A075326 -id:A075326 - cp's OEIS frontend

A299409 Solution (e(n)) of the system of 5 complementary equations in Comments.

10, 26, 45, 62, 78, 94, 114, 130, 146, 162, 180, 198, 214, 230, 248, 266, 282, 298, 317, 334, 350, 366, 386, 402, 418, 434, 451, 470, 486, 502, 520, 538, 554, 570, 589, 606, 622, 638, 658, 674, 690, 706, 725, 742, 758, 774, 792, 810, 826, 842, 861, 878, 894

Offset: 0

Clark Kimberling, Apr 22 2018

Define sequences a(n), b(n), c(n), d(n) recursively, starting with a(0) = 1, b(0) = 2, c(0) = 3:
a(n) = least new;
b(n) = least new;
c(n) = least new;
d(n) = least new;
e(n) = a(n) + b(n) + c(n) + d(n);
where "least new k" means the least positive integer not yet placed.
***
Conjecture: for all n >= 0,
0 <= 17n - 11 - 4 a(n) <= 4
0 <= 17n - 7 - 4 b(n) <= 4
0 <= 17n - 3 - 4 c(n) <= 3
0 <= 17n + 1 - 4 d(n) <= 3
0 <= 17n - 5 - e(n) <= 3
***
The sequences a,b,c,d,e partition the positive integers.  The sequence e can be called the "anti-tetranacci sequence"; see A075326 (anti-Fibonacci numbers) and A265389 (anti-tribonacci numbers).

			n:   0  1   2    3   4   5   6   7   8   9
a:   1  5   9   14  18  22  27  31  35  39
b:   2  6   11  15  19  23  28  32  36  40
c:   3  7   12  16  20  24  29  33  37  41
d:   4  8   13  17  21  25  30  34  38  42
e:  10  26  45  62  78  94 114 130 146 162

Clark Kimberling, Table of n, a(n) for n = 0..1000
Wieb Bosma, Rene Bruin, Robbert Fokkink, Jonathan Grube, Anniek Reuijl, and Thian Tromp, Using Walnut to solve problems from the OEIS, arXiv:2503.04122 [math.NT], 2025.
Robbert Fokkink and Gandhar Joshi, Anti-recurrence sequences, arXiv:2506.13337 [math.NT], 2025. See pp. 2, 11, 18.

Cf. A036554, A299634, A299405, A299637, A299638, A299641.

(* Program 1: sequences a,b,c,d,e generated from the complementary equations *)
z = 200;
mex[list_, start_] := (NestWhile[# + 1 &, start, MemberQ[list, #] &]);
a = {1}; b = {2}; c = {3}; d = {4}; e = {}; AppendTo[e,
Last[a] + Last[b] + Last[c] + Last[d]];
Do[{AppendTo[a, mex[Flatten[{a, b, c, d, e}], 1]],
   AppendTo[b, mex[Flatten[{a, b, c, d, e}], 1]],
   AppendTo[c, mex[Flatten[{a, b, c, d, e}], 1]],
   AppendTo[d, mex[Flatten[{a, b, c, d, e}], 1]],
   AppendTo[e, Last[a] + Last[b] + Last[c] + Last[d]]}, {z}];
Take[a, 100]  (* A299405 *)
Take[b, 100]  (* A299637 *)
Take[c, 100]  (* A299638 *)
Take[d, 100]  (* A299641 *)
Take[e, 100]  (* A299409 *)
(* Program 2: sequence e generated by iterating a morphism *)
morph = Nest[Flatten[# /. Thread[{0, 1, 2, 3} -> {{2, 3, 3, 1}, {2, 3, 2, 1}, {2, 3, 1, 1}, {2, 3, 0, 1}}]] &, {0}, 9];
A299409 = Accumulate[Prepend[Drop[Flatten[morph /. Thread[{0, 1, 2, 3} -> {{1, 1, 2, 4}, {1, 1, 3, 3}, {1, 1, 4, 2}, {1, 1, 5, 1}}]], 1] + 15, 10]];
Take[A299409, 100]  (* Peter J. C. Moses, May 04 2018 *)

A299637 Solution (b(n)) of the system of 5 complementary equations in Comments.

Original entry on oeis.org

2, 6, 11, 15, 19, 23, 28, 32, 36, 40, 44, 49, 53, 57, 61, 66, 70, 74, 79, 83, 87, 91, 96, 100, 104, 108, 112, 117, 121, 125, 129, 134, 138, 142, 147, 151, 155, 159, 164, 168, 172, 176, 181, 185, 189, 193, 197, 202, 206, 210, 215, 219, 223, 227, 232, 236, 240

Offset: 0

Clark Kimberling, Apr 22 2018

Define sequences a(n), b(n), c(n), d(n) recursively, starting with a(0) = 1, b(0) = 2, c(0) = 3:
a(n) = least new;
b(n) = least new;
c(n) = least new;
d(n) = least new;
e(n) = a(n) + b(n) + c(n) + d(n);
where "least new k" means the least positive integer not yet placed.
***
Conjecture: for all n >= 0,
0 <= 17n - 11 - 4 a(n) <= 4
0 <= 17n - 7 - 4 b(n) <= 4
0 <= 17n - 3 - 4 c(n) <= 3
0 <= 17n + 1 - 4 d(n) <= 3
0 <= 17n - 5 - e(n) <= 3
***
The sequences a,b,c,d,e partition the positive integers.  The sequence e can be called the "anti-tetranacci sequence"; see A075326 (anti-Fibonacci numbers) and A265389 (anti-tribonacci numbers).

			n:   0  1   2    3   4   5   6   7   8   9
a:   1  5   9   14  18  22  27  31  35  39
b:   2  6   11  15  19  23  28  32  36  40
c:   3  7   12  16  20  24  29  33  37  41
d:   4  8   13  17  21  25  30  34  38  42
e:  10  26  45  62  78  94 114 130 146 162

Clark Kimberling, Table of n, a(n) for n = 0..1000

Cf. A036554, A299634, A299405, A299638, A299641, A299409.

z = 200;
mex[list_, start_] := (NestWhile[# + 1 &, start, MemberQ[list, #] &]);
a = {1}; b = {2}; c = {3}; d = {4}; e = {}; AppendTo[e,
Last[a] + Last[b] + Last[c] + Last[d]];
Do[{AppendTo[a, mex[Flatten[{a, b, c, d, e}], 1]],
   AppendTo[b, mex[Flatten[{a, b, c, d, e}], 1]],
   AppendTo[c, mex[Flatten[{a, b, c, d, e}], 1]],
   AppendTo[d, mex[Flatten[{a, b, c, d, e}], 1]],
   AppendTo[e, Last[a] + Last[b] + Last[c] + Last[d]]}, {z}];
Take[a, 100]  (* A299405 *)
Take[b, 100]  (* A299637 *)
Take[c, 100]  (* A299638 *)
Take[d, 100]  (* A299641 *)
Take[e, 100]  (* A299409 *)

A299638 Solution (c(n)) of the system of 5 complementary equations in Comments.

Original entry on oeis.org

3, 7, 12, 16, 20, 24, 29, 33, 37, 41, 46, 50, 54, 58, 63, 67, 71, 75, 80, 84, 88, 92, 97, 101, 105, 109, 113, 118, 122, 126, 131, 135, 139, 143, 148, 152, 156, 160, 165, 169, 173, 177, 182, 186, 190, 194, 199, 203, 207, 211, 216, 220, 224, 228, 233, 237, 241

Offset: 0

Clark Kimberling, Apr 22 2018

Define sequences a(n), b(n), c(n), d(n) recursively, starting with a(0) = 1, b(0) = 2, c(0) = 3:
a(n) = least new;
b(n) = least new;
c(n) = least new;
d(n) = least new;
e(n) = a(n) + b(n) + c(n) + d(n);
where "least new k" means the least positive integer not yet placed.
***
Conjecture: for all n >= 0,
0 <= 17n - 11 - 4 a(n) <= 4
0 <= 17n - 7 - 4 b(n) <= 4
0 <= 17n - 3 - 4 c(n) <= 3
0 <= 17n + 1 - 4 d(n) <= 3
0 <= 17n - 5 - e(n) <= 3
***
The sequences a,b,c,d,e partition the positive integers.  The sequence e can be called the "anti-tetranacci sequence"; see A075326 (anti-Fibonacci numbers) and A265389 (anti-tribonacci numbers).

			n:   0  1   2    3   4   5   6   7   8   9
a:   1  5   9   14  18  22  27  31  35  39
b:   2  6   11  15  19  23  28  32  36  40
c:   3  7   12  16  20  24  29  33  37  41
d:   4  8   13  17  21  25  30  34  38  42
e:  10  26  45  62  78  94 114 130 146 162

Clark Kimberling, Table of n, a(n) for n = 0..1000

Cf. A036554, A299634, A299405, A299637, A299641, A299409.

z = 200;
mex[list_, start_] := (NestWhile[# + 1 &, start, MemberQ[list, #] &]);
a = {1}; b = {2}; c = {3}; d = {4}; e = {}; AppendTo[e,
Last[a] + Last[b] + Last[c] + Last[d]];
Do[{AppendTo[a, mex[Flatten[{a, b, c, d, e}], 1]],
   AppendTo[b, mex[Flatten[{a, b, c, d, e}], 1]],
   AppendTo[c, mex[Flatten[{a, b, c, d, e}], 1]],
   AppendTo[d, mex[Flatten[{a, b, c, d, e}], 1]],
   AppendTo[e, Last[a] + Last[b] + Last[c] + Last[d]]}, {z}];
Take[a, 100]  (* A299405 *)
Take[b, 100]  (* A299637 *)
Take[c, 100]  (* A299638 *)
Take[d, 100]  (* A299641 *)
Take[e, 100]  (* A299409 *)

A299641 Solution (d(n)) of the system of 5 complementary equations in Comments.

Original entry on oeis.org

4, 8, 13, 17, 21, 25, 30, 34, 38, 42, 47, 51, 55, 59, 64, 68, 72, 76, 81, 85, 89, 93, 98, 102, 106, 110, 115, 119, 123, 127, 132, 136, 140, 144, 149, 153, 157, 161, 166, 170, 174, 178, 183, 187, 191, 195, 200, 204, 208, 212, 217, 221, 225, 229, 234, 238, 242

Offset: 0

Clark Kimberling, Apr 22 2018

Define sequences a(n), b(n), c(n), d(n) recursively, starting with a(0) = 1, b(0) = 2, c(0) = 3:
a(n) = least new;
b(n) = least new;
c(n) = least new;
d(n) = least new;
e(n) = a(n) + b(n) + c(n) + d(n);
where "least new k" means the least positive integer not yet placed.
***
Conjecture: for all n >= 0,
0 <= 17n - 11 - 4 a(n) <= 4
0 <= 17n - 7 - 4 b(n) <= 4
0 <= 17n - 3 - 4 c(n) <= 3
0 <= 17n + 1 - 4 d(n) <= 3
0 <= 17n - 5 - e(n) <= 3
***
The sequences a,b,c,d,e partition the positive integers.  The sequence e can be called the "anti-tetranacci sequence"; see A075326 (anti-Fibonacci numbers) and A265389 (anti-tribonacci numbers).

			n:   0  1   2    3   4   5   6   7   8   9
a:   1  5   9   14  18  22  27  31  35  39
b:   2  6   11  15  19  23  28  32  36  40
c:   3  7   12  16  20  24  29  33  37  41
d:   4  8   13  17  21  25  30  34  38  42
e:  10  26  45  62  78  94 114 130 146 162

Clark Kimberling, Table of n, a(n) for n = 0..1000

Cf. A036554, A299634, A299405, A299637, A299638, A299409.

z = 200;
mex[list_, start_] := (NestWhile[# + 1 &, start, MemberQ[list, #] &]);
a = {1}; b = {2}; c = {3}; d = {4}; e = {}; AppendTo[e,
Last[a] + Last[b] + Last[c] + Last[d]];
Do[{AppendTo[a, mex[Flatten[{a, b, c, d, e}], 1]],
   AppendTo[b, mex[Flatten[{a, b, c, d, e}], 1]],
   AppendTo[c, mex[Flatten[{a, b, c, d, e}], 1]],
   AppendTo[d, mex[Flatten[{a, b, c, d, e}], 1]],
   AppendTo[e, Last[a] + Last[b] + Last[c] + Last[d]]}, {z}];
Take[a, 100]  (* A299405 *)
Take[b, 100]  (* A299637 *)
Take[c, 100]  (* A299638 *)
Take[d, 100]  (* A299641 *)
Take[e, 100]  (* A299409 *)

A075325 Pair the natural numbers such that the m-th pair is (r, s) where r, s and s-r are the smallest numbers which have not occurred earlier and also are not equal to the difference of any earlier pair: (1, 3), (4, 9), (6, 13), (8, 18), (11, 23), (14, 29), (16, 33), (19, 39), (21, 43), (24, 49), (26, 53), (28, 58), ... Sequence gives first term of each pair.

Original entry on oeis.org

1, 4, 6, 8, 11, 14, 16, 19, 21, 24, 26, 28, 31, 34, 36, 38, 41, 44, 46, 48, 51, 54, 56, 59, 61, 64, 66, 68, 71, 74, 76, 79, 81, 84, 86, 88, 91, 94, 96, 99, 101, 104, 106, 108, 111, 114, 116, 118, 121, 124, 126, 128, 131, 134, 136, 139, 141, 144, 146, 148, 151, 154, 156

Offset: 1

Amarnath Murthy, Sep 16 2002

nonn

Most of the pairs are of the form (r,2r+1) except for the ones like a(4) = (8,18) and a(12) = (28,58) and (38,78) etc. which are of the form (r,2r +2).

			The first pair (1, 3) covers 1, 2, 3. The second pair is (4, 9) covering 4, 5, 9.

Clark Kimberling, Table of n, a(n) for n = 1..10000
Robbert Fokkink and Gandhar Joshi, Anti-recurrence sequences, arXiv:2506.13337 [math.NT], 2025. See p. 4.

Cf. A007814, A075326, A075327.

The sequence formed by listing the differences between the second and first elements of each pair is A047215.

(* Here, the offset for (a(n)) is 0. *)
z = 200;
mex[list_, start_] := (NestWhile[# + 1 &, start, MemberQ[list, #] &]);
a = {}; b = {}; c = {};
Do[AppendTo[a,
   mex[Flatten[{a, b, c}], If[Length[a] == 0, 1, Last[a]]]];
  AppendTo[b, mex[Flatten[{a, b, c}], Last[a]]];
  AppendTo[c, Last[a] + Last[b]], {z}];
Take[a, 100] (* A075325 *)
Take[b, 100] (* A047215 *)
Take[c, 100] (* A075326 *)
Grid[{Join[{"n"}, Range[0, 20]], Join[{"a(n)"}, Take[a, 21]],
  Join[{"b(n)"}, Take[b, 21]], Join[{"c(n)"}, Take[c, 21]]},
 Alignment -> ".",
 Dividers -> {{2 -> Red, -1 -> Blue}, {2 -> Red, -1 -> Blue}}]
(* Peter J. C. Moses, Apr 26 2018 *)

used = vector(500); i = 1; A = vector(80); B = A; C = A; for (n = 1, 80, while (used[i], i++); j = i + 1; while (used[j] || used [i + j], j++); A[n] = i; B[n] = i + j; C[n] = i + i + j; used[i] = 1; used[j] = 1; used[i + j] = 1); A \\ David Wasserman, Jan 16 2005

Let A(n) = A007814(n). Let B(n) = A(n) + 1 if A(n) < 2; B(n) = 0 if A(n)>=2 & A(n) is even; B(n) = 2 if A(n) >= 2 & A(n) is odd. Then a(n) = (5n+B(n)-4)/2. - John Chew (jjchew(AT)math.utoronto.ca), Jun 20 2006

More terms from David Wasserman, Jan 16 2005

A075327 Sum of n-th pair in A075325.

Original entry on oeis.org

4, 13, 19, 26, 34, 43, 49, 58, 64, 73, 79, 86, 94, 103, 109, 116, 124, 133, 139, 146, 154, 163, 169, 178, 184, 193, 199, 206, 214, 223, 229, 238, 244, 253, 259, 266, 274, 283, 289, 298, 304, 313, 319, 326, 334, 343, 349, 356, 364, 373, 379, 386, 394, 403, 409

Offset: 1

Amarnath Murthy, Sep 16 2002

nonn

Cf. A075325, A075326.

More terms from David Wasserman, Jan 16 2005

A297464 Solution (a(n)) of the system of 4 complementary equations in Comments.

Original entry on oeis.org

1, 4, 8, 11, 14, 18, 21, 24, 28, 31, 34, 38, 41, 44, 48, 51, 54, 58, 61, 64, 68, 71, 74, 78, 81, 84, 88, 91, 94, 98, 101, 104, 108, 111, 114, 118, 121, 124, 128, 131, 134, 138, 141, 144, 148, 151, 154, 158, 161, 164, 168, 171, 174, 178, 181, 184, 188, 191

Offset: 0

Clark Kimberling, Apr 19 2018

Define sequences a(n), b(n), c(n), d(n) recursively, starting with a(0) = 1, b(0) = 2, c(0) = 3;:
a(n) = least new;
b(n) = least new;
c(n) = least new;
d(n) = a(n) + b(n) + c(n);
where "least new k" means the least positive integer not yet placed.
***
Conjecture: for all n >= 0,
0 <= 10n - 6 - 3 a(n) <= 2
0 <= 10n - 2 - 3 b(n) <= 3
0 <= 10n + 1 - 3 c(n) <= 3
0 <= 10n - 3 - d(n) <= 2
***
The sequences a,b,c,d partition the positive integers.  The sequence d can be called the "anti-tribonacci sequence"; viz., if sequences a and b are defined as above, and c(n) is defined by c(n) = a(n) + b(n), then the resulting system of 3 complementary sequences gives c = A075326, the "anti-Fibonacci sequence."  See A299409 for the "anti-tetranacci" sequences.

			n:   0    1    2    3    4    5    6    7    8    9
a:   1    4    8   11   14   18   21   24   28   31
b:   2    5    9   12   15   19   22   25   29   32
c:   3    7   10   13   17   20   23   26   30   33
d:   6   16   27   36   46   57   66   75   87   96

Clark Kimberling, Table of n, a(n) for n = 0..1000
Wieb Bosma, Rene Bruin, Robbert Fokkink, Jonathan Grube, Anniek Reuijl, and Thian Tromp, Using Walnut to solve problems from the OEIS, arXiv:2503.04122 [math.NT], 2025. See p. 8.

Cf. A036554, A299634, A297465, A297466, A265389.

z = 400;
mex[list_, start_] := (NestWhile[# + 1 &, start, MemberQ[list, #] &]);
a = {1}; b = {2}; c = {3}; d = {}; AppendTo[d, Last[a] + Last[b] + Last[c]];
Do[{AppendTo[a, mex[Flatten[{a, b, c, d}], 1]],
   AppendTo[b, mex[Flatten[{a, b, c, d}], 1]],
   AppendTo[c, mex[Flatten[{a, b, c, d}], 1]],
   AppendTo[d, Last[a] + Last[b] + Last[c]]}, {z}];
Take[a, 100]  (* A297464 *)
Take[b, 100]  (* A297465 *)
Take[c, 100]  (* A297466 *)
Take[d, 100]  (* A265389 *)

a(n) = a(n-1) + a(n-3) - a(n-4) (conjectured).

d(n) = A275389(n) for n >= 0.

A332788 Positive-pan primes (see Comments).

Original entry on oeis.org

5, 31, 59, 107, 271, 223, 269, 313, 647, 457, 503, 941, 1579, 12919, 3571, 1667, 2897, 3037, 3187, 1993, 3461, 2179, 10141, 5927, 11969, 4957, 13627, 5519, 22787, 3851, 3889, 3929, 15217, 44221, 65867, 21799, 10211, 31727, 24623, 11467, 6983, 7039, 16651, 73351

Offset: 1

Ivan N. Ianakiev, Feb 24 2020

nonn

Take a double-pan balance scale and name the pans "negative" and "positive". At each step, the question is: "Is there an unused prime that would balance the scale if added to the positive pan?" If the answer is yes, add that prime to the positive pan. Otherwise, add the smallest unused prime to the negative pan.

Is the number of primes in the positive pan (P) infinite? If it is, is fractalization of P, i.e., further subdivision into PN and PP subpans, possible (including infinitely)?

			2 and 3 unbalance the scale (and go to the negative pan N), but 5 = 2 + 3 balances it (and goes to the positive pan P).

Michael S. Branicky, Table of n, a(n) for n = 1..10000

Cf. A075326, A332341, A332787.

a[1]=-2;a[n_]:=a[n]=Module[{tab=Table[a[i],{i,1,n-1}],
    totalN=Abs[Total[Select[Table[a[i],{i,1,n-1}],Negative]]],
     totalP=Total[Select[Table[a[i],{i,1,n-1}],Positive]],
    l=NextPrime[Last[Select[Table[a[i],{i,1,n-1}],Negative]],-1],
    m=NextPrime[Abs[Last[Select[Table[a[i],{i,1,n-1}],Negative]]]]},
   If[totalN==totalP,
    If[PrimePi[tab[[-1]]]-PrimePi[Abs[tab[[-2]]]]==1,-NextPrime[tab[[-1]]],
     If[FreeQ[Abs[tab],m],-m,While[! FreeQ[Abs[tab],m],m=NextPrime[m]];-m]],
    If[PrimeQ[totalN-totalP]&&FreeQ[Abs[tab],totalN-totalP],totalN-totalP,
     If[FreeQ[Abs[tab],Abs[l]],l,While[!FreeQ[Abs[tab],Abs[l]],l=NextPrime[l,-1]];l]
]]];Select[a/@Range[370],Positive]

from itertools import islice
from sympy import isprime, nextprime
def agen(): # generator of terms
    used, d, nextp = set(), 0, 2
    while True:
        if d > 0 and d not in used and isprime(d):
            used.add(d); yield d; d = 0
        while nextp in used:
            nextp = nextprime(nextp)
        used.add(nextp); d += nextp
print(list(islice(agen(), 44))) # Michael S. Branicky, May 12 2022

a(41) and beyond from Michael S. Branicky, May 12 2022

A035166 Let d(m) = denominator of Sum_{k=1..m} 1/k^2 and consider f(m) = product of primes which appear to odd powers in d(m); sequence lists m such that f(m) is different from f(m-1).

Original entry on oeis.org

1, 10, 15, 20, 25, 42, 49, 50, 55, 66, 75, 78, 91, 100, 110, 121, 125, 136, 153, 156, 164, 169, 171, 182, 189, 190, 205, 250, 253, 272, 276, 289, 294, 342, 343, 354, 361, 375, 406, 413, 435, 465, 473, 496, 500, 506, 516, 529, 555, 592, 605, 625

Offset: 1

Bill Gosper, Sep 04 2002

The prime 479 first appears in f(m) at m = 2395, ahead of 71, which first appears in f(2485).
The first occurrence of four distinct primes is at m = 2500, with 5^7, 17^3, 71 and 479.
For 1890 < m < 2006, d(m) is a square (f(m)=1). The lone prime in 1875 .. 1890 is 61 and in 2006 .. 2027 it is 59.
It appears that f(m) can differ from f(m-1) in at most one prime.
(f from definition) = A007913, squarefree part. - Reinhard Zumkeller, Jul 06 2012

			f(10) = 5 is the first time f(m) > 1. The 5 persists until it disappears at m = 15.

Reinhard Zumkeller, Table of n, a(n) for n = 1..250

Cf. A075326, A075327, A007407.

import Data.List (findIndices)
a035166 n = a035166_list !! (n-1)
a035166_list = map (+ 1) $ findIndices (/= 0) $ zipWith (-) (tail gs) gs
   where gs = 0 : map a007913 a007407_list
-- Reinhard Zumkeller, Jul 06 2012

for k:1 do (subset(factor_number(denom(harmonic(k,2))), lambda([x],oddp(second(x)))), if old#old:%% then print(k,%%))

d[n_] := Denominator[ HarmonicNumber[n, 2]]; f[n_] := Times @@ Select[ FactorInteger[d[n]], OddQ[#[[2]]]&][[All, 1]]; A035166 = Join[{1}, Select[ Range[1000], f[#] != f[#-1]&]] (* Jean-François Alcover, Feb 26 2016 *)

d(m) = denominator(sum(k=1, m, 1/k^2));
f(m) = my(f=factor(d(m))); for (k=1, #f~, if (!(f[k,2] % 2), f[k,2] = 0)); factorback(f);
isok(m) = if (m==1, 1, f(m) != f(m-1)); \\ Michel Marcus, Sep 06 2023

A249055 a(1)=0; the next term is always the product of the two smallest numbers not yet in the sequence and which have not yet been used.

Original entry on oeis.org

0, 2, 12, 30, 56, 90, 143, 210, 272, 342, 420, 506, 600, 702, 812, 992, 1122, 1260, 1406, 1560, 1722, 1892, 2070, 2256, 2450, 2652, 2862, 3135, 3422, 3660, 3906, 4160, 4422, 4692, 4970, 5256, 5550, 5852, 6162, 6480, 6806, 7140, 7482, 7832, 8372, 8742, 9120, 9506, 9900

Offset: 1

N. J. A. Sloane, Nov 01 2014

nonn

Suggested by A075336 and A249406.

			Start with a(1) = 0. The missing numbers are 1 2 3 4 5 6 ...
Multiply the first two, and we get 2, which is therefore a(2).
Cross 1, 2, and 1*2 = 2 off the missing list.
The first two missing numbers are now 3 and 4, so a(3) = 3*4 = 12.
Cross off 3,4,12 from the missing list.
Repeat!

Cf. A075326, A000045, A249406, A000304, A000301.

M:=50; A:=[0]; miss:=[seq(n,n=1..M^2)]:
for n from 1 to M do t1:=miss[1]*miss[2]; A:=[op(A),t1];
miss:=[seq(miss[i],i=3..nops(miss))];
miss:=remove('x->x=t1',miss);
od:
A;

Typo in definition corrected by Douglas Latimer, Nov 01 2014

cp's OEIS Frontend

A299409 Solution (e(n)) of the system of 5 complementary equations in Comments.

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A299637 Solution (b(n)) of the system of 5 complementary equations in Comments.

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A299638 Solution (c(n)) of the system of 5 complementary equations in Comments.

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A299641 Solution (d(n)) of the system of 5 complementary equations in Comments.

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A075327 Sum of n-th pair in A075325.

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A297464 Solution (a(n)) of the system of 4 complementary equations in Comments.

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A332788 Positive-pan primes (see Comments).

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