A065039 -id:A065039 - cp's OEIS frontend

A067080 If n = ab...def in decimal notation then the left digitorial function Ld(n) = ab...defab...deab...d...ab*a.

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 90, 93, 96, 99, 102, 105, 108, 111, 114, 117, 160, 164, 168, 172, 176, 180, 184, 188, 192, 196, 250, 255, 260, 265, 270, 275, 280, 285, 290, 295, 360, 366, 372

Offset: 1

Amarnath Murthy, Jan 05 2002

This entry should probably start at n=0, just as A067079 does. But that would require a number of changes, so it can wait until the editors have more free time. - N. J. A. Sloane, Nov 29 2014

			Ld(256) = 256*25*2 =12800.
a(31)=floor(31/10^0)*floor(31/10^1)=31*3=93;
a(42)=168 since 42=42(base-10) and so a(42)=42*4(base-10)=42*4=168.

Hieronymus Fischer, Table of n, a(n) for n = 1..10000

Cf. A067079, A065039, A048651, A098844, A132019, A132026, A132038, A000217.
For formulas regarding a general parameter p (i.e. terms floor(n/p^k)) see A132264.
For the product of terms floor(n/p^k) for p=2 to p=12 see A098844(p=2), A132027(p=3)-A132033(p=9), A132263(p=11), A132264(p=12).
For the products of terms 1+floor(n/p^k) see A132269-A132272, A132327, A132328.

a067080 n = if n <= 9 then n else n * a067080 (n `div` 10)
-- Reinhard Zumkeller, Nov 29 2012

Table[d = IntegerDigits[n]; rd = 1; While[ Length[d] > 0, rd = rd*FromDigits[d]; d = Drop[d, -1]]; rd, {n, 1, 75} ]
Table[Times@@NestList[Quotient[#,10]&,n,IntegerLength[n]-1],{n,70}] (* Harvey P. Dale, Dec 16 2013 *)

a(n)=my(t=n);while(n\=10,t*=n); t \\ Charles R Greathouse IV, Nov 20 2012

a(n) = Product_{k=1..length(n)} floor(n/10^(k-1)). - Vladeta Jovovic, Jan 08 2002
From Hieronymus Fischer, Aug 13 2007: (Start)
a(n) = product{0<=k<=floor(log_10(n)), floor(n/10^k)}, n>=1.
Recurrence:
a(n) = n*a(floor(n/10));
a(n*10^m) = n^m*10^(m(m+1)/2)*a(n).
a(k*10^m) = k^(m+1)*10^(m(m+1)/2), for 0a(n) <= b(n), where b(n)=n^(1+floor(log_10(n)))/10^(1/2*(1+floor(log_10(n)))*floor(log_10(n))); equality holds for n=k*10^m, m>=0, 1<=k<10. Here b(n) can also be written n^(1+floor(log_10(n)))/10^A000217(floor(log_10(n))).
Also: a(n) <= 3^((1-log_10(3))/2)*n^((1+log_10(n))/2)=1.332718...*10^A000217(log_10(n)), equality for n=3*10^m, m>=0.
a(n) > c*b(n), where c=0.472362443816572... (see constant A132026).
Also: a(n) > c*2^((1-log_10(2))/2)*n^((1+log_10(n))/2) = 0.601839...*10^A000217(log_10(n)).
lim inf a(n)/b(n) = 0.472362443816572..., for n-->oo.
lim sup a(n)/b(n) = 1, for n-->oo.
lim inf a(n)/n^((1+log_10(n))/2) = 0.472362443816572...*sqrt(2)/2^log_10(sqrt(2)), for n-->oo.
lim sup a(n)/n^((1+log_10(n))/2) = sqrt(3)/3^log_10(sqrt(3)), for n-->oo.
lim inf a(n)/a(n+1) = 0.472362443816572... for n-->oo (see constant A132026).
a(n) = O(n^((1+log_10(n))/2)). (End)

More terms from Robert G. Wilson v, Jan 07 2002

A171397 Write n in base 10, but then read it as if it were written in base 11: if n = Sum_{i >= 0} d_i10^i, with 0 <= d_i <= 9, then a(n) = Sum_{i >= 0} d_i11^i.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 66, 67, 68, 69, 70, 71, 72

Offset: 0

Paul Weisenhorn, Jul 11 2011

This is the sequence of all decimal integers that are created when base 10 numbers are interpreted as base 11 numbers.
Numbers without digit A (=10) in their representation in base 11. Complement of A095778. - François Marques, Oct 20 2020
Original definition: Earliest sequence containing no 11-term arithmetic progression.
In general, if p is prime, the earliest sequence containing no p-term arithmetic progression is created when base (p-1) numbers are interpreted as base p numbers.

			a(53)=58 because 53_11 in base 11 equals 58. - _François Marques_, Oct 20 2020

D. E. Arganbright, Mathematical Modeling with Spreadsheets, ABACUS, Vol. 3, #4(1986), 19-31.

François Marques, Table of n, a(n) for n = 0..10000

Different from A065039. - Alois P. Heinz, Sep 07 2011
CNumbers with at least one digit b-1 in base b : A074940 (b=3), A337250 (b=4), A337572 (b=5), A333656 (b=6), A337141 (b=7), A337239 (b=8), A338090 (b=9), A011539 (b=10), A095778 (b=11).
Numbers with no digit b-1 in base b : A005836 (b=3), A023717 (b=4), A020654 (b=5), A037465 (b=6), A020657 (b=7), A037474 (b=8), A037477 (b=9), A007095 (b=10), this sequence (b=11).

seq(`if`(numboccur (10, convert (n, base, 11))=0, n, NULL), n=0..122);
# second Maple program:
a:= n-> (l-> add(l[i]*11^(i-1), i=1..nops(l)))(convert(n, base, 10)):
seq(a(n), n=0..66);  # Alois P. Heinz, Aug 30 2024

Table[FromDigits[RealDigits[n, 10], 11], {n, 0, 100}] (* François Marques, Oct 20 2020 *)

a(n) = fromdigits(digits(n), 11); \\ Michel Marcus, Oct 09 2020

def A171397(n): return int(str(n),11) # Chai Wah Wu, Aug 30 2024

Edited by N. J. A. Sloane, Aug 31 2024

A067082 If n = abc...def in decimal notation then the right digit sum function = abc...def + bc...def + c...def + ... + def + ef + f.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 50, 52, 54, 56, 58, 60, 62, 64, 66, 68, 60, 62, 64, 66, 68, 70, 72, 74, 76, 78, 70, 72

Offset: 0

Amarnath Murthy, Jan 05 2002

			a(256) = 256 + 56 + 6 = 318.

Cf. A067079, A067080, A065039.

Table[d = IntegerDigits[n]; rd = 0; While[ Length[d] > 0, rd = rd + FromDigits[d]; d = Drop[d, 1]]; rd, {n, 0, 75} ]

a(abcd) = abcd+(abcd-1000a)+(abcd-1000a-100b)+(abcd-1000a-100b-10c).

n*length(n)-Sum_{k=1..length(n)} 10^k*floor(n/10^k). - Vladeta Jovovic, Jan 08 2002

More terms from Robert G. Wilson v, Jan 07 2002

A337143 Numbers k for which there are only 3 bases b (2, k+1 and another one) in which the digits of k contain the digit b-1.

Original entry on oeis.org

5, 6, 8, 9, 12, 16, 18, 28, 37, 81, 85, 88, 130, 150, 262, 810, 1030, 1032, 4132, 9828, 9832, 10662, 10666, 562576, 562578

Offset: 1

François Marques, Sep 14 2020

This sequence is the list of indices k such that A337496(k)=3.
Conjecture: this sequence is finite and full. a(26) > 3.8*10^12 if it exists.
All terms of this sequence increased by 1 are either prime numbers, or prime numbers squared, or 2 times a prime number because if b is a strict divisor of k+1, the digit for the units in the expansion of k in base b is b-1 so it must be 2 or the third base. In fact k+1 could have been equal to 8=2*4 but 7 is not a term of the sequence (7 = 111_2 = 21_3 = 13_4 = 7_8).

			a(7)=18 because there are only 3 bases (2, 19 and 3) which satisfy the condition of the definition (18=200_3) and 18 is the seventh of these numbers.

Cf. Numbers with at least one digit b-1 in base b : A074940 (b=3), A337250 (b=4), A337572 (b=5), A011539 (b=10), A095778 (b=11).
Cf. Numbers with no digit b-1 in base b: A005836 (b=3), A023717 (b=4), A020654 (b=5), A037465 (b=6), A020657 (b=7), A037474 (b=8), A037477 (b=9), A007095 (b=10), A065039 (b=11).
Cf. A337496, A337535, A337536.

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cp's OEIS Frontend

A065438 Complement of A065039.

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A067080 If n = ab...def in decimal notation then the left digitorial function Ld(n) = ab...def*ab...de*ab...d*...*ab*a.

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A171397 Write n in base 10, but then read it as if it were written in base 11: if n = Sum_{i >= 0} d_i*10^i, with 0 <= d_i <= 9, then a(n) = Sum_{i >= 0} d_i*11^i.

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A067082 If n = abc...def in decimal notation then the right digit sum function = abc...def + bc...def + c...def + ... + def + ef + f.

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Mathematica

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Extensions

A337143 Numbers k for which there are only 3 bases b (2, k+1 and another one) in which the digits of k contain the digit b-1.

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A067080 If n = ab...def in decimal notation then the left digitorial function Ld(n) = ab...defab...deab...d...ab*a.

A171397 Write n in base 10, but then read it as if it were written in base 11: if n = Sum_{i >= 0} d_i10^i, with 0 <= d_i <= 9, then a(n) = Sum_{i >= 0} d_i11^i.