keyword:new

A381323

a(n) is the least nondecreasing prime p such that n is the number of primes between p (not included) and p+n*log(p).

+0
0

11, 37, 37, 59, 59, 59, 79, 79, 137, 151, 229, 229, 347, 367, 373, 379, 379, 397, 397, 571, 571, 571, 587, 587, 587, 587, 587, 587, 587, 587, 853, 853, 853, 877, 877, 877, 877, 877, 967, 967, 1009, 1009, 1019, 1021, 1021, 1277, 1297, 1297, 1361, 1361, 1361, 1361, 1361, 1361, 1361

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,1

LINKS

Table of n, a(n) for n=1..55.

FORMULA

a(n) ~ n^(3/2) as n -> oo.

Limit_{N->oo} (Sum_{n=2..N} log(a(n))) / (Sum_{n=2..N} log(n)) = 3/2.

EXAMPLE

a(1) = 11 since 11 is the first prime for which we find the only prime 13 between 11 and 11+1*log(11)=13.39.. .

a(2) = 37 since 37 is the second prime for which we find the two primes 41 and 43 between 37 and 37+2*log(37)=44.22.. .

a(9) = 137 since 137 is the 9th prime (>= a(8)) for which we find 9 primes between 137 and 137+9*log(137)~181.29.

PROG

(PARI) Pdeb=2; for(K=1, 40, forprime(P=Pdeb, 40^2, AR=P+K*log(P); NPR=primepi(AR)-primepi(P); if(NPR==K, print1(P, ", "); Pdeb=P; break())));

CROSSREFS

Cf. A381293.

KEYWORD

nonn,new

AUTHOR

Alain Rocchelli, Feb 21 2025

STATUS

approved

A381488

Pentagonal numbers that are deficient.

+0
0

1, 5, 22, 35, 51, 92, 117, 145, 247, 287, 376, 425, 477, 590, 651, 715, 782, 925, 1001, 1162, 1247, 1335, 1426, 1617, 1717, 2035, 2147, 2501, 2625, 2882, 3015, 3151, 3577, 3725, 4187, 4347, 4845, 5017, 5551, 5735, 6112, 6305, 6501, 6902, 7107, 7315, 7526, 7957

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,2

LINKS

Table of n, a(n) for n=1..48.

EXAMPLE

22 = 2*11 is the 4th pentagonal number and is a deficient number, since it is larger than the sum of its proper divisors (14).

117 = 3^2*13 is the 9th pentagonal number and is a deficient number, since it is larger than the sum of its proper divisors (65).

1001 = 7*11*13 is the 26th pentagonal number and is a deficient number, since it is larger than the sum of its proper divisors (343).

MATHEMATICA

Select[Table[n*(3*n-1)/2, {n, 1, 75}], DivisorSigma[-1, #] < 2 &] (* Amiram Eldar, Feb 25 2025 *)

PROG

(PARI) select(x->sigma(x)<2*x, vector(100, k, k*(3*k-1)/2)) \\ Michel Marcus, Feb 25 2025

CROSSREFS

Intersection of A005100 and A000326.

Cf. A379264.

KEYWORD

nonn,new

AUTHOR

Massimo Kofler, Feb 25 2025

STATUS

approved

A381189

Ulam numbers that are squarefree semiprimes.

+0
0

6, 26, 38, 57, 62, 69, 77, 82, 87, 106, 145, 155, 177, 206, 209, 219, 221, 253, 309, 319, 339, 341, 358, 382, 451, 485, 497, 502, 566, 685, 695, 734, 781, 849, 866, 893, 905, 949, 1018, 1037, 1079, 1081, 1101, 1157, 1167, 1169, 1186, 1191, 1257, 1313, 1355, 1387, 1389

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,1

COMMENTS

Ulam numbers: a(1) = 1; a(2) = 2; for n>2, a(n) = least number > a(n-1) which is a unique sum of two distinct earlier terms.

Number such as 4, 2809, 3481, 6889, etc., are in A378795 but not in this or in a longer sequence.

LINKS

Table of n, a(n) for n=1..53.

EXAMPLE

4 is not a term of this sequence because 4=2^2 is the product of 2 not distinct primes even if 4 is an Ulam number.

6 is a term because 6=2*3 is the product of 2 distinct primes and 6 is an Ulam number.

57 is a term because 57=3*19 is the product of 2 distinct primes and 57 is an Ulam number.

2809 is not a term of a longer sequence because 2809=53^2 is the product of 2 not distinct primes even if 2809 is an Ulam number.

MATHEMATICA

seq[numUlams_] := Module[{ulams = {1, 2}}, Do[AppendTo[ulams, n = Last[ulams]; While[n++; Length[DeleteCases[Intersection[ulams, n - ulams], n/2, 1, 1]] != 2]; n], {numUlams}]; Select[ulams, FactorInteger[#][[;; , 2]] == {1, 1} &]]; seq[160] (* Amiram Eldar, Feb 16 2025, after Jean-François Alcover at A002858 *)

CROSSREFS

Intersection of A006881 and A002858.

Cf. A068820, A378795, A379162, A379532.

KEYWORD

nonn,new

AUTHOR

Massimo Kofler, Feb 16 2025

STATUS

approved

A380882

Centered square numbers which are sphenic numbers.

+0
0

1105, 2665, 3445, 7565, 8845, 14965, 15665, 16745, 17485, 18241, 20605, 22685, 23545, 27145, 28085, 32005, 32513, 35113, 37265, 48985, 50245, 50881, 55445, 56785, 62305, 71065, 74885, 78013, 80401, 81205, 84461, 85285, 88621, 89465, 109045, 111865, 113765, 116645, 118585, 119561

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,1

COMMENTS

a(n) == 1 (mod 4).

LINKS

Table of n, a(n) for n=1..40.

EXAMPLE

1105 = 5 * 13* 17 which is the 24-th centered square number.

18241 = 17 * 29 * 37 which is the 96-th centered square number.

32513 = 13 * 41 * 61 which is 128-th centered square number.

MATHEMATICA

Select[Table[2*n*(n+1) + 1, {n, 0, 250}], FactorInteger[#][[;; , 2]] == {1, 1, 1} &] (* Amiram Eldar, Feb 07 2025 *)

CROSSREFS

Intersection of A007304 and A001844.

Cf. A027862, A371016, A370795.

KEYWORD

nonn,new

AUTHOR

Massimo Kofler, Feb 07 2025

STATUS

approved

A381169

List of twin prime averages (A014574) is partitioned by including as many elements as possible in the n-th partition, L_n, such that any gap in L_n is smaller than the gap between L_n and L_(n-1) but not bigger than the first gap in L_n. a(n) is the number of elements in L_n.

+0
0

1, 1, 1, 1, 1, 1, 2, 2, 2, 3, 2, 2, 1, 1, 2, 1, 6, 3, 2, 2, 2, 1, 1, 5, 2, 2, 2, 3, 1, 2, 2, 2, 2, 3, 2, 2, 1, 2, 4, 2, 2, 2, 2, 5, 2, 2, 1, 1, 1, 3, 2, 2, 1, 3, 3, 2, 1, 4, 2, 3, 2, 2, 1, 2, 2, 3, 3, 1, 3, 2, 1, 2, 1, 1, 2, 3, 3, 1, 1, 2, 2, 3, 2, 2, 1, 5, 2

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,7

COMMENTS

The partition method used here is the same as that in A348168.

Conjecture 1: lim_{n->oo} N_i/n = k_i, where N_i is the number of partitions with i elements and k_i is a constant, with k_2 > k_1 > k_3 > k_4 > .... The values of k_i are the same as those in A348168.

Conjecture 2: lim_{n->oo} Sum_{1..n} a(n)/n = lim_{i->oo} Sum_{1..i} i*k_i = e, or the average partition length approaches 2.71828... as n tends to infinity.

Numbers of twin prime pairs (N) and partitions with 1 through 6 twin prime pairs for n up to 10000000 are given in the table below.

n N N_1 N_2 N_3 N_4 N_5 N_6

-------- -------- ------- ------- ------- ------ ------ ------

1 1 1 0 0 0 0 0

10 15 6 3 1 0 0 0

100 209 30 45 16 5 3 1

1000 2536 286 416 145 64 29 19

10000 26474 2851 4331 1271 544 311 190

100000 271338 28034 43375 12923 5731 3002 1870

1000000 2725126 281837 434234 128190 56563 30074 18171

10000000 27120107 2815831 4352926 1276953 563128 302256 181612

LINKS

Table of n, a(n) for n=1..87.

EXAMPLE

Twin prime pair averages in the first 10 partitions are: [4], [6], [12], [18], [30], [42], [60, 72], [102, 108], [138, 150], and [180, 192, 198]. Thus, a(1) = a(2) = a(3) = a(4) = a(5) = a(6) = 1, a(7) = a(8) = a(9) = 2, and a(10) = 3.

PROG

(Python)

from sympy import isprime, nextprime; L = [4]

def nexttwin(x):

p1 = nextprime(x); t1 = p1 + 2

while isprime(t1) == 0: p1 = nextprime(t1); t1 = p1 + 2

return p1+1

for _ in range(2, 89):

print(len(L), end = ', ')

t0 = L[-1]; t1 = nexttwin(t0); g0 = t1 - t0; M = [t1]; t = nexttwin(t1); g1 = t - t1

while g1 < g0 and t - t1 <= g1: M.append(t); t1 = t; t = nexttwin(t)

L = M

CROSSREFS

Cf. A001097, A014574, A348168.

KEYWORD

nonn,new

AUTHOR

Ya-Ping Lu, Feb 15 2025

STATUS

approved

A380877

Primes p where the prime race 12m+1 versus 12m+7 is tied.

+0
0

2, 3, 5, 13, 17, 433, 457, 461

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,1

COMMENTS

Primes p such that pi_{12,1}(p) = pi_{12,7}(p), where pi_{m,a}(x) is the number of primes <= x which are congruent to a (mod m). For the first 5 billion primes, pi_{12,7}(p) >= pi_{12,1}(p). If exists, a(9) > 122430513841.

LINKS

Table of n, a(n) for n=1..8.

PROG

(Python)

from sympy import nextprime; p, d = 2, 0

while p < 500:

if d == 0: print(p, end = ', ')

p = nextprime(p); r = p%12

if r == 7: d += 1

elif r == 1: d -= 1

CROSSREFS

Cf. A007351, A068228, A068229, A379989, A380333.

KEYWORD

nonn,new

AUTHOR

Ya-Ping Lu, Feb 06 2025

STATUS

approved

A380817

a(n) = A380245(A379343(n)).

+0
0

1, 2, 3, 4, 5, 6, 9, 10, 7, 8, 11, 12, 13, 14, 15, 20, 21, 18, 19, 16, 17, 22, 23, 24, 25, 26, 27, 28, 35, 36, 33, 34, 31, 32, 29, 30, 37, 38, 39, 40, 41, 42, 43, 44, 45, 54, 55, 52, 53, 50, 51, 48, 49, 46, 47, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 77, 78, 75, 76, 73, 74, 71, 72

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,2

COMMENTS

This sequence can be regarded as a triangular array read by rows. Each row is a permutation of a block of consecutive numbers; the blocks are disjoint and every positive number belongs to some block. The length of row n is 4n-3 = A016813(n+1), n > 0.

The sequence can also be regarded as a table read by upward antidiagonals. For n>1, row n joins two consecutive antidiagonals.

The sequence is a self-inverse permutation of the positive integers.

In particular, the initial {a(1),a(2), ... a(A000384(n+1))} is self-inverse.

The sequence is an intra-block permutation of the positive integers.

Generalization of the Cantor numbering method.

LINKS

Boris Putievskiy, Table of n, a(n) for n = 1..9730

Boris Putievskiy, Integer Sequences: Irregular Arrays and Intra-Block Permutations, arXiv:2310.18466 [math.CO], 2023.

Index entries for sequences that are permutations of the natural numbers.

FORMULA

T(n,k) for 1 <= k <= 4n - 3: T(n,k) = A000384(n-1) + P(n,k), P(n, k) = -k + m(n) - 1 if k < m(n) and k mod 2 = 1, P(n, k) = -k + m(n) + 1 if k < m(n) and k mod 2 = 0, P(n, k) = k if k >= m(n), where m(n) = 2*n - 1.

EXAMPLE

Triangle array begins:

k= 1 2 3 4 5 6 7 8 9

n=1: 1;

n=2: 2, 3, 4, 5, 6;

n=3: 9, 10, 7, 8, 11, 12, 13, 14, 15;

For n > 1, each row of triangle array joins two consecutive upward antidiagonals in the table:

1, 3, 6, 8, 15, ...

2, 5, 7, 14, 16, ...

4, 10, 13, 19, 26, ...

9, 12, 18, 25, 31, ...

11, 21, 24, 34, 41, ...

...

Subtracting (n-1)*(2*n-3) from each term in row n produces a permutation of numbers from 1 to 4*n-3:

1,

1, 2, 3, 4, 5;

3, 4, 1, 2, 5, 6, 7, 8, 9;

All permutations are self-inverse.

MATHEMATICA

T[n_, k_]:=(n-1)*(2*n-3)+Module[{m=2*n-1}, If[k<m, If[OddQ[k], -k+m-1, -k+m+1], k]]

Nmax=3; Flatten[Table[T[n, k], {n, 1, Nmax}, {k, 1, 4 n-3}]]

CROSSREFS

Cf. A016813 (row lengths), A000384, A378684, A380245, A379343.

KEYWORD

nonn,tabf,new

AUTHOR

Boris Putievskiy, Feb 04 2025

STATUS

approved

A380815

a(n) = A378684(A379343(n)).

+0
0

1, 5, 2, 4, 3, 6, 12, 9, 14, 7, 11, 10, 13, 8, 15, 23, 20, 25, 18, 27, 16, 22, 21, 24, 19, 26, 17, 28, 38, 35, 40, 33, 42, 31, 44, 29, 37, 36, 39, 34, 41, 32, 43, 30, 45, 57, 54, 59, 52, 61, 50, 63, 48, 65, 46, 56, 55, 58, 53, 60, 51, 62, 49, 64, 47, 66, 80, 77, 82, 75, 84, 73, 86, 71

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,2

COMMENTS

This sequence can be regarded as a triangular array read by rows. Each row is a permutation of a block of consecutive numbers; the blocks are disjoint and every positive number belongs to some block. The length of row n is 4n-3 = A016813(n+1), n > 0.

The sequence can also be regarded as a table read by upward antidiagonals. For n>1, row n joins two consecutive antidiagonals.

The sequence is an intra-block permutation of the positive integers.

Generalization of the Cantor numbering method.

LINKS

Boris Putievskiy, Table of n, a(n) for n = 1..9730

Boris Putievskiy, Integer Sequences: Irregular Arrays and Intra-Block Permutations, arXiv:2310.18466 [math.CO], 2023.

Index entries for sequences that are permutations of the natural numbers.

FORMULA

(a(1),a(2), ... a(A000384(n+1)) is permutation of the positive integers from 1 to A000384(n+1). ord (a(1),a(2), ... a(A000384(n+1)) = 3, where ord is the order of the permutation.

T(n,k) for 1 <= k <= 4n - 3: T(n,k) = A000384(n-1) + P(n,k), P(n, k) = k + m(n) if k < m(n) and k mod 2 = 1, P(n, k) = -k + m(n) if k < m(n) and k mod 2 = 0, P(n, k) = k if k >= m(n) and k mod 2 = 1, P(n, k) = -k + 2 * m(n) if k >= m(n) and k mod 2 = 0, Where m(n) = 2n - 1.

EXAMPLE

Triangle array begins:

k= 1 2 3 4 5 6 7 8 9

n=1: 1;

n=2: 5, 2, 4, 3, 6;

n=3: 12, 9, 14, 7, 11, 10, 13, 8, 15;

ord(1,5,2... 10,15) = 3.

For n > 1, each row of triangle array joins two consecutive upward antidiagonals in the table:

1, 2, 6, 7, 15, ...

5, 3, 14, 8, 27, ...

4, 9, 13, 18, 26, ...

12, 10, 25, 19, 42, ...

11, 20, 24, 33, 41, ...

...

Subtracting (n-1)*(2*n-3) from each term in row n produces a permutation of numbers from 1 to 4*n-3:

1;

4, 1, 3, 2, 5;

6, 3, 8, 1, 5, 4, 7, 2, 9;

MATHEMATICA

T[n_, k_]:=(n-1)*(2*n-3)+Module[{m=2*n-1}, If[k<m, If[OddQ[k], k+m, -k+m], If[OddQ[k], k, -k+2*m]]]

Nmax=3; Flatten[Table[T[n, k], {n, 1, Nmax}, {k, 1, 4 n-3}]]

CROSSREFS

Cf. A016813 (row lengths), A000384, A378684, A379343.

KEYWORD

nonn,tabf,new

AUTHOR

Boris Putievskiy, Feb 04 2025

STATUS

approved

A376214

a(n) = A379343(A379342(n)).

+0
0

1, 2, 3, 4, 5, 6, 7, 10, 9, 8, 11, 14, 13, 12, 15, 16, 21, 18, 19, 20, 17, 22, 27, 24, 25, 26, 23, 28, 29, 36, 31, 34, 33, 32, 35, 30, 37, 44, 39, 42, 41, 40, 43, 38, 45, 46, 55, 48, 53, 50, 51, 52, 49, 54, 47, 56, 65, 58, 63, 60, 61, 62, 59, 64, 57, 66, 67, 78, 69, 76, 71, 74, 73, 72

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,2

COMMENTS

This sequence can be regarded as a triangular array read by rows. Each row is a permutation of a block of consecutive numbers; the blocks are disjoint and every positive number belongs to some block. For n > 0, the length of row n is 4n-3 = A016813(n+1).

The sequence is a self-inverse permutation of the positive integers.

In particular, the initial {a(1),a(2), ... a(A000384(n+1))} is self-inverse.

The sequence is an intra-block permutation of the positive integers.

Generalization of the Cantor numbering method.

LINKS

Boris Putievskiy, Table of n, a(n) for n = 1..9730

Boris Putievskiy, Integer Sequences: Irregular Arrays and Intra-Block Permutations, arXiv:2310.18466 [math.CO], 2023.

Index entries for sequences that are permutations of the natural numbers.

FORMULA

a(n) = A380245(A378684(n)).

T(n,k) for 1 <= k <= 4n - 3: T(n,k) = A000384(n-1) + P(n,k), P(n, k) = k if k < m(n) and k mod 2 = 1, P(n, k) = -k + m(n) + 1 if k < m(n) and k mod 2 = 0, P(n, k) = k if k >= m(n) and k mod 2 = 1, P(n, k) = -k + 3 * m(n) - 1 if k >= m(n) and k mod 2 = 0, where m(n) = 2n - 1.

EXAMPLE

Triangle array begins:

k= 1 2 3 4 5 6 7 8 9

n=1: 1;

n=2: 2, 3, 4, 5, 6;

n=3: 7, 10, 9, 8, 11, 14, 13, 12, 15;

(1,2,3... 12,15) = (1,2,3... 12,15)^(-1).

(1,2,3... 12,15) = (1,5,2... 10,15)(1,3,5, ...7,15). The first permutation is from

Example A380245 and the second from Example A378684.

For n > 1, each row of triangle array joins two consecutive upward antidiagonals in the table:

1, 3, 6, 8, 15, ...

2, 5, 9, 12, 20, ...

4, 10, 13, 19, 26, ...

7, 14, 18, 25, 33, ...

11, 21, 24, 34, 41, ...

...

Subtracting (n-1)*(2*n-3) from each term in row n produces a permutation of numbers from 1 to 4*n-3:

1,

1, 2, 3, 4, 5;

1, 4, 3, 2, 5, 8, 7, 6, 9;

MATHEMATICA

T[n_, k_]:=(n-1)*(2*n-3)+Module[{m=2*n-1}, If[k<m, If[OddQ[k], k, -k+m+1], If[OddQ[k], k, -k+3*m-1]]]

Nmax=3; Flatten[Table[T[n, k], {n, 1, Nmax}, {k, 1, 4 n-3}]]

CROSSREFS

Cf. A016813 (row lengths), A000384, A378684, A380245, A379342, A379343.

KEYWORD

nonn,tabf,new

AUTHOR

Boris Putievskiy, Feb 05 2025

STATUS

approved

A381464

Lexicographically earliest positive integer sequence satisfying a(n) = a(a(n))/n.

+0
0

1, 3, 6, 5, 20, 18, 8, 56, 10, 90, 12, 132, 14, 182, 16, 240, 19, 108, 323, 100, 22, 462, 24, 552, 26, 650, 28, 756, 30, 870, 32, 992, 34, 1122, 36, 1260, 38, 1406, 40, 1560, 42, 1722, 44, 1892, 46, 2070, 48, 2256, 50, 2450, 52, 2652, 54, 2862, 57, 448, 3135, 59, 3422, 61, 3660, 63, 3906, 65, 4160

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,2

COMMENTS

While extending the sequence at a(k) we will check if k equals a previous term in the sequence. If such a term a(m) = k is found a(k) is determined as a(k) = a(m)*m. If no previous term matches k we may choose a(k) = k+c with the least c such that c > 0 and k+c does not equal any previous term in the sequence. It is conjectured that this check is sufficient. Reasoning behind this conjecture:

The greatest common divisor of two consecutive Fibonacci numbers is 1, thus we know that (k-1)^F(m)*k^F(m+1) and (t-1)^F(n)*t^F(n+1) are all different for some m,n > 1 if k and t are chosen such that for m or n < 2 no solution for (k-1)^F(m)*k^F(m+1) = (t-1)^F(n)*t^F(n+1) exist, because this cannot be equal if (k-1)*k and (t-1)*t have different prime numbers as divisors and if the only difference is the exponent of the prime factors, then the distribution of these between (t-1) and t and thus their progression F(n) or F(n+1) is individually distinct. In this sequence we need also to consider the more general case (k-c)^F(m)*k^F(m+1) = (t-1)^F(n)*t^F(n+1) because sometimes we need to set a(k) = k+c. It is conjectured that in this case c is bounded to be < 3.

LINKS

Table of n, a(n) for n=1..65.

Thomas Scheuerle, Scatter plot log(a(n)/n^2) for n = 1..800.

FORMULA

Let b(n, m) be m times recursion into a(n), for example b(3, 2) = a(a(3)).

b(3, m) = A000304(m+1) for m > 0.

b(n, m+2) = b(n, m)*b(n, m+1).

b(5, m) = 4^F(m)*5^F(m+1), where F(m) = A000045(m).

b(k, m) = (k-1)^F(m)*k^F(m+1), for all k where k+1 = a(k).

PROG

(PARI) listA(max_n) = {my(v=[1, 0], t=1); for(k=2, max_n, if(v[k]==0, t=1; if(k+t<#v, while(v[k+t]>0, t++)); v[k]=k+t); v=concat(v, vector(max(0, v[k]+1-#v))); if(v[v[k]]>0, print("The conjecture that a single forward check is sufficient failed at:", k)); v[v[k]]=k*v[k]); v[1..max_n]}

CROSSREFS

Cf. A257794, A358793.

Cf. A000045, A000304.

Cf. A099267 ( a(n) = a(a(n))-n ).

KEYWORD

nonn,easy,new

AUTHOR

Thomas Scheuerle, Feb 24 2025

STATUS

approved