598, 6439, 6337, 12776, 19113, 31889, 512, 3241

Offset: 1

Anthony Sand, Jun 09 2014

Let x(1) = 1, x(2) = 1, then begin the sequence x(i) = no-zero(x(i-2) + x(i-1)), where the function no-zero(n) removes all zero digits from n.

The sequence behaves like a standard Fibonacci sequence until step 15, where x = no-zero(233 + 377) = no-zero(610) = 61. At step 16, x = 377 + 61 = 438. The sequence then proceeds until step 927, where x = no-zero(224 + 377) = no-zero(601) = 61. Therefore at step 928, x = 377 + 61 = 438 and the sequence repeats.

			x(3) = x(1) + x(2) = 1 + 1 = 2.
x(4) = x(2) + x(3) = 1 + 2 = 3.
x(15) = no-zero(x(13) + x(14)) = no-zero(233 + 377) = no-zero(610) = 61.
x(16) = 377 + 61 = 438.

Anthony Sand, Table of n, a(n) for n = 1..927
Index entries for linear recurrences with constant coefficients, order 912.

Cf. A000045, A004719, A242350, A243657, A243658, A306773.

noz:=proc(n) local a,t1,i,j; a:=0; t1:=convert(n,base,10); for i from 1 to nops(t1) do j:=t1[nops(t1)+1-i]; if j <> 0 then a := 10*a+j; fi; od: a; end; # A004719
t1:=[1,1]; for n from 3 to 100 do t1:=[op(t1),noz(t1[n-1]+t1[n-2])]; od: t1; # N. J. A. Sloane, Jun 11 2014

Nest[Append[#, FromDigits@ DeleteCases[IntegerDigits[Total@ #[[-2 ;; -1]] ], ?(# == 0 &)]] &, {1, 1}, 45] (* _Michael De Vlieger, Jun 27 2020 *)
nxt[{a_,b_}]:={b,FromDigits[DeleteCases[IntegerDigits[a+b],0]]}; NestList[nxt,{1,1},50][[All,1]] (* Harvey P. Dale, Sep 12 2022 *)

x(i) = no-zero(x(i-2) + x(i-1)). For example, no-zero(233 + 377) = no-zero(610) = 61.

A370812 a(1) = 1; for n >= 2, a(n) = noz(a(n-1) + 2*n - 1), where noz(n) = A004719(n).

Original entry on oeis.org

1, 4, 9, 16, 25, 36, 49, 64, 81, 1, 22, 45, 7, 34, 63, 94, 127, 162, 199, 238, 279, 322, 367, 414, 463, 514, 567, 622, 679, 738, 799, 862, 927, 994, 163, 234, 37, 112, 189, 268, 349, 432, 517, 64, 153, 244, 337, 432, 529, 628, 729, 832, 937, 144, 253, 364, 477

Offset: 1

Paolo Xausa, May 24 2024

Zeroless analog of the positive squares.

Paolo Xausa, Table of n, a(n) for n = 1..10000

Cf. A000290, A004719, A243657, A243658, A373171, A373172.

Row n = 4 of A373169.

noz[n_] := FromDigits[DeleteCases[IntegerDigits[n], 0]];
Block[{n = 1}, NestList[noz[++n*2 - 1 + #] &, 1, 100]]

A321475 Zeroless factorials (version 2): a(0) = 1, and for any n > 0, a(n) = noz(1 * noz(2 * ... * noz((n-1) * n))), where noz(n) = A004719(n) omits the zeros from n.

Original entry on oeis.org

1, 1, 2, 6, 24, 12, 72, 54, 432, 3888, 3888, 399168, 576, 82728, 879912, 2397168, 337968, 5924736, 8851949568, 143936352, 31644, 92589264, 118459638, 3698784, 1197539136, 2387625984, 954864, 236271168, 3573339984, 238453776, 69587928, 142275168, 33566976

Offset: 0

Rémy Sigrist, Nov 11 2018

This sequence is a variant of A243657 where the multiplications are carried in the opposite order; as (i, j) -> noz(i * j) is not associative in general we obtain another sequence.

Is this sequence bounded?

			For n = 12:
- noz(11 * 12) = noz(132) = 132,
- noz(10 * 132) = noz(1320) = 132,
- noz(9 * 132) = noz(1188) = 1188,
- noz(8 * 1188) = noz(9504) = 954,
- noz(7 * 954) = noz(6678) = 6678,
- noz(6 * 6678) = noz(40068) = 468,
- noz(5 * 468) = noz(2340) = 234,
- noz(4 * 234) = noz(936) = 936,
- noz(3 * 936) = noz(2808) = 288,
- noz(2 * 288) = noz(576) = 576,
- noz(1 * 576) = noz(576) = 576,
- hence a(12) = 576.

Rémy Sigrist, Table of n, a(n) for n = 0..10000

Cf. A000142, A004719, A243657.

noz[n_] := FromDigits[DeleteCases[IntegerDigits[n], 0]];
A321475[n_] := If[n == 0, 1, Block[{k = n}, Nest[noz[--k * #] &, n, n-1]]];
Array[A321475, 50, 0] (* Paolo Xausa, May 20 2024 *)

a(n, base=10) = my (f=max(1, n)); forstep (k=n-1, 2, -1, f = fromdigits(select(sign, digits(f*k, base)), base)); f

a(10^k) = a(10^k - 1) for any k >= 0.

A356757 Omit zero digits from factorial numbers.

Original entry on oeis.org

1, 1, 2, 6, 24, 12, 72, 54, 432, 36288, 36288, 399168, 47916, 622728, 871782912, 137674368, 2922789888, 35568742896, 64237375728, 121645148832, 243292817664, 5199421717944, 11247277776768, 258521673888497664, 624484173323943936, 15511214333985984, 4329146112665635584

Offset: 0

Stefano Spezia, Aug 26 2022

			a(12) = 47916 since 12! = 479001600.

Index entries for sequences related to factorial numbers

Cf. A000142, A004719.
Cf. A004154, A243657, A321475.
Cf. A027869 (number of omitted zero digits), A356758 (number of nonzero digits).

Table[FromDigits[Select[IntegerDigits[n!],Positive]], {n,0,26}]

a(n) = fromdigits(select(x->(x>0), digits(n!))); \\ Michel Marcus, Aug 26 2022

from math import factorial
def a(n): return int(str(factorial(n)).replace("0", ""))
print([a(n) for n in range(27)]) # Michael S. Branicky, Aug 26 2022

a(n) = A004719(A000142(n)).

A371911 Zeroless analog of tribonacci numbers.

Original entry on oeis.org

1, 1, 1, 3, 5, 9, 17, 31, 57, 15, 13, 85, 113, 211, 49, 373, 633, 155, 1161, 1949, 3265, 6375, 11589, 21229, 39193, 7211, 67633, 11437, 86281, 165351, 26369, 2781, 19451, 4861, 2793, 2715, 1369, 6877, 1961, 127, 8965, 1153, 1245, 11363, 13761, 26369, 51493, 91623, 169485, 31261

Offset: 0

Bryle Morga, Apr 11 2024

At n = 208666297 this sequence enters a cycle that has a period of 300056874 and begins: 2847, 26331, 5851, ... (only the first 3 terms of the cycle are needed to reproduce the entire cycle).

This can be compared with the sequence A243063, which enters a cycle that has a (relatively) small period of 912.

			a(9) = Zr(a(8) + a(7) + a(6)) = Zr(17 + 31 + 57) = Zr(105) = 15.

Cf. A000045, A000213, A004719, A242350, A243657, A243658, A243063.

a[0]=a[1]=a[2]=1; a[n_]:=FromDigits[DeleteCases[IntegerDigits[a[n-1]+a[n-2]+a[n-3]],0]]; Array[a,50,0] (* Stefano Spezia, Apr 12 2024 *)

def a(n):
    a, b, c = 1, 1, 1
    for _ in range(n):
        a, b, c = b, c, int(str(a+b+c).replace('0', ''))
    return a

# faster for initial segment of sequence
from itertools import islice
def agen(): # generator of terms
    a, b, c = 1, 1, 1
    while True: yield a; a, b, c = b, c, int(str(a+b+c).replace("0", ""))
print(list(islice(agen(), 50))) # Michael S. Branicky, Apr 13 2024

a(n) = Zr(a(n-1) + a(n-2) + a(n-3)), where the function Zr(k) removes all zero digits from k.

A374265 Minimized zeroless factorials.

Original entry on oeis.org

1, 1, 2, 6, 24, 12, 72, 54, 432, 3888, 3888, 42768, 47916, 62298, 872172, 13968, 221688, 57996, 143928, 134712, 269154, 563994, 1247868, 286344, 877356, 171864, 513324, 1252728, 3414474, 914616, 41868, 119178, 454716, 127188, 527832, 15642, 91332, 192924, 125892, 29718

Offset: 0

Bryle Morga, Jul 02 2024

a(n) is the smallest f(n) such that f(0) = 1 and for i > 0, f(i) = OpNoz_i(i*f(i-1)),where OpNoz_i is a function that either removes zeros or keeps the value unchanged (the choice is made for each value of i).
Removing zeros at every i gives an upper bound a(n) <= A243657(n); is this a strict inequality for n >= 12?
Is this sequence bounded?

			a(12) = 47916 via the path: 1, 1, 2, 6, 24, 12, 72, 504, 4032, 36288, 362880, 3991680, 47916.

Cf. A000142, A004719, A242350, A243063, A243657, A243658, A356757, A374266.

def a(n):
   reach = {1}
   for i in range(1, n+1):
      newreach = set()
      for m in reach:
         newreach.update([m*i, int(str(m*i).replace('0', ''))])
      reach = newreach
   return min(reach)

A374266 Smallest number reachable by a Fibonacci-like iteration where one has the option to either omit or keep zero digits.

Original entry on oeis.org

1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 61, 438, 499, 937, 1436, 2373, 389, 2762, 1657, 1368, 325, 1693, 218, 1911, 2129, 44, 1516, 129, 394, 37, 53, 9, 62, 35, 133, 24, 121, 181, 95, 69, 11, 53, 19, 9, 19, 19, 2, 12, 5, 8, 4, 3, 7, 1, 8, 9, 8, 8, 7, 6, 4

Offset: 1

Bryle Morga, Jul 02 2024

a(n) is the smallest f(n) such that f(1)=f(2)=1 and f(i) = OpNoz_i(f(i-1)+f(i-2)) for 2

Choosing to always remove zero digits at each step gives A243063. This strategy of always choosing to remove zeros is optimal for n < 23. For n >= 23, a(n) < A243063(n), i.e., the optimal path contains a step that keeps zeros.

Removal of zeros preserves the digital root giving the lower bound a(n) >= A030132(n). In fact, for n >= 53, a(n) = A030132(n). It follows that this sequence is eventually periodic with a period of 24.

			a(23) = 1657 via the path: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 1946, 8711, 1657.

Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1).

Cf. A000045, A004719, A030132, A242350, A243657, A243658, A306773, A374265.

def a(n):
    reach = {(1, 1)}
    for _ in range(n-1):
        newreach = set()
        for a, b in reach:
            newreach.update([(b, a+b), (b, int(str(a+b).replace('0', '')))])
        reach = newreach
    return min(reach, key = lambda k:k[0])[0]

a(n) <= A243063(n); Strict inequality for n >= 23.

a(n) = A030132(n) and a(n) = a(n+24) for n >= 53.

A371916 Zeroless analog of tetranacci numbers.

Original entry on oeis.org

1, 1, 1, 1, 4, 7, 13, 25, 49, 94, 181, 349, 673, 1297, 25, 2344, 4339, 85, 6793, 13561, 24778, 45217, 9349, 9295, 88639, 1525, 1888, 11347, 13399, 28159, 54793, 17698, 11449, 11299, 95239, 135685, 253672, 495895, 98491, 983743, 183181, 176131, 1441546, 278461, 279319, 2175457

Offset: 0

Bryle Morga, Apr 12 2024

It is not known whether this sequence cycles, but it is conjectured to cycle just like A243063 and A371911 (have periods of 912 and 300056874, respectively) because the expected growth factor in the number of digits of successive terms is 0.9.

It's been computationally verified that if the sequence does cycle, then s+p > 10^10, where s and p are the starting index and period of the cycle, respectively.

			a(14) = Zr(a(13)+a(12)+a(11)+a(10)) = Zr(1297+673+349+181) = Zr(2500) = 25.

Cf. A000045, A000213, A000288, A004719, A242350, A243657, A243658, A243063.

a[0]=a[1]=a[2]=a[3]=1; a[n_]:=FromDigits[DeleteCases[IntegerDigits[a[n-1]+a[n-2]+a[n-3]+a[n-4]], 0]]; Array[a, 46, 0] (* Stefano Spezia, Apr 12 2024 *)

def a(n):
    a, b, c, d = 1, 1, 1, 1
    for _ in range(n):
        a, b, c, d = b, c, d, int(str(a+b+c+d).replace('0', ''))
    return a

# faster for initial segment of sequence
from itertools import islice
def agen(): # generator of terms
    a, b, c, d = 1, 1, 1, 1
    while True:
        yield a
        a, b, c, d = b, c, d, int(str(a+b+c+d).replace("0", ""))
print(list(islice(agen(), 45))) # Michael S. Branicky, Apr 13 2024

a(n) = Zr(a(n-1)+a(n-2)+a(n-3)+a(n-4)), where the function Zr(k) removes zero digits from k.

Showing 1-10 of 10 results.

cp's OEIS Frontend

A243658 a(0)=0; thereafter a(n) = noz(n+a(n-1)), where noz(n) = A004719(n).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Mathematica

Python

A242350 Multiply a(n-1) by 2 and drop all 0's.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Extensions

A243063 Numbers generated by a Fibonacci-like sequence in which zeros are suppressed.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Formula

A370812 a(1) = 1; for n >= 2, a(n) = noz(a(n-1) + 2*n - 1), where noz(n) = A004719(n).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

A321475 Zeroless factorials (version 2): a(0) = 1, and for any n > 0, a(n) = noz(1 * noz(2 * ... * noz((n-1) * n))), where noz(n) = A004719(n) omits the zeros from n.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A356757 Omit zero digits from factorial numbers.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Python

Formula

A371911 Zeroless analog of tribonacci numbers.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

Python

Python