cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-8 of 8 results.

A262408 Positive integers m such that pi(m^2) = pi(j^2) + pi(k^2) for no 0 < j <= k < m.

Original entry on oeis.org

1, 2, 5, 10, 21, 23, 46, 103, 105, 193, 222, 232, 285, 309, 345, 392, 404, 476, 587, 670, 779, 912, 1086, 1162, 1249, 2508, 2592, 2852, 2964, 3362, 3673, 3895, 4218, 4732, 5452, 6417, 7667, 7759, 8430, 8796, 9606, 11096, 11953, 12014, 12125, 13956, 14474, 15018, 17854, 18861, 18879, 19307, 22843, 28106, 29423, 31576, 37182
Offset: 1

Views

Author

Zhi-Wei Sun, Sep 21 2015

Keywords

Comments

Conjecture: (i) There are infinitely many positive integers m such that pi(m^2) = pi(x^2) + pi(y^2) for some 0 < x <= y < m. Also, the current sequence has infinitely many terms.
(ii) For every n = 4,5,... the equation pi(x^n) + pi(y^n) = pi(z^n) has no integral solution with 0 < x <= y < z.
It is interesting to compare this conjecture with Fermat's Last Theorem. See also A262409 for the equation pi(x^3) + pi(y^3) = pi(z^3).

Examples

			a(3) = 5 since pi(5^2) = 9 cannot be written as pi(j^2) + pi(k^2) with 0 < j <= k < 5. Note that pi(1^2) = 0, pi(2^2) = 2, pi(3^2) = 4 and pi(4^2) = 6 are all even.
		

References

  • Zhi-Wei Sun, Problems on combinatorial properties of primes, in: M. Kaneko, S. Kanemitsu and J. Liu (eds.), Number Theory: Plowing and Starring through High Wave Forms, Proc. 7th China-Japan Seminar (Fukuoka, Oct. 28 - Nov. 1, 2013), Ser. Number Theory Appl., Vol. 11, World Sci., Singapore, 2015, pp. 169-187.

Crossrefs

Programs

  • Mathematica
    f[n_]:=PrimePi[n^2]
    T[n_]:=Table[f[k],{k,1,n}]
    n=0;Do[Do[If[MemberQ[T[m-1],f[m]-f[k]],Goto[aa]],{k,1,m-1}];n=n+1;Print[n," ",m];Label[aa];Continue,{m,1,32000}]

A262443 Positive integers m such that pi(m^2) = pi(j^2)*pi(k^2) for some 0 < j < k < m, where pi(x) denotes the number of primes not exceeding x.

Original entry on oeis.org

8, 11, 14, 19, 20, 36, 38, 45, 66, 87, 91, 115, 139, 143, 152, 155, 201, 220, 227, 279, 357, 383, 391, 415, 418, 452, 476, 480, 489, 496, 500, 514, 521, 524, 549, 552, 557, 588, 595, 632, 653, 676, 706, 708, 749, 753, 761, 766, 820, 846, 863, 877, 922, 1009, 1038, 1041, 1044, 1052, 1057, 1080
Offset: 1

Views

Author

Zhi-Wei Sun, Sep 23 2015

Keywords

Comments

Conjecture: (i) The sequence has infinitely many terms. Also, there are infinitely many positive integers m such that pi(m^2) = pi(j^2)*pi(k^2) for no 0 < j <= k < m.
(ii) For any integer n > 2, the equation pi(x^n)*pi(y^n) = pi(z^n) has no solution with 0 < x <= y < z.

Examples

			 a(1) = 8 since pi(8^2) = pi(64) = 18 = 2*9 = pi(2^2)*pi(5^2) with 0 < 2 < 5 < 8.
a(4) = 19 since pi(19^2) = pi(361) = 72 = 4*18 = pi(3^2)*pi(8^2) with 0 < 3 < 8 < 19.
		

References

  • Zhi-Wei Sun, Problems on combinatorial properties of primes, in: M. Kaneko, S. Kanemitsu and J. Liu (eds.), Number Theory: Plowing and Starring through High Wave Forms, Proc. 7th China-Japan Seminar (Fukuoka, Oct. 28 - Nov. 1, 2013), Ser. Number Theory Appl., Vol. 11, World Sci., Singapore, 2015, pp. 169-187.

Crossrefs

Programs

  • Mathematica
    f[n_]:=PrimePi[n^2]
    T[n_]:=Table[f[k],{k,1,n}]
    Dv[n_]:=Divisors[f[n]]
    Le[n_]:=Length[Dv[n]]
    n=0;Do[Do[If[MemberQ[T[m],Part[Dv[m],i]]&&MemberQ[T[m],Part[Dv[m],Le[m]-i+1]],n=n+1;Print[n," ",m];Goto[aa]],{i,2,(Le[m]-1)/2}];Label[aa];Continue,{m,1,1080}]

A262403 Number of ways to write pi(T(n)) = pi(T(k)) + pi(T(m)) with 1 < k < m < n, where T(x) is the triangular number x*(x+1)/2, and pi(x) is the number of primes not exceeding x.

Original entry on oeis.org

0, 0, 0, 0, 1, 1, 1, 2, 2, 1, 2, 1, 3, 4, 4, 4, 3, 3, 3, 3, 5, 4, 3, 4, 6, 4, 5, 2, 3, 6, 4, 1, 5, 8, 3, 2, 6, 1, 4, 5, 4, 2, 7, 2, 4, 5, 5, 5, 3, 4, 9, 9, 4, 5, 4, 8, 7, 6, 9, 4, 7, 5, 6, 2, 5, 9, 3, 8, 5, 6, 8, 5, 4, 3, 8, 4, 8, 7, 8, 5, 7, 8, 7, 4, 6, 2, 7, 7, 8, 7, 4, 5, 6, 4, 6, 4, 6, 4, 6, 6
Offset: 1

Views

Author

Zhi-Wei Sun, Sep 21 2015

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 4, and a(n) = 1 only for n = 5, 6, 7, 10, 12, 32, 38, 445, 727.
(ii) All those numbers pi(T(n)) (n = 1,2,3,...) are pairwise distinct. Moreover, if sum_{i=j,...,k}1/pi(T(i)) and sum_{r=s,...,t}1/pi(T(r)) with 1 < j <= k and j <= s <= t have the same fractional part but the ordered pairs (j,k) and (s,t) are different, then j = 2, k = 5 and s = t = 4.
Clearly, part (i) is related to addition chains, and the first assertion in part (ii) is an analog of Legendre's conjecture that pi(n^2) < pi((n+1)^2) for all n = 1,2,3,....
See also A262408 and A262409 for related conjectures involving powers.

Examples

			a(5) = 1 since pi(T(5)) = pi(15) = 6 = 2 + 4 = pi(3) + pi(10) = pi(T(2)) + pi(T(4)).
a(6) = 1 since pi(T(6)) = pi(21) = 8 = 2 + 6 = pi(3) + pi(15) = pi(T(2)) + pi(T(5)).
a(7) = 1 since pi(T(7)) = pi(28) = 9 = 3 + 6 = pi(6) + pi(15) = pi(T(3)) + pi(T(5)).
a(10) = 1 since pi(T(10)) = pi(55) = 16 = 2 + 14 = pi(3) + pi(45) = pi(T(2)) + pi(T(9)).
a(12) = 1 since pi(T(12)) = pi(78) = 21 = 3 + 18 = pi(6) + pi(66) = pi(T(3)) + pi(T(11)).
a(32) = 1 since pi(T(32)) = pi(528) = 99 = 9 + 90 = pi(28) + pi(465) = pi(T(7)) + pi(T(30)).
a(38) = 1 since pi(T(38)) = pi(741) = 131 = 32 + 99 = pi(136) + pi(528) = pi(T(16)) + pi(T(32)).
a(445) = 1 since pi(T(445)) = pi(99235) = 9526 = 2963 + 6563 = pi(27028) + pi(65703) = pi(T(232)) + pi(T(362)).
a(727) = 1 since pi(T(727)) = pi(264628) = 23197 = 10031 + 13166 = pi(105111) + pi(141778) = pi(T(458)) + pi(T(532)).
		

References

  • R. K. Guy, Unsolved Problems in Number Theory, 3rd Edition, Springer, 2004. (Cf. Section C6 on addition chains.)
  • Zhi-Wei Sun, Problems on combinatorial properties of primes, in: M. Kaneko, S. Kanemitsu and J. Liu (eds.), Number Theory: Plowing and Starring through High Wave Forms, Proc. 7th China-Japan Seminar (Fukuoka, Oct. 28 - Nov. 1, 2013), Ser. Number Theory Appl., Vol. 11, World Sci., Singapore, 2015, pp. 169-187.

Crossrefs

Programs

  • Mathematica
    f[n_]:=PrimePi[n(n+1)/2]
    T[m_,n_]:=Table[f[k],{k,m,n}]
    Do[r=0;Do[If[MemberQ[T[k+1,n-1],f[n]-f[k]],r=r+1];Continue,{k,2,n-2}];Print[n," ",r];Continue,{n,1,100}]

A262462 Positive integers k with pi(k^3) a square, where pi(x) denotes the number of primes not exceeding x.

Original entry on oeis.org

1, 2, 3, 14, 1122
Offset: 1

Views

Author

Zhi-Wei Sun, Sep 23 2015

Keywords

Comments

Conjecture: (i) The Diophantine equation pi(x^2) = y^2 with x > 0 and y > 0 has infinitely many solutions.
(ii) The only solutions to the Diophantine equation pi(x^m) = y^n with {m,n} = {2,3}, x > 0 and y > 0 are as follows:
pi(89^2) = 10^3, pi(2^3) = 2^2, pi(3^3) = 3^2, pi(14^3) = 20^2 and pi(1122^3) = 8401^2.
(iii) For m > 1 and n > 1 with m + n > 5, the equation pi(x^m) = y^n with x > 0 and y > 0 has no integral solution.
The conjecture seems reasonable in view of the heuristic arguments.
Part (ii) of the conjecture implies that the only terms of the current sequence are 1, 2, 3, 14 and 1122.

Examples

			a(1) = 1 since pi(1^3) = 0^2.
a(2) = 2 since pi(2^3) = 2^2.
a(3) = 3 since pi(3^3) = 3^2.
a(4) = 14 since pi(14^3) = pi(2744) = 400 = 20^2.
a(5) = 1122 since pi(1122^3) = pi(1412467848) = 70576801 = 8401^2.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=IntegerQ[Sqrt[n]]
    f[n_]:=PrimePi[n^3]
    n=0;Do[If[SQ[f[k]],n=n+1;Print[n," ",k]],{k,1,1200}]
    Select[Range[1200],IntegerQ[Sqrt[PrimePi[#^3]]]&] (* Harvey P. Dale, Aug 21 2024 *)

A262698 Positive integers m such that pi(k^3) + pi(m^3) is a cube for some k = 1,...,m, where pi(x) denotes the number of primes not exceeding x.

Original entry on oeis.org

1, 2, 4, 24, 41, 51, 88, 95, 99, 179, 183, 663, 782, 829, 1339, 2054, 2816, 7918, 8474, 13264, 16664, 27415, 39514, 48606, 51145, 187222, 200906, 261980, 353209, 375162, 396967, 400469
Offset: 1

Views

Author

Zhi-Wei Sun, Sep 27 2015

Keywords

Comments

Conjecture: (i) There are infinitely many distinct primes p,q,r such that pi(p^2) + pi(q^2) = r^2.
(ii) The Diophantine equation pi(x^3) + pi(y^3) = z^3 with 0 < x <= y and z >= 0 only has the following 17 solutions: (x,y,z) = (1,1,0), (2,2,2), (3,4,3), (16,24,13), (3,41,19), (37,51,26), (53,88,41), (18,95,41), (45,99,44), (108,179,79), (149,183,87), (8,663,251), (243,782,297), (803,829,385), (100,1339,489), (674,2054,745), (1519,2816,1047).
(iii) The Diophantine equation pi(x^n) + pi(y^n) = z^n with n > 3 and x,y,z > 0 has no solution.
a(26) > 10^5, if it exists. Conjecture (ii) above is false since these further solutions exist: (1339, 7918, 2682), (3360, 8474, 2922), (8443, 13264, 4764), (15590, 16664, 6696), (15883, 27415, 9431), (9719, 39514, 12689), (22265, 48606, 15933), (38606, 51145, 18297). - Giovanni Resta, Jun 14 2020
Further solutions: (79522, 187222, 58554), (65281, 200906, 61833), (222863, 261980, 92917), (226465, 353209, 114585), (41559, 375162, 112168), (244967, 396967, 127399), (291034, 400469, 133443) - Chai Wah Wu, Apr 13 2021

Examples

			a(4) = 24 since pi(16^3) + pi(24^3) = pi(4096) + pi(13824) = 564 + 1633 = 2197 = 13^3.
		

References

  • Zhi-Wei Sun, Problems on combinatorial properties of primes, in: M. Kaneko, S. Kanemitsu and J. Liu (eds.), Number Theory: Plowing and Starring through High Wave Forms, Proc. 7th China-Japan Seminar (Fukuoka, Oct. 28 - Nov. 1, 2013), Ser. Number Theory Appl., Vol. 11, World Sci., Singapore, 2015, pp. 169-187.

Crossrefs

Programs

  • Mathematica
    f[n_]:=PrimePi[n^3]
    CQ[n_]:=IntegerQ[n^(1/3)]
    n=0;Do[Do[If[CQ[f[x]+f[y]],n=n+1;Print[n," ",y];Goto[aa]],{x,1,y}];Label[aa];Continue,{y,1,3000}]
  • PARI
    lista(nn) = {my(c=0, v=vector(nn)); for(m=1, nn, forprime(p=(m-1)^3+1, m^3, c++); v[m]=c; if(sum(k=1, m, ispower(v[k]+v[m], 3)), print1(m, ", "))); } \\ Jinyuan Wang, Jun 13 2020

Extensions

a(18)-a(25) from Giovanni Resta, Jun 14 2020
a(26)-a(29) from Chai Wah Wu, Apr 05 2021
a(30)-a(32) from Chai Wah Wu, Apr 09 2021

A262536 Positive integers z such that pi(x^3+y^3) = pi(z^3) for some 0 < x <= y < z, where pi(m) denotes the number of primes not exceeding m.

Original entry on oeis.org

7, 9, 11, 12, 34, 46, 49, 65, 89, 95, 103, 127, 144, 150, 163, 172, 206, 236, 249, 258, 275, 288, 300, 309, 312, 385, 492, 495, 505, 577, 641, 683, 729, 738, 751, 796, 835, 873, 904, 990, 995, 1010, 1154, 1210, 1297, 1312, 1403, 1458, 1476, 1502, 1544, 1626, 1661, 1731, 1808, 1852, 1985, 1988, 2020, 2059, 2107, 2214, 2304, 2316, 2370, 2448, 2594, 2833, 2840, 2883, 2920, 3073, 3088, 3097
Offset: 1

Views

Author

Zhi-Wei Sun, Sep 24 2015

Keywords

Comments

The sequence has infinitely many terms. In fact, for any integer t > 0, we have (6*t^2)^3 + (6*t^3-1)^3 = (6*t^3+1)^3 - 2 and hence pi((6*t^2)^3+(6*t^3-1)^3) = pi((6*t^3+1)^3) since neither (6*t^3+1)^3 nor (6*t^3+1)^3-1 is prime.
Concerning the equation pi(x^3+y^3) = pi(z^3) with 0 < x <= y < z, there are exactly 70 solutions with z <= 2700. They are (x,y,z) = (5,6,7),(6,8,9),(7,10,11),(9,10,12),(15,33,34),(23,44,46),(24,47,49),(43,58,65),(41,86,89),(47,91,95),(64,94,103),(95,106,127),(71,138,144),(73,144,150),(54,161,163),(135,138,172),(128,188,206),(55,235,236),(135,235,249),(197,212,258),(159,256,275),(142,276,288),(146,288,300),(192,282,309),(161,297,312),(96,383,385),(252,345,385),(390,391,492),(334,438,495),(372,426,505),(426,486,577),(297,619,641),(353,650,683),(242,720,729),(244,729,738),(150,749,751),(602,659,796),(161,833,835),(470,825,873),(566,823,904),(668,876,990),(514,947,995),(744,852,1010),(791,812,1010),(509,1120,1154),(852,972,1154),(236,1207,1210),(216,1295,1297),(459,1293,1312),(915,1259,1403),(484,1440,1458),(488,1458,1476),(300,1498,1502),(368,1537,1544),(511,1609,1626),(420,1652,1661),(1278,1458,1731),(1132,1646,1808),(1033,1738,1852),(1241,1808,1985),(1010,1897,1988),(1582,1624,2020),(294,2057,2059),(237,2106,2107),(732,2187,2214),(575,2292,2304),(577,2304,2316),(1518,2141,2370),(1611,2189,2448),(432,2590,2594).
Recall Fermat's Last Theorem, which asserts that the Diophantine equation x^n + y^n = z^n with n > 2 and x,y,z > 0 has no solution. In 1936 K. Mahler discovered that
(9*t^3+1)^3 + (9*t^4)^3 - (9*t^4+3*t)^3 = 1.
Conjecture: (i) For any integers n > 3 and x,y,z > 0 with {x,y} not equal to {1,z}, we have |x^n+y^n-z^n| >= 2^n-2, unless n = 5, {x,y} = {13,16} and z = 17.
(ii) For any integer n > 3 and x,y,z > 0 with {x,y} not containing z, there is a prime p with x^n+y^n < p < z^n or z^n < p < x^n+y^n, unless n = 5, {x,y} = {13,16} and z = 17.
(iii) For any integers n > 3, x > y >= 0 and z > 0 with x not equal to z, there always exists a prime p with x^n-y^n < p < z^n or z^n < p < x^n-y^n.
We have verified part (i) of the conjecture for n = 4..10 and 0 < x,y,z <= 1700.

Examples

			a(1) = 7 since pi(5^3+6^3) = pi(125+216) = pi(341) = 68 = pi(343) = pi(7^3).
a(2) = 9 since pi(6^3+8^3) = pi(216+512) = pi(728) = 129 = pi(729) = pi(9^3).
a(50) = 1502 since pi(300^3+1498^3) = pi(27000000+3361517992) = pi(3388517992) = 162202081 = pi(3388518008) = pi(1502^3).
		

References

  • Zhi-Wei Sun, Problems on combinatorial properties of primes, in: M. Kaneko, S. Kanemitsu and J. Liu (eds.), Number Theory: Plowing and Starring through High Wave Forms, Proc. 7th China-Japan Seminar (Fukuoka, Oct. 28 - Nov. 1, 2013), Ser. Number Theory Appl., Vol. 11, World Sci., Singapore, 2015, pp. 169-187.

Crossrefs

Programs

  • Mathematica
    pi[n_]:=PrimePi[n]
    n=0;Do[Do[If[pi[x^3+y^3]==pi[z^3],n=n+1;Print[n," ",z];Goto[aa]],{x,1,z-1},{y,x,z-1}];Label[aa];Continue,{z,1,2700}]

A262730 Primes p such that p^2 = pi(x^3) + pi(y^3) for some positive integers x and y, where pi(m) denotes the number of primes not exceeding m.

Original entry on oeis.org

2, 3, 23, 83, 199, 331, 487, 1069, 1289, 1697, 2467, 3463, 3617, 3733, 5153, 5449, 6221, 9203, 9811, 9967, 12473, 13883, 14723, 15791, 16001, 18919, 33589, 33827, 46093, 58321, 59051, 59921, 60289, 71249, 84349, 85133, 88211, 124309, 126047, 126359, 127541, 145679, 146807, 153247, 165233
Offset: 1

Views

Author

Zhi-Wei Sun, Sep 28 2015

Keywords

Comments

Conjecture: (i) The sequence has infinitely many terms.
(ii) There are infinitely many primes p such that p^2 = pi(x^3+y^3) for some positive integers x and y.
See also A262731 for a related conjecture.

Examples

			a(1) = 2 since pi(1^3)+pi(2^3) = 0+4 = 2^2 with 2 prime.
a(3) = 23 since pi(9^3)+pi(14^3) = pi(729)+pi(2744) = 129+400 = 529 = 23^2 with 23 prime.
a(20) = 9967 since pi(841^3)+pi(1109^3) = pi(594823321)+pi(1363938029) = 31068537+68272552 = 99341089 = 9967^2 with 9967 prime.
a(38) = 124309 since pi(5773^3)+pi(5779^3) = pi(192399824917)+pi(193000344139) = 7714808769+7737918712 = 15452727481 = 124309^2 with 124309 prime.
a(45) = 165233 since pi(6924^3)+pi(7148^3) = pi(331948857024)+pi(365219225792) = 13025048890+14276895399 = 27301944289 = 165233^2 with 165233 prime.
		

References

  • Zhi-Wei Sun, Problems on combinatorial properties of primes, in: M. Kaneko, S. Kanemitsu and J. Liu (eds.), Number Theory: Plowing and Starring through High Wave Forms, Proc. 7th China-Japan Seminar (Fukuoka, Oct. 28 - Nov. 1, 2013), Ser. Number Theory Appl., Vol. 11, World Sci., Singapore, 2015, pp. 169-187.

Crossrefs

Programs

  • Mathematica
    f[n_]:=PrimePi[n^3]
    T[1]:={0}
    T[n_]:=Union[T[n-1],{f[n]}]
    n=0;Do[Do[If[f[x]>Prime[y]^2,Goto[aa]];If[MemberQ[T[Prime[y]],Prime[y]^2-f[x]],n=n+1;Print[n," ",Prime[y]];Goto[aa]];Continue,{x,1,Prime[y]}];
    Label[aa];Continue,{y,1,15111}]

A262722 Positive integers m such that pi(k^3+m^3) is a cube for some k = 1..m, where pi(x) denotes the number of primes not exceeding x.

Original entry on oeis.org

1, 41, 56, 74, 103, 157, 384, 491, 537, 868, 1490, 1710, 4322, 4523, 4877, 4942, 5147, 5407, 7564, 17576, 67722, 131455, 220641, 438895, 443475, 553878, 571473, 625611
Offset: 1

Views

Author

Zhi-Wei Sun, Sep 28 2015

Keywords

Comments

Conjecture: (i) There are infinitely many distinct primes p,q,r such that pi(p^2+q^2) = r^2.
(ii) The Diophantine equation pi(x^3+y^3) = z^3 with 0 < x <= y and z > 0 only has the following 13 solutions: (x,y,z) = (1,1,1), (5,41,19), (47,56,29), (28,74,33), (2,103,44), (3,103,44), (6,157,65), (235,384,160), (266,491,198), (91,537,206), (359,868,331), (783,1490,565), (1192,1710,677).
(iii) The Diophantine equation pi(x^n+y^n) = z^n with n > 3 and x,y,z > 0 has no solution.
Part (ii) of the conjecture implies that the current sequence only has 12 terms as shown here.
Conjecture (ii) is false as there are more terms beyond 1710. It is likely the sequence has an infinite number of terms. (x,y,z) for 1710 < y <= 7564: (1429, 4322, 1514), (1974, 4523, 1604), (3361, 4877, 1840), (3992, 4942, 1949), (3253, 5147, 1902), (971, 5407, 1859), (935, 7564, 2563). - Chai Wah Wu, Apr 12 2021
Solutions (x,y,z) for 7564 < y <= 67722: (3484, 17576, 5783), (25184, 67722, 21604). - Chai Wah Wu, Apr 17 2021
Solutions (x,y,z) for 67722 < y <= 625611: (61021, 131455, 41715), (93577, 220641, 68507), (394510, 438895, 155930), (3086, 443475, 131933), (338485, 553878, 175133), (239982, 571473, 172855), (610794, 625611, 228409). - Chai Wah Wu, Apr 26 2021

Examples

			a(2) = 41 since pi(5^3+41^3) = pi(125+68921) = pi(69046) = 6859 = 19^3.
		

References

  • Zhi-Wei Sun, Problems on combinatorial properties of primes, in: M. Kaneko, S. Kanemitsu and J. Liu (eds.), Number Theory: Plowing and Starring through High Wave Forms, Proc. 7th China-Japan Seminar (Fukuoka, Oct. 28 - Nov. 1, 2013), Ser. Number Theory Appl., Vol. 11, World Sci., Singapore, 2015, pp. 169-187.

Crossrefs

Programs

  • Mathematica
    f[x_,y_]:=PrimePi[x^3+y^3]
    CQ[n_]:=IntegerQ[n^(1/3)]
    n=0;Do[Do[If[CQ[f[x,y]],n=n+1;Print[n," ",y];Goto[aa]],{x,1,y}];Label[aa];Continue,{y,1,1800}]

Extensions

a(13)-a(19) from Chai Wah Wu, Apr 12 2021
a(20)-a(21) from Chai Wah Wu, Apr 17 2021
a(22)-a(28) from Chai Wah Wu, Apr 26 2021
Showing 1-8 of 8 results.