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 1-3 of 3 results.

A378501 Integers m such that A379816(m) = A379815(m) + m.

Original entry on oeis.org

3, 5, 6, 7, 10, 11, 13, 14, 15, 17, 19, 21, 22, 23, 24, 26, 27, 28, 29, 30, 31, 33, 34, 35, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 49, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 65, 66, 67, 68, 69, 70, 71, 73, 74, 75, 76, 77, 78, 79, 82, 83, 85, 86, 87
Offset: 1

Views

Author

Michel Marcus, Feb 10 2025

Keywords

Crossrefs

Complement of A380743.

Programs

  • PARI
    isok(m) = for(k=m+1, 2*m, if(issquare(1/m-1/k), return(0))); m>4||m==3; \\ Jinyuan Wang, Feb 11 2025

Extensions

More terms from Jinyuan Wang, Feb 11 2025

A380743 Integers m such that A379816(m) != A379815(m) + m.

Original entry on oeis.org

1, 2, 4, 8, 9, 12, 16, 18, 20, 25, 32, 36, 48, 50, 63, 64, 72, 80, 81, 84, 90, 98, 100, 108, 117, 128, 144, 150, 162, 180, 192, 198, 200, 225, 242, 252, 256, 272, 275, 288, 300, 306, 320, 324, 336, 338, 350, 360, 363, 392, 400, 432, 441, 450, 468, 500, 507, 512, 525, 528, 539
Offset: 1

Views

Author

Michel Marcus, Feb 11 2025

Keywords

Crossrefs

Complement of A378501.

A379816 a(n) is the smallest integer k > n such that sqrt(1/n - 1/k) is a rational number; or 0 if no such k exists.

Original entry on oeis.org

0, 4, 12, 0, 25, 18, 448, 16, 12, 100, 1100, 18, 4225, 112, 240, 18, 289, 36, 549100, 25, 588, 2178, 13248, 72, 45, 676, 108, 126, 142129, 180, 71622400, 64, 396, 612, 1260, 48, 1369, 722, 507, 400, 42025, 294, 521345932, 242, 225, 559728, 108288, 72, 112, 100, 127500, 169, 1755625, 162, 2475, 448, 4332, 568516, 16573100, 150
Offset: 1

Views

Author

Felix Huber, Feb 07 2025

Keywords

Comments

a(1) = a(4) = 0. Proof: See Huber link.
k > n exists for n > 4.
Continues a(61) <= 53872731025, a(62) = 1984, a(63) = 112, a(64) = 72, a(65) = 4225, a(66) = 2178, a(67) <= 159831244588, a(68) = 1156, a(69) = 268272, a(70) = 8820, a(71) <= 859838400, a(72) = 144, a(73) <= 83265625, a(74) = 136900, a(75) = 300, a(76) = 6498, a(77) = 13552, a(78) = 2106, a(79) = 505600, a(80) = 100, a(81) = 108, a(82) = 6724, a(83) = 558092, a(84) = 147, a(85) = 12145225, a(86) = 447458, a(87) = 68208, a(88) = 8712, a(89) = 22250089, a(90) = 100, a(91) = 225450316, a(92)=1150, a(93) = 565068, a(97) <= 3046267249, a(101) = 10201, a(103) <= 5332206050752, a(107) = 99022508. - R. J. Mathar, Feb 21 2025
For nonsquare n, solutions k (not necessarily the smallest ones) are given by k= n*A002350(n)^2 such that 1/n-1/k= (A002349(n)/A002350(n))^2. - R. J. Mathar, Feb 25 2025
For n=p^2, squared odd primes, solutions k (not necessarily the smallest) are constructed by k=p*(p+1)^2/4 such that 1/n -1/k = 1/p^2-1/k = [(p-1)/(p*(p+1))]^2 is a rational square. For other perfect squares n, start from such a solution for a prime factor p|n, n=(p*f)^2, and multiply the 3 terms of both sides of that solution with 1/f^2 to find a solution for n. - R. J. Mathar, Feb 25 2025

Examples

			a(16) = 18 because sqrt(1/16 - 1/17) = sqrt(1/272) is irrational but sqrt(1/16 - 1/18) = sqrt(1/144) = 1/12 is rational.
		

Crossrefs

Programs

  • Maple
    A379816:=proc(n)
        local k;
        if n=1 or n=4 then
            return 0
        else
            for k from n+1 do
                if type(sqrt(1/n-1/k),rational) then
                    return k
                fi
            od
        fi;
    end proc;
    seq(A379816(n),n=1..58);
  • PARI
    a(n) = if ((n==1) || (n==4), return(0)); my(k=n+1); while (!issquare(1/n - 1/k), k++); k; \\ Michel Marcus, Feb 08 2025

Extensions

Terms a(59-60) from R. J. Mathar, Feb 12 2025
Showing 1-3 of 3 results.