A145023 -id:A145023 - cp's OEIS frontend

A145047 Primes p of the form 4k+1 for which s=10 is the least positive integer such that sp-(floor(sqrt(sp)))^2 is a square.

1237, 1621, 1721, 1933, 1949, 1993, 2221, 2237, 2309, 2341, 2473, 2621, 2657, 2789, 2797, 2857, 2953, 3221, 3361, 3533, 3677, 3881, 3889, 3917, 4133, 4457, 4481, 4549, 4813, 4889, 4973, 5153, 5189, 5261, 5441, 5653, 5717, 5813, 6101, 6217, 6301, 6329

Offset: 1

Vladimir Shevelev, Sep 30 2008, Oct 05 2008

nonn

Conjecture: The least positive integer s can take values only from A008784 (see for s=1,2,5,10 sequences A145016, A145022, A145023 and this sequence).

			a(1)=1237 since p=1237 is the least prime of the form 4k+1 for which sp-(floor(sqrt(sp)))^2 is not a square for s=1..9, but 10p-(floor(sqrt(10p)))^2 is a square (for p=1237 it is 49).

Cf. A145016, A145017, A145022, A145023, A008784.

A145048 Primes p of the form 4k+1 for which s=13 is the least positive integer such that sp-(floor(sqrt(sp)))^2 is a square.

Original entry on oeis.org

2749, 2897, 3049, 3529, 3557, 3929, 4073, 4253, 4657, 4817, 5081, 5281, 5417, 5449, 5657, 5693, 5869, 6053, 6121, 6529, 6793, 6833, 7109, 7393, 7541, 7829, 7877, 7993, 8209, 8329, 8377, 8429, 8501, 8741, 8761, 8893, 9001, 9109, 9157, 9209, 9257, 9293

Offset: 1

Vladimir Shevelev, Sep 30 2008

nonn

			a(1)=2749 since p=2749 is the least prime of the form 4k+1 for which sp-(floor(sqrt(sp)))^2 is not a square for s=1..12, but 13p-(floor(sqrt(13p)))^2 is a square (for p=2749 it is 16).

Cf. A145016, A145022, A145023, A145047.

A145049 Primes p of the form 4k+1 for which s=17 is the least positive integer such that sp-(floor(sqrt(sp)))^2 is a square.

Original entry on oeis.org

3037, 3169, 3257, 3769, 4013, 4421, 4793, 4957, 5237, 5297, 5701, 5821, 5881, 6373, 6689, 6761, 6949, 7013, 7213, 7417, 7481, 7549, 7621, 7757, 8389, 8461, 8537, 8681, 8753, 9049, 9133, 9277, 9349, 9733, 10133, 10529, 10601, 11093, 11177, 11257, 11677, 11701

Offset: 1

Vladimir Shevelev, Sep 30 2008

nonn

			a(1)=3037 since p=3037 is the least prime of the form 4k+1 for which sp-(floor(sqrt(sp)))^2 is not a square for s=1..16, but 17p-(floor(sqrt(17p)))^2 is a square (for p=3037 it is 100).

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A145016, A145022, A145023, A145047, A145048.

filter:= proc(p) local s;
  if not isprime(p) then return false fi;
  for s from 1 to 17 do
    if issqr(s*p - floor(sqrt(s*p))^2) then return evalb(s=17) fi
  od;
  false
end proc:
select(filter, [seq(i,i=1..10000,4)]); # Robert Israel, Jan 22 2024

A145236 a(n) is the least positive integer such that if p_n is the n-th prime then (ceiling(sqrt(a(n)p_n)))^2 - a(n)p_n is a perfect square.

Original entry on oeis.org

2, 1, 1, 3, 5, 5, 9, 9, 13, 17, 19, 23, 25, 27, 31, 35, 41, 41, 47, 51, 51, 57, 61, 65, 73, 75, 77, 81, 83, 85, 99, 101, 107, 109, 117, 119, 125, 129, 133, 139, 145, 145, 155, 157, 161, 163, 173, 183, 187, 189, 193, 199, 201, 209, 215, 221, 225, 227, 233, 237, 239, 247

Offset: 1

Vladimir Shevelev, Oct 05 2008, Oct 07 2008

nonn

Conjectures: 1) for n >= 2, the sequence does not decrease; 2) for n > 1, a(n) is odd; 3) a(n) can be equal to a(n+1) only for twins: p_(n+1) - p_n = 2 (although there also exist twins for which a(n) < a(n+1)).

All these conjectures are proved using the formula a(n) = p_n - 2*floor(sqrt(2p_n)) + 2, n > 1. See also A145701 and A145714. - Vladimir Shevelev, Oct 18 2008

Cf. A145016, A145022, A145023, A145047, A145048, A145049, A145050, A145215.

A145236 := proc(n) local p,k,a ; p := ithprime(n) ; for k from 1 do ceil(sqrt(ceil(k*p))) ; a := %^2-k*p ; if issqr(a) then return k ; end if; end do: end proc:
for n from 1 do printf("%d,\n",A145236(n)) ; end do: # R. J. Mathar, Aug 02 2010

a(12)=23 (not 21). - Vladimir Shevelev, Oct 16 2008

Extended by R. J. Mathar, Aug 02 2010

A145050 Primes p of the form 4k+1 for which s=26 is the least positive integer such that sp-(floor(sqrt(s*p)))^2 is a square.

Original entry on oeis.org

6569, 8117, 8689, 9221, 9281, 9829, 10289, 10457, 11597, 11953, 12577, 12721, 13093, 14561, 15737, 15817, 16529, 17041, 17341, 17737, 18089, 18397, 19121, 19997, 20129, 20693, 20789, 21601, 21701, 22093, 22433, 22777, 22877, 23029, 23633, 23833, 24809, 25589

Offset: 1

Vladimir Shevelev, Sep 30 2008, Oct 03 2008

nonn

For all primes of the form 4*k+1 not exceeding 10000 the least integer s takes only values: 1, 2, 5, 10, 13, 17, 26. These values are the first numbers in A145017 (see our conjecture at A145047).

			a(1)=6569 since p=6569 is the least prime of the form 4*k+1 for which s*p-(floor(sqrt(s*p)))^2 is not a square for s=1..25, but 26*p-(floor(sqrt(26*p)))^2 is a square (for p=6569 it is 225).

Cf. A145016, A145017, A145022, A145023, A145043, A145047, A145048, A145049.

More terms from Jinyuan Wang, Jul 16 2025

A145017 Squarefree positive integers k for which k-(floor(sqrt(k)))^2 is a perfect square.

Original entry on oeis.org

1, 2, 5, 10, 13, 17, 26, 29, 34, 37, 53, 58, 65, 73, 82, 85, 97, 101, 109, 122, 130, 137, 145, 170, 173, 178, 185, 194, 197, 205, 221, 226, 229, 241, 257, 265, 281, 290, 293, 298, 305, 314, 349, 362, 365, 370, 377, 386, 397, 401, 409, 442, 445, 457, 466, 485, 493

Offset: 1

Vladimir Shevelev, Sep 29 2008

nonn

If an odd prime p divides a(n) then it has the form 4k+1.

Conjecture. For every n>=1 there exist infinitely many primes p of the form 4k+1 for which for a(n) > 1 we have s*p-(floor(sqrt(s*p)))^2 is not a perfect square for s=1,...,a(n)-1 while a(n)*p-(floor(sqrt(a(n)*p)))^2 is a perfect square. (See A145016(s=1) and A145022, A145023, A145047, A145048, A145049, A145050 correspondingly for s=2, s=5, s=10, s=13, s=17, s=26.) - Vladimir Shevelev, Sep 30 2008

Andrew Howroyd, Table of n, a(n) for n = 1..1000

Cf. A005117, A020893, A145016, A145022, A145023, A145047, A145048, A145049, A145050.

Select[Range@ 500, And[SquareFreeQ@ #, IntegerQ@ Sqrt[# - Floor[Sqrt@ #]^2]] &] (* Michael De Vlieger, Jan 12 2020 *)

is(n)={issquarefree(n) && issquare(n-sqrtint(n)^2)} \\ Andrew Howroyd, Jan 12 2020

Missing a(40) inserted and terms a(42) and beyond from Andrew Howroyd, Jan 12 2020

A145215 a(n) is the minimal prime of the form 4k+1 for which s=A008784(n) is the minimal positive integer such that sa(n)-floor(sqrt(sa(n)))^2 is a square.

Original entry on oeis.org

5, 41, 353, 1237, 2749, 3037, 10369, 6569, 27253, 38561, 14897, 33289, 27917, 171629, 143513, 76081, 37649, 373273, 399181, 63029, 133157, 637601, 425197, 94261, 499321, 910853, 229849, 149837

Offset: 1

Vladimir Shevelev, Oct 05 2008

nonn

See the conjecture in the comment at A145047. In addition, I conjecture that for every such s there exist infinitely many primes of the form 4k+1.

Cf. A145016, A145022, A145023, A145047-A145050, A008487.

f(s)=forprime(p=2,,if(p%4>1 || !issquare(s*p-sqrtint(s*p)^2),next);for(i=1,s-1,if(issquare(i*p-sqrtint(i*p)^2), next(2)));return(p))
S=select(n->if(n%2==0, if(n%4, n/=2, return(0))); n==1||vecmax(factor(n)[, 1]%4)==1, vector(150,i,i));
apply(f, S) \\ Charles R Greathouse IV, Feb 07 2013

a(22) corrected by Charles R Greathouse IV, Feb 07 2013

A245474 a(n) = smallest positive integer s such that sn - floor(sqrt(sn))^2 is a square.

Original entry on oeis.org

1, 1, 1, 3, 1, 1, 6, 7, 1, 1, 1, 11, 3, 1, 14, 3, 1, 1, 2, 19, 1, 21, 22, 23, 6, 1, 1, 3, 7, 1, 3, 31, 2, 33, 1, 35, 1, 1, 38, 6, 1, 2, 42, 43, 11, 1, 46, 47, 3, 1, 1, 3, 2, 1, 6, 55, 14, 57, 1, 59, 6, 2, 62, 7, 1, 1, 66, 67, 1, 69, 35, 71, 2, 1, 2, 3, 19

Offset: 0

Thomas Ordowski, Jul 23 2014

nonn

a(n) <= n for n > 0. If prime p == 3 (mod 4) then a(p) = p.
Conjecture: a(p) < p for prime p == 1 (mod 4).
Outline of proof of conjecture: write p = x^2 + y^2.  Since gcd(x,y) = 1, there are u,v with x*u + y*v = 1, u^2 + v^2 < y^2 + x^2 = p.  Taking s = u^2 + v^2, s*p = (u*y+v*x)^2 + 1^2, and |u*y+v*x| = floor(sqrt(s*p)). - Robert Israel, Aug 04 2014
For the first 100000 primes p == 1 (mod 4), a(p) < sqrt(p)/2. - Robert Israel, Aug 03 2014

Robert Israel, Table of n, a(n) for n = 0..10000

Cf. A145022, A145023, A363340.

A:= proc(n) local s,a;
     for s from 1 do
       a:= floor(sqrt(s*n));
       if issqr(s*n-a^2) then return s fi
     od
end proc:
seq(A(n),n=0..1000); # Robert Israel, Jul 23 2014

a245474[n_Integer] := Catch[
  Do[
   If[IntegerQ[Sqrt[(s*n - Floor[Sqrt[s*n]]^2)]] == True, Throw[s]],
   {s, n}]
  ]; Map[a245474, Range[100]] (* Michael De Vlieger, Aug 03 2014 *)

a(n) = s=1; while(!issquare(s*n-sqrtint(s*n)^2), s++); s \\ Colin Barker, Jul 23 2014

More terms from Colin Barker, Jul 23 2014

A145043 Primes p for which s=2 is the least positive integer such that sp-t(sp) is a triangular number, where t(n) is the maximal triangular number not exceeding n.

Original entry on oeis.org

5, 17, 23, 47, 53, 71, 73, 127, 173, 233, 251, 269, 281, 347, 353, 359, 431, 487, 491, 509, 541, 563, 569, 593, 613, 677, 743, 773, 827, 857, 863, 883, 929, 953, 977, 1013, 1153, 1187, 1283, 1319, 1361, 1373, 1439, 1481, 1583, 1613, 1619, 1709, 1871, 1997

Offset: 1

Vladimir Shevelev, Sep 30 2008

nonn

Cf. A000217, A145032, A145016, A145022, A145023.

maxt(n) = {k = 1; while ((t=k*(k+1)/2) < n, k++); if (t == n, t, k--; k*(k+1)/2);}
lista(nn) = {forprime(p = 2, nn, if (! ispolygonal(p - maxt(p), 3) && ispolygonal(2*p - maxt(2*p), 3), print1(p, ", ");););} \\ Michel Marcus, Jul 25 2014

More terms from Michel Marcus, Jul 25 2014

A145281 a(n) is the least prime such that (ceiling(sqrt(a(n)p_n)))^2 - a(n)p_n is a perfect square, where p_n is the n-th prime.

Original entry on oeis.org

2, 3, 3, 3, 5, 5, 11, 11, 13, 17, 19, 23, 29, 29, 31, 37, 41, 41, 47, 53, 53, 59, 61, 67, 73, 79, 79, 83, 83, 89, 101, 101, 107, 109, 127, 127, 127, 131, 137, 139, 149, 149, 157, 157, 163, 163, 173, 191, 191, 191, 193, 199, 211, 211, 223, 223, 227, 227, 233, 239

Offset: 1

Vladimir Shevelev, Oct 06 2008, Oct 07 2008

nonn

Theorem. p_n - 2*sqrt(2p_n) + 2= A145236(n).
Or a(n) is the least prime q_n <= p_n such that sqrt(p_n) - sqrt(q_n) < sqrt(2) [or (p_n + q_n)/2 < sqrt(p_n*q_n) + 1]. See also our comment to A145300. - Vladimir Shevelev, Oct 09 2008
The above conjecture is true. This means that a(n) is the nearest prime p > p_n - 2*floor(sqrt(2*p_n)) + 2. A considerably more important and deep question is whether p < p_n. The answer does not follow even from the Riemann conjecture about zeros of the zeta function. - Vladimir Shevelev, Oct 17 2008

Cf. A145236, A000040, A145016, A145022, A145023, A145047, A145048, A145049, A045050.

a(n) = {my(p = prime(n)); my(q = 2); while (! issquare(ceil(sqrt(q*p))^2 - q*p), q = nextprime(q+1)); q;} \\ Michel Marcus, Jul 06 2015

More terms from Michel Marcus, Jul 06 2015

cp's OEIS Frontend

A145047 Primes p of the form 4k+1 for which s=10 is the least positive integer such that sp-(floor(sqrt(sp)))^2 is a square.

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A145048 Primes p of the form 4k+1 for which s=13 is the least positive integer such that sp-(floor(sqrt(sp)))^2 is a square.

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A145049 Primes p of the form 4k+1 for which s=17 is the least positive integer such that sp-(floor(sqrt(sp)))^2 is a square.

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A145236 a(n) is the least positive integer such that if p_n is the n-th prime then (ceiling(sqrt(a(n)*p_n)))^2 - a(n)*p_n is a perfect square.

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A145050 Primes p of the form 4*k+1 for which s=26 is the least positive integer such that s*p-(floor(sqrt(s*p)))^2 is a square.

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A145017 Squarefree positive integers k for which k-(floor(sqrt(k)))^2 is a perfect square.

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Mathematica

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A145215 a(n) is the minimal prime of the form 4k+1 for which s=A008784(n) is the minimal positive integer such that s*a(n)-floor(sqrt(s*a(n)))^2 is a square.

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A245474 a(n) = smallest positive integer s such that s*n - floor(sqrt(s*n))^2 is a square.

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Maple

Mathematica

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Extensions

A145043 Primes p for which s=2 is the least positive integer such that sp-t(sp) is a triangular number, where t(n) is the maximal triangular number not exceeding n.

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Extensions

A145236 a(n) is the least positive integer such that if p_n is the n-th prime then (ceiling(sqrt(a(n)p_n)))^2 - a(n)p_n is a perfect square.

A145050 Primes p of the form 4k+1 for which s=26 is the least positive integer such that sp-(floor(sqrt(s*p)))^2 is a square.

A145215 a(n) is the minimal prime of the form 4k+1 for which s=A008784(n) is the minimal positive integer such that sa(n)-floor(sqrt(sa(n)))^2 is a square.

A245474 a(n) = smallest positive integer s such that sn - floor(sqrt(sn))^2 is a square.

A145281 a(n) is the least prime such that (ceiling(sqrt(a(n)p_n)))^2 - a(n)p_n is a perfect square, where p_n is the n-th prime.