Jamie Robert Creasey - cp's OEIS frontend

A344104 a(0) = 10; for n > 0, a(n) is a(n-1) multiplied by the number of 0's so far in the sequence.

10, 10, 20, 60, 240, 1200, 8400, 75600, 831600, 10810800, 183783600, 3491888400, 73329656400, 1686582097200, 43851134527200, 1227831766761600, 36834953002848000, 1289223355099680000, 51568934203987200000, 2372170973383411200000, 123352890615937382400000

Offset: 0

Jamie Robert Creasey, May 09 2021

This sequence is the multiplicative counterpart to A130232, but the initial term 0 is replaced with 10 (0 obviously results in A000004). The data demonstrates that the number of 0's grows rapidly as a(n) increases, because each term is a multiple of 10, thus adding at least one 0 to successive terms.

			To calculate a(5), multiply a(4)=240 by the number of 0's present in itself and previous terms, of which there are 5, thus yielding 1200.
a(6) is 1200 multiplied by 7, which is the number of 0's present so far, thus giving 8400.

Cf. A130224, A130232.

a[0]=10;a[n_]:=a[n]=a[n-1]Count[Flatten[IntegerDigits/@Array[a,n,0]],0];Array[a,20,0] (* Giorgos Kalogeropoulos, May 09 2021 *)

A344104_list, c = [10], 1
for _ in range(20):
    A344104_list.append(A344104_list[-1]*c)
    c += str(A344104_list[-1]).count('0') # Chai Wah Wu, Jun 06 2021

a(n+1) = a(n)*#_0[a(n)...a(0)], where #_0(n) is the number of 0's in n.

A341993 a(0)=0. For n > 0, a(n+1) = 2a(n) if the sum of digits of 2a(n) exceeds that of a(n); otherwise, a(n+1) is the smallest unused nonnegative integer.

Original entry on oeis.org

0, 1, 2, 4, 8, 3, 6, 5, 7, 9, 10, 20, 40, 80, 11, 22, 44, 88, 12, 24, 48, 96, 13, 26, 14, 28, 56, 15, 16, 17, 18, 19, 38, 76, 21, 42, 84, 168, 23, 46, 92, 184, 368, 25, 27, 29, 58, 30, 60, 31, 62, 32, 64, 128, 256, 33, 66, 34, 68, 35, 36, 37, 74, 148, 296, 39

Offset: 0

Jamie Robert Creasey, Feb 25 2021

This sequence is a permutation of the nonnegative integers; the inverse permutation begins 0, 1, 2, 5, 3, 7, 6, 8, 4, 9, 10, ...
There exist areas that feature numbers in runs of three or more in arithmetic progression, such as (5, 7, 9) and (15, 16, 17, 18, 19).
Record values are 0, 1, 2, 4, 8, 9, 10, 20, 40, 80, 88, ...

			We start the sequence with 0. Doubling this integer results in 0, but as the sum of digits of 0 is equal to that of 0, we choose the smallest nonnegative integer not yet used, which is 1. We can double 1 three times before the sum of digits of 2*a(n) (i.e., 16) does not exceed that of a(n) (8). Thus the next term after 8 is the next unused nonnegative integer, 3, after which we resume doubling.

Index entries for sequences that are permutations of the nonnegative integers

Cf. A000079 (powers of 2), A331440 (similar principle, except lesser or equal sum of digits replaced by containing the digit S).

A339951 Similar to A100787, but alternate with addition and multiplication instead. See Comments section for more information.

Original entry on oeis.org

1, 2, 4, 8, 64, 70, 280, 287, 0, 2, 16, 16, 32, 40, 280, 280, 560, 561, 3366, 3367, 20202, 20205, 40410, 40414, 0, 2, 16, 16, 32, 40, 0, 5, 30, 30, 150, 156, 156, 159, 477, 483, 2898, 2901, 8703, 8709, 60963, 60965, 0, 2, 0, 2, 4, 4, 8, 8, 40, 44, 0, 4, 4, 4

Offset: 0

Jamie Robert Creasey, Dec 24 2020

We start with a(0) as 1, placing an addition sign below the digit. As such, we add 1 to a(0) to receive the next term which is 2. This time, place a multiplication sign below the new digit and double a(1) to receive a(2) which is 4. Place add under the next unused digit and add this to a(2), followed by multiply under the next a(3) and repeat.

Unlike A100787, terms within this sequence are not strictly increasing, as the 8th digit in this sequence is a 0 where we apply multiplication. Also, the differences do not reflect the digits of this sequence, due to alternation with multiplication. The records are 1, 2, 4, 8, 64, 70, 280, 287, 280, 560, 561, 3366 ...

			1, 2, 4, 8, 64, 70, 280, 287, 0, 2, 16, ...
+  *  +  *  +*  +*  +*+  *+*  +  *  +*
To calculate a(5), we add 6 to 64, as the 5th digit is a 6 with the add operator underneath. Thus, a(5) is 70.
To calculate a(8), we multiply 287 by 0, as the 8th digit is a 0 with the multiply operator underneath. Thus, a(8) is 0.

Cf. A100787, A107974, A107975, A107976, A107977.

A337016 a(0) = 0. Successive terms are double the previous, then with their digits incremented by 1.

Original entry on oeis.org

0, 1, 3, 7, 25, 61, 233, 577, 2265, 5641, 22393, 55897, 2228105, 5567321, 22245753, 555102617, 2221316345, 55537437101, 222185985313, 5554821081737, 222110753274585, 5553326176510281, 22217763464131673, 555466371039374457, 222110438531898591025

Offset: 0

Jamie Robert Creasey, Nov 21 2020

If 9 appears anywhere in the decimal expansion of 2*a(n-1), we replace that digit with 10 upon incrementing by 1. See a(12) of the Examples section and A216556 for more information.

			To calculate a(12), double 55897 to get 111794, then increment the digits by 1 to get 2228105.
To calculate a(13), double 2228105 to get 4456210, then increment the digits by 1 to get 5567321.

Cf. A000079, A061511, A089718.

a:= proc(n) option remember; `if`(n=0, 0, (l-> parse(cat(seq(
      l[-i]+1, i=1..nops(l)))))(convert(2*a(n-1), base, 10)))
    end:
seq(a(n), n=0..25);  # Alois P. Heinz, Dec 12 2020

NestList[FromDigits[Flatten@ Map[IntegerDigits, IntegerDigits[2 #] + 1]] &, 0, 24] (* Michael De Vlieger, Dec 11 2020 *)

digs(n) = if (n==0, [0], digits(n));
lista(nn) = {a = 0; print1(a, ", "); for (n=1, nn, a = eval(concat(apply(t->Str(t+1), digs(2*a)))); print1(a, ", "););} \\ Michel Marcus, Nov 28 2020

a(n) = A216556(2*a(n-1)), a(0) = 0.

A338767 a(0) = 0; a(n) is obtained by incrementing each digit of a(n-1) by n.

Original entry on oeis.org

0, 1, 3, 6, 10, 65, 1211, 8988, 16171616, 1015101610151015, 11101115111011161110111511101115, 1212121112121216121212111212121712121211121212161212121112121216

Offset: 0

Jamie Robert Creasey, Nov 07 2020

This sequence is the additive counterpart of the digit factorials which, unlike the digit factorials, increases at a faster pace. A061511 and its relatives bear similarities to this sequence, but each of these increase at varying rates depending on the chosen constant. However, unlike these sequences, the constant increases by 1 each time. If digits within a(n-1) exceed 9 when one adds a constant, we ignore carrying and replace the digit with its correct value, thus 9+1 = 10. a(15) has 1024 digits.

			a(5) = {1+5, 0+5} = 65, where {x, y} is the concatenation of x and y.
a(6) = {6+6, 5+6} = 1211.

Cf. A061511-A061750, A061581-A061587, A089718, A337770.

a:= proc(n) option remember; `if`(n=0, 0, (l-> parse(cat(
      seq(n+l[-i], i=1..nops(l)))))(convert(a(n-1), base, 10)))
    end:
seq(a(n), n=0..12);  # Alois P. Heinz, Nov 15 2020

Nest[Append[#1, FromDigits@ Apply[Join, Map[IntegerDigits, IntegerDigits[#1[[-1]] ] + #2]]] & @@ {#, Length@ #} &, {0}, 11] (* Michael De Vlieger, Nov 13 2020 *)

A336743 a(n) is the product of the first n positive evil numbers.

Original entry on oeis.org

1, 3, 15, 90, 810, 8100, 97200, 1458000, 24786000, 446148000, 8922960000, 205228080000, 4925473920000, 132987795840000, 3856646079360000, 115699382380800000, 3818079618566400000, 129814707031257600000, 4673329453125273600000, 182259848671885670400000

Offset: 0

Jamie Robert Creasey, Sep 26 2020

Despite 0 appearing in the evil numbers, it is excluded here for obvious reasons.

This sequence features a mixture of odious and evil numbers, demonstrated in terms such as a(5) and a(8), where 8100 is evil but 24768000 is odious. As such, this shows that despite doubling preserving evilness, this is untrue for all multipliers.

			a(3) = 3*5*6 = 90.
a(8) = 3*5*6*9*10*12*15*17 = 24786000.

Cf. A000142, A000069 (odious), A001969 (evil).

FoldList[Times, 1, Select[Range[40], EvenQ @ DigitCount[#, 2, 1] &]] (* Amiram Eldar, Sep 26 2020 *)

a(n) = Product_{k=2, n+1} A001969(k). - Michel Marcus, Sep 27 2020

A337770 a(0)=1; a(n) is the leading digit of a(n-1) multiplied by n concatenated with the remaining digits of a(n-1).

Original entry on oeis.org

1, 1, 2, 6, 24, 104, 604, 4204, 32204, 272204, 2072204, 22072204, 242072204, 2642072204, 28642072204, 308642072204, 4808642072204, 68808642072204, 1088808642072204, 19088808642072204, 209088808642072204, 4209088808642072204, 88209088808642072204

Offset: 0

Jamie Robert Creasey, Sep 19 2020

This sequence bears similarities to the digit factorials, see A089718. However, unlike the digit factorials, we only multiply the leading digit of a(n-1) by n, instead of all digits present. As such, for indices greater than 4, a(n) includes all the digits from a(n-1), except those resulting from the lead digit of a(n-1) being multiplied by n.

If one attempts this with the last digit of a(n-1) instead, 220 is the largest integer reached by the process. All indices greater than 4 yield the same number, as the last digit of 220 is 0 which, if multiplied by 5, results in itself and, if other digits remain consistent, causes 220 to repeat infinitely.

			As a(4) is 24, a(5) is {2*5, 4} which is 104, where {x, y} is the concatenation of x and y.
a(7) is 4204, a(8) is {4*8, 204} which is 32204.

Harvey P. Dale, Table of n, a(n) for n = 0..468

Cf. A000142, A089718.

nxt[{n_,a_}]:=Module[{ida=IntegerDigits[a]},{n+1,ida[[1]](n+1)10^(Length[ ida]-1)+FromDigits[Rest[ida]]}]; NestList[nxt,{0,1},25][[All,2]] (* Harvey P. Dale, Nov 13 2021 *)

seq(n)={my(v=vector(n+1)); v[1]=1; for(n=1, n, my(t=v[n], b=10^logint(t,10), h=t\b*b); v[n+1] = h*n + (t-h)); v} \\ Andrew Howroyd, Sep 19 2020

A334025 a(0)=0, a(1)=1; and a(n) = {2a(n-2), 2a(n-1)}, where {x,y} is the concatenation of x and y.

Original entry on oeis.org

0, 1, 2, 24, 448, 48896, 89697792, 97792179395584, 179395584195584358791168, 195584358791168358791168391168717582336, 358791168391168717582336391168717582336717582336782337435164672

Offset: 0

Jamie Robert Creasey, Apr 14 2020

This sequence, due to the process of concatenating one number with another, bears similarities to A131293 and other familiar sequences. However, unlike A131293, this sequence increases at a faster rate. It happens due to the multiplier applied to the existing terms, which increases the number of digits present in the successive term drastically (see a(7) and a(8)). a(11) is too large to include here and has 102 digits.

			a(2) = {2*a(2-2), 2*a(2-1)} = {2*0, 2*1} = 02 = 2.
a(5) = {2*a(5-2), 2*a(5-1)} = {2*24, 2*448} = 48896.

Cf. A002605, A061107, A108047, A131293, A008352, A113526, A104458.

a[0] = 0; a[1] = 1; a[n_] := a[n] = FromDigits @ Join[IntegerDigits[2*a[n - 2]], IntegerDigits[2*a[n - 1]]]; Array[a, 11, 0] (* Amiram Eldar, Apr 18 2020 *)

A285470 Numbers k where "2" appears as the second digit of the decimal representation.

Original entry on oeis.org

12, 22, 32, 42, 52, 62, 72, 82, 92, 120, 121, 122, 123, 124, 125, 126, 127, 128, 129, 220, 221, 222, 223, 224, 225, 226, 227, 228, 229, 320, 321, 322, 323, 324, 325, 326, 327, 328, 329, 420, 421, 422, 423, 424, 425, 426, 427, 428, 429, 520, 521, 522, 523, 524, 525, 526, 527, 528, 529, 620, 621, 622, 623, 624, 625, 626, 627

Offset: 1

Jamie Robert Creasey, Apr 19 2017

To find a(n), concatenate the first digit of n with 2 and then the other digits (if any) from n. See example. - David A. Corneth, Jun 12 2017

			a(21) = 221, a(36) = 326.
As the first digit of 983 is 9, and the others are 83, a(983) = 9283. - _David A. Corneth_, Jun 12 2017

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

Cf. A011532 (containing 2), A052404 (without 2), A217394 (starting with 2).

seq(seq(seq(a*10^d + 2*10^(d-1)+c, c=0..10^(d-1)-1),a=1..9),d=1..2); # Robert Israel, Jun 12 2017

Table[FromDigits@ Apply[Join, {{First@ #}, {2}, Rest@ #}] &@ IntegerDigits@ n, {n, 67}] (* Michael De Vlieger, Jun 12 2017 *)
Select[Range[700],NumberDigit[#,IntegerLength[#]-2]==2&] (* Harvey P. Dale, Aug 15 2025 *)

isok(n) = (n>9) && digits(n)[2] == 2; \\ Michel Marcus, Jun 12 2017

a(n) = my(d = digits(n)); fromdigits(concat([d[1], [2], vector(#d-1, i, d[i+1])])) \\ David A. Corneth, Jun 12 2017

nxt(n) = if(isok(n+1), n+1, d = digits(n); t = 9*10^(#d-2); if(d[1]==9,t*=3); n+=t++) \\ David A. Corneth, Jun 12 2017

def a(n): s = str(n); return int(s[0] + "2" + s[1:])
print([a(n) for n in range(1, 68)]) # Michael S. Branicky, Dec 22 2021

From Robert Israel, Jun 12 2017: (Start)
a(10*n+j) = 10*a(n)+j for 0<=j<=9 and n >= 1.
G.f. g(x) satisfies g(x) = 10*(1-x^10)*g(x^10)/(1-x) + (x + 2*x + ... + 9*x^9)*x^10/(1-x^10) + 12*x + 22*x^2 + ... + 92*x^9. (End)

cp's OEIS Frontend

User: Jamie Robert Creasey

A344104 a(0) = 10; for n > 0, a(n) is a(n-1) multiplied by the number of 0's so far in the sequence.

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A341993 a(0)=0. For n > 0, a(n+1) = 2*a(n) if the sum of digits of 2*a(n) exceeds that of a(n); otherwise, a(n+1) is the smallest unused nonnegative integer.

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A339951 Similar to A100787, but alternate with addition and multiplication instead. See Comments section for more information.

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A337016 a(0) = 0. Successive terms are double the previous, then with their digits incremented by 1.

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A338767 a(0) = 0; a(n) is obtained by incrementing each digit of a(n-1) by n.

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A336743 a(n) is the product of the first n positive evil numbers.

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A337770 a(0)=1; a(n) is the leading digit of a(n-1) multiplied by n concatenated with the remaining digits of a(n-1).

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A334025 a(0)=0, a(1)=1; and a(n) = {2*a(n-2), 2*a(n-1)}, where {x,y} is the concatenation of x and y.

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A341993 a(0)=0. For n > 0, a(n+1) = 2a(n) if the sum of digits of 2a(n) exceeds that of a(n); otherwise, a(n+1) is the smallest unused nonnegative integer.

A334025 a(0)=0, a(1)=1; and a(n) = {2a(n-2), 2a(n-1)}, where {x,y} is the concatenation of x and y.