A146561 -id:A146561 - cp's OEIS frontend

A034089 Numbers that are proper divisors of the number you get by rotating digits right once.

102564, 128205, 142857, 153846, 179487, 205128, 230769, 102564102564, 128205128205, 142857142857, 153846153846, 179487179487, 205128205128, 230769230769, 1012658227848, 1139240506329, 102564102564102564

Offset: 1

Erich Friedman

Let p(q) denote the period of the fraction q; then sequence is generated by p(i / (10k-1)), k=2,3,4,5,6,7,8,9; k <= i <= 9 and the concatenations of those periods, e.g., p(7/39)=a(5) p(2/19)=a(17).
Example if k=5: p((5+2)/49)=142857 which is in the sequence as the concatenations 142857142857, 142857142857142857, 142857142857142857142857, etc. - Benoit Cloitre, Feb 02 2002
The i in p(i / (10k-1)) is the last digit of the period, while k is equal to the ratio (right-rotated of p)/p. Thus no concatenation of any different such p's can be in the sequence. There are 8*9/2 = 36 terms which are not concatenation of previous terms, the last one being a(124) = 1525423728813559322033898305084745762711864406779661016949 with 58 digits. The term a(3)=p(7/49) is the only period of length (6) different from the length (42) of the other terms corresponding to the same value of k. - M. F. Hasler, Nov 18 2007
Numbers comprising multiple copies of a single digit, e.g., 111111, are not permitted. - Harvey P. Dale, Mar 08 2013
From Emmanuel Vantieghem, Oct 25 2015: (Start)
Subsequence of A245680.
Every element of the sequence is a multiple of 3.
The leading digit of every element is < 5.
(End)

M. F. Hasler, Table of n, a(n) for n = 1..124

Subsequences of this sequence (with quotient k): A146088 (k=2), A146561 (k=3), A146569 (k=4), A146754 (k=5), A291354 (k=6), A291215 (k=7), A291321 (k=8), A291353 (k=9).

Cf. A092697, A097717.

period(p,q,S=[])=until(setsearch(S,p),S=setunion(S,[p]);p=10*p%q);S=[];until(p==S[1],S=concat(S,p);p=10*p%q);S*10\q /* print list of periods, right-rotated and ratio */ rotquo(n,d)={d=divrem(n,10);d[1]+=d[2]*10^#Str(d[1]);[n,d[1],d[1]/n]} for(k=2,9,for(i=k,9,print1( i/(10*k-1),"\t",rotquo(sum(j=1,#p=period(i,k*10-1),p[j]*10^(#p-j))))) /* build the sequence up to the greatest period */ A034089()={local(S=[],p); for(k=2,9,for(i=k,9,S=concat(S,sum(j=1,#p=period(i,k*10-1),p[j]*10^(#p-j))))); S=vecsort(S); for(i=1,#S, for(c=2,58\p=#Str(S[i]), S=concat(S,S[i]*(10^(c*p)-1)/(10^p-1)) )); vecsort(S)} \\ M. F. Hasler, Nov 18 2007

Edited, corrected and extended by M. F. Hasler, Nov 18 2007

A146088 Numbers k with the property that shifting the rightmost digit of k to the left end doubles the number.

Original entry on oeis.org

0, 105263157894736842, 157894736842105263, 210526315789473684, 263157894736842105, 315789473684210526, 368421052631578947, 421052631578947368, 473684210526315789, 105263157894736842105263157894736842, 157894736842105263157894736842105263

Offset: 0

N. J. A. Sloane, based on correspondence from William A. Hoffman III (whoff(AT)robill.com), Apr 10 2009

The sequence is infinite, since repeating 105263157894736842 any number of times (e.g. 105263157894736842105263157894736842) gives another number with the same property.
A number N = 10n+m is in the sequence iff 2N = m*10^d+n, where d is the number of digits of n = [N/10]. This is equivalent to 19n = m(10^d-2), i.e. 10^d=2 (mod 19) and n = m(10^d-2)/19, m=2..9 (to ensure that n has d digits). Thus for each d = 18j-1, j=1,2,3... we have exactly 8 solutions which are the j-fold repetition of one among {a(1),...,a(8)}. - M. F. Hasler, May 04 2009
Normally lists have offset 1, but there are good reasons to make an exception in this case. - N. J. A. Sloane, Dec 24 2012

			The sequence starts with a(0)=0 because rotating a lone 0 does double 0. That initial trivial term was not given an index of 1 when it was added, so that the index of other terms of A146088 would not change and invalidate delicate prior cross-references within OEIS (e.g., A217592) or outside of it.
a(4) = 263157894736842105 because 2*a(4) = 526315789473684210.

Gerard P. Michon, Deriving A146088 from linear decadic equations.

Cf. A092697, A146561, A146569, A146754, A217592.

Subsequence of A034089 (except for the initial 0).

a[n_] := (m = Mod[n - 1, 8] + 2; d = Floor[(n + 7)/8]*18 - 1; ((10/19)*(10^d - 2) + 1)*m); Table[a[n], {n, 0, 10}] (* Jean-François Alcover, Jan 16 2013, after M. F. Hasler *)

A146088(n) = ((10^((n+7)\8*18-1)-2)/19*10+1)*((n-1)%8+2)
/* or a more experimental approach: */ for(d=1,99, Mod(10,19)^k-2 & next; for(m=2,9, print1(",",m*(10^k-2)/19,m))) \\\\ M. F. Hasler, May 04 2009

a(n) = ((10^d-2)/19*10+1)m, where m=(n-1)%8+2 is the trailing digit and d=(n+7)\8*18-1 is the number of other digits. - M. F. Hasler, May 04 2009

a(8k+i) = A217592(9k+i+1)/2 for i=1..8 with any k.

More terms from M. F. Hasler, May 04 2009

a(0) = 0 added by Gerard P. Michon, Oct 29 2012

A146754 Numbers m with the property that shifting the rightmost digit of m to the left end multiplies the number by 5.

Original entry on oeis.org

142857, 142857142857, 142857142857142857, 142857142857142857142857, 142857142857142857142857142857, 142857142857142857142857142857142857, 102040816326530612244897959183673469387755, 122448979591836734693877551020408163265306, 142857142857142857142857142857142857142857

Offset: 1

N. J. A. Sloane, based on correspondence from William A. Hoffman III (whoff(AT)robill.com), Apr 10 2009

From Seiichi Manyama, Aug 22 2017: (Start)
For k >= 1, (10^(6*k) - 1)/7 is a term.
For 5 <= a <= 9 and k >= 1, a*(10^(42*k) - 1)/49 is a term. (End)
For consistency with A146088 (similar for ratio 2), where an initial a(0) = 0 has been added, the same could be considered here. It would be compatible with the formulas given above (with k = 0). - M. F. Hasler, May 03 2025

			From _Seiichi Manyama_, Aug 22 2017: (Start)
a(1) = b1*10 + 7 with b1 = 14285, and 5*a(1) = 714285 = 7*10^5 + b1.
a(7) = b7*10 + 5 with b7 = 10204081632653061224489795918367346938775, and
  5*a(7) = 510204081632653061224489795918367346938775 = 5*10^41 + b7. (End)

Robert Israel, Table of n, a(n) for n = 1..258
Wikipedia, Parasitic number.

Cf. A146088 (k=2), A146561 (k=3), A146569 (k=4), this sequence (k=5), A291354 (k=6), A291215 (k=7), A291321 (k=8), A291353 (k=9).
All these are subsequences of A034089 (except for an initial 0 in some of them).
Cf. A092697, A097717.

f:= proc(d) # solutions with d+1 digits
    local b,R,a;
    R:= NULL;
    for b from ceil(49*10^(d-1)/(10^d - 1)) to 9 do
       a:= (10^d-5)*b/49;
       if a::integer then R:= R, 10*a+b fi
    od;
   R
end proc:
map(f, [$1..42]); # Robert Israel, Nov 05 2024

A146569 Numbers m with the property that shifting the rightmost digit of m to the left end multiplies the number by 4.

Original entry on oeis.org

0, 102564, 128205, 153846, 179487, 205128, 230769, 102564102564, 128205128205, 153846153846, 179487179487, 205128205128, 230769230769, 102564102564102564, 128205128205128205, 153846153846153846, 179487179487179487

Offset: 0

N. J. A. Sloane, based on correspondence from William A. Hoffman III (whoff(AT)robill.com), Apr 10 2009

a(13) <= 102564102564102564. - Donovan Johnson, Jun 06 2009

The condition is equivalent to constraining the numbers to be of the form 10*m+d with a k-digit number m and a nonzero digit d such that 4*(10*m+d) = 10^k * d + m, i.e., 39*m = (10^k - 4)*d. Checking modulo 13, this implies k = 5 (mod 6). Also, m >= 10^(k-1) implies d >= 4. Each such k and d leads to a solution. - Hagen von Eitzen, Jun 26 2009

Wikipedia, Parasitic number.

Cf. A146088 (k=2), A146561 (k=3), this sequence (k=4), A146754 (k=5), A291354 (k=6), A291215 (k=7), A291321 (k=8), A291353 (k=9).
All these are subsequences of A034089 (except for an initial 0 in some of these).
Cf. A092697, A097717.

a(n) = local(r=(n-1)%6+1,k=(n-r)/6);floor((r+3)/39*10^(6*(k+1))) \\ Hagen von Eitzen, Jun 26 2009

If n = 6*k + r with 1 <= r <= 6, then a(n) = (10^(6*k) - 1)/(10^6 - 1)*a(r) as well as a(n) = floor((r + 3)/39*10^(6*(k+1))). - Hagen von Eitzen, Jun 26 2009

a(7)-a(12) from Donovan Johnson, Jun 06 2009
More terms from Hagen von Eitzen, Jun 26 2009
a(0) = 0 prefixed for consistency with A146088 by M. F. Hasler, May 03 2025

A291215 Numbers m with the property that shifting the rightmost digit of m to the left end multiplies the number by 7.

Original entry on oeis.org

1014492753623188405797, 1159420289855072463768, 1304347826086956521739, 10144927536231884057971014492753623188405797, 11594202898550724637681159420289855072463768, 13043478260869565217391304347826086956521739, 101449275362318840579710144927536231884057971014492753623188405797

Offset: 1

Seiichi Manyama, Aug 21 2017

With x = (10^21 - 7)/69 = 14492753623188405797, we have
  a(1) = 7*x*10 + 7, a(2) = 8*x*10 + 8, a(3) = 9*x*10 + 9.
For consistency with A146088 (similar for ratio k=2) and others, where an initial a(0) = 0 has been added, the same could be considered here. It would be compatible with the formula given for a(3k). - M. F. Hasler, May 03 2025

			b = 101449275362318840579.
a(1) = b*10 + 7,
7*a(1) = 7101449275362318840579 = 7*10^21 + b.

Robert Israel, Table of n, a(n) for n = 1..135
Wikipedia, Parasitic number.

Cf. A146088 (k=2), A146561 (k=3), A146569 (k=4), A146754 (k=5), A291354 (k=6), this (k=7), A291321 (k=8), A291353 (k=9).
All these are subsequences of A034089 (except for an initial 0 in some of them).
Cf. A092697, A097717.

seq(seq(y*((10^(22*k)-1)/69),y=7..9),k=1..6); # Robert Israel, Aug 22 2017

From Robert Israel, Aug 22 2017: (Start)
a(3k-2) = 7(10^(22k)-1)/69.
a(3k-1) = 8(10^(22k)-1)/69.
a(3k) = 9(10^(22k)-1)/69.
a(n+6) = (10^22+1) a(n+3) - 10^22 a(n).
G.f.: x*(1304347826086956521739*x^2 + 1159420289855072463768*x + 1014492753623188405797)/(10^22*x^6 - (10^22+1)*x^3 + 1). (End)

A291321 Numbers m with the property that shifting the rightmost digit of m to the left end multiplies the number by 8.

Original entry on oeis.org

1012658227848, 1139240506329, 10126582278481012658227848, 11392405063291139240506329, 101265822784810126582278481012658227848, 113924050632911392405063291139240506329, 1012658227848101265822784810126582278481012658227848

Offset: 1

Seiichi Manyama, Aug 22 2017

Let x = (10^12 - 8)/79 = 12658227848. Then a(1) = 8*x*10 + 8, a(2) = 9*x*10 + 9.

For consistency with A146088 (similar for ratio k=2) and others, where an initial a(0) = 0 has been added, the same could be considered here. It would be compatible with the formula given for a(2k). - M. F. Hasler, May 03 2025

			a(1) = b*10 + 8 with b = 101265822784, and 8*a(1) = 8101265822784 = 8*10^12 + b.

Seiichi Manyama, Table of n, a(n) for n = 1..152
Wikipedia, Parasitic number.

Cf. A146088 (k=2), A146561 (k=3), A146569 (k=4), A146754 (k=5), A291354 (k=6), A291215 (k=7), this sequence (k=8), A291353 (k=9).
All these are subsequences of A034089 (except for an initial 0 in some of them).
Cf. A092697, A097717.

a(n)=10^((n+1)\2*13)\79*(9-n%2) \\ Charles R Greathouse IV, Aug 23 2017

a(2*k - 1) = 8*(10^(13*k) - 1)/79.

a(2*k) = 9*(10^(13*k) - 1)/79.

Edited by M. F. Hasler, May 03 2025

A291353 Numbers m with the property that shifting the rightmost digit of m to the left end multiplies the number by 9.

Original entry on oeis.org

10112359550561797752808988764044943820224719, 1011235955056179775280898876404494382022471910112359550561797752808988764044943820224719

Offset: 1

Seiichi Manyama, Aug 23 2017

For consistency with A146088 (similar for ratio k=2) and others, where an initial a(0) = 0 has been added, the same could be considered here. It would be compatible with the formula given for a(n). - M. F. Hasler, May 03 2025

Seiichi Manyama, Table of n, a(n) for n = 1..22
Wikipedia, Parasitic number.

Cf. A146088 (k=2), A146561 (k=3), A146569 (k=4), A146754 (k=5), A291354 (k=6), A291215 (k=7), A291321 (k=8), this sequence (k=9).
All these are subsequences of A034089 (except for an initial 0 in some of them).
Cf. A092697, A097717.

a(n) = 9*(10^(44*n) - 1)/89.

A291354 Numbers m with the property that shifting the rightmost digit of m to the left end multiplies the number by 6.

Original entry on oeis.org

1016949152542372881355932203389830508474576271186440677966, 1186440677966101694915254237288135593220338983050847457627, 1355932203389830508474576271186440677966101694915254237288

Offset: 1

Seiichi Manyama, Aug 23 2017

For consistency with A146088 (similar for ratio k=2) and others, where an initial a(0) = 0 has been added, the same could be considered here. It would be compatible with the formula given for a(4k). - M. F. Hasler, May 03 2025

Seiichi Manyama, Table of n, a(n) for n = 1..68
Wikipedia, Parasitic number.

Cf. A146088 (k=2), A146561 (k=3), A146569 (k=4), A146754 (k=5), this sequence (k=6), A291215 (k=7), A291321 (k=8), A291353 (k=9).
All these are subsequences of A034089 (except for an initial 0 in some of them).
Cf. A092697, A097717.

a(4*k - 3) = 6*(10^(58*k) - 1)/59.
a(4*k - 2) = 7*(10^(58*k) - 1)/59.
a(4*k - 1) = 8*(10^(58*k) - 1)/59.
a(4*k)     = 9*(10^(58*k) - 1)/59.

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A034089 Numbers that are proper divisors of the number you get by rotating digits right once.

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A146088 Numbers k with the property that shifting the rightmost digit of k to the left end doubles the number.

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A146754 Numbers m with the property that shifting the rightmost digit of m to the left end multiplies the number by 5.

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A146569 Numbers m with the property that shifting the rightmost digit of m to the left end multiplies the number by 4.

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A291215 Numbers m with the property that shifting the rightmost digit of m to the left end multiplies the number by 7.

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A291321 Numbers m with the property that shifting the rightmost digit of m to the left end multiplies the number by 8.

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A291353 Numbers m with the property that shifting the rightmost digit of m to the left end multiplies the number by 9.

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A291354 Numbers m with the property that shifting the rightmost digit of m to the left end multiplies the number by 6.

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