cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-10 of 54 results. Next

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

Original entry on oeis.org

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

Views

Author

N. J. A. Sloane, Jun 11 2014

Keywords

Comments

Zeroless analog of triangular numbers.

Crossrefs

Row n = 3 of A373169.

Programs

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

A243657 Zeroless factorials: a(0)=1; thereafter a(n) = noz(n*a(n-1)), 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, 42768, 513216, 667188, 934632, 141948, 2271168, 3869856, 6965748, 132349212, 264698424, 555866694, 1222967268, 28128247164, 67577931936, 16894482984, 439256557584, 1185992754768, 332779713354, 965611687266, 289683561798, 89819415738
Offset: 0

Views

Author

N. J. A. Sloane, Jun 11 2014

Keywords

Comments

Zeroless analog of factorial numbers.
A very crude calculation suggests a(n) may grow like n^10. I don't really believe that, but certainly a(n) grows very slowly in comparison with n!. - N. J. A. Sloane, Sep 03 2014

Crossrefs

Programs

  • Maple
    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:=[1]; for n from 1 to 50 do t1:=[op(t1),noz(n*t1[n])]; od: t1;
  • Mathematica
    nxt[{n_,a_}]:={n+1,FromDigits[DeleteCases[IntegerDigits[a(n+1)],0]]}; NestList[nxt,{0,1},40][[;;,2]] (* Harvey P. Dale, Feb 13 2024 *)
  • Python
    from itertools import count, islice
    def noz(n): return int(str(n).replace("0", ""))
    def agen(): # generator of terms
        yield (an:=1)
        yield from (an:=noz(n*an) for n in count(1))
    print(list(islice(agen(), 32))) # Michael S. Branicky, Jul 02 2024

A306569 a(n) is the largest value obtained by iterating x -> noz(x * n) starting from 1 (where noz(k) = A004719(k) omits the zeros from k) if any, otherwise a(n) = -1.

Original entry on oeis.org

1, 765257552, 4858337151, 62987487698944, 14281197489647865625, 16756687971893376, 956884714362661, 292452281337511936, 11897243269649199, 1, 824281567746336491, 13552472793415699584, 841944776182612378933, 9434962871842528764976
Offset: 1

Views

Author

Rémy Sigrist, Feb 24 2019

Keywords

Comments

Is every term positive?

Examples

			a(2) = max(A242350) = 765257552.
		

Crossrefs

See A306567 for the additive variant.

Programs

  • PARI
    See Links section.

Formula

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

A373169 Square array read by ascending antidiagonals: T(n,k) = noz(T(n,k-1) + (k-1)*(n-2) + 1), with T(n,1) = 1, n >= 2, k >= 1, where noz(n) = A004719(n).

Original entry on oeis.org

1, 1, 2, 1, 3, 3, 1, 4, 6, 4, 1, 5, 9, 1, 5, 1, 6, 12, 16, 6, 6, 1, 7, 15, 22, 25, 12, 7, 1, 8, 18, 28, 35, 36, 19, 8, 1, 9, 21, 34, 45, 51, 49, 27, 9, 1, 1, 24, 4, 55, 66, 7, 64, 36, 1, 1, 11, 18, 46, 29, 81, 91, 29, 81, 46, 2, 1, 12, 3, 43, 75, 6, 112, 12, 54, 1, 57, 3
Offset: 2

Views

Author

Paolo Xausa, May 27 2024

Keywords

Comments

Row n is the zeroless analog of the positive n-gonal numbers.

Examples

			The array begins:
  n\k|  1  2   3   4   5    6    7    8    9   10  ...
  ----------------------------------------------------
   2 |  1, 2,  3,  4,  5,   6,   7,   8,   9,   1, ... = A177274
   3 |  1, 3,  6,  1,  6,  12,  19,  27,  36,  46, ... = A243658 (from n = 1)
   4 |  1, 4,  9, 16, 25,  36,  49,  64,  81,   1, ... = A370812
   5 |  1, 5, 12, 22, 35,  51,   7,  29,  54,  82, ... = A373171
   6 |  1, 6, 15, 28, 45,  66,  91,  12,  45,  82, ... = A373172
   7 |  1, 7, 18, 34, 55,  81, 112, 148, 189, 235, ...
   8 |  1, 8, 21,  4, 29,   6,  43,  86, 135,  19, ...
   9 |  1, 9, 24, 46, 75, 111, 154,  24,  81, 145, ...
  10 |  1, 1, 18, 43, 76, 117, 166, 223, 288, 361, ...
  ...      |                                     \______ A373170 (main diagonal)
        A004719 (from n = 2)
		

Crossrefs

Cf. rows 2..6: A177274, A243658, A370812, A373171, A373172.
Cf. A373170 (main diagonal).

Programs

  • Mathematica
    noz[n_] := FromDigits[DeleteCases[IntegerDigits[n], 0]];
    A373169[n_, k_] := A373169[n, k] = If[k == 1, 1, noz[A373169[n, k-1] + (k-1)*(n-2) + 1]];
    Table[A373169[n - k + 1, k], {n, 2, 15}, {k, n - 1}]
  • PARI
    noz(n) = fromdigits(select(sign, digits(n)));
    T(n,k) = if (k==1, 1, noz(T(n,k-1) + (k-1)*(n-2) + 1));
    matrix(7,7,n,k,T(n+1,k)) \\ Michel Marcus, May 30 2024

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

Views

Author

Paolo Xausa, May 24 2024

Keywords

Comments

Zeroless analog of the positive squares.

Crossrefs

Programs

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

A373171 a(1) = 1; for n >= 2, a(n) = noz(a(n-1) + 3*n - 2), where noz(n) = A004719(n).

Original entry on oeis.org

1, 5, 12, 22, 35, 51, 7, 29, 54, 82, 113, 147, 184, 224, 267, 313, 362, 414, 469, 527, 588, 652, 719, 789, 862, 938, 117, 199, 284, 372, 463, 557, 654, 754, 857, 963, 172, 284, 399, 517, 638, 762, 889, 119, 252, 388, 527, 669, 814, 962, 1113, 1267, 1424, 1584
Offset: 1

Views

Author

Paolo Xausa, May 28 2024

Keywords

Comments

Zeroless analog of the positive pentagonal numbers.

Crossrefs

Row n = 5 of A373169.

Programs

  • Mathematica
    noz[n_] := FromDigits[DeleteCases[IntegerDigits[n], 0]];
    Block[{n = 1}, NestList[noz[++n*3 - 2 + #] &, 1, 100]]
  • PARI
    noz(n) = fromdigits(select(sign, digits(n))); \\ A004719
    lista(nn) = my(va=vector(nn)); for (n=1, nn, va[n] = if (n==1, 1, noz(va[n-1] + 3*n - 2))); va; \\ Michel Marcus, Jun 03 2024

A373172 a(1) = 1; for n >= 2, a(n) = noz(a(n-1) + 4*n - 3), where noz(n) = A004719(n).

Original entry on oeis.org

1, 6, 15, 28, 45, 66, 91, 12, 45, 82, 123, 168, 217, 27, 84, 145, 21, 9, 82, 159, 24, 19, 18, 111, 28, 129, 234, 343, 456, 573, 694, 819, 948, 181, 318, 459, 64, 213, 366, 523, 684, 849, 118, 291, 468, 649, 834, 123, 316, 513, 714, 919, 1128, 1341, 1558, 1779
Offset: 1

Views

Author

Paolo Xausa, May 28 2024

Keywords

Comments

Zeroless analog of the positive hexagonal numbers.

Crossrefs

Row n = 6 of A373169.

Programs

  • Mathematica
    noz[n_] := FromDigits[DeleteCases[IntegerDigits[n], 0]];
    Block[{n = 1}, NestList[noz[++n*4 - 3 + #] &, 1, 100]]
    nxt[{n_,a_}]:={n+1,FromDigits[DeleteCases[IntegerDigits[a+4n+1],0]]}; NestList[nxt,{1,1},60][[;;,2]] (* Harvey P. Dale, Jul 08 2024 *)
  • PARI
    noz(n) = fromdigits(select(sign, digits(n))); \\ A004719
    lista(nn) = my(va=vector(nn)); for (n=1, nn, va[n] = if (n==1, 1, noz(va[n-1] + 4*n - 3))); va; \\ Michel Marcus, Jun 03 2024

A306567 a(n) is the largest value obtained by iterating x -> noz(x + n) starting from 0 (where noz(k) = A004719(k) omits the zeros from k).

Original entry on oeis.org

9, 99, 27, 99, 96, 99, 63, 99, 81, 91, 99, 195, 94, 295, 93, 291, 113, 189, 171, 992, 159, 187, 187, 483, 988, 475, 153, 281, 181, 273, 279, 577, 297, 997, 567, 369, 333, 363, 351, 994, 219, 465, 357, 663, 459, 461, 423, 192, 441, 965, 399, 999, 437, 126, 551
Offset: 1

Views

Author

Rémy Sigrist, Feb 24 2019

Keywords

Comments

For any n > 0, a(n) is well defined:
- the set of zeroless numbers (A052382) contains arbitrarily large gaps,
- for example, for any k > 0, the interval I_k = [10^k..(10^(k+1)-1)/9-1] if free of zeroless numbers,
- let i be such that #I_i > n,
- let b_n be defined by b_n(0) = 0, and for any j > 0, b_n(j) = noz(b_n(j-1) + n),
- as b_n starts below 10^i and cannot cross the gap constituted by I_i,
- b_n is bounded (and eventually periodic), QED.

Examples

			For n = 1:
- noz(0 + 1) = 1,
- noz(1 + 1) = 2,
- noz(2 + 1) = 3,
  ...
- noz(7 + 1) = 8,
- noz(8 + 1) = 9,
- noz(9 + 1) = noz(10) = 1,
- hence a(1) = 9.
		

Crossrefs

See A306569 for the multiplicative variant.

Programs

  • PARI
    \\ See Links section.

Formula

Empirically, for any k >= 0:
- a( 10^k) = 9 * 10^k + (10^k-1)/9,
- a(2 * 10^k) = 99 * 10^k + 2 * (10^k-1)/9,
- a(3 * 10^k) = 27 * 10^k + 3 * (10^k-1)/9,
- a(4 * 10^k) = 99 * 10^k + 4 * (10^k-1)/9,
- a(5 * 10^k) = 96 * 10^k + 5 * (10^k-1)/9,
- a(6 * 10^k) = 99 * 10^k + 6 * (10^k-1)/9,
- a(7 * 10^k) = 63 * 10^k + 7 * (10^k-1)/9,
- a(8 * 10^k) = 99 * 10^k + 8 * (10^k-1)/9,
- a(9 * 10^k) = 81 * 10^k + 9 * (10^k-1)/9.

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

Views

Author

Rémy Sigrist, Nov 11 2018

Keywords

Comments

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?

Examples

			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.
		

Crossrefs

Programs

  • Mathematica
    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 *)
  • PARI
    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

Formula

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

A321480 Zeroless analog of triangular numbers (version 2): a(0) = 0, 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

0, 1, 3, 6, 1, 15, 3, 28, 9, 18, 19, 39, 6, 28, 15, 12, 1, 9, 99, 37, 39, 177, 64, 69, 39, 19, 72, 99, 37, 12, 69, 64, 87, 12, 289, 27, 54, 82, 39, 42, 19, 6, 57, 37, 27, 54, 82, 12, 69, 64, 69, 12, 64, 27, 27, 82, 12, 87, 289, 69, 39, 289, 72, 99, 64, 57, 24
Offset: 0

Views

Author

Rémy Sigrist, Nov 11 2018

Keywords

Comments

This sequence is a variant of A243658 where the additions are carried in the opposite order; as (i, j) -> noz(i + j) is not associative in general we obtain another sequence.
This sequence is conjectured to be bounded. This could be explained by the fact that the zeros appearing in the last steps of the calculation (when adding small values) erode the number of digits of the intermediate sums.
The distinct values among the first 1000000 terms are: 0, 1, 3, 6, 9, 12, 15, 18, 19, 24, 27, 28, 37, 39, 42, 54, 57, 64, 69, 72, 82, 84, 87, 99, 177, 289.

Examples

			For n = 16:
- noz(15 + 16) = noz(31) = 31,
- noz(14 + 31) = noz(45) = 45,
- noz(13 + 45) = noz(58) = 58,
- noz(12 + 58) = noz(70) = 7,
- noz(11 + 7) = noz(18) = 18,
- noz(10 + 18) = noz(28) = 28,
- noz(9 + 28) = noz(37) = 37,
- noz(8 + 37) = noz(45) = 45,
- noz(7 + 45) = noz(52) = 52,
- noz(6 + 52) = noz(58) = 58,
- noz(5 + 58) = noz(63) = 63,
- noz(4 + 63) = noz(67) = 67,
- noz(3 + 67) = noz(70) = 7,
- noz(2 + 7) = noz(9) = 9,
- noz(1 + 9) = noz(10) = 1,
- hence a(16) = 1.
		

Crossrefs

Programs

  • Mathematica
    noz[n_] := FromDigits[DeleteCases[IntegerDigits[n], 0]];
    A321480[n_] := Block[{k = n}, Nest[noz[--k + #] &, n, Max[0, n-1]]];
    Array[A321480,100,0] (* Paolo Xausa, Apr 17 2024 *)
  • PARI
    a(n, base=10) = { my (t=n); forstep (k=n-1, 1, -1, t = fromdigits(select(sign, digits(t+k, base)), base)); t } \\ corrected by Rémy Sigrist, Apr 17 2024

Extensions

a(10), a(20), a(30), a(40), a(50) and a(60) corrected by Paolo Xausa, Apr 17 2024
Showing 1-10 of 54 results. Next