A182134 -id:A182134 - cp's OEIS frontend

A246810 a(n) is the smallest number m such that np(m) = n, where np(m) is number of primes p such that prime(m) < p < prime(m)^(1 + 1/m).

1, 5, 12, 17, 25, 55, 83, 169, 207, 206, 384, 953, 1615, 2192, 2197, 3024, 3023, 10709, 10935, 29509, 29508, 62736, 62735, 94333, 94332, 196966, 314940, 608777, 1258688, 1767259, 2448975, 2448973, 7939362, 9373136, 9373134, 16854966, 16854967

Offset: 1

Farideh Firoozbakht and Jahangeer Kholdi, Oct 10 2014

nonn

Firoozbakht's conjecture says that for every n, there exists at least one prime p where, prime(n) < p < prime(n)^(1 + 1/n). Hence if Firoozbakht's conjecture is true, then there is no m such that np(m) = 0.
Conjecture: For every positive integer n, a(n) exists.
a(65) > 10^12. - Robert Price, Nov 12 2014

			a(6) = 55 since the number of primes p such that prime(55) < p < prime(55)^(1 + 1/55) is 6 and 55 is the smallest number with this property.

Robert Price, Table of n, a(n) for n = 1..64
A. Kourbatov, Verification of the Firoozbakht conjecture for primes up to four quintillion, arXiv:1503.01744 [math.NT], 2015
A. Kourbatov, Upper bounds for prime gaps related to Firoozbakht's conjecture, J. Int. Seq. 18 (2015) 15.11.2
Nilotpal Kanti Sinha, On a new property of primes that leads to a generalization of Cramer's conjecture, arXiv:1010.1399v2 [math.NT], 2010.
Wikipedia, Firoozbakht's conjecture.

Cf. A000040, A182134, A246781, A246782, A246783, A246787.

np[n_]:=(b=Prime[n]; Length[Select[Range[b+1, b^(1 + 1/n)],PrimeQ]]); a[n_]:=(For[m=1, np[m] !=n, m++]; m);
Do[Print[a[n]], {n, 37}]

A245722 Row products of table A244365.

Original entry on oeis.org

3, 5, 7, 11, 221, 323, 437, 23, 899, 1147, 1517, 82861, 107113, 2491, 3127, 241133, 21182921, 347261, 33984931, 478661, 583573, 7387, 8633, 107972737, 13710311357, 135745657, 1317919, 12317, 14351, 16637, 2494633, 428448457, 490995677, 3532343, 645328247

Offset: 1

Reinhard Zumkeller, Nov 18 2014

nonn

a(n) = Product_{k=1..A182134(n)} A244365(n,k);
A020639(a(n)) = A000040(n+1);
A006530(a(n)) = A245396(n);
A001221(a(n)) = A001222(a(n)) = A182134(n); A008966(a(n)) = 1;
A001221(GCD(a(n),a(n+1))) = A182134(n) - 1.

Reinhard Zumkeller, Table of n, a(n) for n = 1..1000

Cf. A244365, A245396, A182134, A020639, A006530, A001221, A001222, A008966.

Haskell
```
a245722 = product . a244365_row
```

A246790 Indices of record values in A246785.

Original entry on oeis.org

2, 7, 15, 28, 53, 90, 177, 213, 216, 393, 395, 1628, 2206, 2212, 7075, 10364, 10727, 10954, 43444, 46099, 62758, 94347, 196989, 214629, 214631, 608803, 889740, 1353804, 2010048, 2449005, 7939395, 9373169, 16855002, 16855008, 41084086, 80005684, 113726074

Offset: 1

Farideh Firoozbakht, Oct 24 2014

nonn

The record values up to n = 56000: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 14, 15, 16, 17, 18, 19, 20, 21. Note that 13 is not a record value.
See A252474 for additional record values.
a(55) > 3.5*10^11. - Robert Price, Dec 17 2014

			a(13) = 2206 and A246785(2206) = 14. Note that 13 is not a record value.

Robert Price, Table of n, a(n) for n = 1..54

Cf. A000040, A182134, A246785, A252474.

a(21)-a(54) from Robert Price, Dec 17 2014

A246783 Smallest number m such that all the n numbers np(m+k-1), 0 < k < n+1 are equal, where np(t) is number of primes p with prime(t) < p < prime(t)^(1+1/t).

Original entry on oeis.org

1, 1, 1, 1, 67, 67, 67, 67, 67, 54412, 161342, 161342, 1214143, 9915018, 9915018, 68964006, 68964006, 810832784, 19867608968, 52415066804, 119937255921, 272007811177

Offset: 1

Farideh Firoozbakht, Oct 13 2014

np(n) = A182134(n).
Conjecture: For every n, a(n) exists.
The np values for distinct terms of the sequence are: 1, 5, 11, 11, 16, 14, ... . It is interesting that for n = 1, 5, 11 & 14, np(a(n)) = n. What is the next term of this sequence (1, 5, 11, 14, ?, ... .)?
The sequence giving the np values of this sequence is A245101.
The sequence {1, 5, 11, 11, 16, 14} commented on above is expanded in A245098.
a(23) > 10^12. - Robert Price, Nov 12 2014
The next term of the sequence (1, 5, 11, 14, ?, ... .) commented on above is >22 and corresponds to t>10^12. - Robert Price, Nov 12 2014

			a(9) = 67, since all the nine numbers np(67), np(67+1), np(67+2), ..., np(67+8) are equal and 67 is the smallest such number. Note that np(67) = 5.

Cf. A000040, A182134.

np[n_]:=Length[Select[Range[Prime[n]+1,Prime[n]^(1+1/n)], PrimeQ]];a[n_]:=(For[m=1,Length[Union[Table[np[m+k-1],{k,n}]]]!=1,m++];m);Do[Print[a[n]],{n,15}]

a(16)-a(22) from Robert Price, Nov 12 2014

A246792 Smallest number m such that for 0 < k < n+1, np(m+k-1) = np(m)-k+1, where np(t) is number of primes p with prime(t) < p < prime(t)^(1 + 1/t).

Original entry on oeis.org

1, 7, 25, 25, 181, 208, 208, 1867, 14345, 19609, 40918, 40918, 620326, 2552265, 2552265, 7225612, 7225612, 16679492, 33772734, 33772734, 33772734, 620326386, 1516416904, 1516416904, 4764006481, 5272314878, 21423652192

Offset: 1

Farideh Firoozbakht, Oct 16 2014

nonn

np(m) = A182134(m).
According to the definition, numbers np(a(n)), np(a(n)+1), ..., np(a(n)+n-2) and np(a(n)+n-1) are n consecutive numbers in descending order.
a(34) > 10^12. - Robert Price, Dec 07 2014
See A251736 for the corresponding values of np.

			a(15) = 2552265, since np(2552265) = 24, np(2552265+1) = 23 , ..., np(2552265+13) = 11, np(2552265+14) = 10 are 15 consecutive numbers in descending order.

Robert Price, Table of n, a(n) for n = 1..33

Cf. A000040, A182134, A246791, A251736.

np[t_] := np[t] = Length[Select[Range[Prime[t]+1,Prime[t]^(1+1/t)],PrimeQ]]; a[1]=1; a[n_] := a[n] = (For[m = a[n-1],c = Table[np[m+k-1],{k,n}]; c !=  Reverse[Range[Min[c], Max[c]]], m++]; m); Do[Print[a[n]],{n,15}]

np(n) = primepi(prime(n)^(1+1/n))-n;
isok(m, n) = {for (k=1, n, if (np(m+k-1) != np(m)-k+1, return (0));); return (1);}
a(n) = {m = 1; while (! isok(m, n), m++); m;} \\ Michel Marcus, Dec 07 2014

a(18)-a(33) from Robert Price, Dec 07 2014

A246791 Smallest number m such that for 0 <= k < n, np(m+k) = np(m)+k, where np(t) is number of primes p with prime(t) < p < prime(t)^(1 + 1/t).

Original entry on oeis.org

1, 4, 15, 136, 2128, 15453, 479403, 1184231, 10975072, 27112368, 175600366, 2304656281, 14896902677, 59331462112

Offset: 1

Farideh Firoozbakht, Oct 16 2014

np(m) = A182134(m).
According to the definition, numbers np(a(n)), np(a(n)+1), ..., np(a(n)+n-2), np(a(n)+n-1) are n consecutive numbers in ascending order.
a(15) > 10^12. - Robert Price, Nov 19 2014
See A247116 for the corresponding values of np.

			a(8) = 1184231 since np(1184231) = 17, np(1184231+1) = 18, ..., np(1184231+6) = 23, np(1184231+7) = 24 are 8 consecutive numbers and 1184231 is the smallest number with this property.

Cf. A000040, A182134, A246792, A247116.

np[n_] := np[n] = Length[Select[Range[Prime[n]+1, Prime[n]^(1 + 1/n)], PrimeQ]]; a[1]=1; a[n_] := a[n] = (For[m = a[n-1], c = Table[np[m+k], {k,0,n-1}]; c != Range[Min[c], Max[c]], m++]; m); Do[Print[a[n]],{n, 8}]

np(t) = primepi(prime(t)^(1 + 1/t)) - t;
ok(m, n) = {for (k=0, n-1, if (np(m+k) != np(m)+k, return(0));); return (1);}
a(n) = {m = 1; while (! ok(m, n), m++); m;} \\ Michel Marcus, Nov 25 2014

a(9)-a(14) from Robert Price, Nov 19 2014

A246794 a(n) = A246785(A005669(n)).

Original entry on oeis.org

1, 1, 1, 2, 2, 3, 4, 4, 3, 5, 7, 5, 5, 4, 4, 3, 4, 4, 4, 5, 5, 5, 4, 7, 11, 8, 7, 8, 5, 10, 3, 4, 6, 9, 6, 8, 12, 8, 4, 10, 6, 9, 10, 7, 9, 4, 6, 7, 10, 7, 8, 5, 8, 10, 9, 8, 8, 4, 6, 7, 8, 9, 2, 10, 7, 6, 9, 8, 6, 4, 4, 7, 4, 6

Offset: 2

Farideh Firoozbakht, Oct 25 2014

Conjecture: For every n, a(n) is positive.

a(64) = 2 and A246795(64) = 30 hence A182134(49749629143526 - k) = k for 1 < k < 31, where 49749629143526 = A005669(64).

Cf. A000040, A002386, A005669, A182134, A246785, A246795.

A246795 a(n) = A246793(A005669(n)).

Original entry on oeis.org

1, 1, 2, 3, 5, 5, 7, 8, 9, 8, 12, 10, 11, 12, 9, 12, 15, 10, 11, 15, 10, 17, 13, 16, 20, 17, 21, 19, 19, 24, 13, 23, 18, 21, 17, 26, 26, 29, 19, 23, 23, 25, 27, 27, 29, 24, 26, 25, 33, 28, 32, 32, 31, 36, 35, 37, 42, 33, 28, 35, 30, 36, 30, 34, 44, 29, 32, 34, 33, 30, 40, 38, 34, 44

Offset: 2

Farideh Firoozbakht, Oct 24 2014

This sequence evaluates the largest m's defined in A246793 for the index of primes where largest gaps occurs.
Conjecture: For every n, a(n) is positive.
a(75) = 44 and A246794(75) = 6 hence A182134(34952141021660495 - k) = k for 5 < k < 45, where 34952141021660495 = A005669(75).

Cf. A000040, A002386, A005669, A182134, A246785, A246794.

A252474 Record values in A246785.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 32, 33, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 52, 53, 55, 57, 59, 60, 61

Offset: 1

Robert Price, Dec 17 2014

For all known terms of this sequence, a(n+1) < a(n)+3; is this true for all terms? - Farideh Firoozbakht, Dec 20 2014
Integers not in this sequence begin: 13, 31, 34, 51, 54, 56, 58.
The comment above could also be phrased as "no two consecutive numbers in the complement" or "no gaps larger than one". But one sees that the missing numbers become more frequent towards larger values, so the answer to the question might well be "no". - Robert Price, Jan 06 2015

Cf. A182134, A246785, A246790.

A252475 Indices of record values in A246793.

Original entry on oeis.org

2, 7, 15, 21, 30, 61, 90, 177, 216, 395, 965, 1628, 2206, 2212, 3040, 10727, 10954, 29529, 62758, 94357, 196992, 314967, 608805, 1258717, 1767289, 2449005, 7939395, 9373169, 16855002, 16855004, 32881951, 41084088, 83715358, 90288095, 151449068, 315082046

Offset: 1

Robert Price, Dec 17 2014

nonn

a(51) > 3.5*10^11.

Robert Price, Table of n, a(n) for n = 1..50

Cf. A000040, A182134, A246785, A252474, A246793.

cp's OEIS Frontend

A246810 a(n) is the smallest number m such that np(m) = n, where np(m) is number of primes p such that prime(m) < p < prime(m)^(1 + 1/m).

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A245722 Row products of table A244365.

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A246790 Indices of record values in A246785.

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A246783 Smallest number m such that all the n numbers np(m+k-1), 0 < k < n+1 are equal, where np(t) is number of primes p with prime(t) < p < prime(t)^(1+1/t).

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A246792 Smallest number m such that for 0 < k < n+1, np(m+k-1) = np(m)-k+1, where np(t) is number of primes p with prime(t) < p < prime(t)^(1 + 1/t).

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A246791 Smallest number m such that for 0 <= k < n, np(m+k) = np(m)+k, where np(t) is number of primes p with prime(t) < p < prime(t)^(1 + 1/t).

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A246794 a(n) = A246785(A005669(n)).

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A246795 a(n) = A246793(A005669(n)).

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A252474 Record values in A246785.

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