Tamas Sandor Nagy has authored 81 sequences. Here are the ten most recent ones:
A387287
Primes in the order of their first appearance among the factors of the averages of twin prime pairs.
Original entry on oeis.org
2, 3, 5, 7, 17, 23, 11, 19, 47, 13, 29, 103, 107, 137, 43, 59, 41, 71, 31, 67, 139, 283, 149, 313, 37, 347, 373, 397, 443, 113, 467, 271, 181, 281, 577, 593, 199, 157, 653, 131, 101, 89, 241, 83, 251, 379, 773, 787, 167, 109, 907, 163, 73, 1033, 53, 223, 1117
Offset: 1
a(1) = 2 because 2 appeared first as a prime factor of the average of a twin prime pair, namely of 4 = 2*2 = 2^2, the average of 3 and 5, the first twin prime pair.
a(2) = 3 because 3 appeared next as a prime factor of the average of a twin prime pair, here 6 = 2*3, of the twin primes 5 and 7.
a(3) = 5 because 5 appeared next as a prime factor of the average of a twin prime pair, this time of 30 = 2*3*5, between 29 and 30. The averages 12 and 18 are skipped as their factors, 2 and 3, already appeared.
a(5) = 17 following a(4) = 7, skipping the primes 11 and 13 in the order of appearances.
-
P:= select(isprime, {seq(i,i=3..10^4,2)}):
TPA:= map(`+`, P intersect map(`-`,P,2),1):
TPA:= sort(convert(TPA,list)):
R:= NULL: S:= {}:
for t in TPA do
V:= numtheory:-factorset(t) minus S;
if nops(V) > 1 then printf("t = %d: %a\n",t,V) fi;
R:= R, op(sort(convert(V,list)));
S:= S union V;
od:
R; # Robert Israel, Aug 25 2025
-
With[{m = Select[Prime[Range[1000]], PrimeQ[# + 2] &] + 1}, DeleteDuplicates[Flatten[FactorInteger[#][[;; , 1]] & /@ m]]] (* Amiram Eldar, Aug 25 2025 *)
A387095
a(1) = 1 for the single prime 3; for n>=2, a(n) is the number of primes between 2^n and 2^(n+1) whose pairs lay symmetrically at each side of the center 3*2^(n-1) of that interval.
Original entry on oeis.org
1, 2, 2, 4, 4, 6, 8, 22, 26, 42, 92, 128, 218, 416, 750, 1300, 2342, 4136, 7440, 13572, 24820, 45420, 82922, 152964, 282626, 522354, 972388, 1809744, 3379508, 6318652, 11855790, 22277960, 41917480
Offset: 1
Intervals: Primes Centers Count
in A092570: of intervals: a(n):
[2,4] 3 3 1
[4,8] 5 7 6 2
[8,16] 11 13 12 2
[16,32] 17 19 29 31 24 4
[32,64] 37 43 53 59 48 4
[64,128] 79 83 89 103 109 113 96 6
[128,256] 151 157 173 191 193 211 227 233 192 8
...
In row 5 for the interval 2^5 = 32 to 2^6 = 64, the prime pair 37 and 59 lay symmetrically to each side of the center of that interval, 3*2^4 = 48 as abs(48-37) = abs(48-59) = 11, and their sum 37 + 59 = 96, which is 3*2^5. So are 43 and 53 in similarly symmetrical positions in that range, with their sum being 96 also.
-
a[n_] := Module[{c = 0, r = r1 = 3*2^(n-1)}, While[(r1 = NextPrime[r1, -1]) > 2^n, If[PrimeQ[2*r - r1], c += 2]]; c]; a[1] = 1; Array[a, 20] (* Amiram Eldar, Aug 16 2025 *)
A386256
Smallest semiprime p1*p2 such that p2 mod p1 = n and no prime is used more than once in the sequence.
Original entry on oeis.org
6, 35, 377, 407, 817, 391, 3649, 3131, 4841, 4331, 11461, 5293, 7729, 8051, 12031, 25217, 34417, 29503, 24931, 33389, 26051, 57479, 78227, 44377, 68557, 15707, 78119, 64829, 197401, 77059, 166633, 71371, 140579, 86147, 96427, 109237, 84907, 142523, 213341, 158801
Offset: 1
a(4) = 407 = 11 * 37 because 37 mod 11 = 4, and neither of these primes were used before in the sequence as a(1) = 2 * 3, a(2) = 5 * 7, and a(3) = 13 * 29, and so 11 and 37 are the earliest possible primes to satisfy the condition.
a(5) = 817 = 19 * 43 because 43 mod 19 = 5. Smaller candidate primes such as 13 and 31 would have been suitable, but 13 was already used for a(3) = 377 = 13 * 29. Therefore 19 and 43 are the earliest possible primes to satisfy the condition.
-
q[k_, n_, ps_] := Module[{f = FactorInteger[k], p1, p2}, If[f[[;; , 2]] != {1, 1}, {}, p1 = f[[1, 1]]; p2 = f[[2, 1]]; If[Mod[p2, p1] == n && ! MemberQ[ps, p1] && ! MemberQ[ps, p2], {p1, p2}, {}]]];
seq[nmax_] := Module[{ps = {}, s = {}, k, p}, Do[k = 6; While[(p = q[k, n, ps]) == {}, k++]; AppendTo[s, Times @@ p]; ps = Join[ps, p], {n, 1, nmax}]; s]; seq[40] (* Amiram Eldar, Aug 14 2025 *)
nsp[n_Integer] := nsp[n] = Block[{sp = n + 1}, While[ PrimeOmega[sp] != 2, sp++]; sp]; a[n_] := Block[{sp = 4}, While[fi = Flatten[ Table[ #[[1]], {#[[2]]}] & /@ FactorInteger[ sp]]; Mod[ fi[[2]], fi[[1]]] != n || MemberQ[p, fi[[1]]] || MemberQ[p, fi[[2]]], sp = nsp[sp]]; AppendTo[p, fi[[1]]]; AppendTo[p, fi[[2]]]; sp]; p = {}; Do[Print[{n, f[n]}], {n, 50}] (* Robert G. Wilson v, Aug 20 2025 *)
A386520
Column sums of the triangle in A386755.
Original entry on oeis.org
1, 5, 13, 13, 31, 35, 57, 61, 85, 85, 111, 99, 235, 89, 353, 173, 171, 341, 343, 229, 489, 423, 415, 435, 661, 525, 535, 559, 1161, 427, 931, 653, 1201, 787, 941, 885, 1629, 537, 1443, 1839, 1723, 931, 1119, 1525, 2415, 741, 2257, 2327, 1947, 2005, 2767, 1131, 3181, 1055, 3131, 2147
Offset: 1
Triangle whose columns are summed.
m/n| 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
----------------------------------------------------------------
1 | 1
2 | 1
3 | 2 1
4 | 2 1
5 | 3 2 1
6 | 3 2 1
7 | 3 2 1
8 | 3 2 1
9 | 4 3 2 1
10 | 4 3 2 1
11 | 5 4 3 2 1
12 | 5 4 3 2 1
13 | 5 4 3 2 1
14 | 5 4 3 2 1
15 | 5 4 3 2 1
16 | 5 4 3 2 1
17 | 6 5 4 3 2 1
18 | 6 5 4 3 2 1
19 | 6 5 4 3 2 1
20 | 6 5 4 3 2 1
...
The completed column for n=5 is definitely fully visible here because in column 6 for n=6 the divisor k=6 already appeared. That means that column 5 cannot have more divisors in it under the last k=5 in row 17 because in that row only k=7 may follow k=6 in theory, but 7 does not divide 5. So, all similarly proven, definitely fully visible completed columns in this sample array are readily summable by sight. E.g. column 5: a(5) = 1 + 5 + 5 + 5 + 5 + 5 + 5 = 31.
Cf.
A007952 (row number where k=n first appears).
-
\\ uses row(n) from A386755
a(n) = my(ok=1, k=1, last=-1, s=0, r); while(ok, r=row(k); if (#r >= n, s+=r[n]); k++; if (#r>=n, if ((last==n) && (r[n]==0), ok = 0, last = r[n]))); s; \\ Michel Marcus, Aug 02 2025
-
\\ uses row(n) from A386755
lista(nn) = my(ok=1, k=1, vlast=vector(nn,i,-1), vs=vector(nn)); while(ok, my(r=row(k)); for (i=1, nn, if (#r>=i, vs[i]+=r[i])); k++; my(nbok=0); for (i=1, nn, if (#r>=i, if ((vlast[i]==i) && (r[i]==0), nbok++, vlast[i] = r[i]))); if (nbok == nn, ok = 0);); vs; \\ Michel Marcus, Aug 02 2025
A386755
Triangle read by rows, where row terms are filled depending on divisibility of n. See comments.
Original entry on oeis.org
1, 0, 1, 0, 2, 1, 0, 2, 0, 1, 0, 0, 3, 2, 1, 0, 0, 3, 2, 0, 1, 0, 0, 3, 0, 0, 2, 1, 0, 0, 3, 0, 0, 2, 0, 1, 0, 0, 0, 4, 0, 3, 0, 2, 1, 0, 0, 0, 4, 0, 3, 0, 2, 0, 1, 0, 0, 0, 0, 5, 0, 0, 4, 3, 2, 1, 0, 0, 0, 0, 5, 0, 0, 4, 3, 2, 0, 1, 0, 0, 0, 0, 5, 0, 0, 4, 3, 0, 0, 2, 1
Offset: 1
The triangle begins:
1;
0, 1;
0, 2, 1;
0, 2, 0, 1;
0, 0, 3, 2, 1;
0, 0, 3, 2, 0, 1;
0, 0, 3, 0, 0, 2, 1;
0, 0, 3, 0, 0, 2, 0, 1;
0, 0, 0, 4, 0, 3, 0, 2, 1;
0, 0, 0, 4, 0, 3, 0, 2, 0, 1;
...
.
An example for the step by step construction of a particular row, let it be row n=20: We start with k=1 at column 20, and find that k=1 divides c=20. So we enter k=1 into the array in that column. Next, let now k=2, and we look for the greatest c that is less than 20, and which is divisible by k=2. That c is 18 in column 18, so we enter 2 in that column. We increase k by 1 to k=3, and similarly seek out the greatest c again that is less than 18, and which is divisible by 3. This number is 15, and so we enter 3 in column 15. And so on, we test divisibility with k=4, k=5, and k=6 to find that these k's fit under c=12, c=10 and c=6, respectively.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
0 0 0 0 0 6 0 0 0 5 0 4 0 0 3 0 0 2 0 1
-
row(n) = my(v=vector(n), m=n); for(k=1, n, my(keepm = m); while(m%k, m--); if (m == 0, keepm=m, v[m] = k; m--);); v; \\ Michel Marcus, Aug 01 2025
A386364
Irregular triangle read by rows. The prime numbers corresponding to the distinct prime chains in A385871. Also a permutation of the primes.
Original entry on oeis.org
2, 3, 7, 5, 11, 23, 13, 17, 19, 37, 29, 53, 31, 41, 73, 43, 47, 83, 137, 59, 61, 67, 71, 79, 131, 89, 97, 157, 101, 163, 103, 167, 107, 173, 109, 113, 127, 139, 149, 233, 353, 151, 179, 277, 181, 281, 421, 613, 191, 193, 197, 199, 211, 223, 337, 227, 229, 239, 241, 251, 257, 263, 269, 271
Offset: 1
The irregular triangle begins:
2, 3, 7;
5, 11, 23;
13;
17;
19, 37;
29, 53;
31;
41, 73;
...
-
f(n) = my(m=1, x); while((x=(prime(n) + 2*(n - 1) + m - prime(n + m)))>= 0, if (x==0, return(m+n)); m++); \\ A385871
lista(nn) = my(list = List(), vp=vector(nn)); for (n=1, nn, if (!vp[n], my(v=Vec(prime(n)), m=n); vp[n] = 1; while(m = f(m), v=concat(v, prime(m)); if (m<=nn, vp[m]=1)); listput(list, v););); Vec(list); \\ Michel Marcus, Jul 19 2025
A385871
a(n) is the number of primes in the prime chain to which prime(n) belongs. Details are in the Comments.
Original entry on oeis.org
3, 3, 3, 3, 3, 1, 1, 2, 3, 2, 1, 2, 2, 1, 3, 2, 1, 1, 1, 1, 2, 2, 3, 1, 2, 2, 2, 2, 1, 1, 1, 2, 3, 1, 3, 1, 2, 2, 2, 2, 2, 4, 1, 1, 1, 1, 1, 2, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 2, 4, 1, 1, 1, 3, 2, 1, 1, 2, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 2, 1, 1, 3, 2
Offset: 1
The counting walk from 5:
2 3 5 7 11 13 17 19 23
7 <- 6 <- 5
-> 8 -> 9 -> 10 -> 11
^
|
Match
The counting walk from 11:
2 3 5 7 11 13 17 19 23
15 <- 14 <- 13 <- 12 <- 11
-> 16 -> 17 -> 18 -> 19 -> 20 -> 21 -> 22 -> 23
^
|
Match
Since 5 cannot be reached from any lesser prime, and no greater prime can be reached either from 23 by this method, 5, 11, and 23 belong to a common prime chain of length 3, allowing a(3), a(5), and a(9) terms the value of 3 each.
The counting walk from 13:
2 3 5 7 11 13 17 19 23 29 31 37
18 <- 17 <- 16 <- 15 <- 14 <- 13
-> 19 -> 20 -> 21 -> 22 -> 23 -> 24 -> 25 -> 26 -> 27 -> 28 -> 29
^
|
No match
as the faster growing
prime sequence here already
overtook the counting sequence
without reaching equal values
at any second point.
.
Therefore 13 forms a chain with one solitary link only, and so a(6) = 1.
.
In another example, the primes 2, 3, and 7 are in a prime chain of length 3 because
2 = prime(1), so k = 1
and
prime(1) + 2*(1 - 1) + m = prime(1 + m)
2 + 2*0 + m = prime(1 + m)
2 + m = prime(1 + m)
Solved for m by search: m = 1
and since
2 + 1 = prime(1 + 1)
3 = prime(2) -> so far, 3 is linked to 2 in a prime chain.
Furthermore:
3 = prime(2), so let now k = 2
and
prime(2) + 2*(2 - 1) + m = prime(2 + m)
3 + 2*1 + m = prime(2 + m)
5 + m = prime(2 + m)
Solved for m by search: m = 2
and since
5 + 2 = prime(2 + 2)
7 = prime(4) -> so far, 7 is linked to 2 and 3 in a prime chain.
Checking for further members to the 2-3-7 prime chain:
prime(4) + 2*(4 - 1) + m = prime(4 + m)
7 + 2*3 + m = prime(4 + m)
7 + 6 + m = prime(4 + m)
13 + m = prime(4 + m)
No such m > 0 is found, so m and prime(4 + m) cannot exist to satisfy the equation.
The 2-3-7 prime chain has no more members, therefore they form a chain of length 3, and so the terms a(1), a(2), and a(4) each equal to 3.
The prime chains to which the first few primes belong:
2 {2, 3, 7},
3 {2, 3, 7},
5 {5, 11, 23},
7 {2, 3, 7},
11 {5, 11, 23},
13 {13},
17 {17},
19 {19, 37},
23 {5, 11, 23},
29 {29, 53},
31 {31},
37 {19, 37},
41 {41, 73},
43 {43},
47 {47, 83, 137},
53 {29, 53},
...
A385615
Star numbers corresponding to the point numbers in A385330.
Original entry on oeis.org
1, 2, 2, 3, 2, 3, 3, 4, 2, 4, 3, 4, 2, 3, 4, 5, 3, 4, 2, 5, 4, 5, 3, 5, 4, 2, 3, 5, 6, 4, 5, 6, 3, 4, 6, 2, 5, 6, 4, 5, 3, 6, 5, 4, 6, 7, 2, 3, 5, 7, 6, 4, 7, 5, 6, 3, 7, 4, 6, 2, 5, 7, 6, 4, 7, 5, 3, 6, 7, 8, 5, 4, 6, 7, 2, 8, 3, 5, 8, 7, 6, 4, 8, 7, 5, 6, 8
Offset: 1
A385330
The point numbers encountered by a rotating marker following the process described in the Comments.
Original entry on oeis.org
1, 1, 2, 1, 1, 2, 3, 1, 2, 2, 1, 3, 1, 2, 4, 1, 3, 1, 2, 2, 2, 3, 1, 4, 3, 1, 2, 5, 1, 4, 1, 2, 3, 1, 3, 2, 2, 4, 2, 3, 1, 5, 4, 3, 6, 1, 1, 2, 5, 2, 1, 4, 3, 1, 2, 3, 4, 1, 3, 2, 2, 5, 4, 2, 6, 3, 1, 5, 7, 1, 4, 3, 6, 1, 1, 2, 2, 5, 3, 2, 1, 4, 4, 3, 1, 2, 5
Offset: 1
The sequence can be written as an irregular triangle, read by rows, with row n corresponding to laying down the n-star. The first few rows are (star-number, point-number):
(1,1);
(2,1), (2,2)*;
(3,1), (2,1), (3,2), (3,3);
(4,1), (2,2), (4,2), (3,1), (4,3), (2,1), (3,2), (4,4);
(5,1), (3,3), (4,1), (2,2), (5,2), (4,2), (5,3), (3,1), (5,4), (4,3), (2,1), (3,2), (5,5);
...
* The label (2,2) overwrites the (1,1) label (the only known occurrence of an overwrite).
A385404
Numbers that can be split into two at any place between their digits such that the resulting numbers are always a nonprime on the left and a prime on the right.
Original entry on oeis.org
12, 13, 15, 17, 42, 43, 45, 47, 62, 63, 65, 67, 82, 83, 85, 87, 92, 93, 95, 97, 123, 143, 147, 153, 167, 183, 423, 443, 447, 453, 467, 483, 497, 623, 637, 643, 647, 653, 667, 683, 697, 813, 817, 823, 843, 847, 853, 867, 873, 883, 913, 917, 923, 937, 943, 947, 953, 967, 983, 997
Offset: 1
637 is a term because when it is split in two in all possible ways, it first results in 63, a nonprime, and 3, a prime. When split in the second and final possible way, it results in 6, a nonprime, and 37, a prime.
-
q[n_] := !MemberQ[IntegerDigits[n], 0] && AllTrue[Range[IntegerLength[n]-1], PrimeQ[QuotientRemainder[n, 10^#]] == {False, True} &]; Select[Range[10, 1000], q] (* Amiram Eldar, Jun 27 2025 *)
-
from sympy import isprime
def ok(n): return '0' not in (s:=str(n)) and len(s) > 1 and all(not isprime(int(s[:i])) and isprime(int(s[i:])) for i in range(1, len(s)))
print([k for k in range(1000) if ok(k)]) # Michael S. Branicky, Jun 27 2025
-
# uses import and function ok above
from itertools import count, islice, product
def agen(): # generator of terms
tp = list("23579") # set of left-truncatable primes
for d in count(2):
tpnew = []
for f in "123456789":
for e in tp:
if isprime(int(s:=f+e)):
tpnew.append(s)
if ok(t:=int(f+e)):
yield t
tp = tpnew
if len(tp) == 0:
return
afull = list(agen())
print(afull[:60]) # Michael S. Branicky, Jun 27 2025
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