A028871 -id:A028871 - cp's OEIS frontend

A241537 Cubes c such that c + 1234567890 is prime where 1234567890 is the first pandigital number with digits in order.

1, 50653, 79507, 456533, 571787, 1295029, 1685159, 1771561, 2248091, 2685619, 3307949, 4173281, 7880599, 9393931, 10218313, 10793861, 11697083, 17373979, 18191447, 22665187, 30664297, 47045881, 70444997, 111284641, 146363183, 151419437, 156590819, 192100033

Offset: 1

K. D. Bajpai, Apr 25 2014

nonn

			50653 = 37^3 and appears in the sequence because 50653 + 1234567890 = 1234618543, which is prime.
79507 = 43^3  and appears in the sequence because 79507 + 1234567890 = 1234647397, which is prime.
64000 = 40^3 but not included in the sequence since 64000 + 1234567890 = 1234631890 = (2)*(5)*(29389)*(4201), which is not prime.

K. D. Bajpai, Table of n, a(n) for n = 1..10000

Cf. A000040, A002496, A028871, A050289, A056899, A234812, A235640, A240587.

KD := proc() local a,c; c:=n^3;a:=c+1234567890; if isprime(a) then RETURN (c); fi; end: seq(KD(), n=1..1000);

lst={}; Do[c=n^3; If[PrimeQ[c+1234567890], AppendTo[lst,c]], {n,1,1000}]; lst
(*For the b-file*)  m=0; c=n^3; a=c+1234567890; Do[If[PrimeQ[a],m++; Print[m," ",c]], {n,1,4*10^5}]

s=[]; for(n=1, 1000, c=n^3; if(isprime(c+1234567890), s=concat(s, c))); s \\ Colin Barker, Apr 25 2014

A241538 Squares s such that s + 1234567890 is prime.

Original entry on oeis.org

1, 169, 1681, 6889, 8281, 11881, 24649, 27889, 41209, 57121, 58081, 67081, 80089, 101761, 124609, 175561, 185761, 201601, 212521, 332929, 380689, 413449, 461041, 508369, 534361, 609961, 625681, 654481, 683929, 693889, 822649, 829921, 833569, 1014049, 1018081

Offset: 1

K. D. Bajpai, Apr 25 2014

1234567890 is the first pandigital number with digits in order.

			169 = 13^2 and appears in the sequence because 169 + 1234567890 = 1234568059, which is prime.
1681 = 41^2  and appears in the sequence because 1681 + 1234567890 = 1234569571, which is prime.
625 = 25^2 but is not included in the sequence since 625 + 1234567890 = 1234568515 = (5)*(246913703), which is not prime.

K. D. Bajpai, Table of n, a(n) for n = 1..10000

Cf. A000040, A002496, A028871, A050289, A056899, A234812, A235640, A240587.

KD := proc() local a,s; s:=n^2;a:=s+1234567890; if isprime(a) then RETURN (s); fi; end: seq(KD(), n=1..2000);

A241538 = {}; Do[s = n^2; If[PrimeQ[s + 1234567890], AppendTo[A241538, s]], {n, 2000}]; A241538
(* For the b-file *) c = 0; s = n^2; a = s + 1234567890; Do[If[PrimeQ[a], c++; Print[c, " ", s]], {n, 4*10^5}] (* Bajpai *)
Select[Range[1000]^2, PrimeQ[# + 1234567890] &] (* Alonso del Arte, Apr 25 2014 *)

A271725 T(n,k) is an array read by rows, with n > 0 and k=1..4, where row n gives four prime numbers in increasing order with locations in right angles of each concentric square drawn on a distorted version of the Ulam spiral.

Original entry on oeis.org

3, 7, 17, 19, 13, 23, 37, 41, 307, 359, 401, 419, 13807, 14159, 14401, 14519, 41413, 42023, 42437, 42641, 6317683, 6325223, 6330257, 6332771, 22958473, 22972847, 22982437, 22987229, 39081253, 39100007, 39112517, 39118769, 110617807, 110649359, 110670401, 110680919

Offset: 1

Michel Lagneau, Apr 13 2016

nonn

See the illustration for more information.
Conjecture: there is an infinity of concentric squares having a prime number in each right angle. The number 5 is the center of all the squares.
It seems that the drawing of an infinite number of concentric squares having a prime number in each corner is impossible in an Ulam spiral. But with a slight distortion of this space, the problem becomes possible.
The illustration (see the link) shows the new version of a spiral with two remarkable orthogonal diagonals containing four classes of prime numbers given by the sequences A125202, A121326, A028871 and A073337 supported by four line segments. These intersect at a single point represented by the prime number 5.
The sequence of the corresponding length of the sides is {s(k)} = {2, 4, 18, 118, 204, 2514, 4792, 6252, 10518, 14032, 16752, 17598, ...}
The primes are defined by the polynomials: [4*m^2-10*m+7, (2*m-1)^2-2, 4*m^2+1, 4*(m+1)^2-6*(m+1)+1]. The sequence of the corresponding m is {b(k)} = {2, 3, 10, 60, 103, 1258, 2397, 3127, 5260, 7017, 8377, 8800, 10375, 11518, 11523, 12498, 15415, 15888, ...} with the relation b(k) = 1 + s(k)/2.
The array begins:
      3,     7,    17,    19;
     13,    23,    37,    41;
    307,   359,   401,   419;
  13807, 14159, 14401, 14519;
  41413, 42023, 42437, 42641;
  ...
Construction of the spiral (see the illustration in the link):
                .  .  .  .  .  .  .  .  .  .  .  .
               .  42  41  40  39  38  37   .  .  .
                                       |
               .  43  20  19  18  17  36  35  .  .
                                   |
               .  .   21   6   5  16  15  34  .  .
                               |
               .  .   22   7   4   3  14  33  .  .
               .  .   23   8   1   2  13  32  .  .
               .  .   24   9  10  11  12  31  .  .
               .  .   25  26  27  28  29  30  .  .
                   .  .  .  .  .  .  .  .  .  .  .
The first squares of center 5 having a prime number in each vertex are:
    19  18  17      41  40  39  38  37
     6   5  16      20  19  18  17  36
     7   4  3       21   6   5  16  15   . . . .
                    22   7   4   3  14
                    23   8   1   2  13

Michel Lagneau, Illustration

Cf. A028871, A073337, A121326, A125202, A200975.

for n from 1 to 10000 do :
  x1:=4*n^2-10*n+7:x2:=(2*n-1)^2-2:
  x3:=4*(n+1)^2-6*(n+1)+1:x4:=4*n^2+1:
   if isprime(x1) and isprime(x2) and isprime(x3) and isprime(x4)
    then
     printf("%d %d %d %d %d \n",n,x1,x2,x4,x3):
    else
    fi:
od:

A293620 Numbers k such that f(k), f(k+1) and f(k+2) are all primes, where f(k) = (2k+1)^2 - 2 (A073577).

Original entry on oeis.org

1, 2, 16, 58, 149, 177, 534, 681, 954, 1045, 1052, 1255, 1367, 1563, 2046, 2074, 2515, 2557, 2564, 2788, 3586, 3593, 3908, 4062, 4552, 5252, 5371, 5385, 6400, 6729, 7443, 7478, 9305, 9375, 9942, 10355, 10411, 10726, 10740, 11286, 11545, 11559, 11832, 11965

Offset: 1

Amiram Eldar, Oct 13 2017

nonn

Sierpiński proved that under Schinzel's hypothesis H this sequence is infinite.
Sierpiński showed that the only quadruple of consecutive primes of the form (2k+1)^2 - 2 are for k = 1 (i.e., 1 and 2 are the only consecutive terms in this sequence).
Numbers k such that the 3 consecutive integers k, k+1 and k+2 belong to A088572. - Michel Marcus, Oct 13 2017

			The first triples are: k = 1: (7, 23, 47), k = 2: (23, 47, 79), k = 16: (1087, 1223, 1367).

Amiram Eldar, Table of n, a(n) for n = 1..10000
Wacław Sierpiński, Remarque sur la distribution de nombres premiers, Matematički Vesnik, Vol. 2(17), Issue 31 (1965), pp. 77-78.
Eric Weisstein's World of Mathematics, Near-Square Prime.
Wikipedia, Schinzel's hypothesis H.

Cf. A088572, A008865, A028870, A028871, A073577, A088572.

Select[Range[10^4], AllTrue[{(2#+1)^2-2, (2#+3)^2-2, (2#+5)^2-2},PrimeQ] &]
SequencePosition[Table[If[PrimeQ[(2k+1)^2-2],1,0],{k,12000}],{1,1,1}][[;;,1]] (* Harvey P. Dale, Feb 09 2025 *)

f(n) = 4*n^2 + 4*n - 1;
isok(n) = isprime(f(n)) && isprime(f(n+1)) && isprime(f(n+2)); \\ Michel Marcus, Oct 13 2017

A163159 Fibonacci numbers F such that F^2-2 is prime.

Original entry on oeis.org

2, 3, 5, 13, 21, 55, 89, 233, 987, 1597, 5702887, 1836311903, 99194853094755497, 26925748508234281076009, 184551825793033096366333, 468340976726457153752543329995929, 30010821454963453907530667147829489881, 1188518561323126046432205871807859915657177

Offset: 1

Vladimir Joseph Stephan Orlovsky, Jul 21 2009

nonn

In condensed representation: The A000045(i) at i = 3, 4, 5, 7, 8, 10, 11, 13, 16, 17, 34, 46,... [R. J. Mathar, Jul 25 2009]

			2^2-2=2. 3^2-2=7. 5^2-2=23.

Cf. A000045, A028871, A163158

f[n_]:=Fibonacci[n]; f2[n_]:=f[n]^2-2; lst={};Do[If[PrimeQ[f2[n]],AppendTo[lst, f[n]]],{n,6!}];lst
Select[Fibonacci[Range[400]],PrimeQ[#^2-2]&] (* Harvey P. Dale, Oct 21 2011 *)

{A000045(i): A008865(A000045(i)) in A000040}. [R. J. Mathar, Jul 25 2009]

More terms from Harvey P. Dale, Oct 21 2011

A348425 Squares whose second arithmetic derivative is a square.

Original entry on oeis.org

0, 1, 4, 49, 529, 2209, 6241, 27889, 28561, 35344, 49729, 128881, 192721, 250000, 431649, 528529, 703921, 1181569, 1495729, 1610361, 1868689, 3411409, 4870849, 5755201, 9138529, 11390625, 12250000, 13830961, 13845841, 15737089, 22648081, 25391521, 31618129

Offset: 1

Marius A. Burtea, Oct 18 2021

nonn

For prime numbers of the form p = k^2 - 2 (A028871) the number m = p^2 is a term because m'' = (p^2)'' = (2*p*p')' = (2*p)'= p + 2*p' = p + 2 = k^2.
If m is a term in A028873 then p = m^2 - 3 is prime and k = p^4 is a term. Indeed: k' = 4*p^3 and k'' = 4*p^3 + 12*p^2 = 4*p^2*(p + 3) = 4*p^2*m^2.
If m is a term in A201787  then p = 5*m^2 - 6 is prime and k = p^6 is a term. Indeed: k' = 6*p^5 and k'' = 5*p^5 + 30*p^4 = 5*p^4*(p + 6) = 25*p^4*m^2.

			4'' = 4' = 4 so 4 is a term.
49'' = 14' = 9 so 49 is a term.

Cf. A000290, A003415, A028871, A028873, A068346, A201787, A266890, A192192, A328244.

f:=func; [n:n in [s*s:s in [0.. 5623]] | IsSquare(Floor(f(Floor(f(n)))))];

d[0] = d[1] = 0; d[n_] := n * Plus @@ ((Last[#]/First[#]) & /@ FactorInteger[n]); Select[Range[0, 6000]^2, IntegerQ @ Sqrt[d[d[#]]] &] (* Amiram Eldar, Oct 18 2021 *)

ad(n) = if (n<1, 0, my(f = factor(n)); n*sum(k=1, #f~, f[k, 2]/f[k, 1])); \\ A003415
lista(nn) = {for (n=0, nn, if (issquare(ad(ad(n^2))), print1(n^2, ", ")); ); } \\ Michel Marcus, Oct 30 2021

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A241537 Cubes c such that c + 1234567890 is prime where 1234567890 is the first pandigital number with digits in order.

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A241538 Squares s such that s + 1234567890 is prime.

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A271725 T(n,k) is an array read by rows, with n > 0 and k=1..4, where row n gives four prime numbers in increasing order with locations in right angles of each concentric square drawn on a distorted version of the Ulam spiral.

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Maple

A293620 Numbers k such that f(k), f(k+1) and f(k+2) are all primes, where f(k) = (2k+1)^2 - 2 (A073577).

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A163159 Fibonacci numbers F such that F^2-2 is prime.

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A348425 Squares whose second arithmetic derivative is a square.

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Magma

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