A240588 -id:A240588 - cp's OEIS frontend

A241811 a(1) = 1, a(2) = 0; for n >= 3, a(n) = least number not included earlier that divides the concatenation of all previous terms.

1, 0, 2, 3, 11, 71, 29, 9, 683, 67, 7, 743, 739, 1933, 23, 161, 21, 37, 19, 17, 119, 49, 332534262883, 13, 39, 13739483941387, 83, 111, 79853560395691, 5431567, 70610371, 69, 51, 4112497, 28384496881337963, 353, 77, 1531, 42787, 63, 27, 41, 709, 33, 81, 487, 139697

Offset: 1

Paolo P. Lava, Apr 29 2014

			a(1)=1 and a(2)=0. a(1) U a(2) = 10 and its divisors are 1, 2, 5, 10. Therefore 2 is the least number not yet present in the sequence which divides 10. Again, a(1) U a(2) U a(3) = 102 and its divisors are 1, 2, 3, 6, 17, 34, 51, 102. Therefore a(4)=3, etc.

Chai Wah Wu, Table of n, a(n) for n = 1..68

Cf. A096097, A096098, A240588.

with(numtheory):
T:=proc(t) local x, y; x:=t; y:=0; while x>0 do x:=trunc(x/10); y:=y+1; od; end:
P:=proc(q) local a,b,c,k,n; b:=10; print(1); print(0); c:=[0,1];
for n from 1 to q do a:=sort([op(divisors(b))]); for k from 2 to nops(a) do
if not member(a[k],c) then c:=[op(c),a[k]]; b:=a[k]+b*10^T(a[k]); print(a[k]); break;
fi; od; od; end: P(30);

a(23)-a(28) from Zak Seidov, May 08 2014

a(29)-a(47) from Giovanni Resta, Aug 15 2019

A250745 Start with a(1) = 1; then a(n) = smallest number, not already in the sequence, such that a(n) divides concat(a(1), a(2), ..., a(n)).

Original entry on oeis.org

1, 2, 3, 5, 10, 4, 8, 6, 11, 20, 13, 7, 9, 12, 15, 18, 14, 25, 30, 24, 16, 32, 40, 29, 50, 100, 26, 52, 39, 21, 28, 35, 42, 17, 34, 51, 23, 46, 27, 36, 45, 43, 19, 38, 68, 48, 60, 75, 90, 54, 56, 58, 22, 44, 33, 55, 97, 125, 200, 64, 80, 69, 66, 88, 70, 41, 82

Offset: 1

Paolo P. Lava, Nov 27 2014

Like A171785 but without the constraint a(n) > a(n-1).
Among the first 1000 terms, a(n) = n for n = 1, 2, 3, 15, 170, 577, 759, and the numbers not yet found are 149, 298, 347, 401, 447, 454, 457, 467, 487, 509, etc.
Is this sequence a rearrangement of the natural numbers?

			a(1) = 1;
a(2) = 2 -> 12 /2 = 6;
a(3) = 3 -> 123 / 3 = 41;
Then we cannot use 4 as the next term because 1234 / 4 = 617 / 2.
a(4) = 5 -> 1235 / 5 = 247;
Again, 4, 6, 7, 8 and 9 cannot be used as the next term.
a(5) = 10 -> 123510 / 10 = 12351;
a(6) = 4 -> 1235104 / 4 = 308776;
a(7) = 8 -> 12351048 / 8 = 1543881; etc.

Paolo P. Lava, Table of n, a(n) for n = 1..1000

Cf. A171785, A240588, A241811, A249398, A249399, A250746, A250747.

with(numtheory); P:=proc(q) local a,b,k,n; a:=0; b:={};
for k from 1 to q do for n from 1 to q do if nops({n} intersect b)<1
then if type((a*10^(1+ilog10(n))+n)/n,integer)
then a:=a*10^(1+ilog10(n))+n; b:= b union {n}; print(n); break;
fi; fi; od; od; end: P(10^5);

A250746 Start with a(0) = 0; then a(n) = smallest number > a(n-1) such that a(n) divides concat(a(n), a(n-1), ..., a(0)).

Original entry on oeis.org

0, 1, 2, 3, 5, 10, 15, 18, 19, 35, 42, 51, 55, 70, 85, 93, 95, 106, 155, 217, 310, 745, 1210, 1342, 3355, 5185, 6222, 6330, 9495, 10413, 11115, 12070, 13774, 34435, 41322, 61983, 68870, 1601065116264571, 2217993924228622, 2324778503347862, 2325380783693255

Offset: 0

Paolo P. Lava, Nov 27 2014

This sequence is infinite. - Robert G. Wilson v, Dec 09 2014

			a(0) = 0;
a(1) = 1 -> 10 / 1 = 10;
a(2) = 2 -> 210 / 2 = 105;
a(3) = 3 -> 3210 / 3 = 1070;
Now we cannot use 4 as the next term because 43210 / 4 = 21605 / 2.
a(4) = 5 -> 32105 / 5 = 6421; etc.

Robert G. Wilson v, Table of n, a(n) for n = 0..46

Cf. A171785, A240588, A241811, A249398, A249399, A250745, A250747.

with(numtheory); P:=proc(q) local a,k,n; print(0); print(1); a:=10;
for n from 2 to q do if type((n*10^(1+ilog10(a))+a)/n,integer)
then a:=n*10^(1+ilog10(a))+a; print(n);
fi; od; end: P(10^9);

f[lst_List] := Block[{k = lst[[-1]] + 1, id = FromDigits@ Flatten@ IntegerDigits@ Reverse@ lst}, While[ Mod[ id, k] > 0, k++]; Append[lst, k]]; Nest[f, {0}, 36] (* or *)
f[lst_List] := Block[{mn = lst[[-1]], id = FromDigits@ Flatten@ IntegerDigits@ Reverse@ lst}, d = Divisors@ id; Append[lst, Min@ Select[d, # > mn &]]]; Nest[f, {0, 1}, 36] (* Robert G. Wilson v, Dec 08 2014 *)

a(37)-a(40) from Robert G. Wilson v, Dec 08 2014

A250747 Start with a(0) = 0; then a(n) = smallest number not already in the sequence such that a(n) divides concat(a(n), a(n-1), ..., a(0)).

Original entry on oeis.org

0, 1, 2, 3, 5, 10, 6, 9, 13, 26, 15, 18, 30, 431, 73, 67, 134, 7, 14, 21, 35, 29, 58, 127, 27, 39, 43, 70, 11, 22, 19, 38, 95, 190, 2748070932534311, 2768821759897, 5537643519794, 787, 191, 382, 955, 17, 31, 45, 54, 90, 101, 202, 303, 57, 114, 47, 55, 33, 66

Offset: 0

Paolo P. Lava, Nov 28 2014

Like A250746, but without the constraint a(n) > a(n-1).

			a(0) = 0;
a(1) = 1 -> 10 / 1 = 10;
a(2) = 2 -> 210 / 2 = 105;
a(3) = 3 -> 3210 / 3 = 1070;
Now we cannot use 4 as the next term because 43210 / 4 = 21605 / 2.
a(4) = 5 -> 32105 / 5 = 6421;
Again, we cannot use 4, 6, 7, 8 or 9.
a(5) = 10 -> 1053210 / 10 = 105321.
We still cannot use 4, but 6 is ok.
a(6) = 6 -> 61053210 / 6 = 10175535. Etc.

Jon E. Schoenfield, Table of n, a(n) for n = 0..165

Cf. A171785, A240588, A241811, A249398, A249399, A250745, A250746.

with(numtheory); P:=proc(q) local a,b,k,n; print(0); print(1); a:=10; b:={0,1};
for k from 1 to q do for n from 1 to q do if nops({n} intersect b)<1
then if type((n*10^(1+ilog10(a))+a)/n,integer)
then a:=n*10^(1+ilog10(a))+a; b:= b union {n}; print(n); break; fi; fi;
od; od; end: P(10^5);

More terms from Jon E. Schoenfield, Nov 29 2014

A330291 a(1) = 1; for n > 1, a(n) is the smallest prime divisor of the number formed by the concatenation of a(1) to a(n-1) that has not previously appeared in the sequence.

Original entry on oeis.org

1, 1, 11, 101, 3, 17, 65358901, 7, 109, 10193590070216413001, 359, 4007, 149, 43787, 169031777, 67, 13, 14767, 30323, 151651, 1194421, 31008087256045370298219860945197217288544368471059741857398977350440454626775436666667, 1087, 27457, 19301, 23, 22349

Offset: 1

Scott R. Shannon, Dec 09 2019

The next term a(28) requires the factorization of a composite 195 digit number 49715...700001. Note that for the second term 1 is considered to be prime.

			a(3) = 11 as the concatenation of a(1) and a(2) = '11' and 11, the smallest prime divisor of 11, has not appeared in the sequence.
a(4) = 101 as the concatenation of a(1)..a(3) is '1111' and 101 is the smallest prime divisor of 1111 which has not appeared in the sequence. Note that 11 also divides 1111 but a(3) = 11.
a(6) = 17 as the concatenation of a(1)..a(5) is '11111013' and 17 is the smallest prime divisor of 11111013 which has not appeared in the sequence. Note that 9 also divides 11111013 and has not appeared but only prime divisors are considered.

Tyler Busby, Table of n, a(n) for n = 1..57

Cf. A000040, A020639, A240588, A330290, A330293.

A330290 a(1) = 1; for n > 1, a(n) = the smallest divisor of the number formed by the concatenation of a(1) to a(n-1) that has not previously appeared in the sequence.

Original entry on oeis.org

1, 1, 11, 101, 3, 9, 7, 61, 1821477701, 311, 449, 89, 29, 37, 121, 349, 1047, 73, 2741, 59, 5527, 21, 487, 3679724250117316595527, 137, 257, 33, 99, 27, 47, 17, 19, 13, 39, 63, 23, 557, 53, 159, 117, 351, 81, 3439913, 677, 83, 197, 151, 453, 211, 1033, 239296537198126662281, 167, 501

Offset: 1

Scott R. Shannon, Dec 09 2019

The next term a(54) requires the factorization of a composite 177 digit number 22177...67001.

			a(3) = 11 as the concatenation of a(1) and a(2) = '11' and 11, the smallest divisor of 11, has not appeared in the sequence.
a(4) = 101 as the concatenation of a(1)..a(3) is '1111' and 101 is the smallest divisor of 1111 which has not appeared in the sequence. Note that 11 also divides 1111 but a(3) = 11.
a(6) = 9 as the concatenation of a(1)..a(5) is '11111013' and 9 is the smallest divisor of 11111013 which has not appeared in the sequence. Note that 3 also divides 11111013 but a(5) = 3.

Cf. A020639, A000040, A000005, A330291, A330293, A240588.

A330293 a(1) = 1, a(2) = 2; for n > 2, a(n) = the smallest prime divisor of the number formed by the concatenation of a(1) to a(n-1) that has not previously appeared in the sequence.

Original entry on oeis.org

1, 2, 3, 41, 7, 653, 331, 2536483, 191, 176081, 18307, 2143406938831, 101, 73, 3541, 439, 5665417, 37, 17302849, 86113, 11, 878390431, 2969, 1385625388248048145493629820571541645230648738185397486740279040908468652182116663161996667, 59, 30956837, 181, 151, 159833, 1629097816565791058167, 293, 2063, 3251, 31219483, 13

Offset: 1

Scott R. Shannon, Dec 09 2019

The next term a(36) requires the factorization of a composite 246 digit number 18604...12467.

			a(3) = 3 as the concatenation of a(1) and a(2) = '12' and 3 is the smallest prime divisor of 12 that has not appeared in the sequence.
a(4) = 41 as the concatenation of a(1)..a(3) is '123' and 41 is the smallest prime divisor of 123 which has not appeared in the sequence. Note that 3 also divides 123 but a(3) = 3.
a(6) = 653 as the concatenation of a(1)..a(5) is '123417' and 653 is the smallest prime divisor of 123417 has not appeared in the sequence. Note that 9 also divides 123417 and has not appeared but only prime divisors are considered.

Cf. A020639, A000040, A000005, A330290, A330291, A240588.

A235623 Numbers n for which in the prime power factorization of n!, the numbers of exponents 1 and >1 are equal.

Original entry on oeis.org

0, 1, 4, 7, 8, 9, 13, 19, 20, 21

Offset: 1

Vladimir Shevelev, Apr 20 2014

Number n is in the sequence, if and only if pi(n) = 2*pi(n/2), where pi(x) is the number of primes<=x. Indeed, all primes from interval (n/2, n] appear in prime power factorization of n! with exponent 1, while all primes from interval (0, n/2] appear in n! with exponents >1. However, it follows from Ehrhart's link that, for n>=22, pi(n) < 2*pi(n/2). Therefore, a(9)=21 is the last term of the sequence.

m is in this sequence if and only if the number of prime divisors of [m/2]! equals the number of unitary prime divisors of m! - Peter Luschny, Apr 29 2014

			21! = 2^20*3^9*5^4*7^3*11*13*17*19. Here 4 primes with exponent 1 and 4 primes with exponents >1, so 21 is in the sequence.

Eugene Ehrhart, On prime numbers, Fibonacci Quarterly 26:3 (1988), pp. 271-274.

Cf. A056171, A177329, A177333, A177334, A240537, A240588, A240606, A240619, A240620, A240668, A240669, A240670, A240672, A240695, A240751, A240755, A240764, A240905, A240906, A241123, A241124, A241139, A241148, A241289.

with(numtheory): a := proc(n) factorset(n!); factorset(iquo(n,2)!);
`if`(nops(%% minus %) = nops(%), n, NULL) end: seq(a(n), n=0..30); # Peter Luschny, Apr 28 2014

isok(n) = {f = factor(n!); sum(i=1, #f~, f[i,2] == 1) == sum(i=1, #f~, f[i,2] > 1);} \\ Michel Marcus, Apr 20 2014

A240229 a(n) is the shortest concatenation of the Fibonacci numbers F(1), F(2), ..., divisible by F(n) = A000045(n), n >= 1. a(n) = 0 if there is no such concatenation.

Original entry on oeis.org

1, 1, 112, 11235, 11235, 112, 1123581321345589144233, 11235, 11235813213455891442333776109871597258, 11235813213455891442333776109871597258441816765, 1123581321345589144233377610987159725844181676, 1123581321345589144233377610987159725844181676510946177112865746368

Offset: 1

Wolfdieter Lang, May 10 2014

The corresponding numbers a(n)/F(n) are 1, 1, 56, 3745, 2247, 14, 86429332411199164941, 535, 330465094513408571833346356172694037, 204287512971925298951523201997665404698942123, 12624509228602125216105366415586064335327884, 7802648064899924612731788965188609207251261642437126229950456572, ...
The author's opinion is that this is an example of a not-so-interesting sequence. I call this a WOTS (waste of time sequence). But because I had to write a program to test similar proposed sequences I thought I would apply it to this prominent example.
The next entry a(13) has 324 digits for the divisibility by F(13) = 233 with a(13)/F(13) a 321 digit composite. The given a(n) are all nonprimes.
Question: is there an n with a(n) = 0?

			a(3) = 112 because neither 1 nor 11 are divisible by F(3) = 2, but 112, the concatenation of F(1), F(2) and F(3) is.

Cf. A096098, A096097, A240588.

See the name.

A302687 a(1) = 1; a(2) = 2; then a(n) is the smallest number > a(n-1) such that a(n) divides concat(a(1), a(2), ..., a(n-1)).

Original entry on oeis.org

1, 2, 3, 41, 43, 129, 9567001, 21147541, 22662659, 23817877, 24837187, 28850377, 28872229, 37916473, 48749751, 70416307, 439229167, 834385607, 2270365163, 2278377431, 3751789547, 4433933101, 4810754611, 14432263833, 15632412757, 30530543651, 42441819717, 65591903199, 65857498407

Offset: 1

Daniel Sterman, Apr 11 2018

			a(3) = 3, which makes the concatenation of the first three terms: 123. After 3, the next-highest factor of 123 is 41, so a(4) = 41. The concatenation of the first four terms is then 12341. After 41, the next-highest factor of 12341 is 43, so a(5) = 43.

Compare A240588, in which each term does not need to strictly increase as long as it has not yet appeared in the sequence.
Compare also A171785, in which each term must divide the concatenation of all terms in the sequence including itself.
In A029455, each term divides the concatenation of all smaller positive integers.
In A110740, each term divides the concatenation of all strictly smaller positive integers.

A[1]:= 1: A[2]:= 2: C:= 1:
for n from 3 to 20 do
  C:= A[n-1]+C*10^(ilog10(A[n-1])+1);
  A[n]:= min(select(`>`,numtheory:-divisors(C),A[n-1]))
od:
seq(A[i],i=1..20); # Robert Israel, Apr 12 2018

a(16)-a(20) from Robert Israel, Apr 12 2018

a(21)-a(29) from Daniel Suteu, Apr 12 2018

cp's OEIS Frontend

A241811 a(1) = 1, a(2) = 0; for n >= 3, a(n) = least number not included earlier that divides the concatenation of all previous terms.

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A250745 Start with a(1) = 1; then a(n) = smallest number, not already in the sequence, such that a(n) divides concat(a(1), a(2), ..., a(n)).

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A250746 Start with a(0) = 0; then a(n) = smallest number > a(n-1) such that a(n) divides concat(a(n), a(n-1), ..., a(0)).

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A250747 Start with a(0) = 0; then a(n) = smallest number not already in the sequence such that a(n) divides concat(a(n), a(n-1), ..., a(0)).

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A330291 a(1) = 1; for n > 1, a(n) is the smallest prime divisor of the number formed by the concatenation of a(1) to a(n-1) that has not previously appeared in the sequence.

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A330290 a(1) = 1; for n > 1, a(n) = the smallest divisor of the number formed by the concatenation of a(1) to a(n-1) that has not previously appeared in the sequence.

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A330293 a(1) = 1, a(2) = 2; for n > 2, a(n) = the smallest prime divisor of the number formed by the concatenation of a(1) to a(n-1) that has not previously appeared in the sequence.

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A235623 Numbers n for which in the prime power factorization of n!, the numbers of exponents 1 and >1 are equal.

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A240229 a(n) is the shortest concatenation of the Fibonacci numbers F(1), F(2), ..., divisible by F(n) = A000045(n), n >= 1. a(n) = 0 if there is no such concatenation.

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